Midterm 1 Review

01:57:39
https://www.youtube.com/watch?v=qUYej7xU8bo

Resumo

TLDRDum ĉi tiu kunveno, ni diskutis diversajn kemiajn konceptojn en prepariĝo por la unua ekzameno. La instruisto klarigis temojn pri la perioda tabelo inkluzive de radiuso, jonigo de energio, elektronegativeco kaj siaj periodaj tendencoj. Ni ankaŭ reviziis kiel desegni Lewis-strukturojn kun minimumaj formalaj ŝargoj kaj kiel apliki la bohran modelon por elektra translokigo en atomoj de hidrogeno. La kunveno enhavis ekzemplajn ekzamenajn demandojn kiuj testis niajn konojn pri kalkulado de subatomaj partikloj, kvantuma nombrosistemo, kaj ligoj inter atomoj. La partoprenantoj povis demandi kaj ricevi klarigojn en realtempa diskuto, kio helpas en pli bona konceptado de la materialo prezentita.

Conclusões

  • 🔍 Koncepto pri perioda tabelo gravas por kompreni radiuson kaj ionigan energion.
  • 📚 Revizio pri Bohr-modelo rilate al hidrogenaj ekscitoj estis farita.
  • 🧪 Akcento estis pri minimumigado de formala ŝargo en Lewis-strukturoj.
  • ❓ Realtempaj demandoj estis pritraktitaj por klarigi konfundajn punktojn.
  • ⚛️ Traktado pri kvantumaj nombroj estis inkluzivita por interna atomstruktura kompreno.
  • 🔗 Diskuto pri molekulaj ligoj kaj iliaj elektronaj karakteroj estis farita.
  • 🔠 Nomoformado de kovalentaj kaj acidaj kunmetaĵoj estis klarigita.
  • 💡 Subatomaj partikloj kaj iliaj kalkuloj restis kerna temo.
  • 🔄 Komprenado de variaĵoj ene periodaj tendencoj estis prifokusa.
  • 📝 Prepariĝo por venonta ekzameno estis la ĉefa celo de la kunveno.

Linha do tempo

  • 00:00:00 - 00:05:00

    La prelego komenciĝas kun la prezentanto kontrolante la funkciecon de la teamsoftvaro kaj klarigante la strukturon de la sesio. Subitaj demandoj devus esti rezervitaj ĝis fino de ĉiu problemo ĉar ili povas esti respondotaj dum la solvo de aliaj taskoj. Laŭvole, oni povas uzi manlevon aŭ tajpadon por demandi, kaj la sesio okazos tute virtuale por la resto de la semestro.

  • 00:05:00 - 00:10:00

    La unua problemo esploras masleĝon, demonstrante la difinitan konsiston aŭ proporcian leĝon, kie oni substrekas ke natrio estas konstanta per masprocento en natria klorido el diversaj fontoj. Oni klarigas, ke ĉi tio koncernas ununuran substancon prefere ol kemia reago.

  • 00:10:00 - 00:15:00

    Pluverkanta problemo eksploras subatomajn partiklojn en mangana jono, kondukante partoprenantojn tra la kalkulo de neŭtronoj kaj elektronoj en speciala izotopa kaj jona konteksto. La metodo uzas masnumeron kaj atomnumeron por informi la procezon.

  • 00:15:00 - 00:20:00

    La sekva problemo traktas energion kaj frekvencon de fotono uzante ĝian ondolongon en la infraruĝa spektro. Uzante la konstantajn valorojn por h (Planck-konstanto) kaj c (lumrapido), oni solvas por ambaŭ postulataj mezuroj.

  • 00:20:00 - 00:25:00

    Problemo kovras la minimuman energion aŭ laborfunkcion postulatan por la eltira efiko fotolektra en metaloj. Oni uzas la saman fundamentan ekvacion kiel en la antaŭa frekvenca problemaro, sed emfazas la konceptan komprenadon de fotolektra efiko.

  • 00:25:00 - 00:30:00

    Diskutante la Bohr-modelon, oni klarigas kiel kalkuli la frekvencon de fotono bezonata por eksciti elektronon de unu enerĝinivelo al alia en hydrogeno, kun instrukcioj pri ĝusta uzo de la ekvacio kaj konvertado de mezurunuoj.

  • 00:30:00 - 00:35:00

    Oni devas kompreni la elektron-konfiguron de mangano uzante la periodan tabelon por plenigi subshell-ojn laŭ energia ordo. Oni substrekas la uzon de la n-1 regulo por transiraj metaloj, montrante la bezonon sekvi ĝustan proceduron por tiaj elementoj.

  • 00:35:00 - 00:40:00

    Demandate pri la nombro de orbitaloj ĉe specifa energinivelo, la solvo uzas la rilaton 2l+1 por kalkuli ilin en subnivelo difinita de donitaj kvantumaj nombroj. Diskuto temas pri la uzado de kvantumnombroj kaj ilia interpreto laŭ energiaj niveloj.

  • 00:40:00 - 00:45:00

    Malkovru la maksimuman eblan nombron de elektronoj en specifita ŝelo kaj subshell surbaze de kvantumaj nombroj. Tio implicas analizon de kapacito per la formulo 2n^2 por maksimumo en ŝelo kaj 2(2l+1) por subshell.

  • 00:45:00 - 00:50:00

    Uzu konceptajn sciojn de kvantumnombroj por identigi taŭgajn valorojn en orbitaloj, inkluzive de subshelloj kaj specifaj orbitaloj, klarigante la regulojn kaj rilatojn inter magnitudoj kiel eblaj naŭncaj esploroj de loka spacokovro en atoma skalo.

  • 00:50:00 - 00:55:00

    Analizo pri la grunda stato de atomo postulas aplikon de diversaj principoj kiel Hund kaj Paul por aserti la plej stabilan elektronaranĝon en orbitaloj, uzante detalojn provizitajn en la problema diagrama reprezentado.

  • 00:55:00 - 01:00:00

    Reveni al bazaj prediktoj bazitaj sur la perioda tabelo por klasifikado kaj ŝarĝo de elementoj hala kaj metalaj karakterizaĵoj kun referenco al iliaj pozicioj kaj komunumaj propraĵoj en formadleĝoj de reago.

  • 01:00:00 - 01:05:00

    Analizas periodajn tendencojn elektita radiuso kaj ioniga energio, atenta konsidero al distingoj en grupniveloj kaj sia influo sur prediktoj de kompareblaj fizikaj priskriboj tra elementoj kiel similaj punktoj de molekuligita aspekto.

  • 01:05:00 - 01:10:00

    Enkonduko al la sekcio pri jonigitaj periodaj tendencoj, emfaza noto sur energio de jonizado, kaj tiuj grupoj kun rimarkindaj esceptoj, kiel trovitaj en la kvina grupo kie anormale grandaj energetikaj saltegoj okazas.

  • 01:10:00 - 01:15:00

    Oni esploras la polusecon de kemiaj ligoj ekzamenante la diferencan elektronegativecon inter elementparoj, disvolvante metodon por kontroli la polusecon surbaze de la relativaj pozicioj en la perioda tabelo, sen limigo al numeraj ekvivalentoj.

  • 01:15:00 - 01:20:00

    Ankaŭ en kelkfoje la disjunticece, kompara analizo de atomaj kaj jonaj radioj uzante la bazajn konojn de elektrona strukturo kaj periodaj tendencoj al mezaj valentinoj kaj jonstruktura ekspansio.

  • 01:20:00 - 01:25:00

    La solvo de formula predikto implikas la uzon de klinika logiko kaj scio pri periodaj tendencoj de metalo kaj nemetalo por produkti ekvilibran chemian formon respondas la elekto al fiksitaj elekto kaj agnoskas la sisteman enkondukon.

  • 01:25:00 - 01:30:00

    Analizinte Lewis-strukturan por formala ŝargo, oni enprofundiĝas en konceptoj necesaj por ĝuste decidi pri elektronarangoj kaj rekonstruo, planita celis forigon de neta spaca neeskapo sur centraj elementoj.

  • 01:30:00 - 01:35:00

    Trajto pri la nomado de kovalentaj kombinaĵoj uzas la regulojn pri prefiksoj kaj oksoanionoj, helpante edukantojn kapti la esencajn fonojn por intelekta demandokurso en la preciza prognozo de kunmeta formulo.

  • 01:35:00 - 01:57:39

    Post la kompletigo de revizio, donita demandoj retaksas kaj resumas konceptojn en periodaj tendencoj kaj la uzon de elektronegativacaj diferencoj por helpi kompreni kemio-dezajnojn en logika aŭtokratia formo.

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Mapa mental

Mind Map

Perguntas frequentes

  • Kian enhavon enhavas la kunveno?

    La kunveno enhavas klarigojn kaj ekzercajn demandojn pri kemio, inkluzive de temoj kiel kvantumaj nombroj, Lewis-strukturoj, periodaj tendencoj, kaj kemiaj nomadoj.

  • Kiuj konceptoj de kemio estis traktataj en la kunveno?

    Traktataj konceptoj inkluzivas kvantumajn nombrojn, periodajn tendencojn, bohran modelon, Lewis-strukturojn, kemiajn nomojn, kaj elektra negoco de ligoj.

  • Kiel oni determinis la plej grandan ionigan energion?

    La plej granda ioniga energio estis determinita konsiderante la poziciojn de la elementoj en la perioda tabelo kaj la specifajn esceptojn inter grupoj 2 kaj 5.

  • Kiel la kunveno traktis la temon de Lewis-strukturoj?

    La kunveno diskutis kiel desegni Lewis-strukturojn atentante valentelektronojn kaj minimumigi formalajn ŝargojn.

  • Kio estis la ĉefa fokuso de la diskuto pri kvantumaj nombroj?

    La fokuso estis kiel determini la aĝustajn kvantumajn nombrojn por elektrona orbitralo kaj la rilato de tiuj nombroj al perioda tabelo.

  • Kiel oni determinas la formalan ŝargon en Lewis-strukturo?

    Oni determinas per subtraho de la nombro de bendoj kaj senligaj elektronoj en orbitralo de la valentelektronoj de la elemento.

  • Kiel la seminaro traktis nomadon de kemiaj kunmetaĵoj?

    Ĝi klarigis kiel uzi prefiksojn por nomi kovalentajn kunmetaĵojn kaj uzadon de anjonaj nomoj por determini acidajn formulojn.

  • Kiel oni analizis ligojn rummoritaj en la ekzamenaj demandoj?

    Ligoj estis analizitaj laŭ ilia elektroninklinio por determini ilian ionan karakteron kaj polusecon.

  • Ĉu temis pri specialaj demandoj pri unu elektra negativigeco?

    Jes, temis pri komparado de la elektronegativigeco inter diversaj elementoj por taksi ligan polusecon.

  • Kio estis aparta rimarkinda temo traktata dum la kunveno?

    Rimarkinda temo estis la klarigo pri kiel la elektra negativo kaj la istimularoj de la periodoj influas ligan karakteron.

