transition metals KEYdoc

By Yolanda Sims,2014-11-16 16:59
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transition metals KEYdocmetal,Metal

     Name: _________________________________________ Date:____________________________

    Optional Homework 8 do not turn in!!

    1.) Write the electron configurations for the following species:

    11022626210629a. Ag 1s2s2p3s3p4s3d4p5s4d 5s4d

    010226262106+1b. Ag 1s2s2p3s3p4s3d4p5s4d

    010226262106+2 c. Cd1s2s2p3s3p4s3d4p5s4d

    014622626210621062147+3d. Ir 1s2s2p3s3p4s3d4p5s4d5p6s4f5d 6s4f5d

    2.) Find the charge of all species in each of the following coordination compounds, if it is a polyatomic

    ion, I just want the overall charge of that ion, not individual oxidation numbers of every element!:

    HINT: you may need to go back and review your polyatomic ion charges!!

    a. K[CuCl] 2

    K: +1

    Cu: +1

    Cl: - 1

    b. (NH)[Ni(CN)] 424

    NH: +1 4

    Ni: +2

    CN: -1

    c. [Ti(HO)](SO) 26243

    Ti: +3

    HO: 0 2

    SO: -2 4

    d. Al[V(CN)] 463

    Al: +3

    V: +2

    CN: -1

    3.) Identify the complex ion and the ligands in the compound K[Fe(CN)CO]. Find the oxidation 35

    number of the metal ion in complex ion: be sure to label the inner sphere (ligands bonded

    covalently to the metal) and outer sphere ligands (counter ions) in the complex!

K[Fe(CN)CO]: K is an outer sphere ligand 35

     CN is an inner sphere ligand (-1)

     CO is an inner sphere ligand (0)

     Iron is the transition metal oxidation state = +2

    4.) Find the charge on the nitrosyl ligand (NO) in the Co(III) compound: Na[Co(CN)NO] 25

Na[Co(CN)NO] Na = +1 x 2 = +2 25

     CN = -1 x 5 = -5

     Co = +3

    So the NO ligand must be neutral!

    5.) If the coordination complex [Cr(NH)Cl]Cl dissociated into its ions, what ions would be formed? 352

    Write a balanced equation showing this dissociation!

    [Cr(NH)Cl]Cl ? ??? 352

    +2-1[Cr(NH)Cl]Cl ? [Cr(NH)Cl] + 2Cl35235 (aq) (aq)

    6.) Using the ideas of Lewis and resonance structures, determine which of the following ligands can

    participate in linkage isomerism (meaning can they cause a compound to be a linkage isomer).

    DRAW the resonance structures in order to explain! Remember that a linkage isomer can bond to

    the metal ion from two distinctly different places on the molecule. And in order to have a bond

    you need electrons!!!

    -1a. NO 2


     There are lone pairs present on

    both the N and the O so the species can bond through either the N or O which means this species can be a linkage isomer

    b. SO 2


     There are lone pair electrons

    on the S and the O in the molecule, therefore this species can participate in linkage isomerism

    -1c. NO 3




     There are only

    lone pair electrons on the oxygens in this species, therefore, the only way for this ion to bond is through the oxygen. Since all of the oxygens are equivalent to one another, there is no linkage isomerism possible

    7.) Define/explain the crystal field splitting energy ()?

    Crystal field splitting occurs when incoming ligands “invade” the space of certain d orbitals.

    This causes those d orbitals to become higher in energy than their non-interfered with

    counterparts. Depending on the shape (octahedral, tetrahedral, or square planar) the d

    orbitals will split apart differently. The crystal field splitting energy is the new different in

    energy between the higher energy d orbitals and the lower energy d orbitals as a result of

    the split.


    9.) Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high

    temperatures, as described by the following balanced equation. How many grams of N are formed 2

    when 18.1 g of NH are reacted with 90.4 g of CuO? What starting material, if any is left over? How 3

    many grams of that material are left over?