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Legendas
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Rolagem automática:
  • 00:00:02
    um it is two o'clock
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    and I believe the meeting is now
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    recording
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    so let's hope that there's no issues
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    with that so that I can post this later
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    on so before I begin could I get one
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    person to confirm for me that they can
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    hear me
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    very good okay I have not used teams
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    very often lately so I don't want this
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    to be a total disaster so thank you for
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    clarifying and then the next thing to
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    confirm is once I get over to it are you
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    able to see the screen that I'm sharing
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    yes but you can see okay perfect
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    all right so we'll get right into it
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    um it's gonna be obviously some uh
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    differences between this and a normal
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    review session they will for out for the
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    rest of semester be in person uh
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    what I'll ask is if you have questions
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    I'm going to ideally save them to the
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    end of each problem in part because a
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    lot of times you'll have a question pop
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    in your head that I will then answer as
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    I go through the rest of the problems
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    that'll save us some redundant questions
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    and then also uh just easier for me to
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    monitor on teams if I'm going you know
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    just back and forth once every question
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    uh you can shout them out that's in some
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    ways prefer because then I don't have to
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    look for you to you know raise your hand
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    or look in the you know typed messages
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    but at the same time
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    um
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    if you uh want to you know just rather
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    just raise your hand or type it in I'll
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    try to catch all of those questions
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    about maybe I'll get a little bit behind
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    on those but I'll try to keep up with it
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    um so however you choose to answer ask
  • 00:01:38
    questions is fine and whenever you
  • 00:01:40
    choose to do it or whatever you choose
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    to ask about so you can ask about the
  • 00:01:43
    specific problem that I that I go over
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    and you're also welcome to ask questions
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    about anything that you think is related
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    to that topic that you want some
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    clarification on so it's pretty
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    open-ended and obviously this is for
  • 00:01:55
    your benefit primarily so whatever we
  • 00:01:57
    discuss today is totally fine with me
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    all right so if there's no procedural
  • 00:02:01
    questions that you have
  • 00:02:04
    [Music]
  • 00:02:04
    um
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    get the habit of checking regularly so
  • 00:02:07
    it looks like we're good
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    and I will get right into the first
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    problem so this goes back to the very
  • 00:02:13
    beginning of the course
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    um on mass laws
  • 00:02:18
    so we're asking you to identify in this
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    problem which Mass law is demonstrated
  • 00:02:21
    by the fact that sodium chloride mined
  • 00:02:23
    in God Rich Ontario which has the
  • 00:02:25
    largest salt mine in the world and
  • 00:02:27
    sodium chloride mine and Kura Pakistan
  • 00:02:30
    which is the second largest salt mine
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    both consists of 39.3 percent sodium by
  • 00:02:34
    Mass
  • 00:02:36
    um so for these types of problems
  • 00:02:39
    you know we're looking at
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    a situation that's presented to us
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    and so what we can do is you know go
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    through the answer choices and kind of
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    give a brief description of what each of
  • 00:02:51
    these Mass laws means and then try to
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    relate that to the situation that we're
  • 00:02:54
    given here so Choice a which is law of
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    conservation of mass that's probably the
  • 00:02:59
    most straightforward one that tells us
  • 00:03:00
    that matter is neither created nor
  • 00:03:02
    destroyed during a chemical reaction
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    and in this problem here we're not
  • 00:03:08
    talking about a chemical reaction
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    so this does not seem relevant to the
  • 00:03:13
    situation we're presented we're talking
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    about mass per sense of elements but
  • 00:03:16
    we're not talking about a chemical
  • 00:03:18
    reaction converting one substance to
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    another so that we can probably
  • 00:03:21
    eliminate that one
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    for B which is definite composition B
  • 00:03:25
    and C are the two that we
  • 00:03:27
    kind of most often mix up of these three
  • 00:03:30
    Mass laws and so the first one which is
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    definite composition or sometimes called
  • 00:03:35
    definite proportion you might see it
  • 00:03:37
    that way on your exam but either way
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    it'll have the word definite in it so
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    that's going to be talking about the
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    elemental mass percentage of a single
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    compound
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    and so this is saying that you know
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    any substance
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    typically a compound but it also applies
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    to elements themselves which are 100 of
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    a single element but any substance has
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    specific mass ratios of elements
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    all right now this one sounds like what
  • 00:04:10
    we're describing here because we're
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    talking about a substance sodium
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    chloride
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    yeah we're told that it has 39.3 sodium
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    by mass now we do want to make sure that
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    we can distinguish this one from
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    multiple proportions which is the one
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    that's going to be
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    closely related and sometimes a little
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    bit confused with definite composition
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    so the key distinction is that if we're
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    talking about law of multiple
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    proportions we're talking about
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    relationships involving two or more
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    different substances so I'm not going to
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    write out the whole law of multiple
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    proportions because it's kind of wordy
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    but when you're trying to identify it in
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    a question like this multiple
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    proportions requires that you're talking
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    about two or more different substances
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    first of all and they have to be formed
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    from the same elements
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    so in this problem here we're only
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    talking about a single substance sodium
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    chloride
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    so it's not multiple substances it's
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    multiple it's it's one single substance
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    sodium chloride from different sources
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    but not different substances so the one
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    that's going to work best for this one
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    will be Choice B
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    um now if we go through we should also
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    eliminate Choice D I haven't done that
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    yet Dalton's law we'll learn about this
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    later in the semester this is not
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    actually a mass law
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    not to be confused with Dalton's atomic
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    theory which is a theory that explains
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    the mass laws but Dalton's law which is
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    a separate law dealing with gases is not
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    something that we're going to talk about
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    until much later so the one that fits
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    best is Choice B because we're dealing
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    with the composition of a single
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    substance which would not depend on the
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    source of that substance all right so
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    does anybody have questions on that
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    problem or on the topics in general
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    related to mass laws
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    okay so we are silenced so I will move
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    on to the next one and keep us going
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    be mindful of our time okay so the next
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    question here deals with
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    um atomic structure the subatomic
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    particles that are present in atoms and
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    ions and it wants to know how many
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    neutrons and electrons are present in a
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    55 manganese plus three ion okay so the
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    first thing to decompose is what's going
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    to be
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    um you know the the anatomy of this
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    atomic symbol so we gave you an atomic
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    symbol here that has a symbol and two
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    numbers attached to us let's talk about
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    what those exactly mean so this is 55
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    mn3 plus so when it's written in this
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    way
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    what we have to recall is that the
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    number that's on the top left if it's
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    given to us
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    that's going to be the mass number
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    and that's going to tell us the number
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    of protons plus the number of neutrons
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    basically the sum of the subatomic
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    particles that are in the nucleus
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    okay so that's what the mass number
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    tells us and then this number up here
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    which won't always be there but this if
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    it's given is going to indicate the
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    charge telling us that this is actually
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    an ion not a neutral atom
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    okay so that's going to be the two
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    pieces of this that are critical now
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    what we're going to typically start with
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    for problems like this is identifying
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    the number of protons first and the
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    reason that that one is usually the most
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    straightforward to get is because it
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    comes directly from the atomic number
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    now in this atomic symbol here we didn't
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    give you the atomic number but that's
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    going to be directly related to the
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    symbol of the element which is MN so to
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    find the atomic number Z which would go
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    down here if it was given but it's not
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    given to us in this problem we'll go on
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    the sort of bottom left of the symbol
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    then we have to go to the periodic table
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    to identify that element so I'll flip
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    over to the periodic table here which
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    you should still be able to see it's in
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    the same shared window I believe and
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    we're going to find MN so this is one of
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    the first 36 so you should be familiar
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    this is called manganese and it's right
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    here middle of the 3D block
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    okay so it's but the key part of this
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    which I'll try to zoom in on a little
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    bit
  • 00:08:09
    in case that's small for you guys
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    is that manganese is right here the
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    number above that is the number 25.
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    okay so that's going to tell us the
  • 00:08:20
    atomic number of that element all right
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    so when we go back to over here we know
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    because it's MN and the atomic number on
  • 00:08:28
    the periodic table is 25 that's going to
  • 00:08:30
    be the number of protons in the nucleus
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    now we're not actually asked for that
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    number here but we need it to get the
  • 00:08:35
    other two so now that we have
  • 00:08:38
    the number of protons and we know that
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    the mass number is 55 55 is equal to the
  • 00:08:43
    number of protons
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    so this is going to be the mass number
  • 00:08:47
    which is sometimes given as a plus the
  • 00:08:50
    number of neutrons and then with a
  • 00:08:51
    little bit of algebra we can find out
  • 00:08:53
    that the number of neutrons or Brewing
  • 00:08:55
    is n0 is 55 minus 25 which is 30. so we
  • 00:08:59
    get the number of neutrons by difference
  • 00:09:00
    so we can eliminate already three of the
  • 00:09:03
    answer choices because they have the
  • 00:09:05
    wrong number of neutrons if we trusted
  • 00:09:07
    ourselves to do that
  • 00:09:08
    all right and then finally for the
  • 00:09:10
    number of electrons that is going to
  • 00:09:12
    relate to the charge so if this was a
  • 00:09:15
    neutral atom the number of protons the
  • 00:09:18
    number of electrons would have to be
  • 00:09:19
    equal they would both be 25 but we're
  • 00:09:21
    telling you that there's a three plus
  • 00:09:22
    charge here so the three plus charge
  • 00:09:24
    comes about because we have three fewer
  • 00:09:27
    electrons than protons so we have 25
  • 00:09:29
    protons
  • 00:09:32
    but there's going to be three fewer than
  • 00:09:34
    that for the number of electrons because
  • 00:09:35
    of the plus three charge
  • 00:09:40
    so that's why we're going to subtract 3
  • 00:09:42
    from this and we're going to leave us
  • 00:09:43
    with 22 as the number of electrons so 30
  • 00:09:47
    for neutrons 22 for electrons and that's
  • 00:09:50
    going to be Choice B here okay so the
  • 00:09:53
    number of neutrons requires us to have
  • 00:09:55
    the mass number and the atomic number
  • 00:09:57
    and the number of electrons requires us
  • 00:09:59
    to consider any charge that might be
  • 00:10:01
    present which in this case there was one
  • 00:10:04
    all right so that's again subatomic
  • 00:10:06
    particles
  • 00:10:07
    um and you know knowing all those ways
  • 00:10:09
    of finding the different numbers of
  • 00:10:11
    protons neutrons electrons for both
  • 00:10:13
    atoms and ions are all going to be a
  • 00:10:15
    fair game for you guys so do we have any
  • 00:10:17
    questions on that problem or that topic
  • 00:10:24
    all right you guys are easy so far but
  • 00:10:26
    please feel free with just to chime in
  • 00:10:29
    with any questions you have
  • 00:10:30
    um at any point
  • 00:10:32
    all right on number three
  • 00:10:34
    all right so as you can kind of tell
  • 00:10:36
    from the format of this exam and if
  • 00:10:39
    you're keeping track of sort of where we
  • 00:10:41
    are in the textbook those were really
  • 00:10:43
    the only two
  • 00:10:45
    questions that dealt with chapter one
  • 00:10:48
    um so we're already now on to chapter
  • 00:10:49
    two which starts with electromagnetic
  • 00:10:52
    radiation
  • 00:10:54
    all right and so this problem here we
  • 00:10:57
    have an infrared Photon has a wavelength
  • 00:11:00
    of
  • 00:11:01
    1234 nanometers what are the energy and
  • 00:11:04
    frequency of the photon so this problem
  • 00:11:06
    sort of deals with both relationships
  • 00:11:09
    that we would have involving photons and
  • 00:11:12
    their wavelengths and frequencies and
  • 00:11:14
    energies so there are two different
  • 00:11:16
    equations
  • 00:11:17
    um and you need to know in a specific
  • 00:11:19
    problem which one to use in this case
  • 00:11:20
    I'm using them both just to sort of show
  • 00:11:22
    them both to you in reality you know
  • 00:11:25
    most of the exam questions would require
  • 00:11:27
    you to use one or the other but not
  • 00:11:29
    necessarily both at the same time
  • 00:11:31
    but nonetheless what those are is the
  • 00:11:34
    first one we talked about which is just
  • 00:11:36
    dealing with the wavelength and
  • 00:11:38
    frequency relationship is that the
  • 00:11:40
    product of those two new which is the
  • 00:11:41
    frequency Lambda which is the wavelength
  • 00:11:43
    equals speed of light C and that's going
  • 00:11:46
    to be a constant on your periodic table
  • 00:11:48
    so you don't need to memorize that
  • 00:11:49
    number even though you'll use it a lot
  • 00:11:52
    and then the other relationship is for
  • 00:11:54
    the energy of a photon so the energy is
  • 00:11:57
    equal to a different constant H flux
  • 00:11:59
    constant multiplied by
  • 00:12:02
    new which is again the frequency and
  • 00:12:04
    then we can combine these two equations
  • 00:12:07
    because Lambda is equal to C sorry Nu is
  • 00:12:10
    equal to C divided by Lambda and so if I
  • 00:12:12
    combining those two we get an
  • 00:12:13
    alternative form HC over Lambda
  • 00:12:16
    so there's really two equations and then
  • 00:12:18
    a third one that emerges by combining
  • 00:12:20
    those two algebraically and in this
  • 00:12:23
    problem here we're going to end up using
  • 00:12:24
    both of them
  • 00:12:26
    um and so the first thing we want is
  • 00:12:28
    well we can do it either order but we
  • 00:12:30
    want the energy and the frequency so
  • 00:12:32
    whichever one you choose to do first but
  • 00:12:33
    energy is listed first so let's do that
  • 00:12:35
    we're going to use this form of the
  • 00:12:37
    equation here energy equals HC over
  • 00:12:39
    Lambda because wavelength Lambda is what
  • 00:12:42
    we're given in this problem 12 34
  • 00:12:43
    nanometers okay so that's what we're
  • 00:12:46
    going to use first or with this I choose
  • 00:12:49
    to do it that's totally fine so we're
  • 00:12:50
    going to go HC over Lambda
  • 00:12:53
    and again to remind you H and C are both
  • 00:12:56
    fundamental physical constants and
  • 00:12:59
    they're on your periodic table so again
  • 00:13:00
    look at the bottom here this is the
  • 00:13:02
    exact same periodic table you'll have
  • 00:13:04
    an exam
  • 00:13:05
    H is here 6.626 times 10 to the minus 34
  • 00:13:09
    C is the last one 3 times 10 to the
  • 00:13:12
    eighth so those are those two constants
  • 00:13:14
    that we used a lot in chapter two
  • 00:13:16
    problems and we're going to use both of
  • 00:13:17
    them in this equation so H which is
  • 00:13:19
    6.626
  • 00:13:22
    times 10 to the minus 34 and the units
  • 00:13:25
    for that are joules seconds
  • 00:13:26
    and then C is the speed of light in
  • 00:13:28
    meters per second now the one thing we
  • 00:13:30
    have to be careful about when using both
  • 00:13:32
    of these equations is that we have the
  • 00:13:34
    correct units especially for Lambda so
  • 00:13:37
    in this we need SI units all these
  • 00:13:39
    constants are in terms of SI units and
  • 00:13:41
    particularly we see that the speed of
  • 00:13:43
    light is meters per second so that means
  • 00:13:45
    our wavelength which is a measure of
  • 00:13:46
    distance or length has to be given to us
  • 00:13:49
    in meters or after we have to use it in
  • 00:13:52
    meters to be able to put it in the
  • 00:13:53
    equation so that's the one unit
  • 00:13:55
    conversion will have to do in this
  • 00:13:56
    problem which is that the wavelength in
  • 00:13:58
    meters we're starting with 1234
  • 00:14:01
    nanometers
  • 00:14:03
    and then Nano is 10 to the ninth so
  • 00:14:05
    there's 10 to the minus 9 technically
  • 00:14:07
    but there's two ways of writing that but
  • 00:14:08
    it's going to be one meter is 10 to the
  • 00:14:11
    ninth nanometers if you prefer positive
  • 00:14:13
    exponents or alternatively 10 to the
  • 00:14:15
    minus 9 meters over one nanometer works
  • 00:14:17
    fine as well mathematically they're the
  • 00:14:19
    same
  • 00:14:19
    so when we multiply that out we get
  • 00:14:21
    1.234 times 10 to the minus 6 meters
  • 00:14:26
    and so that's the number we're going to
  • 00:14:28
    put in here for Lambda in the correct
  • 00:14:30
    units 1.234
  • 00:14:33
    times 10 to the minus 6
  • 00:14:36
    and then so when we multiply that across
  • 00:14:38
    we get
  • 00:14:40
    1.61
  • 00:14:42
    times 10 to the minus 19 joules
  • 00:14:45
    um
  • 00:14:46
    so that's the number we get for energy
  • 00:14:48
    so that's going to match choices A and B
  • 00:14:50
    so again if we trust ourselves for the
  • 00:14:53
    first part by elimination we can get rid
  • 00:14:55
    of c and d
  • 00:14:57
    and then finally
  • 00:14:58
    um
  • 00:14:59
    we have to calculate the frequency as
  • 00:15:01
    well and so for that we're going to use
  • 00:15:03
    this first equation here and solve for
  • 00:15:07
    NU which is the frequency which is just
  • 00:15:08
    C divided by Lambda
  • 00:15:11
    and so that's going to be
  • 00:15:14
    speed of light is the only constant in
  • 00:15:16
    this equation 3 times 10 to the eighth
  • 00:15:18
    divided by the same wavelength still in
  • 00:15:21
    meters so make sure the units match up
  • 00:15:27
    and when we divide those two we get 2.43
  • 00:15:34
    times 10 to the 14.
  • 00:15:37
    and the units for that it's inverse
  • 00:15:39
    seconds but more commonly written as
  • 00:15:40
    hearse HZ all right so that's new and
  • 00:15:44
    then that's our energy up there and so
  • 00:15:46
    the answer choice that has both of those
  • 00:15:48
    would be a
  • 00:15:50
    um now in problems you know in the exam
  • 00:15:53
    we we have a lot of problems like this
  • 00:15:55
    where you have to calculate numbers and
  • 00:15:57
    you know given the timing of the exam
  • 00:15:59
    you have uh 75 minutes to do 20
  • 00:16:03
    Questions you're looking at you know
  • 00:16:04
    3.75 minutes per question
  • 00:16:07
    um you're not going to probably have
  • 00:16:08
    time to do every problem
  • 00:16:11
    multiple times you're gonna have to you
  • 00:16:13
    know move through and be efficient to
  • 00:16:15
    make sure the time's not an issue for
  • 00:16:16
    most people time is not an issue I
  • 00:16:17
    should say but you know trying to do the
  • 00:16:19
    exam twice and a lot of time might be
  • 00:16:21
    might be a challenge so what's helpful
  • 00:16:23
    in numerical problems like this
  • 00:16:24
    especially if you don't have time to
  • 00:16:26
    completely check your work and
  • 00:16:27
    recalculate everything is to make sure
  • 00:16:29
    at least the order of magnitude for the
  • 00:16:31
    answer is reasonable
  • 00:16:33
    and so you know what's nice to remember
  • 00:16:36
    is that if we're calculating the energy
  • 00:16:37
    of a photon and we're in you know
  • 00:16:40
    somewhere in the typical region of the
  • 00:16:41
    spectrum that's you know UV divisible to
  • 00:16:45
    maybe infrared is where we're going to
  • 00:16:47
    most often be looking at in the context
  • 00:16:49
    of chemistry your Photon energy should
  • 00:16:51
    be somewhere on the order of you know 10
  • 00:16:53
    to the minus 18 10 to the minus 19.
  • 00:16:55
    somewhere within that negative
  • 00:16:57
    exponential range so if you calculate an
  • 00:16:59
    energy that's you know a huge number
  • 00:17:01
    like positive exponent or way way
  • 00:17:04
    smaller than this like C and D or then
  • 00:17:07
    you probably made a math mistake and
  • 00:17:08
    then similarly with frequency typical
  • 00:17:11
    frequencies for electromagnetic
  • 00:17:12
    radiation in the parts of the spectrum
  • 00:17:14
    we work with are you know around
  • 00:17:17
    gigahertz range you know 10 of the 14th
  • 00:17:20
    10 to the 15th Hertz or so
  • 00:17:23
    um
  • 00:17:24
    and and so that's gonna or terahertz
  • 00:17:26
    even but anyways it's going to be a
  • 00:17:28
    typically a large positive exponent on
  • 00:17:30
    that order and again if you get a
  • 00:17:31
    negative exponent for frequency that's
  • 00:17:34
    almost certainly a math mistake that you
  • 00:17:36
    made so like choices B and D here have
  • 00:17:39
    really small numbers for the frequency
  • 00:17:41
    that's not a number you'd expect to get
  • 00:17:44
    for anything that we're going to
  • 00:17:45
    calculate so that might cue you in that
  • 00:17:47
    you missed it up if you've got one of
  • 00:17:48
    those because a lot of times if you
  • 00:17:50
    haven't if you have a multiple choice
  • 00:17:52
    question that's numerical
  • 00:17:54
    um you know some of the answer choices
  • 00:17:57
    are going to be common mistakes that
  • 00:17:58
    people make so they're ones that you
  • 00:18:00
    they're not just random numbers they're
  • 00:18:02
    ones that you would be likely to pick if
  • 00:18:04
    you made one of the more common mistakes
  • 00:18:06
    that that comes up a lot of the
  • 00:18:08
    numerical questions you actually have to
  • 00:18:09
    type your answer in in which case you
  • 00:18:11
    can't really eliminate answer choices
  • 00:18:13
    like we did here but still knowing
  • 00:18:15
    approximately the appropriate order of
  • 00:18:18
    magnitude for a typical number that you
  • 00:18:20
    would calculate a problem like this can
  • 00:18:22
    help you easily identify if you made an
  • 00:18:24
    obvious mistake
  • 00:18:26
    all right so that's just a little bit of
  • 00:18:27
    word of advice and on some of the
  • 00:18:28
    miracle things so do we have any
  • 00:18:30
    questions on on that so there's someone
  • 00:18:32
    in the chat send some things let me
  • 00:18:33
    check what it says
  • 00:18:36
    will the test do the calculator oh
  • 00:18:39
    okay will the test do the calculation
  • 00:18:41
    using the exact C value around it to 3
  • 00:18:44
    times 10 to the eighth
  • 00:18:46
    um it doesn't matter so you are right
  • 00:18:48
    that the value of C is given to us on
  • 00:18:52
    the periodic table with like I don't
  • 00:18:54
    know it's like five digits at least like
  • 00:18:56
    2.9972 or something like that I forget
  • 00:18:58
    you can look it up
  • 00:19:00
    um but if you round to 3 times 10 to the
  • 00:19:02
    eighth that's fine we typically give you
  • 00:19:04
    for numerical questions about a you know
  • 00:19:07
    one to two percent window of where we'll
  • 00:19:10
    accept the correct answer
  • 00:19:13
    um and that depends on how many steps
  • 00:19:14
    the question is and stuff like that so
  • 00:19:16
    we kind of decide that depending on the
  • 00:19:18
    nature of the question but it's at least
  • 00:19:19
    one percent leeway and
  • 00:19:22
    2.997 versus 3 is like a much smaller
  • 00:19:25
    difference than one percent so rounding
  • 00:19:27
    to three times ten to the eighth is not
  • 00:19:28
    going to cause any problems so I've
  • 00:19:31
    gotten the thumbs up from the questioner
  • 00:19:32
    so I think that means I've
  • 00:19:34
    satisfactorily answered it if anybody
  • 00:19:35
    else has anything want to clarify please
  • 00:19:37
    let me know
  • 00:19:39
    okay so that's perfectly fine to send
  • 00:19:41
    questions in that way we had a question
  • 00:19:43
    in the chat that I was able to get to
  • 00:19:45
    reasonably quickly so feel free to do it
  • 00:19:47
    in that format if you prefer all right
  • 00:19:49
    so on to number four then
  • 00:19:52
    all right so this is
  • 00:19:54
    um
  • 00:19:55
    still chapter two material and we have
  • 00:19:58
    the element hafnium here HF is the is
  • 00:20:01
    the symbol it's pretty far down the
  • 00:20:02
    periodic table it is observed that light
  • 00:20:05
    with wavelength less than 318 nanometers
  • 00:20:08
    is required to eject an electron from
  • 00:20:10
    the surface via the photoelectric effect
  • 00:20:13
    what is the threshold energy work
  • 00:20:16
    function for hafnium and joules now it
  • 00:20:18
    actually turns out that in in this um in
  • 00:20:22
    this problem here we're going to use the
  • 00:20:23
    same equation that we just did
  • 00:20:26
    um
  • 00:20:26
    but it may not be obvious if that's the
  • 00:20:28
    case so first let's talk about the
  • 00:20:30
    concept that this problem deals with
  • 00:20:31
    because that is still relevant to this
  • 00:20:33
    exam and still helpful to think about in
  • 00:20:35
    this problem so we're talking about
  • 00:20:36
    photoelectric effect
  • 00:20:38
    um that's kind of an experiment where
  • 00:20:40
    you have a metal surface in this case
  • 00:20:41
    it's half neon but you can do it with
  • 00:20:43
    any metal
  • 00:20:44
    so indicating this is a metal surface
  • 00:20:47
    you shine light on the surface
  • 00:20:50
    with some frequency new and some energy
  • 00:20:53
    H times Nu and when the product of H Nu
  • 00:20:57
    is greater than some minimum value which
  • 00:21:00
    we call the the threshold energy or the
  • 00:21:03
    work function so if that which is that
  • 00:21:06
    that's equal the work function or
  • 00:21:08
    threshold energy is a synonym for that
  • 00:21:10
    both of those terms are given here when
  • 00:21:12
    that energy is high enough you kick off
  • 00:21:14
    electrons from the surface
  • 00:21:19
    um and you can measure the kinetic
  • 00:21:20
    energy of these electrons and how fast
  • 00:21:21
    they are relates also to the energy of
  • 00:21:23
    the photon but the key part is that if
  • 00:21:25
    the energy of your photon is is too
  • 00:21:28
    small then nothing will happen you won't
  • 00:21:30
    kick off any electrons but if the energy
  • 00:21:32
    is greater than the work function then
  • 00:21:33
    you'll start ejecting electrons so here
  • 00:21:36
    what we're looking for is we're actually
  • 00:21:37
    giving you
  • 00:21:38
    the threshold wavelength so
  • 00:21:42
    we can also just Define that in terms of
  • 00:21:44
    a wavelength 318 nanometers
  • 00:21:48
    so that's our threshold wavelength
  • 00:21:50
    telling us that that's the maximum
  • 00:21:52
    wavelength where an electron will be
  • 00:21:54
    injected we want to calculate what is
  • 00:21:56
    the energy that's associated with that
  • 00:21:57
    so we're just going to use the same
  • 00:21:59
    equation we did in the last problem
  • 00:22:00
    which is that the threshold energy is
  • 00:22:02
    going to be related to the threshold
  • 00:22:04
    wavelength by HC divided by lambda zero
  • 00:22:07
    where lambda zero is the threshold
  • 00:22:09
    wavelength and you notice that they're
  • 00:22:11
    inversely related to each other so the
  • 00:22:13
    photon has to have an energy that's
  • 00:22:14
    larger than e0 which then means it has
  • 00:22:17
    to have a corresponding wavelength
  • 00:22:19
    that's smaller than Lambda 0 because of
  • 00:22:21
    that inverse relationship so but we're
  • 00:22:23
    just going to use the equation to
  • 00:22:25
    calculate the threshold energy or the
  • 00:22:26
    work function e0
  • 00:22:28
    and we're just going to plug things in
  • 00:22:30
    like we did before so we have to do the
  • 00:22:33
    same two constants H and C
  • 00:22:36
    and again
  • 00:22:38
    I always do four four you know three
  • 00:22:40
    decimal places for age but
  • 00:22:42
    rounding to three times in a day if
  • 00:22:45
    that's fine for C
  • 00:22:47
    and then we have to put in Lambda in the
  • 00:22:49
    correct units of meters so same stories
  • 00:22:51
    last one I'm not going to go through the
  • 00:22:53
    conversion again in great detail but
  • 00:22:55
    it's 318 nanometers which we have to
  • 00:22:58
    divide by 10 to the ninth to get it into
  • 00:23:00
    meters which is 3.18 times 10 to the
  • 00:23:02
    minus 7.
  • 00:23:04
    all right and so that's just going to be
  • 00:23:07
    the same equation and what we get is a
  • 00:23:09
    value of 6.25 times 10 to the minus 19.
  • 00:23:15
    joules all right again it's on that
  • 00:23:16
    order of 10 to the minus 19 if it was
  • 00:23:18
    positive like a or really negative like
  • 00:23:22
    C it's probably a mistake somewhere but
  • 00:23:24
    the one that matches would be Choice d
  • 00:23:27
    and that would be the minimum energy
  • 00:23:29
    then for ejecting an electron which
  • 00:23:31
    corresponds to the maximum wavelength of
  • 00:23:34
    318 nanometers so it was set up a little
  • 00:23:36
    bit differently but it is the same
  • 00:23:37
    equation but it is still useful to
  • 00:23:39
    review the concept of photoelectric
  • 00:23:41
    effect because there could be conceptual
  • 00:23:43
    questions related to that as well
  • 00:23:45
    all right do we have any questions
  • 00:23:46
    coming in
  • 00:23:48
    so far no
  • 00:23:50
    so I will transition into number five
  • 00:23:53
    then keep us moving along
  • 00:23:57
    all right so number five is
  • 00:23:59
    uh we're dealing with the Bohr model now
  • 00:24:01
    so
  • 00:24:02
    calculate the frequency of a photon
  • 00:24:04
    needed to promote an electron from
  • 00:24:07
    hydrogen's ground state into the third
  • 00:24:08
    shell
  • 00:24:10
    report the logarithm of your answer to
  • 00:24:12
    two decimal places I don't know if we
  • 00:24:13
    have questions set up like this anymore
  • 00:24:15
    we this is
  • 00:24:17
    sort of old school format but basically
  • 00:24:18
    in some in some forms of the exam
  • 00:24:21
    questions which I don't know if they
  • 00:24:23
    actually exist on the current test bank
  • 00:24:25
    but we would ask you to report the log
  • 00:24:27
    just as a you know easier way of
  • 00:24:29
    reporting a answer that has a large or
  • 00:24:32
    small exponent exponential to it so
  • 00:24:35
    don't get too bogged down in that but if
  • 00:24:36
    it asks you to do it you have to learn
  • 00:24:37
    how to take the log on your calculator
  • 00:24:38
    so make sure you know how to do that
  • 00:24:41
    um but the important thing is to know
  • 00:24:43
    how to actually do the problem okay so
  • 00:24:45
    for this problem here anytime you see
  • 00:24:47
    terminology dealing with you know
  • 00:24:49
    hydrogen promoting electrons
  • 00:24:52
    different shells ground state all that
  • 00:24:54
    stuff you should immediately Zone in on
  • 00:24:56
    the Bohr equation
  • 00:24:58
    and this is not given to you on the
  • 00:25:01
    periodic table this constant here but
  • 00:25:02
    this is the more in my opinion the more
  • 00:25:04
    useful form of the equation there is one
  • 00:25:06
    that's a little bit different that some
  • 00:25:07
    of you might be familiar with called
  • 00:25:08
    ridberg equation but I always do this
  • 00:25:11
    one because this one is useful in any
  • 00:25:12
    situation and then we're going to have
  • 00:25:15
    that constant out front and then it's
  • 00:25:17
    going to be 1 over NF squared where NF
  • 00:25:19
    is the final value of n during the
  • 00:25:22
    electronic excitation or relaxation and
  • 00:25:25
    then ni both of those squared
  • 00:25:28
    where that's the initial value of n that
  • 00:25:30
    it starts at so that's what we're going
  • 00:25:32
    to use First in these types of problems
  • 00:25:34
    so when you're doing a problem that
  • 00:25:35
    deals with a Bohr model you're typically
  • 00:25:38
    going to start with this equation
  • 00:25:39
    occasionally you'll have to work
  • 00:25:40
    backwards in these but the more common
  • 00:25:42
    situation is the one like here where you
  • 00:25:45
    want to get the frequency of a photon
  • 00:25:47
    associated with this event promoting an
  • 00:25:51
    electron from the ground state to the
  • 00:25:52
    third shell which is an absorption event
  • 00:25:54
    but we're going to typically start with
  • 00:25:56
    this equation first and then work our
  • 00:25:57
    way to what the question is asking for
  • 00:25:59
    so we're going to calculate Delta e
  • 00:26:01
    initially
  • 00:26:02
    and we need the two values of n to do
  • 00:26:04
    that
  • 00:26:08
    all right so the two values of n the
  • 00:26:09
    final value of n third shell that tells
  • 00:26:12
    us that an f equals three
  • 00:26:15
    but then we didn't give you directly the
  • 00:26:17
    value of the initial n value but anytime
  • 00:26:19
    you see ground state
  • 00:26:21
    hydrogen's ground state that tells you
  • 00:26:23
    that you're talking about n equals one
  • 00:26:24
    so our initial value of n where it
  • 00:26:26
    starts is n equals one which is the
  • 00:26:28
    ground state so those are the two n
  • 00:26:30
    values we're going to put in
  • 00:26:32
    so one over three squared 1 over 1
  • 00:26:34
    squared the two values of n
  • 00:26:37
    and so what we get is a Delta E A change
  • 00:26:39
    in energy for that electron of 1.94
  • 00:26:42
    times 10 to the minus 18 joules
  • 00:26:45
    and it makes sense that it's positive
  • 00:26:47
    because we're going from a lower energy
  • 00:26:49
    to a higher energy orbit or shell
  • 00:26:52
    but that's not where we want to stop
  • 00:26:54
    here because we don't want the Delta e
  • 00:26:56
    or we don't even want we don't want the
  • 00:26:57
    energy of the photon we want the
  • 00:26:58
    frequency of the photon that causes this
  • 00:27:00
    to happen so then for that we have to
  • 00:27:02
    recognize that
  • 00:27:04
    the energy of the photon that would be
  • 00:27:06
    involved in this is the absolute value
  • 00:27:08
    of delta e
  • 00:27:10
    whether it's absorption or emission so
  • 00:27:12
    if you get the wrong sign for Delta e
  • 00:27:13
    it's okay as long as if you're just
  • 00:27:15
    trying to find the energy or the
  • 00:27:16
    wavelength or the frequency of the
  • 00:27:18
    photon but we get and that's going to be
  • 00:27:20
    equal then to
  • 00:27:22
    H times Nu or HC or Lambda so once we do
  • 00:27:25
    this first part of a Bohr model problem
  • 00:27:28
    of calculating Delta e using this
  • 00:27:30
    equation typically or not not always but
  • 00:27:33
    but in a lot of cases after that you're
  • 00:27:35
    going to then use one of the equations
  • 00:27:38
    for Photon energy to be able to relate
  • 00:27:40
    either that to whether that's either the
  • 00:27:42
    frequency or the wavelength that's
  • 00:27:44
    exactly what we're doing here so we want
  • 00:27:45
    to calculate frequency so we're going to
  • 00:27:47
    use H Nu and we're solving for you which
  • 00:27:49
    is the frequencies that's going to be
  • 00:27:51
    the energy of the photon
  • 00:27:53
    divided by H which is Planck's constant
  • 00:27:56
    so the energy of the photon is the
  • 00:27:58
    absolute value of delta e so it's just
  • 00:27:59
    that number that's the energy of the
  • 00:28:01
    photon that would cause the electron to
  • 00:28:03
    be promoted between those two levels so
  • 00:28:06
    it's 1.94 times 10 to the minus 18
  • 00:28:08
    joules and then we're going to divide
  • 00:28:10
    that by H which is playing constant
  • 00:28:13
    6.626
  • 00:28:15
    times 10 to the minus 34
  • 00:28:19
    dual seconds
  • 00:28:21
    and what we get is two point keeping at
  • 00:28:24
    least one extra significant figure for
  • 00:28:25
    now
  • 00:28:26
    times 10 to the 15th
  • 00:28:29
    2.928 times 10 to the 15th Hertz
  • 00:28:33
    um
  • 00:28:34
    in the the current format of the exam
  • 00:28:36
    the most common way for you to have to
  • 00:28:38
    enter that would be
  • 00:28:41
    2.928 E15
  • 00:28:44
    and the setup that I have here which may
  • 00:28:47
    show up in some of the questions but I
  • 00:28:49
    did not look through it closely to see
  • 00:28:50
    if that's the case so this would be one
  • 00:28:52
    way to answer enter the answer but here
  • 00:28:54
    we're instructed to take the log so if
  • 00:28:56
    you're asked to do that follow the
  • 00:28:57