    2 NH (g) + 3 CuO (s) ; N (g) + 3 Cu (s) + 3 HO (g) 322

    Ahhhh, a good ol’ limiting reagent problem!

    mol1 mol N218.1 g NH x x = 0.531 mol N 322 mol NH17.031 g3

    1 mole CuO1 mol N2 = 0.379 mol N Limiting Reagent!90.4 g CuO x x 23 mol CuO79.545 g CuO

    28.013 g N2 x = 10.6 g N produced0.379 mol N221 mol N2

Since CuO is the limiting reagent, NH is in excess. To determine this amount of excess, determine 3

    how much extra NH, will react with the starting amount of CuO and then subtract the amount 3;;used from the amount of NH you started with. 3

    1 mole CuO2 mol NH17.04 grams NH3390.4 g CuO x x x 12.9 grams NHused3 79.545 g CuO3 mol CuO1 mole NH3

    We started with 18.1 grams of NH and we used up 12.9 grams of it, that leaves 5.2 grams of NH 33


    left over!

    10.) Given the following information, draw the Lewis structure, determine the molecular shape, VSEPR

    shape, hybridization around the central atom, and predicted bond angle for each species.

    a. PH 3

P: 1 x 5 = 5

    H: 1 x 3 = 3

    -1 8 valence e


    o3H tetrahedral shape for both angle 109.5 sp

    b. CF 4

    C: 1 x 4 = 4

    F: 4 x 7 = 28

    -1 32 valence e



    Fo3 tetrahedral for both angle: 109.5 sp

    -2c. SO 4

    S: 1 x 6 = 6

    O: 4 x 6 = 24

    -1 30 + 2 = 32 valence e




    Oo3 tetrahedral for both angle: 109.5 sp

    d. PF 5

    P: 1 x 5 = 5

    F: 5 x 7 = 35

    -1 40 valence e



    FFooo3 trigonal bipyramidal for both angles: 180, 90, and 120 spd

    e. KrF 2

    Kr: 1 x 8 = 8

    F: 2 x 7 = 14

    -1 22 valence e

    FKrFo3 V: trigonal bipyramidal angle: 180 spd

     M: linear (between the atoms)

    -1f. TeF 5

    Te:1 x 6 = 6

    F: 5 x 7 = 35

    -1 41 + 1 = 42 valence e






    oo32 V: octahedral angle: 90 and 180 spd

     M: square based pyramid

    11.) Does the Q for the formation of 1 mole of NO from its elements differ from the Qof the cc

    decomposition of 1 mole of NO into its elements? Explain and give the relationship between the

    two Qvalues. c

    ? N + ? O ? 1NO 22


    Q = 1/21/2[N] [O]22

     1NO ? ? N + ? O 22

    1/21/2[N] [O]22

    Q = [NO]

    The two Q values are inverses of one another! When the equation is reversed, the products

    become the reactants and the reactants become the products, so their location in the Q

    expression changes from denominator to numerator so the Q value is “upside” down or


12.) Balance the reaction and write the Q c

    a. ____2____NaHCO ________NaCO + ________CO + _______HO 3 (s)23 (s)2 (g)2 (g)

    Q = [CO][HO] 22

13.) Given the following Q expressions, write the balanced chemical equation: c

    22a. Q = [CO] [HO] 22

    3 [CH][O]242

The Q expression is products over reactants and the exponents are the coefficients used in the

    balanced equation:

    CH + 3O ? 2CO + 2HO 24222

    -5-7__C__ 9. What is the correct answer to the following expression: 3.33 x 10 + 8.13 x 10?

    -5a. 3 x 10

    -5b. 3.4 x 10

    -5c. 3.41 x 10

    -5d. 3.411 x 10

    -5e. 3.4113 x 10

    __B__ 10. Rank the subatomic particles from lowest to highest mass?

    a. electrons < neutrons < protons

    b. electrons < protons ( neutrons

    c. neutrons < electrons < protons

    d. electrons = protons < neutrons

    _C__ 11. A substance that contains unpaired electrons is attracted to a magnetic field. This

     substance is said to be ________.

    a. Ionic

    b. Ferromagnetic

    c. Paramagnetic

    d. Polar

    e. Diamagnetic

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