    directions and you take the log of 2.928
  • 00:29:01
    times 10 to the 15th
  • 00:29:03
    so make sure you know the log function
  • 00:29:05
    on your calculator in case that comes up
  • 00:29:06
    and it will later in the course and
  • 00:29:08
    that's 15.47
  • 00:29:11
    so
  • 00:29:12
    pay close attention to answer format
  • 00:29:15
    um
  • 00:29:16
    I mean if you miss a problem because of
  • 00:29:19
    some stupid formatting thing and it's
  • 00:29:21
    obvious to me that you did it correctly
  • 00:29:22
    I can you know I can manually correct
  • 00:29:24
    that when appropriate but it's easier
  • 00:29:27
    for you easier for me less stressful for
  • 00:29:29
    all of us if you're able to just follow
  • 00:29:32
    the directions the first time and put
  • 00:29:33
    the answer in the format we're looking
  • 00:29:34
    for
  • 00:29:35
    um but anyway whatever the answer format
  • 00:29:37
    is this is how you would approach the
  • 00:29:39
    problem and the key thing with Bohr
  • 00:29:41
    model problems is is to know what all
  • 00:29:43
    the numbers mean that you calculate and
  • 00:29:45
    and not to either stop too early or use
  • 00:29:48
    the wrong equation in the Second Step so
  • 00:29:50
    it's often multiple steps and you know
  • 00:29:53
    knowing what to do with the numbers and
  • 00:29:55
    knowing which steps are required for
  • 00:29:56
    that particular problem is going to be
  • 00:29:58
    key so read the question closely figure
  • 00:30:00
    out what exactly is looking for and
  • 00:30:01
    that's going to guide you to what you
  • 00:30:03
    need to do okay do we have any questions
  • 00:30:06
    on on that problem
  • 00:30:09
    well we actually have a lot of people
  • 00:30:10
    here today it's like close to 40 I think
  • 00:30:13
    um we usually get that many for the
  • 00:30:15
    in-person reviews so
  • 00:30:17
    maybe I'm onto something but I don't
  • 00:30:19
    particularly like teams this is probably
  • 00:30:20
    the only one we're going to do this way
  • 00:30:23
    um all right on to number six Still
  • 00:30:25
    Still chapter two we're talking about
  • 00:30:28
    electron configurations
  • 00:30:29
    all right so chapter six I'm sorry
  • 00:30:31
    number six which is from chapter two
  • 00:30:33
    what is the ground state electron
  • 00:30:35
    configuration for manganese
  • 00:30:38
    so this is going to be using the
  • 00:30:40
    periodic table to figure out
  • 00:30:43
    um
  • 00:30:45
    you know the the correct number of
  • 00:30:47
    electrons and also how they're arranged
  • 00:30:50
    so the first thing is when we're doing
  • 00:30:52
    problems like these there will
  • 00:30:53
    occasionally be times when we have to be
  • 00:30:55
    able to
  • 00:30:56
    connect the name of an element to its
  • 00:30:58
    symbol that is something that you're
  • 00:31:00
    expected to be able to do for the first
  • 00:31:02
    36 elements on the periodic table
  • 00:31:05
    um a lot of them are very common but
  • 00:31:06
    some of them not as much manganese is
  • 00:31:08
    probably not the most common one and
  • 00:31:11
    people get it mixed up with magnesium is
  • 00:31:13
    pretty similar so manganese which has
  • 00:31:15
    the letter n in it uh first before G is
  • 00:31:18
    going to be MN
  • 00:31:20
    so we have to recognize the symbols we
  • 00:31:22
    have to find this on the periodic table
  • 00:31:23
    we actually saw this one earlier in the
  • 00:31:25
    exam although GSR won't come up twice on
  • 00:31:28
    your real exam but anyways manganese is
  • 00:31:30
    MN and as we also saw earlier
  • 00:31:33
    you might have to zoom out because we
  • 00:31:34
    need to see more of the periodic table
  • 00:31:36
    for this question
  • 00:31:38
    um
  • 00:31:42
    there we go okay it's number 25 already
  • 00:31:44
    circled because we encountered it
  • 00:31:46
    earlier
  • 00:31:47
    all right but this question is dealing
  • 00:31:49
    with the electron configuration so when
  • 00:31:51
    you have a multiple choice question
  • 00:31:52
    dealing with electron configurations
  • 00:31:54
    which most of them are the first thing
  • 00:31:57
    to look at is does the answer Choice
  • 00:31:59
    even have the right number of electrons
  • 00:32:00
    in it we can usually or sometimes anyway
  • 00:32:02
    we can eliminate answer choices so if
  • 00:32:06
    manganese is atomic number 25 and it's
  • 00:32:08
    not charged in this case it's just
  • 00:32:09
    manganese
  • 00:32:11
    then the correct answer needs to have 25
  • 00:32:13
    electrons in it
  • 00:32:15
    um it turns out all these do so I didn't
  • 00:32:17
    help you out on this particular one but
  • 00:32:19
    a lot of times one of the one or more of
  • 00:32:21
    the instrument traces will have the raw
  • 00:32:22
    number of electrons so manganese is 25
  • 00:32:25
    the core electrons would be from argon
  • 00:32:27
    in all the given configurations so
  • 00:32:29
    that's 18 electrons there plus whatever
  • 00:32:30
    comes after that
  • 00:32:32
    [Music]
  • 00:32:33
    um
  • 00:32:34
    so all four of these have 25 electrons
  • 00:32:36
    in the configuration 18 from argon plus
  • 00:32:39
    seven more that come after it so we
  • 00:32:41
    can't eliminate any of them on that
  • 00:32:43
    basis alone sometimes one or more of the
  • 00:32:45
    answer choices has completely the wrong
  • 00:32:47
    number of electrons in it so you know
  • 00:32:49
    it's wrong no matter which subshells and
  • 00:32:51
    things are populated but we can't get
  • 00:32:52
    rid of any of those answer choices here
  • 00:32:54
    so we actually have to go through and
  • 00:32:56
    figure out then where
  • 00:32:58
    you know what the actual configuration
  • 00:33:00
    is so argon is the noble gas that comes
  • 00:33:03
    before manganese on the periodic table
  • 00:33:05
    so we're identifying configurations we
  • 00:33:07
    typically start with the noble gas that
  • 00:33:10
    comes before the element MN is 25 the
  • 00:33:13
    noble gas before that is argon which is
  • 00:33:15
    18. so we're going to just write that in
  • 00:33:16
    practice now after that we get over to
  • 00:33:18
    here number 19. we're in the fourth row
  • 00:33:20
    of the periodic table one two three four
  • 00:33:22
    so that's going to be the 4S block 4s1
  • 00:33:25
    4s2
  • 00:33:26
    and then this whole block of the
  • 00:33:28
    periodic table remember is the d block
  • 00:33:31
    um sounds like something related to
  • 00:33:33
    prison but it's it's not as part of the
  • 00:33:34
    periodic table this is n minus one D so
  • 00:33:37
    after 4S we go to 3D and then we're
  • 00:33:39
    going to count over one two three four
  • 00:33:42
    five to take us to manganese so the last
  • 00:33:44
    electrons we put into the configuration
  • 00:33:46
    for MN are going to be 3d5 so the
  • 00:33:49
    correct configuration would be argon is
  • 00:33:51
    a core 4s2 next and then 3d5 and the
  • 00:33:56
    only one that has that is Choice a again
  • 00:33:59
    I guess I like Choice a on this exam
  • 00:34:02
    um
  • 00:34:03
    but nonetheless Choice B has the wrong
  • 00:34:05
    subshell for p Choice C has 4D instead
  • 00:34:09
    of 3D Choice D has 3s instead of 4S so
  • 00:34:12
    it's only this one here that's correct
  • 00:34:14
    and matches with Choice a
  • 00:34:17
    all right any questions on that problem
  • 00:34:18
    or electron configurations
  • 00:34:22
    yes absolutely why is it four instead of
  • 00:34:25
    three since it's on a four fourth row
  • 00:34:28
    um because for the the d block of the
  • 00:34:30
    periodic table it's n minus one the
  • 00:34:33
    first row of the D of the of the D Block
  • 00:34:37
    is actually 3D so we're in the fourth
  • 00:34:40
    row of the periodic table but it goes
  • 00:34:41
    from 4S to then 3D so it's 3D 4D 5D
  • 00:34:45
    um for for the S and P blocks it is
  • 00:34:47
    exactly the row that you're in so this
  • 00:34:49
    is you know 1s 2s 3s4s and so on this is
  • 00:34:54
    2p 3p 4p but for the middle part it's
  • 00:34:57
    going to be n minus one so it'll be 3D
  • 00:34:59
    405d and down here even though we don't
  • 00:35:01
    use those as much that's going to be n
  • 00:35:03
    minus 2 so this would be
  • 00:35:05
    four after the 5f okay so it is a little
  • 00:35:09
    bit different in those parts of the
  • 00:35:10
    periodic table that's just the order
  • 00:35:11
    that those subshells come in and you
  • 00:35:13
    don't want to skip 3D 3D is the first d
  • 00:35:16
    sub shell and so that's why it's
  • 00:35:18
    um that part of the periodic table
  • 00:35:21
    okay thank you yep any other questions
  • 00:35:26
    all right back to it
  • 00:35:29
    number seven
  • 00:35:34
    all right this question deals with
  • 00:35:38
    um the number of orbitals the
  • 00:35:40
    relationships involving quantum numbers
  • 00:35:42
    uh how many orbitals are there in the
  • 00:35:44
    sub level where n equals five and L
  • 00:35:48
    equals four so for problems like this
  • 00:35:50
    you have to distinguish whether we're
  • 00:35:53
    talking about
  • 00:35:54
    a shell or a subshell or a level or a
  • 00:35:58
    sub levels are sometimes called and you
  • 00:36:00
    need to distinguish are we talking about
  • 00:36:02
    number of orbitals or number of
  • 00:36:03
    electrons and just you know be very
  • 00:36:06
    clear about what the question is asking
  • 00:36:07
    so here we're looking for how many
  • 00:36:10
    orbitals and we're telling you it's a
  • 00:36:12
    sub level and even if we didn't tell you
  • 00:36:13
    it was a sub level we're giving you an n
  • 00:36:16
    and an L value so we give you both of
  • 00:36:18
    those n and L together that defines a
  • 00:36:20
    subshell
  • 00:36:21
    all right so let's let's go over all the
  • 00:36:23
    different relationships then that are
  • 00:36:24
    going to be related to problems like
  • 00:36:25
    this so if you're in a shell or a
  • 00:36:29
    sometimes it's called a level but
  • 00:36:30
    usually a shell
  • 00:36:31
    a shell would only be defined by a value
  • 00:36:33
    of n
  • 00:36:35
    and in that case the number of orbitals
  • 00:36:38
    is just equal to N squared
  • 00:36:44
    so in this problem if we said how many
  • 00:36:46
    orbitals are in the fifth shell or in
  • 00:36:49
    the Shell where n equals five and that's
  • 00:36:51
    all we gave you you would use N squared
  • 00:36:52
    5 squared but here we're giving you n
  • 00:36:55
    and L defining a subshell or a sub level
  • 00:36:57
    so for defining a subshell or a sub
  • 00:36:59
    level
  • 00:37:04
    that's where we're giving you n and L
  • 00:37:06
    together or sometimes we just give you L
  • 00:37:08
    but don't specify n but if you have the
  • 00:37:10
    value of L you know that you're talking
  • 00:37:12
    about a subshell of some type in that
  • 00:37:15
    case the number of orbitals
  • 00:37:17
    which
  • 00:37:19
    for some sub shells you can get that
  • 00:37:20
    from the periodic table also by counting
  • 00:37:22
    how many columns there are but in
  • 00:37:24
    general it's given as 2L plus one which
  • 00:37:26
    is what we'll want to use here
  • 00:37:28
    so that's going to be what we're going
  • 00:37:30
    to use if we look for the number of
  • 00:37:32
    orbitals in a subshell so with n equals
  • 00:37:35
    five there would be 25 orbitals but then
  • 00:37:37
    how many of those are in the L equals 4
  • 00:37:39
    sublevel that's going to be using 2L
  • 00:37:41
    plus one so to answer this question here
  • 00:37:44
    it's going to be that the number of
  • 00:37:45
    orbitals
  • 00:37:48
    foreign
  • 00:37:52
    plus one two times four plus one is nine
  • 00:37:55
    so that'd be Choice D here
  • 00:37:58
    um and again it's because we gave you n
  • 00:38:01
    and L together or anytime you have L
  • 00:38:03
    that's the relationship you're going to
  • 00:38:04
    want to use for the number of orbitals
  • 00:38:06
    now there is some related relationships
  • 00:38:08
    that don't come up here but just to make
  • 00:38:10
    sure we're complete and covering them
  • 00:38:12
    in a given shell
  • 00:38:14
    the maximum number of electrons
  • 00:38:17
    is now 2N squared because there's up to
  • 00:38:20
    two electrons per orbital so N squared
  • 00:38:22
    is number of orbitals two n squares the
  • 00:38:24
    number of electrons possible and then in
  • 00:38:27
    a subshell you just double the 2L plus
  • 00:38:29
    one to figure out the maximum number of
  • 00:38:30
    electrons
  • 00:38:32
    again this doesn't tell you how many
  • 00:38:33
    electrons will be in that subshell it
  • 00:38:35
    tells you how many electrons can be and
  • 00:38:37
    it's two times two out plus one because
  • 00:38:39
    there's up to two electrons per orbital
  • 00:38:42
    so those relationships are helpful
  • 00:38:43
    because you you know we could also
  • 00:38:46
    go through the exercise of figuring out
  • 00:38:48
    all the different quantum number pop
  • 00:38:50
    combinations but that is usually more
  • 00:38:53
    tedious and more prone to mistakes than
  • 00:38:55
    just using these relatively simple
  • 00:38:57
    relationships but we have to remember
  • 00:38:58
    them we have to know when they're
  • 00:39:00
    appropriate to use all right any
  • 00:39:02
    questions on on that problem or quantum
  • 00:39:04
    numbers
  • 00:39:06
    foreign
  • 00:39:13
    so this is another one with a two-part
  • 00:39:15
    answer not necessarily because you would
  • 00:39:17
    have
  • 00:39:19
    um
  • 00:39:19
    questions that are always two parts on
  • 00:39:21
    the test but just because I so I can
  • 00:39:23
    review more of the concepts I suppose so
  • 00:39:25
    anyway in this question what is the
  • 00:39:27
    maximum number of electrons allowed in
  • 00:39:28
    the N equals five shell and how many of
  • 00:39:31
    these can reside in a 5f subshell so
  • 00:39:34
    this this sort of you know
  • 00:39:36
    he uses the same relationships as the
  • 00:39:38
    last one but now we're talking about
  • 00:39:40
    maximum number of electrons not number
  • 00:39:42
    of orbitals
  • 00:39:43
    um
  • 00:39:44
    so
  • 00:39:46
    we're going to use the 2N squared or for
  • 00:39:49
    this one because we're looking for
  • 00:39:51
    first how many are in the N equals five
  • 00:39:53
    shell so in a shell
  • 00:39:57
    the number of electrons maximum is equal
  • 00:40:00
    to 2N squared
  • 00:40:02
    and so that's going to be 2 times 5
  • 00:40:05
    squared which is 50. so there's 50
  • 00:40:08
    electrons possible maximum in the fifth
  • 00:40:11
    shell and then how many of those are in
  • 00:40:14
    the 5f subshell
  • 00:40:16
    now the other thing that this requires
  • 00:40:17
    us to review is these letter
  • 00:40:19
    designations for subshells
  • 00:40:22
    um so we see this a lot where we don't
  • 00:40:24
    use a number for L we use a
  • 00:40:26
    corresponding letter so L can be 0 1 2 3
  • 00:40:30
    4 and so on
  • 00:40:34
    and those correspond to the letters s p
  • 00:40:38
    d f and then pretty much alphabetical
  • 00:40:40
    after that although we typically only
  • 00:40:42
    use those first four spdnf are the most
  • 00:40:45
    common ones because those occur in real
  • 00:40:47
    atoms so here we're talking about a 5f
  • 00:40:50
    subshell so F tells us that L equals
  • 00:40:53
    three
  • 00:40:54
    and so that's the value of L we're going
  • 00:40:55
    to use where the number of electrons
  • 00:40:59
    is 2 times 2L plus 1 is the maximum
  • 00:41:01
    number of electrons again not not how
  • 00:41:03
    many there necessarily will be but
  • 00:41:04
    that's the maximum
  • 00:41:06
    l equals 3 so 2 times 3 plus 1 is 7.
  • 00:41:12
    and then double that is 14. all right so
  • 00:41:15
    50 electrons possible in n equals 5 14
  • 00:41:18
    of those can go into 4f and so that
  • 00:41:21
    matches answer Choice C here
  • 00:41:24
    um
  • 00:41:24
    and so that's again very similar to the
  • 00:41:28
    relationships we use on the last problem
  • 00:41:30
    um the differences here we're talking
  • 00:41:31
    what number of electrons and because we
  • 00:41:33
    can have up to two electrons per orbital
  • 00:41:35
    we double those those expressions to get
  • 00:41:39
    the number of electrons that are
  • 00:41:40
    possible
  • 00:41:41
    all right similar topics so any any
  • 00:41:43
    questions on either of those
  • 00:41:46
    uh so the the five measures that number
  • 00:41:51
    right for any map
  • 00:41:53
    well we used it in sort of the first
  • 00:41:54
    part of the question because it was
  • 00:41:57
    the first part dealt with how many were
  • 00:41:59
    total in the N equals five shell but for
  • 00:42:01
    five F that 5 so represents the value of
  • 00:42:03
    n
  • 00:42:04
    but when we're finding the number for
  • 00:42:06
    electrons that are in the subshell we
  • 00:42:07
    don't need the value of n we just need
  • 00:42:09
    the value of L
  • 00:42:11
    okay
  • 00:42:13
    anything else
  • 00:42:16
    okay
  • 00:42:18
    moving on
  • 00:42:21
    because this Friday
  • 00:42:23
    all right another quantum number
  • 00:42:24
    question so you're getting the idea
  • 00:42:26
    um hopefully from this exam review that
  • 00:42:30
    quantum numbers figure prominently I
  • 00:42:33
    don't I don't write the exams I helped
  • 00:42:36
    kind of put them together over the years
  • 00:42:37
    but I don't really
  • 00:42:38
    write that myself so whoever you know
  • 00:42:41
    put this exam together was personally
  • 00:42:43
    responsible likes quantum numbers
  • 00:42:44
    because there are there's like three
  • 00:42:45
    questions on that so be ready
  • 00:42:48
    um so I just want to ask for which of
  • 00:42:49
    the following is a valid set of quantum
  • 00:42:51
    numbers for a 4D orbital
  • 00:42:54
    all right so here we have to again
  • 00:42:57
    recognize now we're talking about three
  • 00:42:59
    different quantum numbers because an
  • 00:43:01
    individual orbital also has a third
  • 00:43:03
    quantum number associated with the m sub
  • 00:43:05
    l
  • 00:43:06
    so let's quickly review the rules for
  • 00:43:08
    quantum numbers in the relationship so n
  • 00:43:10
    can be any positive integer
  • 00:43:13
    those are the possible values of n the
  • 00:43:15
    shell number
  • 00:43:17
    and then the values of L that you can
  • 00:43:19
    have are related to that
  • 00:43:22
    they start at zero
  • 00:43:24
    and go up to n minus one
  • 00:43:27
    so the biggest value of L is n minus one
  • 00:43:30
    and then once you know the value of l
  • 00:43:33
    once that specified the possible values
  • 00:43:35
    for M sub L go all the way from negative
  • 00:43:37
    L to plus L including zero
  • 00:43:42
    in increments of one so those are the
  • 00:43:45
    possible values where we introduce M sub
  • 00:43:47
    L here which we hadn't really dealt with
  • 00:43:48
    before so in this question here we're
  • 00:43:50
    talking about a 4D orbital and as came
  • 00:43:54
    up just on the the clarification that
  • 00:43:56
    was saw a little bit ago when you have
  • 00:43:58
    that as a designation for a subshell
  • 00:44:01
    which is a set of orbitals or for a
  • 00:44:03
    specific orbital in that subshell as we
  • 00:44:05
    do here remember that the first number
  • 00:44:08
    there is n
  • 00:44:09
    so in 4D n equals 4 and then the value
  • 00:44:13
    of L comes from the letter which can be
  • 00:44:16
    zero one two or three
  • 00:44:18
    4 n equals 4 and we have letters that
  • 00:44:21
    are SPD and F so we're talking about L
  • 00:44:23
    equals two all right so what we have is
  • 00:44:26
    n equals 4 and L equals two and now we
  • 00:44:30
    have to figure out
  • 00:44:32
    so that's going to be the first two
  • 00:44:33
    numbers so with that we can eliminate
  • 00:44:36
    Choice C because it has the wrong L
  • 00:44:38
    value
  • 00:44:40
    and choice e which has the wrong L value
  • 00:44:42
    so both of those we can eliminate
  • 00:44:44
    because they have the wrong L value well
  • 00:44:46
    so we're looking for four and two is the
  • 00:44:48
    first two but now we have to figure out
  • 00:44:49
    which of these remaining three answer
  • 00:44:51
    choices has a valid third quantum number
  • 00:44:54
    M sub l so M sub L we have not talked
  • 00:44:56
    about yet it doesn't have as much
  • 00:44:58
    physical significance of those related
  • 00:45:00
    to the orientation of an orbital but the
  • 00:45:03
    key is that is minus L to plus l so for
  • 00:45:05
    an L equals two subshell the allowed
  • 00:45:08
    values of M sub L go from negative two
  • 00:45:10
    to positive two so negative two negative
  • 00:45:12
    one zero one and two are all the
  • 00:45:16
    possible values of M sub L it'd be five
  • 00:45:18
    orbitals each with a different M sub L
  • 00:45:20
    value so these are the ones that are
  • 00:45:22
    possible for M sub L has to be between
  • 00:45:23
    negative two and plus two and so what we
  • 00:45:26
    see is that choice B has three which is
  • 00:45:28
    too big
  • 00:45:29
    Choice D has 1.5 those values are never
  • 00:45:33
    half integer values they're always
  • 00:45:34
    they're always whole numbers and so the
  • 00:45:37
    only one that has a valid set would be
  • 00:45:40
    Choice a n equals four L equals two
  • 00:45:42
    which we specified from it being a four
  • 00:45:44
    D orbital and then the values of M sub L
  • 00:45:47
    can be anywhere between negative two and
  • 00:45:48
    plus two so this one works out that
  • 00:45:50
    would designate one of those four D
  • 00:45:52
    orbitals out of the five that are in
  • 00:45:54
    that subshell
  • 00:45:56
    all right so do we have any questions
  • 00:45:58
    then on that one which also dealt with
  • 00:46:00
    quantum numbers
  • 00:46:04
    what do you mean that you can only go up
  • 00:46:06
    to n minus one
  • 00:46:08
    so that's going to be sort of the limit
  • 00:46:10
    of L and so if you have
  • 00:46:13
    so here we're in the fourth shell n
  • 00:46:15
    equals four that's the number there and
  • 00:46:18
    that means the possible values for l in
  • 00:46:20
    a n equals four shell would be zero one
  • 00:46:23
    two or three n minus 1 which is four or
  • 00:46:26
    sorry which is three in this case is the
  • 00:46:28
    biggest value of L that you can have so
  • 00:46:30
    that means in an N equals four shell you
  • 00:46:33
    would have 4S 4p 4D and four F sub
  • 00:46:37
    shells in this case we're specifying it
  • 00:46:39
    as the 4D subshells that refers to L
  • 00:46:42
    equals two so L equals three is possible
  • 00:46:44
    but because we're talking about a d
  • 00:46:46
    subshell here that's the value of L that
  • 00:46:47
    we're specifying so it's all the
  • 00:46:49
    different possible values of L are
  • 00:46:51
    bounded by n minus one starting at zero
  • 00:46:54
    bounded by n minus one so depending on
  • 00:46:56
    what n is so all these quantum numbers
  • 00:46:58
    are related to each other in kind of
  • 00:47:00
    this you know top to bottom hierarchy
  • 00:47:02
    where the value of n dictates which
  • 00:47:04
    values of L are possible the value of L
  • 00:47:07
    did case which possible values of M sub
  • 00:47:09
    value have
  • 00:47:10
    um hopefully that clarifies that a
  • 00:47:12
    little bit
  • 00:47:13
    okay one more question sure
  • 00:47:16
    so we had a four of f it would be M's
  • 00:47:22
    level would be negative three to
  • 00:47:24
    positive three yeah so if you had four F
  • 00:47:26
    the only difference would be that you're
  • 00:47:28
    specifying in that case an L value of
  • 00:47:30
    three
  • 00:47:32
    and with L equals three then the answer
  • 00:47:34
    value has now become negative three to
  • 00:47:36
    plus three so it would also have more
  • 00:47:38
    about possible values of M so well as we
  • 00:47:41
    talked about on previous questions
  • 00:47:42
    that's because there are two L plus one
  • 00:47:43
    orbitals and so the each orbital has a
  • 00:47:46
    unique M sub L value so the higher the
  • 00:47:49
    value of L the more M sub L values are
  • 00:47:51
    possible as well
  • 00:47:53
    okay thank you
  • 00:47:55
    what would it be if it were uh 4S or S
  • 00:48:00
    would be
  • 00:48:03
    l equals zero
  • 00:48:06
    so n equal four because that's the
  • 00:48:09
    number tells you l equals zero for S
  • 00:48:11
    orbitals
  • 00:48:12
    and then when you have l equals zero the
  • 00:48:14
    only M sub L value is also zero there's
  • 00:48:16
    a single orbital in that subshell and it
  • 00:48:19
    has an M sub L equals zero plus out a
  • 00:48:21
    minus L well there's no such thing as
  • 00:48:23
    negative zero or plus zero which is zero
  • 00:48:25
    so that's the only value you can have
  • 00:48:27
    that one
  • 00:48:29
    okay thank you yep
  • 00:48:31
    anything else
  • 00:48:34
    all right doing okay on time
  • 00:48:36
    um but we can still take more questions
  • 00:48:38
    whenever they come up
  • 00:48:39
    on to number 10
  • 00:48:42
    um
  • 00:48:43
    all right so this one asked for which of
  • 00:48:45
    the following orbital diagrams
  • 00:48:48
    reflects an atom in the ground state
  • 00:48:52
    um so ground state means the lowest
  • 00:48:54
    energy configuration for an atom and we
  • 00:48:57
    use the periodic table to predict where
  • 00:49:00
    the electrons go but this also then
  • 00:49:02
    takes the additional step of looking at
  • 00:49:04
    the relative spins of those electrons
  • 00:49:06
    and which individual orbitals are
  • 00:49:08
    occupied okay so for these types of
  • 00:49:10
    problems when we're dealing with the
  • 00:49:12
    ground state of a multi-electron atom
  • 00:49:14
    there's three rules that we have to make
  • 00:49:16
    sure are followed
  • 00:49:19
    so the first one which is what the
  • 00:49:21
    periodic table kind of allows us to
  • 00:49:23
    follow is called the offbao principle
  • 00:49:25
    that's where we fill
  • 00:49:27
    lowest to highest subshells in terms of
  • 00:49:29
    energies and so if you go in the order
  • 00:49:31
    of the periodic table which starts at 1s
  • 00:49:33
    and then goes you know 2s to 2p to 3S to
  • 00:49:37
    3p to 4S to 3D to 4p and so on when you
  • 00:49:41
    go in that order through the um to the
  • 00:49:44
    periodic table that's allowing you to
  • 00:49:45
    follow the off file principle filling
  • 00:49:47
    the subshells from lowest to highest
  • 00:49:49
    energy
  • 00:49:50
    um and that's called offbao now the
  • 00:49:52
    other ones that are important are
  • 00:49:56
    poly exclusion
  • 00:50:02
    which tells us that no two electrons
  • 00:50:06
    have the same quantum numbers
  • 00:50:09
    now in terms of how that manifests
  • 00:50:11
    itself in a diagram like this it means
  • 00:50:14
    that an orbital can have no more than
  • 00:50:16
    two electrons in it and when two
  • 00:50:19
    electrons are in the same orbital
  • 00:50:22
    they need to be one spin up and one spin
  • 00:50:24
    down they have to have different M's the
  • 00:50:26
    best quantum numbers for that to work so
  • 00:50:28
    that's the sort of manifestation of Paul
  • 00:50:30
    exclusion and the third one which we
  • 00:50:32
    have to consider in some of these as
  • 00:50:33
    well is Hun's rule
  • 00:50:36
    um which is the one where you maximize
  • 00:50:38
    the number of unpaired electrons you
  • 00:50:40
    singlely if you have degenerate orbitals
  • 00:50:42
    in particular you singly fill them first
  • 00:50:45
    before you go through and start pairing
  • 00:50:47
    up electrons and that means you want as
  • 00:50:49
    many unpaired electrons as possible
  • 00:50:52
    so we have to think about all of these
  • 00:50:53
    for
  • 00:50:55
    this problem here so we go to Choice a
  • 00:50:57
    here
  • 00:50:59
    um if it follows the off file principle
  • 00:51:01
    because 1s is filled 2s is filled and
  • 00:51:04
    then we have the last electrons in 2p so
  • 00:51:06
    that's the correct order
  • 00:51:07
    um one S2s and then 2p
  • 00:51:10
    um but and then poly exclusion is
  • 00:51:12
    followed none of the orbitals have more
  • 00:51:13
    than two electrons so we're good there
  • 00:51:15
    but if you look at Hun's rule here we
  • 00:51:18
    see that our first electron is spin up
  • 00:51:20
    which means the second one should also
  • 00:51:22
    be or at the very least you have to be
  • 00:51:23
    the same they either both be down or
  • 00:51:25
    both be up to follow Hun's Rule and to
  • 00:51:27
    maximize this so so this one here we can
  • 00:51:30
    eliminate because it violates Hun's rule
  • 00:51:32
    does not maximize the number of unpaired
  • 00:51:34
    electrons
  • 00:51:35
    it singly occupies the orbitals which is
  • 00:51:38
    correct but it also has you also have to
  • 00:51:40
    arrange the spins to be the same as each
  • 00:51:42
    other being both up or both down when
  • 00:51:44
    you do it like this so this one violates
  • 00:51:46
    Hun's rules
  • 00:51:47
    because those two electrons are
  • 00:51:48
    opposites fin
  • 00:51:50
    if we go to answer Choice B here what we
  • 00:51:54
    notice is that we have an empty spot
  • 00:51:56
    here in 2s so we fill one s but then we
  • 00:51:59
    don't completely fill 2s but we're
  • 00:52:01
    starting to put electrons in 2p already
  • 00:52:02
    so this one violates the optile
  • 00:52:05
    principle you have to completely fill
  • 00:52:06
    one subshell before you go through and
  • 00:52:10
    put electrons into higher energy
  • 00:52:12
    subshells so we can't start putting
  • 00:52:14
    electrons into 2p unless 2s is already
  • 00:52:16
    filled so this one violates that first
  • 00:52:19
    one called the offbought principle
  • 00:52:21
    if you look at Choice D here sorry sorry
  • 00:52:24
    C we're not quite to D yet we have
  • 00:52:27
    Alpha principles followed 1s filled 2s
  • 00:52:30
    filled and then we start putting
  • 00:52:31
    electrons into p
  • 00:52:33
    no violation of all the exclusion we
  • 00:52:35
    have two electrons at most per orbital
  • 00:52:37
    and when they when we do there's their
  • 00:52:39
    opposite spins and then we see that the
  • 00:52:41
    two electrons they're not next to each
  • 00:52:43
    other they don't have to be they just
  • 00:52:44
    have to be in two different orbitals
  • 00:52:46
    both with the same spin so this is
  • 00:52:48
    totally fine even though conventionally
  • 00:52:50
    we put them right next to each other
  • 00:52:51
    they can be in any combination of three
  • 00:52:53
    of two of those three orbitals with the
  • 00:52:55
    same Spin and that still follows the
  • 00:52:57
    rules so this one follows all the rules
  • 00:53:00
    so we believe that that's going to be
  • 00:53:01
    our answer Choice here
  • 00:53:03
    let's check that there's a problem with
  • 00:53:05
    d so D has
  • 00:53:07
    the correct ordering of the subshell is
  • 00:53:09
    One S then 2s and then we start adding
  • 00:53:11
    to 2p but the problem here is that we
  • 00:53:14
    put the second electron into the same
  • 00:53:16
    orbital which is another violation of
  • 00:53:19
    hund's rule so none of these violated
  • 00:53:21
    poly Exclusion Principle but they either
  • 00:53:23
    violated off file principle or hund's
  • 00:53:25
    rule for the ones that don't work
  • 00:53:28
    all right so we can eliminate a b and d
  • 00:53:31
    by violating one of those rules Choice C
  • 00:53:34
    is the one that follows all of them
  • 00:53:37
    okay so those are
  • 00:53:39
    um taking just a little bit more
  • 00:53:40
    detailed look at how the electrons
  • 00:53:42
    arrange themselves in orbitals which is
  • 00:53:44
    related to electron configuration but
  • 00:53:46
    takes the additional steps of specifying
  • 00:53:49
    the population of individual orbitals
  • 00:53:51
    and whether the electrons have up spin
  • 00:53:54
    or down spin
  • 00:53:55
    and so that equals plus one half or
  • 00:53:57
    minus one half
  • 00:53:58
    okay any questions on that one
  • 00:54:03
    um I have a question yep
  • 00:54:05
    if or not if would be be sorta an
  • 00:54:10
    example because on one of the last
  • 00:54:11
    homework I asked
  • 00:54:13
    um to show which one would be an excited
  • 00:54:16
    gra or an excited state using this would
  • 00:54:19
    be sort of be an example if we said that
  • 00:54:22
    one of those electrons moved from the 2s
  • 00:54:25
    to the 2p
  • 00:54:27
    yes B would be an excited state
  • 00:54:30
    um we're not going to ask you a lot
  • 00:54:31
    about excited States and multi-electron
  • 00:54:32
    atoms but yeah if you take if you take
  • 00:54:35
    one electron from a ground state
  • 00:54:36
    configuration and move it up to a higher
  • 00:54:39
    energy subshell that's how you would
  • 00:54:41
    generate an excited state so that's
  • 00:54:42
    exactly what B is because you can think
  • 00:54:44
    of it as taking starting with C which is
  • 00:54:46
    correct and then taking one electron and
  • 00:54:49
    moving it over
  • 00:54:51
    to the higher energy subshell and
  • 00:54:53
    flipping the spin but basically just
  • 00:54:54
    taking one electron and promoting it so
  • 00:54:56
    that would be an exciting State not a
  • 00:54:58
    ground state
  • 00:54:59
    um so that is correct observation
  • 00:55:02
    thank you
  • 00:55:04
    all right anything else
  • 00:55:08
    all right we've
  • 00:55:10
    crossed we've gotten to the top of the
  • 00:55:12
    hill so now let's do the second half
  • 00:55:14
    of this review
  • 00:55:17
    all right still dealing with
  • 00:55:19
    periodic table
  • 00:55:21
    um I can't remember if we've
  • 00:55:22
    transitioned into chapter three or not
  • 00:55:24
    but we're close
  • 00:55:26
    um I think we're not there yet
  • 00:55:27
    temperatures a lot of stuff
  • 00:55:29
    um yeah we're still not playing chapter
  • 00:55:31
    three yet but anyway
  • 00:55:32
    um this one deals with predicting the
  • 00:55:34
    charges of ions and the classifications
  • 00:55:36
    of an element so again multiple Concepts
  • 00:55:39
    in the same question mainly just for the
  • 00:55:40
    purpose of
  • 00:55:42
    making the review more useful so which
  • 00:55:44
    of the following show the correct
  • 00:55:45
    classification and expected monoatomic
  • 00:55:48
    ion charge for the given element
  • 00:55:51
    um and so classification would be you
  • 00:55:53
    know things like metal non-metal
  • 00:55:55
    halogen main group metal all those
  • 00:55:57
    different terms we learned that sort of
  • 00:56:00
    specify
  • 00:56:01
    the place on the periodic table
  • 00:56:04
    and then
  • 00:56:05
    um monoatomic ion charge that's where we
  • 00:56:07
    predict the charge also based on where
  • 00:56:09
    it's located in the periodic table so
  • 00:56:11
    let's go through the three answer
  • 00:56:13
    choices in our you know multiple choice
  • 00:56:15
    question is possibly more than one of
  • 00:56:17
    those as being correct so for number one
  • 00:56:20
    which is potassium
  • 00:56:22
    we have to recognize that potassium is
  • 00:56:25
    the atomic symbol K
  • 00:56:26
    one of the least obvious ones but it is
  • 00:56:29
    in those first 36. so now we have to
  • 00:56:31
    find that on the periodic table and and
  • 00:56:33
    specify
  • 00:56:34
    which classification is in and what the
  • 00:56:37
    charge of that would be if it Formed an
  • 00:56:39
    ion so when we find K on the periodic
  • 00:56:42
    table it's number 19 over here on the
  • 00:56:44
    left
  • 00:56:45
    um and so the classification that's
  • 00:56:48
    given in the answer choices main group
  • 00:56:50
    metal and that would be valid
  • 00:56:52
    classification for this one because
  • 00:56:53
    everything on the left side of the
  • 00:56:55
    periodic table is a metal those in the
  • 00:56:57
    first two columns in particular would be
  • 00:56:59
    classified as main group metals and they
  • 00:57:01
    have more specific names as well but in
  • 00:57:03
    general they're called main group Metals
  • 00:57:04
    when they're on the S or the P block so
  • 00:57:07
    you know those two columns first and
  • 00:57:09
    second as well as some of the metals
  • 00:57:11
    that are you know over here
  • 00:57:13
    on the p-block like aluminum and gallium
  • 00:57:15
    and lead and bismuth and things like
  • 00:57:18
    that some of those are metalloids but
  • 00:57:20
    all those things that are in the p-block
  • 00:57:22
    that are metals are also called main
  • 00:57:24
    group Metals so that's a fine
  • 00:57:26
    classification but then when we predict
  • 00:57:28
    the ionic charge for potassium it's
  • 00:57:30
    going to want to lose enough electrons
  • 00:57:32
    to get to the noble gas configuration
  • 00:57:34
    since it's in the First Column it only
  • 00:57:36
    has to lose one electron to achieve
  • 00:57:38
    argon configuration so we would expect
  • 00:57:41
    potassium and everything else in the
  • 00:57:43
    First Column to form a plus one charge
  • 00:57:46
    but the answer Choice was given to us as
  • 00:57:48
    plus two so that one is incorrect
  • 00:57:51
    [Music]
  • 00:57:51
    um
  • 00:57:53
    because the charge is given wrong
  • 00:57:56
    um so we don't want number one as part
  • 00:57:58
    of answer choices but now we have to
  • 00:58:00
    look at the next two so just write out
  • 00:58:03
    that so potassium is a main group metal
  • 00:58:06
    more specifically it's an alkaline metal
  • 00:58:08
    but that General classification is fine
  • 00:58:10
    but we will predict the charge to be
  • 00:58:12
    plus one which was not correctly given
  • 00:58:14
    in the answer of choice number two which
  • 00:58:16
    is aluminum the symbol is Al so this one
  • 00:58:20
    is not the most obvious from its
  • 00:58:23
    location of the periodic table we do
  • 00:58:24
    want to be familiar with it so a l is
  • 00:58:26
    number 13 here
  • 00:58:28
    you know so the first thing is it a you
  • 00:58:30
    know a metal or a non-metal or or what
  • 00:58:32
    is this classification the staircase
  • 00:58:34
    goes right here
  • 00:58:36
    um
  • 00:58:37
    and even though it's close to the
  • 00:58:38
    staircase it is to the left and it is
  • 00:58:40
    classified as a metal and I don't think
  • 00:58:42
    that's surprising to anybody that's you
  • 00:58:44
    know dealt with aluminum foil it's a
  • 00:58:46
    silvery shiny
  • 00:58:49
    flexible substance conduct electricity
  • 00:58:51
    all those properties we associate with
  • 00:58:52
    Metals which will formalize more towards
  • 00:58:55
    the end of the course all associated
  • 00:58:56
    with aluminum so clearly aluminum is a
  • 00:58:58
    metal even though it's positioned on the
  • 00:59:00
    periodic table is kind of on the
  • 00:59:01
    borderline so it is a main group metal
  • 00:59:03
    also because it's in the P block
  • 00:59:04
    anything that's in the s or the P block
  • 00:59:06
    that's a metal would be called main
  • 00:59:08
    group and then to predict the charge of
  • 00:59:10
    aluminum this one is not again the most
  • 00:59:12
    obvious but there's one you're expecting
  • 00:59:14
    to know the valence configuration for
  • 00:59:16
    aluminum is 3s2 3p1 and so if it's able
  • 00:59:20
    to form a noble gas configuration it
  • 00:59:22
    does that by losing three electrons it
  • 00:59:24
    loses its three p electron both of the
  • 00:59:27
    3s electrons and that gives it then the
  • 00:59:29
    neon configuration so aluminum is number
  • 00:59:31
    13. it's a metal so it tends to lose
  • 00:59:33
    electrons by losing three it gets to
  • 00:59:36
    that noble gas so the charge for an
  • 00:59:38
    aluminum ion in an ionic compound would
  • 00:59:40
    always be expected to be plus three so
  • 00:59:42
    Al is also a main group metal
  • 00:59:45
    and that one doesn't really have a more
  • 00:59:47
    specific classification than that and it
  • 00:59:49
    would expect to be a plus three so those
  • 00:59:51
    are all given correctly in the answer of
  • 00:59:53
    choice
  • 00:59:54
    all right then finally for selenium
  • 00:59:57
    selenium is s e
  • 01:00:00
    and so when we find that on the periodic
  • 01:00:02
    table which is number 34
  • 01:00:04
    um it's clearly to the right of the
  • 01:00:06
    staircase so it is a non-metal
  • 01:00:08
    um it's not on it's not one of the metal
  • 01:00:10
    volumes it's not right on the staircase
  • 01:00:11
    and then in terms of the charge of the
  • 01:00:15
    ion because it's a non-metal we expect
  • 01:00:17
    it to be more likely to form anions
  • 01:00:19
    which is what it does so here it's in
  • 01:00:22
    the 6A column so it's two away from the
  • 01:00:25
    noble gas so it can gain two electrons
  • 01:00:27
    to get to the KR Krypton configuration
  • 01:00:30
    so we respected to form a minus two
  • 01:00:33
    anion if it was involved in ionic
  • 01:00:35
    bonding all right so selenium is a
  • 01:00:38
    non-metal
  • 01:00:40
    and because it's non-metal exp if we
  • 01:00:43
    expected to form anions and it's based
  • 01:00:45
    on this column retractable minus two is
  • 01:00:47
    the most likely anion that it would form
  • 01:00:49
    so that one is also stated correctly so
  • 01:00:51
    both 2 and 3 are correct so e would be
  • 01:00:54
    the answer choice we would choose so
  • 01:00:57
    that again relates to using the periodic
  • 01:00:59
    table to make predictions about ions and
  • 01:01:02
    also classifications both of those kind
  • 01:01:04
    of came up in this so if you have any
  • 01:01:06
    questions on that
  • 01:01:09
    yes so
  • 01:01:11
    um would you only
  • 01:01:13
    um
  • 01:01:14
    okay so so whenever you're trying to get
  • 01:01:17
    to a noble gas you go to the left so you
  • 01:01:20
    take away electrons so um when would you
  • 01:01:23
    go to the right and like or can you even
  • 01:01:25
    add electrons or no yeah you can do
  • 01:01:28
    either so if you're in the if you're on
  • 01:01:30
    the left of the periodic table when
  • 01:01:33
    you're we're dealing with Metals in that
  • 01:01:35
    case you would go to the left losing
  • 01:01:38
    electrons to get to the nearest noble
  • 01:01:40
    gas you're always going to get to the
  • 01:01:41
    one that's closest I suppose is a way to
  • 01:01:42
    think about it so for these first two
  • 01:01:44
    columns here as well as for you know
  • 01:01:46
    things like aluminum you would lose
  • 01:01:48
    electrons that's what metals tend to do
  • 01:01:51
    to get to the nearest noble gas for
  • 01:01:53
    things that are on the right side of the
  • 01:01:54
    periodic table they're close to noble
  • 01:01:56
    gas but in the other direction then you
  • 01:01:58
    would add electrons so you'd add one
  • 01:02:00
    electron if you're in the halogen column
  • 01:02:02
    here group seven to get to a minus one
  • 01:02:04
    anion or two electrons I don't know if
  • 01:02:06
    you can actually see my pointer on teams
  • 01:02:08
    by the way I've got to mention that
  • 01:02:10
    that's one limitation of Microsoft teams
  • 01:02:11
    is that at least in the past my laser
  • 01:02:14
    pointer didn't work on teams my
  • 01:02:15
    highlighter so if you can't see that I
  • 01:02:18
    apologize it's a limitation of doing it
  • 01:02:19
    online that I can never figure out how
  • 01:02:21
    to solve but anyway if you're you can
  • 01:02:23
    kind of follow my gestation you can see
  • 01:02:25
    it or not
  • 01:02:26
    yeah we can oh really okay so maybe
  • 01:02:29
    teams improved which is rare
  • 01:02:32
    um all right so yeah for the non-metals
  • 01:02:34
    these last three columns or so when
  • 01:02:36
    you're to the right of the staircase the
  • 01:02:38
    the expectations that they gain
  • 01:02:40
    electrons and would form stable anions
  • 01:02:42
    they wouldn't lose electrons to perform
  • 01:02:43
    cations in that case
  • 01:02:46
    what about the ones in the middle
  • 01:02:49
    uh those are hard to predict so they
  • 01:02:51
    they behave like Metals so when they if
  • 01:02:54
    you're talking about like the transition
  • 01:02:55
    metals here
  • 01:02:56
    um
  • 01:02:58
    they would be they behave like you might
  • 01:03:00
    also they tend to form cations and not
  • 01:03:02
    anions
  • 01:03:03
    um but there's no easy way to predict
  • 01:03:05
    what the charges will be there's use
  • 01:03:06
    they're usually variable
  • 01:03:09
    um most of these have more than one
  • 01:03:11
    possible charge you should be able to
  • 01:03:12
    figure out what that charge is if we
  • 01:03:14
    give you the formula of a compound where
  • 01:03:16
    you have an anion charge built into that
  • 01:03:18
    we might have problems like that later
  • 01:03:20
    um but you can't predict it from the
  • 01:03:21
    periodic table what the charge would be
  • 01:03:23
    and there's often more than one
  • 01:03:24
    possibility
  • 01:03:26
    the ones the only ones from the middle
  • 01:03:27
    that you should probably know would be
  • 01:03:29
    zinc is always plus two
  • 01:03:34
    which again is not obvious from his
  • 01:03:36
    location but that's that's one that's
  • 01:03:38
    pretty much invariance and then silver
  • 01:03:40
    is this one's not as
  • 01:03:43
    always I suppose but for our purposes is
  • 01:03:46
    always plus one
  • 01:03:48
    um so those are not clear based on the
  • 01:03:50
    location and probably not going to come
  • 01:03:52
    up anyway
  • 01:03:53
    um but the middle ones in general you
  • 01:03:55
    can't really predict what their charges
  • 01:03:56
    would be and there's usually more than
  • 01:03:58
    one that's possible
  • 01:04:00
    all right any
  • 01:04:03
    anything else related to that
  • 01:04:06
    all right so the other thing that we can
  • 01:04:08
    do with the periodic table is make
  • 01:04:10
    predictions about certain properties
  • 01:04:12
    which we call periodic trends
  • 01:04:14
    um
  • 01:04:15
    and that's what this question deals with
  • 01:04:17
    really all of them atomic radius
  • 01:04:19
    ionization energy or electron affinity
  • 01:04:21
    and we're comparing
  • 01:04:22
    oxygen and nitrogen to each other
  • 01:04:25
    all right so we have three terms we're
  • 01:04:27
    going to pair here radius
  • 01:04:30
    ionization energy
  • 01:04:33
    in electron affinity
  • 01:04:34
    now in some level these
  • 01:04:37
    sort of work opposite of each other so
  • 01:04:39
    the the directions that the radius
  • 01:04:43
    decreases correspond to the directions
  • 01:04:45
    in the periodic table where the other
  • 01:04:47
    two increase although radius is a pretty
  • 01:04:49
    smooth Trend ionization energy and
  • 01:04:52
    electron affinity have some exceptions
  • 01:04:55
    um
  • 01:04:56
    all right but if we look at radius first
  • 01:04:58
    that's usually the easier one because
  • 01:04:59
    there's not really exceptions we have to
  • 01:05:01
    worry about
  • 01:05:01
    we have to recall is that as we go
  • 01:05:06
    left to right in the same row of the
  • 01:05:08
    periodic table
  • 01:05:10
    where adding electrons to the same
  • 01:05:13
    subshell so we're adding electrons to
  • 01:05:15
    subshells that are roughly the same
  • 01:05:16
    distance from the nucleus to start with
  • 01:05:19
    but we're also adding more protons to
  • 01:05:21
    the nucleus as we go left to right and
  • 01:05:23
    so that effect wins out which means that
  • 01:05:24
    as we go left to right across a single
  • 01:05:27
    row of the periodic table the radius
  • 01:05:29
    will decrease it gets smaller as we go
  • 01:05:31
    left to right so when we're comparing
  • 01:05:33
    nitrogen and oxygen
  • 01:05:35
    we see that oxygen is to the right of
  • 01:05:37
    nitrogen so oxygen should be smaller so
  • 01:05:40
    in terms of radius we'll do that one
  • 01:05:42
    first and write it out so we don't
  • 01:05:43
    forget it nitrogen should be bigger than
  • 01:05:46
    oxygen because it's to the left and for
  • 01:05:49
    radius there's not really any exceptions
  • 01:05:51
    to that that you have to worry about
  • 01:05:52
    radius would then increase as we go top
  • 01:05:54
    to bottom is that we're adding subshells
  • 01:05:57
    that are further and further away from
  • 01:05:58
    the nucleus but here we're comparing two
  • 01:06:00
    that are next to each other left to
  • 01:06:02
    right so the one to the right would be
  • 01:06:03
    the smaller one
  • 01:06:04
    now the next Trend that this question
  • 01:06:06
    asks about is ionization energy for
  • 01:06:09
    ionization energy electron affinity we
  • 01:06:11
    do need to remember
  • 01:06:13
    some exceptions to the trend so let me
  • 01:06:16
    sort of try to sketch those out as best
  • 01:06:19
    as I can here so for ionization energy
  • 01:06:21
    first
  • 01:06:23
    as we go across the periodic table the
  • 01:06:25
    main groups of the periodic table so one
  • 01:06:27
    and two which are the first two columns
  • 01:06:29
    and then
  • 01:06:30
    three four five six seven eight which
  • 01:06:33
    are all on the P block on the right side
  • 01:06:35
    of the periodic table as we go across
  • 01:06:37
    left to right what we'll find is that
  • 01:06:41
    they tend to increase left to right
  • 01:06:43
    but we have a sort of
  • 01:06:46
    um jumps up a lot here at group two
  • 01:06:51
    and then
  • 01:06:53
    comes back down in group three ocean be
  • 01:06:55
    kind of in between those two
  • 01:06:57
    and then it goes up at group four and
  • 01:07:00
    then for ionization energy we have a big
  • 01:07:01
    jump at group five
  • 01:07:03
    and then group six comes back down a
  • 01:07:05
    little bit and then it goes up to seven
  • 01:07:07
    and way up for eight
  • 01:07:09
    all right but the important part is that
  • 01:07:11
    it generally increases left to right but
  • 01:07:13
    you have exceptions here at group two
  • 01:07:14
    and group five which for ionization
  • 01:07:17
    energy are are higher than you would
  • 01:07:19
    expect and that's because in group two
  • 01:07:21
    and group five you have
  • 01:07:23
    either an S2 or an S2 P3 valence
  • 01:07:27
    configuration and those for reasons that
  • 01:07:30
    are hard to get into in a general
  • 01:07:31
    chemistry class but those are
  • 01:07:32
    particularly stable so you don't want to
  • 01:07:34
    lose electrons if you already have S2 or
  • 01:07:37
    if you have already P3 and so that makes
  • 01:07:39
    the ionization energy which is how much
  • 01:07:41
    energy you need to pull off an electron
  • 01:07:42
    abnormally high for those two groups so
  • 01:07:45
    as it relates to this problem here
  • 01:07:48
    nitrogen is group five and oxygen is
  • 01:07:51
    group six so based on this argument we
  • 01:07:54
    expect that even though the general
  • 01:07:55
    trend is increasing left to right as we
  • 01:07:58
    go from group five to group six we see
  • 01:08:00
    the decrease that's where one of our
  • 01:08:01
    exceptions occurs and so for ionization
  • 01:08:04
    energy we expect nitrogen to be bigger
  • 01:08:06
    than oxygen
  • 01:08:07
    even those to the left but that's where
  • 01:08:09
    that exception occurs and then for
  • 01:08:11
    electron affinity we have the same
  • 01:08:13
    exception but it occurs in the opposite
  • 01:08:16
    Direction so it's still generally
  • 01:08:18
    increasing left to right
  • 01:08:20
    for electron affinity over across the
  • 01:08:22
    main groups of the periodic table
  • 01:08:27
    but here what we find is that for group
  • 01:08:30
    two and group 5 electron Infinity is
  • 01:08:32
    abnormally low so we have group one and
  • 01:08:35
    then it's going to actually dip at group
  • 01:08:37
    two
  • 01:08:38
    going to go back up for group three
  • 01:08:41
    up for group four dip it group five
  • 01:08:44
    and go up and up and we don't really
  • 01:08:46
    talk about electron Infinities for group
  • 01:08:48
    eight they're essentially zero as well
  • 01:08:50
    so again it's not a smooth Trend but it
  • 01:08:52
    does generally increase left to right
  • 01:08:53
    but at group two group five we have the
  • 01:08:56
    exceptions and so that's that's again
  • 01:08:57
    what we're talking about with
  • 01:09:00
    nitrogen and oxygen is group fives and
  • 01:09:02
    group six
  • 01:09:04
    um and so this one you don't have to
  • 01:09:06
    necessarily remember the exception but
  • 01:09:08
    nitrogens would be abnormally low and
  • 01:09:10
    they would increase again as we go to
  • 01:09:12
    the right to oxygen so for electron
  • 01:09:13
    affinity nitrogen is smaller than oxygen
  • 01:09:16
    all right so those are the three
  • 01:09:18
    comparisons we made so see which of
  • 01:09:20
    these are correct so the atomic radius
  • 01:09:22
    of oxygen is larger that's incorrect the
  • 01:09:25
    ionization of energy ionization energy
  • 01:09:27
    of oxygen is larger that is also
  • 01:09:30
    incorrect
  • 01:09:32
    the electron affinity of oxygen is
  • 01:09:33
    larger that one is the one that would be
  • 01:09:35
    true
  • 01:09:37
    and so the only one that works is Choice
  • 01:09:39
    C here that the electron affinity of
  • 01:09:41
    oxygen is larger than nitrogen yeah okay
  • 01:09:44
    so those Trends are
  • 01:09:47
    you're going to come up from time to
  • 01:09:48
    time and
  • 01:09:50
    ionization energy Electro Infinity are
  • 01:09:52
    just a little bit tricky because of the
  • 01:09:54
    exceptions if you remember that the
  • 01:09:55
    exceptions occur at group two and group
  • 01:09:57
    five and that they're going to be in
  • 01:09:59
    opposite directions from each other
  • 01:10:01
    ionization energy abnormally High
  • 01:10:03
    electron affinity abnormally low in
  • 01:10:05
    those columns that should help you with
  • 01:10:07
    any of these comparisons we ask you to
  • 01:10:09
    do so does anybody have any points of
  • 01:10:12
    clarification on that one
  • 01:10:17
    so we should remember the the trends for
  • 01:10:20
    the test right
  • 01:10:21
    yes
  • 01:10:23
    um it would always be qualitative you
  • 01:10:25
    know just comparisons not really knowing
  • 01:10:27
    exact numbers obviously but
  • 01:10:30
    um you do need to be familiar with the
  • 01:10:31
    trends and especially for ionization and
  • 01:10:32
    electron affinity where those small
  • 01:10:34
    exceptions do occur
  • 01:10:36
    thank you
  • 01:10:39
    okay
  • 01:10:41
    ready for number 13. okay this is
  • 01:10:43
    another ionization energy one so this
  • 01:10:45
    deals with the same concept so now that
  • 01:10:47
    we've discussed that we can see if that
  • 01:10:49
    again
  • 01:10:50
    okay so which atom has the largest
  • 01:10:53
    iodization energy we have a bunch of
  • 01:10:55
    them chosen here
  • 01:10:56
    nitrogen oxygen carbon phosphorus
  • 01:10:59
    germanium
  • 01:11:00
    um
  • 01:11:01
    so not all these are in the same row so
  • 01:11:04
    these three here
  • 01:11:06
    are in the same row they're all 2p
  • 01:11:09
    elements then for phosphorus it's a 3p
  • 01:11:11
    element and germanium I think is also a
  • 01:11:13
    3p let me show you remember that
  • 01:11:15
    correctly
  • 01:11:16
    GE is over here which is sorry even
  • 01:11:19
    lower 4p
  • 01:11:21
    um so we have three different rows of
  • 01:11:23
    the periodic table we're talking about
  • 01:11:25
    um
  • 01:11:27
    so in general the ionization energy
  • 01:11:30
    increases up not only left to right if
  • 01:11:33
    you're talking about the last one but
  • 01:11:34
    also bottom to Tops as you go further
  • 01:11:37
    and further to the top right that's
  • 01:11:38
    where the ionization energies get larger
  • 01:11:40
    so you know germanium phosphorus
  • 01:11:43
    nitrogen oxygen carbon we would expect
  • 01:11:46
    that these two P Elements which are
  • 01:11:48
    higher up on the periodic table and
  • 01:11:51
    further to the right than most of the
  • 01:11:52
    other choices p and GE should be the
  • 01:11:55
    higher ones so if we're looking for the
  • 01:11:57
    largest ionization energy we can from
  • 01:11:58
    these ones eliminate
  • 01:12:00
    germanium and phosphorus because they're
  • 01:12:03
    so far down in the lower row and pretty
  • 01:12:06
    far to the left although the carbons
  • 01:12:07
    further to the left but
  • 01:12:09
    um usually there's a pretty big increase
  • 01:12:11
    as you go bottom bottom to top so it's
  • 01:12:13
    going to be one of those three carbon
  • 01:12:15
    nitrogen and oxygen that would be
  • 01:12:16
    identified as the highest based on their
  • 01:12:18
    location and we again have to then think
  • 01:12:20
    about the trend as we go left to right
  • 01:12:22
    which relates to
  • 01:12:24
    exactly what we talked about in the last
  • 01:12:25
    problem so if you have carbon nitrogen
  • 01:12:27
    and oxygen which are groups four five
  • 01:12:30
    and six remember that ionization energy
  • 01:12:32
    is abnormally high for group five so it
  • 01:12:35
    doesn't get so high that it like
  • 01:12:36
    completely blows out the rest of those
  • 01:12:39
    in Europe it's going to be higher than
  • 01:12:40
    group six so if I drew this accurately
  • 01:12:42
    you would expect that of these three
  • 01:12:45
    elements group four has the lowest
  • 01:12:47
    ionization energy group five is then
  • 01:12:49
    abnormally High which makes it a little
  • 01:12:50
    bit higher than group six which is kind
  • 01:12:52
    of in between those two
  • 01:12:54
    all right so that's going to be the the
  • 01:12:56
    trend that's relevant also to here
  • 01:12:59
    and that leaves us with nitrogen again
  • 01:13:02
    is the highest one because of that
  • 01:13:04
    Spike we see in the group five elements
  • 01:13:08
    um so similar to the last one but this
  • 01:13:10
    one also asked us to consider as well
  • 01:13:13
    the trend of going to different rows of
  • 01:13:15
    the periodic table
  • 01:13:17
    um which was how we eliminated those
  • 01:13:19
    those other two answer choices all right
  • 01:13:21
    any more questions on ionization energy
  • 01:13:24
    or periodic trends
  • 01:13:29
    all right so we are finally into the
  • 01:13:32
    chapter three part of the test
  • 01:13:34
    which deals with bonding and sort of a
  • 01:13:39
    more basic level than we are going to
  • 01:13:40
    cover in chapter four right now
  • 01:13:42
    and so the first question deals with
  • 01:13:45
    the bond that has the
  • 01:13:47
    smallest ionic character among the
  • 01:13:49
    following
  • 01:13:51
    um
  • 01:13:52
    so
  • 01:13:54
    for problems like this we have to
  • 01:13:55
    recognize the terminology and this is
  • 01:13:57
    one of the concepts that
  • 01:13:59
    in dealing with you know student
  • 01:14:01
    questions over the years seems to be one
  • 01:14:03
    that is particularly challenging and
  • 01:14:05
    it's it requires a little bit of you
  • 01:14:07
    know higher level thinking because what
  • 01:14:10
    we're dealing with here is
  • 01:14:12
    when we say smallest ionic character the
  • 01:14:14
    first thing is the terminology that
  • 01:14:16
    means we're looking for the one that's
  • 01:14:17
    least polar so when we have bonds
  • 01:14:19
    between two atoms
  • 01:14:21
    they're sort of two extremes and there's
  • 01:14:24
    you know everything else in the middle
  • 01:14:25
    so if you have the same atom bonded to
  • 01:14:27
    itself like hydrogen bonding to hydrogen
  • 01:14:31
    this would be a what we call a perfect
  • 01:14:33
    covalent bond the two electrons in that
  • 01:14:36
    Bond are shared exactly equally between
  • 01:14:38
    the two hydrogen atoms and that would be
  • 01:14:40
    a nonpolar covalent
  • 01:14:43
    and then if we go to The Other Extreme
  • 01:14:46
    where it's a metal with a non-metal like
  • 01:14:48
    sodium and chloride we typically
  • 01:14:50
    classify this bonding as ionic where we
  • 01:14:54
    don't share electrons anymore but rather
  • 01:14:55
    the non-metal steals electrons from the
  • 01:14:58
    metal makes a cation and anion that then
  • 01:15:00
    are attracted ionically then everything
  • 01:15:03
    else in the middle would be classified
  • 01:15:04
    as a sort of polar covalent bond so it's
  • 01:15:07
    something like HCL
  • 01:15:10
    and so when you have this unequal
  • 01:15:12
    sharing of electrons either you know the
  • 01:15:14
    extreme where
  • 01:15:17
    um you know it's ionic or somewhere in
  • 01:15:18
    the middle which is a polar covalent
  • 01:15:20
    bond the sharing of those electrons is
  • 01:15:23
    dictated by What's called the
  • 01:15:24
    electronegativity of the atoms another
  • 01:15:26
    periodic property that we learned which
  • 01:15:29
    also increases as we go left to right
  • 01:15:32
    and also as we go bottom to top so same
  • 01:15:34
    direction as ionization energy and
  • 01:15:36
    electron affinity but this time we're
  • 01:15:38
    talking about the property of an atom
  • 01:15:39
    when it's bonded to something else and
  • 01:15:41
    the nice thing about electronic activity
  • 01:15:43
    is that the trend is very smooth you
  • 01:15:44
    don't have to really worry about
  • 01:15:45
    exceptions like we did for the other
  • 01:15:47
    ones but the point is as we move left to
  • 01:15:49
    right or top to bottom or bottom to top
  • 01:15:51
    we increase electronegativity and so in
  • 01:15:54
    like HCL you're going to have a partial
  • 01:15:55
    negative charge on chlorine a partial
  • 01:15:58
    positive and hydrogen because chlorine
  • 01:16:00
    is more electronegative
  • 01:16:02
    and so what this means is that if you
  • 01:16:04
    have
  • 01:16:07
    higher ionic character
  • 01:16:10
    that correlates with a larger difference
  • 01:16:13
    in electronegativity
  • 01:16:15
    so the same atom bonded to itself the
  • 01:16:18
    difference in electronic activity is
  • 01:16:19
    zero and that's when you have a
  • 01:16:20
    perfectly covalent bond or sometimes two
  • 01:16:22
    different atoms have so similar in
  • 01:16:25
    electronegativity values that they're
  • 01:16:26
    basically nonpolar when they're bonded
  • 01:16:28
    together in the ionic extreme you have a
  • 01:16:31
    very large difference in
  • 01:16:32
    electronegativity metals have very low
  • 01:16:34
    electronegativity non-metals have
  • 01:16:37
    relatively high and so that's a big
  • 01:16:39
    difference you get ionic bonding and
  • 01:16:40
    then polar covalent is somewhere in
  • 01:16:42
    between where there's a significant
  • 01:16:43
    difference in electronegativity hydrogen
  • 01:16:46
    is 2.1 and chlorine I think is three so
  • 01:16:51
    there's a you know a sizable difference
  • 01:16:52
    in negativity but it's not so big that
  • 01:16:55
    you would classify it as ionic so that's
  • 01:16:57
    kind of the key concept of problems like
  • 01:16:58
    this when we say smallest ionic
  • 01:17:00
    character that means we're looking for
  • 01:17:01
    least polar which means we're looking
  • 01:17:03
    for the smallest difference in
  • 01:17:05
    electronegativity
  • 01:17:08
    now if I was being nice to you on this
  • 01:17:10
    problem
  • 01:17:11
    what I would have done is given you one
  • 01:17:13
    answer choice where it's the same atom
  • 01:17:15
    bonded to itself and if that was true
  • 01:17:17
    that would have to be the least polar
  • 01:17:18
    because that's completely non-polar
  • 01:17:20
    unfortunately we don't have that as any
  • 01:17:22
    of our options we have four answer
  • 01:17:23
    choices here and all of them have two
  • 01:17:25
    different you know two different atoms
  • 01:17:26
    bonded together so what we can do then
  • 01:17:28
    is we can
  • 01:17:30
    you know we can kind of compare them one
  • 01:17:33
    at a time because we're guessing pairs
  • 01:17:34
    to each other so if we compare the first
  • 01:17:36
    two Co and CCL
  • 01:17:38
    when you're doing comparisons like this
  • 01:17:40
    it's helpful to you know pick two that
  • 01:17:43
    have something in common so we have for
  • 01:17:45
    the first two they both have carbon in
  • 01:17:46
    them so now we have to figure out which
  • 01:17:48
    is the bigger electronegativity
  • 01:17:50
    difference carbon oxygen or carbon
  • 01:17:52
    chlorine
  • 01:17:53
    so in the periodic table
  • 01:17:56
    if you look where those are
  • 01:17:59
    oxygen is to the right of carbon so it's
  • 01:18:02
    it's definitely more electronegative so
  • 01:18:04
    that's going to be the more or less
  • 01:18:04
    negative atom and then chlorine is over
  • 01:18:07
    here we have to figure out which is more
  • 01:18:08
    electronegative oxygen or chlorine and
  • 01:18:11
    it's not obvious from the position of
  • 01:18:12
    the periodic table because
  • 01:18:15
    um
  • 01:18:17
    you know we're going down into the right
  • 01:18:18
    which is sort of two opposing Trends so
  • 01:18:20
    it's helpful to remember some benchmarks
  • 01:18:22
    and one Benchmark is that you know
  • 01:18:24
    fluorine is the most
  • 01:18:27
    the highest electronegativity is for
  • 01:18:28
    fluorine and then oxygen in terms
  • 01:18:32
    electronic activity is second most so
  • 01:18:34
    oxygen is going to be more or less
  • 01:18:35
    negative than everything except for
  • 01:18:36
    fluorine so that means the difference
  • 01:18:38
    between carbon and oxygen should be
  • 01:18:40
    larger than the difference between
  • 01:18:41
    carbon and chlorine because carbon is
  • 01:18:43
    less electronegative than both but
  • 01:18:45
    oxygen has a bigger value so that
  • 01:18:47
    difference between CNO is bigger and
  • 01:18:50
    again you don't necessarily want to
  • 01:18:51
    memorize all the numbers
  • 01:18:53
    um
  • 01:18:54
    but carbon is 2.4 I've done these
  • 01:18:57
    numbers enough that they're kind of
  • 01:18:58
    stuck in my head now oxygen is 3.5 and
  • 01:19:01
    chlorine is 3.0 so the difference
  • 01:19:03
    between oxygen and carbon is bigger than
  • 01:19:06
    the difference between carbon and
  • 01:19:07
    chlorine and remembering that oxygen is
  • 01:19:08
    the second most electronegative element
  • 01:19:10
    is helpful for some of these types of
  • 01:19:12
    problems when we fluorine is more
  • 01:19:14
    electronegative okay so that takes care
  • 01:19:16
    we're looking for the smallest ionic
  • 01:19:17
    character so we can eliminate Co
  • 01:19:20
    because that would have a bigger ionic
  • 01:19:22
    character than CCL so it can't be the
  • 01:19:24
    smallest overall
  • 01:19:25
    for the next comparison we can look at
  • 01:19:27
    CCL versus CF and this one's actually a
  • 01:19:29
    little bit easier comparison
  • 01:19:32
    um
  • 01:19:34
    and again because both of these chlorine
  • 01:19:36
    and fluorine are more or less negative
  • 01:19:38
    than carbon
  • 01:19:39
    but fluorine is the most electronegative
  • 01:19:42
    element it's clearly more or less
  • 01:19:43
    negative than chlorine because it's
  • 01:19:44
    sitting right above it and so that
  • 01:19:46
    difference between C and F is going to
  • 01:19:48
    be bigger than the difference between C
  • 01:19:50
    and cl so for all the bonds involving
  • 01:19:52
    carbon carbon fluorine would be the most
  • 01:19:55
    polar covalent bond because fluorine is
  • 01:19:57
    the most electronegative elements that
  • 01:19:59
    difference between C and F is the
  • 01:20:01
    biggest you're going to be able to get
  • 01:20:02
    among non-metals involving carbon so
  • 01:20:04
    anyway that tells us that this Bond CF
  • 01:20:07
    is more polar than CCL we're looking for
  • 01:20:09
    the least polar in this case we can
  • 01:20:11
    eliminate that one also so then all
  • 01:20:13
    that's left is to compare CCL
  • 01:20:16
    and sicl
  • 01:20:20
    and again these are two bonds that share
  • 01:20:23
    a common element Chlorine so we just
  • 01:20:25
    need to figure out what's the bigger
  • 01:20:27
    difference carbon versus chlorine
  • 01:20:28
    silicon versus chlorine so we're looking
  • 01:20:30
    for the ones that are further apart as
  • 01:20:32
    we go
  • 01:20:33
    bottom left to top right and so what we
  • 01:20:36
    notice is that both of these are to the
  • 01:20:38
    left of chlorine so they're least less
  • 01:20:40
    or negative but we get silicon is below
  • 01:20:43
    carbon number 14 here is below number
  • 01:20:46
    six which is carbon and because we're
  • 01:20:48
    dropping down the periodic table silicon
  • 01:20:50
    is less negative than carbon meaning
  • 01:20:53
    that the difference between silicon and
  • 01:20:55
    chlorine is greater again we're looking
  • 01:20:56
    for the difference in the two two values
  • 01:20:58
    not the absolute value here and so
  • 01:21:00
    silicon versus chlorine is a bigger
  • 01:21:03
    difference than carbon versus chlorine
  • 01:21:05
    and so what that leaves us with is we
  • 01:21:08
    would predict that this one is more
  • 01:21:10
    polar silicon chlorine because of the
  • 01:21:12
    bigger difference so the one that came
  • 01:21:14
    out smallest and all of these
  • 01:21:16
    comparisons was CCL and so that's going
  • 01:21:19
    to be the one that we would pick as our
  • 01:21:21
    answer Choice
  • 01:21:22
    um and again if you you don't you're not
  • 01:21:24
    going to necessarily have the numbers in
  • 01:21:25
    front of you but just to convince you
  • 01:21:27
    that this is true we said that carbon
  • 01:21:29
    has less negativity you have 2.4 silicon
  • 01:21:31
    is 1.8 so that different is between
  • 01:21:34
    silicon and chlorine a difference of 1.2
  • 01:21:36
    is bigger than the difference between
  • 01:21:38
    carbon and chlorine so when you're
  • 01:21:40
    talking about Bond polarity or ionic
  • 01:21:42
    character
  • 01:21:43
    it's a little bit higher level of
  • 01:21:45
    thinking because you can't just think
  • 01:21:46
    about a single number you have to think
  • 01:21:48
    about electronic activity for both
  • 01:21:50
    elements and which one of those pairs
  • 01:21:52
    has the biggest difference in their
  • 01:21:54
    electronegativity values and that's
  • 01:21:56
    what's going to then relate to the the
  • 01:21:57
    polarity or the ionic character of that
  • 01:21:59
    Bond
  • 01:22:00
    all right so any questions on bond
  • 01:22:02
    polarity
  • 01:22:05
    more or less negativity I have a
  • 01:22:07
    question yep absolutely so um why is HCl
  • 01:22:11
    an ionic bond but not a covalent because
  • 01:22:14
    this is an age like positive one and cl
  • 01:22:17
    negative one
  • 01:22:19
    well so it is a polar covalent bond
  • 01:22:22
    um now the reason for that is even
  • 01:22:26
    though hydrogen is in a weird spot on
  • 01:22:28
    the periodic table all the way on the
  • 01:22:29
    left it is still classified as a
  • 01:22:30
    non-metal so hydrogens in this column
  • 01:22:33
    with a bunch of metals because it only
  • 01:22:35
    has one valence electron but in reality
  • 01:22:37
    it is a non-metal so whenever two
  • 01:22:39
    nonmetals are bonded together we would
  • 01:22:41
    typically classify as being a polar
  • 01:22:43
    covalent bond rather than ionic it does
  • 01:22:45
    behave a little bit like an ionic
  • 01:22:47
    compound when you dissolve it in water
  • 01:22:48
    and we're going to talk about that later
  • 01:22:49
    in the semester but for the purposes of
  • 01:22:51
    classifying the bond and when you just
  • 01:22:54
    have HCL hydrogen chloride by itself it
  • 01:22:57
    would still be typically classified as
  • 01:22:58
    covalent even though there's not a true
  • 01:23:00
    distinction between ionic and covalent
  • 01:23:02
    typically if it's two non-metals we
  • 01:23:04
    would always call it a polar equivalent
  • 01:23:06
    Bond
  • 01:23:07
    okay that makes sense so then um if it's
  • 01:23:10
    too so an ionic bond will never be a
  • 01:23:13
    metal and a metal or a non-metal and a
  • 01:23:15
    non-metal right
  • 01:23:16
    that's correct an ionic bond would be a
  • 01:23:18
    metal with a non-metal okay metal metal
  • 01:23:21
    bonding is a totally different category
  • 01:23:23
    that we would just talk about a little
  • 01:23:24
    bit at the end of the course but in
  • 01:23:26
    terms of the ionic bonds there's always
  • 01:23:28
    one non-metal with a metal all right
  • 01:23:31
    perfect thank you so much all right
  • 01:23:33
    anything else
  • 01:23:35
    okay
  • 01:23:38
    next question then also deals with
  • 01:23:40
    radius which we talked about earlier for
  • 01:23:42
    atoms but this one deals now with ionic
  • 01:23:44
    radius as well so we're comparing oxygen
  • 01:23:48
    to chlorine and then O2 minus and cl
  • 01:23:50
    minus
  • 01:23:52
    um
  • 01:23:53
    and so again
  • 01:23:55
    you know comparisons are going to be
  • 01:23:57
    useful by just doing two at a time not
  • 01:23:59
    trying to look at all four of them and
  • 01:24:01
    pick it out right off the bat
  • 01:24:04
    um so there's going to be you know
  • 01:24:06
    different combinations we can choose but
  • 01:24:08
    a pretty easy comparison would be to
  • 01:24:10
    compare a neutral atom to an ion of the
  • 01:24:14
    same element so either a cation or anion
  • 01:24:16
    and so if we go to oxygen versus O2
  • 01:24:19
    minus
  • 01:24:21
    the comparison is pretty straightforward
  • 01:24:24
    because one is a neutral one's an anion
  • 01:24:26
    and anytime you add electrons to
  • 01:24:28
    something to make an anion it's going to
  • 01:24:30
    get much bigger and that's because you
  • 01:24:32
    increase electron electronic polishes
  • 01:24:34
    putting these two extra electrons in
  • 01:24:36
    into the same subshell but now you have
  • 01:24:38
    more electrons you know crammed into the
  • 01:24:40
    same volume of space they repel each
  • 01:24:41
    other strongly and then they want to
  • 01:24:42
    spread out and that makes the whole
  • 01:24:44
    thing bigger so between oxygen and
  • 01:24:46
    oxygen two minus the anion we expect O2
  • 01:24:48
    minus to be much bigger
  • 01:24:50
    so that's a general thing you can easily
  • 01:24:52
    remember is that
  • 01:24:53
    anions are much larger than neutral
  • 01:24:57
    atoms I'm going to put a you know
  • 01:24:58
    underline on much because it is a pretty
  • 01:25:00
    big effect you know just adding one or
  • 01:25:02
    two electrons really blows out the
  • 01:25:04
    radius in a pretty big way so again you
  • 01:25:06
    don't have to remember numbers or know
  • 01:25:07
    the numbers but it's it's significant
  • 01:25:10
    very significant all right so that means
  • 01:25:13
    that we're looking for the largest one
  • 01:25:14
    here so O2 minus is clearly larger than
  • 01:25:16
    oxygen so we can get rid of that one and
  • 01:25:18
    then similarly when we compare you know
  • 01:25:21
    CL and cl minus
  • 01:25:26
    what we have is again an atom and an
  • 01:25:29
    anion so even adding one electron as we
  • 01:25:31
    did here is enough to really increase
  • 01:25:33
    the radius it'll always increase the
  • 01:25:34
    radius when you add electrons
  • 01:25:36
    conversely it will decrease the radius
  • 01:25:38
    if you take them away but we're only
  • 01:25:39
    dealing with anions in this problem so
  • 01:25:41
    again CL minus much bigger than CL for
  • 01:25:44
    the by the same argument
  • 01:25:46
    so we're looking for the biggest ones we
  • 01:25:48
    can get rid of the two neutrals but now
  • 01:25:49
    we have to compare two anions to each
  • 01:25:52
    other CL minus and O2 minus and when
  • 01:25:55
    we're comparing anions to each other or
  • 01:25:57
    you know anions and cations we can't
  • 01:26:00
    just focus on the periodic table only
  • 01:26:02
    because we've added more electrons or
  • 01:26:04
    with cations are taking some away
  • 01:26:06
    so we do need to think about where those
  • 01:26:08
    valence electrons are because it may not
  • 01:26:10
    always be indicated from the periodic
  • 01:26:12
    table because for ions again you're
  • 01:26:14
    you're changing the number of electrons
  • 01:26:16
    so if we have CL minus versus O2 minus
  • 01:26:18
    we can think about
  • 01:26:20
    you know where do those
  • 01:26:23
    um electrons reside where are the
  • 01:26:24
    valence electrons in both of those so CL
  • 01:26:28
    here is number 17.
  • 01:26:31
    it's in the 3p block of the periodic
  • 01:26:33
    table third row P block and when we go
  • 01:26:36
    to CL minus we're just adding one more
  • 01:26:37
    electron to the 3p to get to argon
  • 01:26:40
    configuration but the point is the
  • 01:26:42
    valence electrons and cl minus are in
  • 01:26:44
    the 3p subshell
  • 01:26:46
    versus oxygen we're up here in 2p so
  • 01:26:49
    oxygen is 2p4 and then we add two more
  • 01:26:52
    electrons also to the 2p so the valence
  • 01:26:55
    electrons in the oxide anion would be in
  • 01:26:58
    the 2p so when you're comparing charged
  • 01:27:01
    species you also want to think about
  • 01:27:02
    where those electrons are for so for cl
  • 01:27:05
    minus they're in the 3p subshell the
  • 01:27:07
    outermost electrons for oxygen they're
  • 01:27:10
    2p
  • 01:27:11
    and as you get to
  • 01:27:13
    higher principal quantum numbers n
  • 01:27:15
    equals two versus n equals three or as
  • 01:27:17
    you go higher and higher that tends to
  • 01:27:19
    lead to larger and larger radius so we
  • 01:27:21
    would expect the three p valence
  • 01:27:23
    electrons for chloride to be further
  • 01:27:25
    away from the nucleus than the two P
  • 01:27:27
    electrons for oxide so we would predict
  • 01:27:29
    that CL minus is bigger even though the
  • 01:27:31
    charges are different
  • 01:27:33
    if you go to a whole different subshell
  • 01:27:35
    further away from the nucleus you expect
  • 01:27:36
    that to expand the radius and so we
  • 01:27:39
    would expect Co minus D to be the
  • 01:27:41
    biggest overall
  • 01:27:42
    so that other part of this comparison is
  • 01:27:45
    that when you're comparing two anions
  • 01:27:49
    or I guess I could just say ions in
  • 01:27:52
    general whether it's cations or anions
  • 01:27:58
    if they're in the same row of the
  • 01:28:00
    periodic table
  • 01:28:02
    look for the one that has the higher
  • 01:28:04
    charge the more positive charge you have
  • 01:28:06
    the smaller it is the more negative you
  • 01:28:08
    charge you have
  • 01:28:13
    the bigger it is
  • 01:28:15
    but if they're in different rows like we
  • 01:28:17
    had here
  • 01:28:23
    the lower you go on the periodic table
  • 01:28:25
    the higher the period number the higher
  • 01:28:28
    the row number
  • 01:28:30
    that's going to be the bigger that you
  • 01:28:31
    get so that was a relevant one for this
  • 01:28:33
    one because we're talking about two ions
  • 01:28:36
    that are in different rows of the
  • 01:28:37
    periodic table so the one that's further
  • 01:28:39
    down in the 3p would expect to be bigger
  • 01:28:41
    than the one that's in 2B if they were
  • 01:28:43
    in the same row then the one that has
  • 01:28:45
    the larger charge or would be we would
  • 01:28:47
    base it on charge the more negative
  • 01:28:49
    charge the bigger the more positive
  • 01:28:51
    charge the smaller but in different rows
  • 01:28:53
    we have to think about which subshell
  • 01:28:55
    would be further or closer to the
  • 01:28:57
    nucleus
  • 01:28:58
    all right any questions on that
  • 01:29:04
    okay
  • 01:29:06
    so then the last few questions deal with
  • 01:29:09
    different aspects of ionic and covalent
  • 01:29:11
    compounds either predicting formulas
  • 01:29:13
    naming them and so on so here we have
  • 01:29:16
    predict the formula of the compound that
  • 01:29:19
    forms when calcium and chlorine react so
  • 01:29:22
    for problems like this we've already
  • 01:29:23
    talked a little bit about predicting the
  • 01:29:26
    ions that would form from an element
  • 01:29:27
    positive ions for Metals getting to the
  • 01:29:31
    noble gas configuration negative ions
  • 01:29:33
    for nonmetals to get to the nearest
  • 01:29:35
    noble gas configuration that's to the
  • 01:29:37
    right of it so that's going to be the
  • 01:29:39
    first step in a problem like this we
  • 01:29:40
    want to predict the formula of an ionic
  • 01:29:42
    compound this would be ionic because
  • 01:29:45
    calcium is a metal
  • 01:29:49
    and over here number 20 chlorine is a
  • 01:29:53
    non-metal over here number 17.
  • 01:29:56
    and so it's a metal with a non-metal so
  • 01:29:58
    it's ionic so the first thing we have to
  • 01:29:59
    do is predict the charges of each of
  • 01:30:01
    those so calcium is number 20 and it's
  • 01:30:03
    in group 2A here so all of the elements
  • 01:30:07
    in this second column periodic table we
  • 01:30:09
    will predict would form a plus two
  • 01:30:10
    cation it loses two electrons to get to
  • 01:30:13
    the nearest noble gas configuration so
  • 01:30:16
    that's going to be plus two and then
  • 01:30:17
    chlorine which is over here in number
  • 01:30:19
    seven number 17 in the group 7A
  • 01:30:23
    all of the elements in this column would
  • 01:30:25
    gain one electron to form
  • 01:30:28
    a negatively charged and a negative one
  • 01:30:30
    anion to get to the noble gas
  • 01:30:32
    configuration in this case also Argon so
  • 01:30:35
    that's how we would predict their
  • 01:30:36
    charges based on the location of the
  • 01:30:38
    periodic table the metal forms of cation
  • 01:30:40
    the non-metal form is an anion so
  • 01:30:42
    calcium being in group 2A
  • 01:30:49
    we predict would have a plus two charge
  • 01:30:51
    when it forms an ionic compound
  • 01:30:54
    and then chlorine which is in group 7A
  • 01:30:58
    on the far right of the periodic table
  • 01:31:01
    we would predict has a minus one charge
  • 01:31:03
    when it forms an ionic compound forms an
  • 01:31:06
    anion so those are the two charges and
  • 01:31:08
    then from there we just have to balance
  • 01:31:10
    them to get the correct formula we have
  • 01:31:11
    to make sure that the positive and the
  • 01:31:12
    negative charges cancel each other out
  • 01:31:14
    so we have calcium two plus
  • 01:31:17
    and cl minus and again the sort of
  • 01:31:21
    low thought way of doing this is just to
  • 01:31:23
    take the charges and cross them over to
  • 01:31:25
    make them as subscripts but either way
  • 01:31:27
    you have to make sure that we have the
  • 01:31:28
    same number of negative charges and
  • 01:31:29
    positive charges so what we need then is
  • 01:31:32
    one calcium two plus paired with two CL
  • 01:31:35
    minuses to balance the plus two charge
  • 01:31:38
    from the metal with two negative one
  • 01:31:40
    charges from the non-metal so cacl2 is
  • 01:31:43
    the correct formula for this compound
  • 01:31:45
    which is called calcium chloride and
  • 01:31:47
    that would be answer Choice a given here
  • 01:31:50
    all right any questions on predicting
  • 01:31:53
    ions or ionic formulas
  • 01:31:59
    all right very good
  • 01:32:02
    four more to go this is usually exam
  • 01:32:05
    reviews always take a while but
  • 01:32:06
    hopefully they're
  • 01:32:08
    useful for you guys all right number 17.
  • 01:32:11
    in the Lewis structure for carbon
  • 01:32:13
    monoxide what is the formal charge on
  • 01:32:15
    the oxygen atom all right so when we're
  • 01:32:18
    doing these
  • 01:32:19
    um
  • 01:32:21
    we have to first get the formula of a
  • 01:32:22
    compound be able to draw it so carbon
  • 01:32:25
    monoxide is a covalent compounds we use
  • 01:32:27
    prefixes to indicate no prefix on the
  • 01:32:30
    first element means there's one of them
  • 01:32:32
    mono means one as well so it's just Co
  • 01:32:35
    is the formula for carbon monoxide
  • 01:32:38
    so a deceptively simple Lewis structure
  • 01:32:39
    just two atoms involved but we still
  • 01:32:41
    have to make sure that we count the
  • 01:32:43
    electrons correctly and that we follow
  • 01:32:45
    the octet rule and all those things okay
  • 01:32:47
    so let's count the electrons first for
  • 01:32:49
    carbon and oxygen valence electrons what
  • 01:32:52
    we care about for Lewis structures
  • 01:32:54
    all right so when we go to the periodic
  • 01:32:56
    table
  • 01:32:57
    in these main group elements the group
  • 01:32:59
    number directly tells you the number of
  • 01:33:01
    valence electrons so carbon is in group
  • 01:33:03
    four meaning it has four valence
  • 01:33:06
    electrons oxygen is in group six so I
  • 01:33:08
    would have six valence electrons so four
  • 01:33:11
    from carbon to six from oxygen for a
  • 01:33:13
    total of ten one carbon in the formula
  • 01:33:20
    one oxygen in the formula
  • 01:33:25
    and a total of 10 electrons then when we
  • 01:33:27
    add those up
  • 01:33:28
    okay so then the next thing we have to
  • 01:33:31
    do is
  • 01:33:33
    draw the Lewis structure
  • 01:33:35
    this is only two atoms so again it seems
  • 01:33:38
    like it should be easy but we have to
  • 01:33:40
    make sure we complete the octet rules so
  • 01:33:42
    if we follow the same approach that we
  • 01:33:44
    usually do
  • 01:33:45
    it's kind of arbitrary which one you
  • 01:33:47
    call the central atom because there's
  • 01:33:48
    only two of them but we start with a
  • 01:33:50
    single bond between them and then we're
  • 01:33:51
    going to complete the octet on one of
  • 01:33:53
    the two it doesn't really matter which
  • 01:33:55
    one you complete the octet on first
  • 01:33:57
    although conventionally we would
  • 01:33:58
    complete the act of the more or less
  • 01:34:00
    negative atom first which is oxygen but
  • 01:34:02
    either way you're going to complete the
  • 01:34:03
    octets of one of them
  • 01:34:05
    so we've completed the octet and oxygen
  • 01:34:07
    that gives us 10 or sorry this eight
  • 01:34:09
    electrons but we have 10 total so that
  • 01:34:11
    leaves a lone pair on the other one
  • 01:34:13
    again whichever order you did that in
  • 01:34:15
    whether you completed the octeton and
  • 01:34:17
    carbon first or oxygen first you would
  • 01:34:19
    end up with one lone pair on the other
  • 01:34:20
    one but this is clearly not the finished
  • 01:34:22
    structure because one of the two atoms
  • 01:34:25
    in this case carbon as I've drawn it
  • 01:34:27
    only has four valence electrons around
  • 01:34:29
    it
  • 01:34:31
    it has two electrons in this Bond here
  • 01:34:34
    two more from a lone pair but that's
  • 01:34:36
    only four and everything that's in the
  • 01:34:38
    2p block of the periodic table starting
  • 01:34:40
    with carbon carbon oxygen nitrogen
  • 01:34:42
    fluorine
  • 01:34:43
    would always want to have eight valence
  • 01:34:45
    electrons in a stable Lewis structure so
  • 01:34:47
    as long as four it needs eight we can't
  • 01:34:49
    just add more electrons because we've
  • 01:34:51
    used all 10 so the way that we do that
  • 01:34:53
    is with multiple bonding so we need two
  • 01:34:55
    four more electrons around carbon so we
  • 01:34:57
    need to draw two more bonds to it with
  • 01:34:59
    two electrons each so we're going to
  • 01:35:00
    take
  • 01:35:01
    two lone pairs from oxygen and turn them
  • 01:35:04
    into multiple bonds making a total of a
  • 01:35:07
    triple bond
  • 01:35:08
    the one Bond we already had plus the two
  • 01:35:10
    new ones to complete the octet so what
  • 01:35:12
    that leaves us with is a structure like
  • 01:35:14
    this where we have a triple bond between
  • 01:35:16
    the two atoms and one lone pair on each
  • 01:35:20
    one and if we check ourselves what we
  • 01:35:21
    have is a total of 10 electrons two four
  • 01:35:24
    six eight ten two lone pairs six more in
  • 01:35:28
    the bonds total of ten and we completed
  • 01:35:30
    that on both atoms six from the bonds
  • 01:35:33
    for each of them two more from each lone
  • 01:35:36
    pair so that's total eight for each atom
  • 01:35:37
    so this follow exact rule has the
  • 01:35:39
    correct number of electrons and then
  • 01:35:41
    finally what this asks us to do
  • 01:35:44
    is calculate the formal charge on the
  • 01:35:46
    oxygen atom so we can calculate the
  • 01:35:48
    formal charge on both of them just to
  • 01:35:49
    practice but we need oxygen for this one
  • 01:35:52
    so carbon and oxygen
  • 01:35:54
    the way that we calculate formal charge
  • 01:35:56
    for carbon is we take the number of
  • 01:35:58
    valence electrons which was four for the
  • 01:36:00
    atom
  • 01:36:01
    we subtract the number of bonds which is
  • 01:36:03
    three has a triple bond and then we
  • 01:36:05
    subtract the number of non-bonding
  • 01:36:08
    electrons the total number of lone pair
  • 01:36:10
    electrons which is two one pair so four
  • 01:36:13
    minus three minus two is minus one
  • 01:36:15
    so this has a minus one formal charge
  • 01:36:17
    and then for oxygen
  • 01:36:20
    it's in group six so six is the number
  • 01:36:22
    of valence electrons it also has three
  • 01:36:24
    bonds same triple bond that shares with
  • 01:36:27
    carbon also has two non-bonding
  • 01:36:29
    electrons so six minus three minus two
  • 01:36:31
    is plus one
  • 01:36:33
    my tablet is starting to lag just as we
  • 01:36:35
    get to the end so hopefully we can make
  • 01:36:36
    it all right so we have a negative one
  • 01:36:38
    and a plus one the one for oxygen is
  • 01:36:41
    plus one so that would be the answer in
  • 01:36:43
    this question
  • 01:36:44
    plus one and this is again a weird Lewis
  • 01:36:47
    structure even though it's deceptively
  • 01:36:48
    simple with only two atoms because it's
  • 01:36:51
    a neutral molecule but you still have a
  • 01:36:54
    plus and negative formal charge and
  • 01:36:56
    moreover the positive formal charge is
  • 01:36:58
    on the more electronegative atom which
  • 01:36:59
    is especially weird but there's no other
  • 01:37:01
    way to do it if you want to follow the
  • 01:37:03
    octet rule and that rule is the most
  • 01:37:04
    important so this is the only way to
  • 01:37:06
    draw the Lewis structure of carbon
  • 01:37:08
    monoxide that gets all of that correct
  • 01:37:10
    follows the ACT rule has the right
  • 01:37:12
    number of electrons and so on
  • 01:37:14
    all right so any questions on that one
  • 01:37:15
    or Lewis structures in general
  • 01:37:26
    are we still recording good
  • 01:37:29
    I saw a message pop up about recording
  • 01:37:31
    and I panicked that it stopped for some
  • 01:37:33
    reason
  • 01:37:34
    all right so we'll go on to the last
  • 01:37:35
    three so the next one deals with
  • 01:37:37
    nomenclature naming compounds
  • 01:37:40
    um these ones are ionic compounds in
  • 01:37:42
    particular so wants to know which of the
  • 01:37:44
    following name formula pairs is
  • 01:37:46
    incorrect
  • 01:37:48
    um so we have strontium carbonate
  • 01:37:50
    Cobalt 2 nitride calcium phosphate zinc
  • 01:37:54
    sulfide
  • 01:37:55
    and potassium chlorite
  • 01:38:00
    did I screw this up and make all of them
  • 01:38:02
    correct no okay I thought I made a
  • 01:38:05
    mistake all right so the first one is
  • 01:38:07
    strontium carbonate srco3
  • 01:38:11
    um so if we go to the prta again we have
  • 01:38:14
    to figure out the charges of the ions
  • 01:38:15
    and make sure that they're balanced and
  • 01:38:17
    all that stuff
  • 01:38:18
    um
  • 01:38:20
    so Sr is the metal on the formula it's
  • 01:38:24
    not one you'd have to know off top of
  • 01:38:25
    your head but here you have enough
  • 01:38:26
    context to find it because the the
  • 01:38:28
    symbol is given in the answer choice and
  • 01:38:31
    so estar is in group two so in a form of
  • 01:38:33
    plus two cation
  • 01:38:36
    and then what we have to remember
  • 01:38:39
    sort of I guess by memorization more
  • 01:38:41
    than anything is that carbonate which is
  • 01:38:44
    CO3
  • 01:38:46
    uh this is really starting to drag I
  • 01:38:48
    hope we can get through this let me save
  • 01:38:49
    this I don't lose it
  • 01:38:51
    that's why I hate teams it doesn't
  • 01:38:54
    eventually it will slow down your tablet
  • 01:38:57
    oh maybe it's because I not I'm not
  • 01:38:59
    plugged in low on battery give me a
  • 01:39:02
    second
  • 01:39:04
    I thought I had this plugged in all
  • 01:39:06
    right let's see if that changes it all
  • 01:39:07
    right so anyways strontium plus two
  • 01:39:09
    there we go just running out of battery
  • 01:39:11
    and didn't decide not to warn me just
  • 01:39:13
    slowed down and then we have CO3 which
  • 01:39:16
    is carbonate and we have to remember
  • 01:39:18
    from our list of polyatomic anions that
  • 01:39:20
    carbonate is two minus
  • 01:39:23
    all right and so for this one plus two
  • 01:39:27
    cation minus two and ions so one of each
  • 01:39:30
    in the formula would be correct
  • 01:39:33
    all right and um balance the charge of
  • 01:39:35
    just one of each so that one is correct
  • 01:39:37
    if we go to Cobalt 2 nitride
  • 01:39:41
    um CO2 n so to figure out if this is the
  • 01:39:44
    right formula we need again think about
  • 01:39:45
    the charges of each ion in transition
  • 01:39:48
    metal compounds the charge is given in
  • 01:39:50
    parentheses with Roman numerals so
  • 01:39:52
    Cobalt two Cobalt is a transition metal
  • 01:39:55
    right there means it's a plus two charge
  • 01:39:57
    on the metal
  • 01:39:58
    and then nitride nitrogen is here in
  • 01:40:01
    group five it would want to gain three
  • 01:40:03
    electrons to get to a normal gas
  • 01:40:06
    and so what that means is that we've
  • 01:40:07
    spent nitrogen to have a minus three
  • 01:40:10
    charge as an anion so the two ions
  • 01:40:12
    involved are
  • 01:40:15
    Cobalt two plus that's directly given to
  • 01:40:17
    us in the name and then N3 minus which
  • 01:40:19
    we predicted from the periodic table
  • 01:40:22
    so if we cross these over and make them
  • 01:40:24
    into subscripts to get the formula the
  • 01:40:26
    correct formula for this should be
  • 01:40:32
    Cobalt three and two not Cobalt 2N so
  • 01:40:36
    this is the wrong formula here or the
  • 01:40:38
    wrong name for that formula this formula
  • 01:40:40
    would be
  • 01:40:41
    I guess Cobalt 1.5 and I try which is a
  • 01:40:44
    weird number but anyway the the name or
  • 01:40:46
    the formula is wrong but it
  • 01:40:48
    um because from Cobalt 2 nitrile we will
  • 01:40:50
    predict a formula of CO3 N2 to balance
  • 01:40:53
    the charge from the plus two cation and
  • 01:40:55
    the minus three anion that gives us plus
  • 01:40:57
    six and minus six so that's the wrong
  • 01:40:59
    formula so that should be the one that
  • 01:41:00
    we would pick as our answer Choice let's
  • 01:41:03
    verify that the rest of these are
  • 01:41:04
    correct
  • 01:41:06
    um
  • 01:41:06
    calcium as we saw earlier so I'm not
  • 01:41:09
    going to flip back to the periodic table
  • 01:41:10
    with ca2 Plus
  • 01:41:12
    and then phosphate we have to remember
  • 01:41:14
    is po43 minus so that's the formula in
  • 01:41:18
    the charge for the cation of the anion
  • 01:41:19
    and so when we make that into
  • 01:41:22
    a formula what it comes out to is
  • 01:41:25
    ca3 and then two phosphates po4 taken
  • 01:41:30
    twice so that one is correct
  • 01:41:33
    um zinc sulfide we talked about this one
  • 01:41:35
    earlier that
  • 01:41:37
    um
  • 01:41:38
    zinc is always plus two even though it's
  • 01:41:41
    not obvious in the periodic table it's
  • 01:41:42
    in the transition metals so you might
  • 01:41:44
    think that is variable but zinc we
  • 01:41:46
    should remember is always plus two even
  • 01:41:48
    as a transition metal so zinc plus two
  • 01:41:51
    and then sorry I should go back to
  • 01:41:53
    period table sulfur here is in the sixth
  • 01:41:56
    column group 6A so two minus to get the
  • 01:41:59
    noble gas so we have 2 minus anion with
  • 01:42:02
    a two plus cation
  • 01:42:04
    so we just need one of each to make zinc
  • 01:42:06
    sulfide
  • 01:42:08
    it wouldn't probably be wrong to say
  • 01:42:10
    zinc 2 sulfide anytime you're dealing
  • 01:42:12
    with transition metals if you want to
  • 01:42:14
    put the Roman numerals in parentheses
  • 01:42:15
    it's okay so it wouldn't be wrong to say
  • 01:42:18
    zinc two soft five we don't necessarily
  • 01:42:19
    need to specify it here because zinc is
  • 01:42:21
    always plus two
  • 01:42:22
    but either way that one's correct and
  • 01:42:24
    then finally potassium chlorite
  • 01:42:26
    K is K plus it's the First Column of the
  • 01:42:29
    periodic table chlorette we have to
  • 01:42:31
    remember is clo2 minus all of these
  • 01:42:34
    oxynions of chlorine hypochlorite
  • 01:42:37
    chlorite chlorate and perchlorate they
  • 01:42:40
    have different numbers of oxygen atoms
  • 01:42:41
    but they're all minus one but anyway
  • 01:42:43
    plus one minus one means we just need
  • 01:42:45
    one of each so kclo2 is the correct
  • 01:42:48
    formula it's the only one that had a
  • 01:42:49
    mistake was Choice B which is what we're
  • 01:42:52
    looking for the one that's incorrect
  • 01:42:55
    all right do we have any questions on
  • 01:42:58
    naming ionic compounds
  • 01:43:03
    okay just two more to get through
  • 01:43:06
    um the next one is covalent which in
  • 01:43:07
    some levels is a little bit more
  • 01:43:10
    straightforward except for the assets
  • 01:43:12
    are a little bit
  • 01:43:14
    I only have three answer choices oh okay
  • 01:43:17
    so this is just one we're just giving
  • 01:43:18
    the formula
  • 01:43:20
    um again you don't have to do one on an
  • 01:43:22
    exam question most likely but we're just
  • 01:43:24
    doing three to give you three examples
  • 01:43:25
    of how we name these things all right so
  • 01:43:28
    the first one here is dichlorine Hep
  • 01:43:30
    oxide so for
  • 01:43:32
    binary covalent compounds two non-metal
  • 01:43:35
    elements in the formula chlorine and
  • 01:43:37
    oxygen going from name to formula is
  • 01:43:40
    very straightforward as long as you
  • 01:43:41
    remember the prefixes so the names
  • 01:43:45
    have these prefixes in them that
  • 01:43:46
    directly tell you the number of each
  • 01:43:48
    atom in the formula so dichlorine
  • 01:43:52
    that's going to be
  • 01:43:53
    cl2 and then hepta means seven you leave
  • 01:43:58
    out the a when it's with oxygen but
  • 01:43:59
    still heptamine seven so cl2o7 would be
  • 01:44:03
    the formula of this compound so going
  • 01:44:05
    named a formula or vice versa for binary
  • 01:44:07
    covalence is usually pretty easy to do
  • 01:44:09
    as long as you remember which prefixes
  • 01:44:11
    correspond with which numbers
  • 01:44:14
    the next two are acids which are a
  • 01:44:16
    little bit
  • 01:44:17
    more tricky to find the formulas of now
  • 01:44:21
    the key with acids though is you need to
  • 01:44:22
    figure out which anion is involved in
  • 01:44:26
    forming that acid and that comes from
  • 01:44:28
    the first part of the name and then once
  • 01:44:30
    you have the charge and the formula the
  • 01:44:31
    anion you just pair it with enough
  • 01:44:33
    hydrogens to balance out that charge one
  • 01:44:35
    plus charge for each hydrogen okay so we
  • 01:44:38
    have hydrocyanic acid
  • 01:44:42
    um
  • 01:44:42
    cyanide is this comes from the cyanide
  • 01:44:45
    anion which we have to remember is CN
  • 01:44:47
    minus
  • 01:44:49
    so cyanic comes from cyanide as the
  • 01:44:52
    anion
  • 01:44:53
    formula for Cyanide is CN minus it's one
  • 01:44:55
    of the polyatomic ions on the list that
  • 01:44:58
    you have to know and so if we want to
  • 01:45:00
    make an acid out of this we have a
  • 01:45:01
    negative one charge so we need to add
  • 01:45:03
    one hydrogen to the formula to balance
  • 01:45:06
    that negative one charge so it becomes
  • 01:45:07
    hcn
  • 01:45:10
    all right so anytime you make an acid
  • 01:45:11
    you have to figure out the charge of the
  • 01:45:13
    anion
  • 01:45:14
    either based on his location on the
  • 01:45:16
    periodic table or if it's polyatomic
  • 01:45:18
    based on what the charge for that anion
  • 01:45:19
    is that you have to remember and then
  • 01:45:21
    you balance that charge with enough
  • 01:45:22
    hydrogen atoms to do that so HCM will be
  • 01:45:25
    this one the reason is hydrocyanic acid
  • 01:45:28
    we we use hydro anytime it's not an Oxo
  • 01:45:31
    anion so this ion is polyatomic but it
  • 01:45:33
    doesn't have oxygen in it so anytime it
  • 01:45:35
    doesn't have oxygen in it we still give
  • 01:45:37
    it the hydro prefix that we would use
  • 01:45:39
    for other types of acids also so that's
  • 01:45:42
    hydrocyanic acid hcn
  • 01:45:44
    and then chloric acid we have to
  • 01:45:47
    remember for acids that have ick in the
  • 01:45:50
    name or
  • 01:45:51
    ous is the other one if it's an ick acid
  • 01:45:54
    it comes from an eight anion
  • 01:45:57
    and if it's an acid that has ous so if
  • 01:46:00
    it was chlorous instead that comes from
  • 01:46:03
    an ite
  • 01:46:06
    anion
  • 01:46:07
    so ick corresponds with h from the anion
  • 01:46:10
    us ous corresponds with ice from the
  • 01:46:13
    anion so because this is chloric acid
  • 01:46:15
    that tells us that it comes from the
  • 01:46:17
    chlorate anion because IC changes to ate
  • 01:46:21
    in acid names
  • 01:46:23
    and so chlorate we have to then remember
  • 01:46:25
    again more polyatomic ions definitely
  • 01:46:27
    make sure you're
  • 01:46:29
    up to speed on those is clo3 minus
  • 01:46:32
    that's the chlorate anion formula and
  • 01:46:35
    once again anytime we're forming an acid
  • 01:46:36
    we just need enough h plus hydrogen to
  • 01:46:39
    balance the charge well there's only one
  • 01:46:40
    negative charge so this one also just
  • 01:46:42
    needs one hydrogen in the formula to
  • 01:46:45
    make it hclo3
  • 01:46:48
    okay so you for so for naming acids it's
  • 01:46:51
    probably the most complicated naming
  • 01:46:54
    that we had to be honest it does require
  • 01:46:56
    you to know the names and charges that
  • 01:46:58
    ions and the anions as well that are
  • 01:47:01
    then used to form that acid
  • 01:47:03
    all right any questions on nomenclature
  • 01:47:06
    of covalent compounds either binary
  • 01:47:08
    compounds or acids
  • 01:47:12
    all right
  • 01:47:14
    very last question then on this review
  • 01:47:15
    is another one dealing with Lewis
  • 01:47:17
    structures
  • 01:47:20
    um in this one we're going to specify
  • 01:47:21
    that we want to draw the Lewis structure
  • 01:47:23
    of this compound here
  • 01:47:27
    that minimizes formal charges
  • 01:47:31
    and we want to know how many double
  • 01:47:32
    Bonds are there and how many lone pairs
  • 01:47:35
    are on the central selenium atom
  • 01:47:37
    respectively so again an ambiguous
  • 01:47:39
    formulas like this will often specify
  • 01:47:41
    the sensor atom so if you read the
  • 01:47:43
    question all the way through you'll see
  • 01:47:45
    that se should be at the center and then
  • 01:47:48
    what we're going to do is we're going to
  • 01:47:49
    draw the Lewis structure we're going to
  • 01:47:50
    minimize the formal charge and then
  • 01:47:52
    we're going to figure out how to answer
  • 01:47:55
    the questions how many double bonds and
  • 01:47:56
    how many lone pairs are there okay so
  • 01:47:59
    let's count the electrons first we have
  • 01:48:01
    chlorine selenium and oxygen in the
  • 01:48:04
    formula
  • 01:48:05
    so we're going to go to the periodic
  • 01:48:06
    table and locate those elements and
  • 01:48:09
    figure out how many valence electrons
  • 01:48:10
    they each have
  • 01:48:12
    chlorine is group seven so there's going
  • 01:48:14
    to be seven valence electrons from each
  • 01:48:15
    chlorine selenium is group six so
  • 01:48:18
    they're going to be six from that one
  • 01:48:20
    oxygen is also group six so six more
  • 01:48:23
    from oxygen so six six and seven are the
  • 01:48:26
    valence electron counts for those three
  • 01:48:27
    elements
  • 01:48:29
    there's two chlorines in the formula
  • 01:48:31
    each with seven electrons as our to our
  • 01:48:34
    valence count
  • 01:48:35
    selenium there's one in the formula and
  • 01:48:37
    it contributes six and oxygen there's
  • 01:48:40
    one also determining six
  • 01:48:42
    all right so 14 plus 6 plus 6 is 26
  • 01:48:46
    valence electrons
  • 01:48:49
    those electrons that we want to include
  • 01:48:50
    in this Lewis structure on the way to
  • 01:48:53
    minimizing the formal charge and so
  • 01:48:56
    we'll start with our Central atom
  • 01:48:58
    single bonds to each of the outer atoms
  • 01:49:01
    in whatever
  • 01:49:02
    arrangement we put them in and we'll
  • 01:49:04
    complete the octets on the outer atoms
  • 01:49:06
    first just like we always do
  • 01:49:08
    so six more electrons go onto each outer
  • 01:49:11
    atom to give them a total of eight two
  • 01:49:13
    from the bond plus the six more that
  • 01:49:15
    we've just added
  • 01:49:17
    and what we've done now is we've
  • 01:49:19
    completed
  • 01:49:20
    three octets all three outer atoms have
  • 01:49:23
    a complete our test so three times eight
  • 01:49:24
    is twenty-four the total number that we
  • 01:49:27
    have available is 26 and so we need to
  • 01:49:30
    have two more electrons on the central
  • 01:49:31
    atom as lone pair
  • 01:49:34
    all right so if we were asking you to
  • 01:49:37
    draw the Lewis structure that follows
  • 01:49:39
    the octet rule this is where we would
  • 01:49:42
    stop because we have exactly eight
  • 01:49:44
    electrons around each atom eight from
  • 01:49:46
    the outer atoms which we took care of by
  • 01:49:49
    adding all the lone Pairs and then by
  • 01:49:50
    putting this lone pair on the central
  • 01:49:51
    atom that also has exactly eight but
  • 01:49:54
    what we're asked to do in this problem
  • 01:49:56
    here
  • 01:49:58
    is to minimize formal charges and so
  • 01:50:01
    when we calculate formal charges in this
  • 01:50:03
    compound
  • 01:50:07
    for the three atoms in the formula
  • 01:50:11
    for chlorine it's in group seven so I
  • 01:50:13
    had seven valence electrons
  • 01:50:15
    it has one Bond
  • 01:50:17
    and each chlorine has six non-bonding
  • 01:50:20
    electrons right now the two chlorines
  • 01:50:21
    are identical so their formal charges
  • 01:50:23
    both come out to zero typically if you
  • 01:50:25
    have outer atom halogen atoms are going
  • 01:50:28
    to get a single Bond and that's it so
  • 01:50:30
    those ones are fine but then if we go to
  • 01:50:31
    selenium and oxygen as is drawn right
  • 01:50:33
    now so selenium is a group six element
  • 01:50:36
    has one Bond or sorry three bonds one
  • 01:50:39
    two three one to each outer atom
  • 01:50:42
    and it has the two non-bonding electrons
  • 01:50:45
    on it so six minus three minus two is
  • 01:50:48
    plus one
  • 01:50:49
    and then for oxygen which is also group
  • 01:50:52
    six it has one Bond just a single bond
  • 01:50:55
    to the central atom and it has six
  • 01:50:57
    non-bonding electrons around it so
  • 01:51:00
    that's minus one so in this structure
  • 01:51:02
    that we've drawn which follows the octet
  • 01:51:04
    rule we we have a plus one and a minus
  • 01:51:08
    one form we'll charge right next to each
  • 01:51:09
    other and we're asking you to minimize
  • 01:51:11
    formal charges here so to do that we
  • 01:51:14
    have to expand the X head of selenium
  • 01:51:15
    but we can because it's in the what
  • 01:51:17
    fourth row of the periodic table way
  • 01:51:19
    down there so if we have a plus one a
  • 01:51:21
    minus one next to each other the way
  • 01:51:22
    that we get rid of that
  • 01:51:24
    is we take off a lone pair
  • 01:51:27
    and make it into a double bond
  • 01:51:29
    so we remove one of the lone pairs from
  • 01:51:31
    oxygen turn it into a double bond
  • 01:51:33
    between the two and now when we
  • 01:51:35
    calculate formal charges
  • 01:51:38
    I'm just going to erase what I did and
  • 01:51:40
    recalculate them here so I don't take up
  • 01:51:42
    more space
  • 01:51:44
    um for selenium it now has two bonds
  • 01:51:47
    because it has or it knows four bonds
  • 01:51:49
    because then you have the double bond
  • 01:51:50
    plus the two single bonds and it still
  • 01:51:52
    has its two non-binding electrons so
  • 01:51:55
    that comes out to zero and then this
  • 01:51:57
    oxygen has now a double bond
  • 01:52:00
    two bonds and then the four nonline
  • 01:52:02
    electrons that it's left with also comes
  • 01:52:04
    out to zero so this structure here which
  • 01:52:07
    has a selenium oxygen double bond is the
  • 01:52:10
    one that minimizes formal charge now
  • 01:52:12
    every single atom in the structure has
  • 01:52:13
    zero formal charge which is ideal where
  • 01:52:16
    you'd like to get to for a neutral atom
  • 01:52:18
    or neutral molecule if we're asking you
  • 01:52:20
    to minimize formal charge so that's the
  • 01:52:22
    one that has and so we have how many
  • 01:52:23
    double bonds we have one double bond
  • 01:52:26
    foreign
  • 01:52:28
    pairs on the central atom well whether
  • 01:52:31
    we minimize formal charges or not we
  • 01:52:33
    have that one lone pair so there's one
  • 01:52:34
    of each one one double bond and one lone
  • 01:52:37
    pair on the central atom so Choice B is
  • 01:52:39
    the correct one here so for problems
  • 01:52:42
    like this
  • 01:52:43
    you need to be able to draw Lewis
  • 01:52:44
    structures very well both for this exam
  • 01:52:47
    and for the next one as it comes out for
  • 01:52:49
    the stuff we're doing in chapter four
  • 01:52:51
    and you know for Lewis structure
  • 01:52:53
    questions you have to make sure you read
  • 01:52:54
    the question closely and if it matters
  • 01:52:56
    whether octet rule or you know formal
  • 01:53:00
    charges should be considered will
  • 01:53:01
    typically specify that as we did here
  • 01:53:04
    all right so that takes us to the end of
  • 01:53:06
    the 20 questions so any any questions on
  • 01:53:09
    that one or anything else that you think
  • 01:53:11
    might come up on the first exam that you
  • 01:53:13
    want me to cover before I adjourn this
  • 01:53:16
    this meeting
  • 01:53:19
    all right
  • 01:53:20
    go for it
  • 01:53:23
    um
  • 01:53:24
    I was having a little trouble on the
  • 01:53:26
    homework questions where they ask you
  • 01:53:28
    how many unpaired electrons a particular
  • 01:53:31
    atom has
  • 01:53:32
    uh especially in the like transition
  • 01:53:35
    metals
  • 01:53:36
    if
  • 01:53:37
    yeah so for those ones it's going to be
  • 01:53:41
    kind of related to
  • 01:53:43
    we didn't do an exact question like that
  • 01:53:45
    but it's related to this concept here of
  • 01:53:47
    following Hun's rule that we talked
  • 01:53:49
    about way up here
  • 01:53:51
    in number what was it nine or ten number
  • 01:53:54
    10 here
  • 01:53:55
    so you want to sort of consider diagrams
  • 01:53:57
    like this so if we give you a transition
  • 01:53:59
    metal
  • 01:54:00
    let's say we do
  • 01:54:02
    iron and we want to know how many
  • 01:54:04
    unpaired electrons there are
  • 01:54:06
    so the starting point for that is going
  • 01:54:08
    to be the electron configuration
  • 01:54:11
    so for iron it's going to be
  • 01:54:14
    Argon
  • 01:54:15
    sorry 4s2
  • 01:54:20
    3d6 let me see if I got that right but
  • 01:54:22
    that's right for iron
  • 01:54:25
    if I didn't get that right I have to
  • 01:54:26
    turn in my inorganic chemistry card
  • 01:54:29
    um so argon is the core and then we have
  • 01:54:32
    4s2 3D one two three four five six so
  • 01:54:36
    for questions that deal with unpaired
  • 01:54:38
    electrons you have to write the
  • 01:54:40
    configuration out first
  • 01:54:42
    now in terms of then what you do from
  • 01:54:43
    there you only need to consider
  • 01:54:45
    subshells that are partially filled to
  • 01:54:47
    identify how many unpaired electrons
  • 01:54:49
    there are anything that's completely
  • 01:54:50
    filled all the electrons have to be
  • 01:54:52
    paired to be able to follow the poly
  • 01:54:54
    Exclusion Principle so all the Argon
  • 01:54:56
    core electrons will be paired and then
  • 01:54:58
    4s2 that's a completely filled subshell
  • 01:55:01
    so you could draw it out but it's going
  • 01:55:02
    to be two paired electrons but then when
  • 01:55:05
    we get to 3d6 that's where there could
  • 01:55:06
    be unpaired electrons that we have to
  • 01:55:08
    figure out how many there are so in the
  • 01:55:10
    3D subshell
  • 01:55:12
    again either using the periodic table or
  • 01:55:15
    using relationships we learned earlier
  • 01:55:17
    l equals 2 for 3D so 2L plus 1 equals
  • 01:55:21
    five orbitals
  • 01:55:23
    or alternatively we know that on the
  • 01:55:25
    periodic table
  • 01:55:27
    the 3D part of the pair table is one two
  • 01:55:30
    three four five six seven eight nine ten
  • 01:55:32
    elements wide meaning that there's up to
  • 01:55:35
    10 electrons or five orbitals present so
  • 01:55:38
    you have to know there's five orbitals
  • 01:55:39
    in a 3D subshell
  • 01:55:43
    and now we just have to follow Hun's
  • 01:55:45
    rule to be able to populate these with
  • 01:55:47
    the six electrons that are in there so
  • 01:55:50
    it's going to go one two three four five
  • 01:55:53
    we individually fill first and then we
  • 01:55:55
    have to pair one up in six so these two
  • 01:55:58
    are paired but the other four would all
  • 01:56:00
    be unpaired in this case so this would
  • 01:56:01
    have four unpaired electrons in the
  • 01:56:03
    ground state
  • 01:56:05
    um and when you follow Hun's Rule and do
  • 01:56:07
    it in that way
  • 01:56:09
    um so that's anytime you're dealing with
  • 01:56:11
    number of unpaired electrons that's kind
  • 01:56:12
    of the level of analysis you want to get
  • 01:56:15
    to for transitional Metals the key is
  • 01:56:16
    that there's going to be you know five a
  • 01:56:19
    set of five D orbitals as you know in
  • 01:56:22
    the valence and
  • 01:56:24
    um you need to fill those with the
  • 01:56:25
    correct number of electrons one at a
  • 01:56:27
    time first before you pair them up
  • 01:56:31
    thank you all right okay good anything
  • 01:56:34
    else related to exam one
  • 01:56:39
    all right um I will have my normal
  • 01:56:42
    availability on Tuesday for you know
  • 01:56:45
    in-person office hours and I'll probably
  • 01:56:47
    be there for a lot of the afternoon as
  • 01:56:48
    well so
  • 01:56:49
    um you know if you need to stop by after
  • 01:56:51
    the regular hours there's a good chance
  • 01:56:52
    I'll be there you can clarify with me on
  • 01:56:55
    other times
  • 01:56:56
    um and you're welcome to send me
  • 01:56:58
    questions via email or any other form of
  • 01:57:01
    communication that you can find me at uh
  • 01:57:04
    so good luck on the first exam and um uh
  • 01:57:08
    you know I encourage you to do the
  • 01:57:10
    practice exam in Blackboard that should
  • 01:57:11
    have and mushrooms opened up yet but
  • 01:57:13
    it'll be open up by tonight if it's not
  • 01:57:15
    already and
  • 01:57:17
    um I will see you guys next week either
  • 01:57:19
    in class or outside of class if you have
  • 01:57:21
    any last questions okay so thank you for
  • 01:57:25
    all thank you all for coming and those
  • 01:57:27
    of you that are watching the recording
  • 01:57:28
    um
  • 01:57:29
    same to you and I'll I'll see you guys
  • 01:57:31
    in a little bit
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