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20: Transition Metals

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    These are homework exercises to accompany the Textmap created for "General Chemistry: Principles and Modern Applications " by Petrucci et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.


    Draw and write out the electronic configurations of these 6 transition metal ions

    1. Ti4+
    2. V2+
    3. Cu
    4. Mn2+
    5. Cr3+


    1. Ti4+ [Ar]
    2. V2+ [Ar] 3d^3
    3. Cu [Ar]3d^10 4s^1
    4. Mn2+ [Ar]3d^5
    5. Cr3+ [Ar] 3d^3


    Write the electron configuration for the following ions: Co2+, Ni2+, Zn2+, Cu2+ and Ag+


    1. Co2+: [Ar]3d7
    2. Ni2+: [Ar] 3d8
    3. Zn2+: [Ar] 3d10
    4. Cu2+: [Ar] 3d9
    5. Ag+: [Kr]4d10




    By means of orbital diagrams, writing electron configurations for the following transition element atoms and ions: a) Zr; b) Nb3+; c) Mo2+; d)Tc4+; e)Tc2+; and f)Ru3+


    1. [Kr]5s2 4d2
    2. [Kr]4d2
    3. [Kr]4d3
    4. [Kr]4d3
    5. [Kr]4d5
    6. [Kr]4d5

    To review electron configurations of transition metals, please visit "Filling Transition Metal Orbitals".



    Write electron configurations and orbital diagrams for the following transition element atom and ions:

    1. Ag+
    2. Fe2+
    3. V
    4. Mn3+
    5. Zn2+
    6. Cu+


    More Review


    Transition metals have both s and d electrons which are relatively easily removed. Depending on the reaction the metal loses different numbers of electrons.


    Describe the following trends of the periodic table

    1. Atomic radii
    2. Ionization energy
    3. Electronegativity


    1. atomic radii decreases across the periodic table and increases down the periodic table.
    2. Ionization energy increases across the periodic table and decreases down the periodic table.
    3. Electronegativity increases across the periodic table and decreases down the periodic table.


    Explain the differences between the group 1 elements and transition elements based on

    1. oxidation states
    2. colors of compounds
    3. magnetic properties
    4. formation of complexes


    1. group 1 elements all have +1 oxidation states but the transition elements can have many different oxidation states
    2. all alkali metals have their own flame colors but transition metals form colorful compounds due to d-d electronic transitions
    3. transition metals are paramagnetic when the it has unpaired electrons in the d orbital and diamagnetic when all electrons are paired. The group 1 elements are not magnetic
    4. group 1 elements form ionic compounds and transition elements form complexes compounds because of the many pairing possibilities in the d orbital

    For review on characteristics of group 1 elements, read "Group 1: The Alkali Metals" and for review of transition elements, go to "Transition Metals and Coordination Complexes".


    Determine and balance the equation (the aluminum is oxidized):

    \[Al_{(s)} + MoO_{3(s)} → ?\]


    \[2Al_{(s)} + MoO_{3(s)} \rightarrow Mo_{(s)} + Al_2O_{3(s)}\]


    Write the half reaction for the production of Mn2+ from MnO2, a reduction half reaction in an acidic environment


    \[MnO_{2(s)} + 4H^+_{(aq)} + 2e^- \rightarrow Mn^{2+}_{(aq)} + 2H_2O_{(l)}\]


    Find the electron potentials of reactions (a) and (b) of the diagram below.



    1. \(PbO_{2(s)} + 2H^+_{(aq)} + 2e^- → PbO_{(s)} + H_2O_{(l)}\)
    2. \(Pb^{2+}_{(aq)} + 2e^- → Pb_{(s)}\)


    Show the possible Lewis structure for each ligands

    1. CO32-
    2. C2O42-
    3. NH3


    Predict the molecular formulas for the following metal carbonyls:

    1. chromium
    2. ruthenium
    3. palladium


    Carbonyls provide two electrons. Using the number of electrons the metal has, you can determine how many carbonyls can be used to make the total number of electrons of the metal carbonyl equal to the number of electrons as the next closest nobel gas.

    1. Cr(CO)6
    2. Ru(CO)5
    3. Pd(CO)4


    Write the balanced reaction for the thermal decomposition of \(ZnCO_{3(s)}\).


    \[ZnCO_{3(s)} \rightarrow ZnO_{(s)} +CO_{2(g)}\]

    More Review


    What geometric structure Mer and far isomers and explain why other structure cannot have this form of isomers?


    only octahedral can have mer and fac isomers because mer isomers nedd a 90°-90°-180° bond angle and mer isomers need 90°-90°-90°angle. This requires the geometric structure to have 6 ligands attached to it.


    Compete and balance the following reaction: Cr2O3(s) +Al(s) →

    Cr2O3(s) +2Al(s) → Al2O3(s) +Cr(l)


    (a) reduction: Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l) E= +1.33V

    oxidation: 6Fe3+(aq) + 6e- → 6Fe2+(aq) E= 0.771V

    net: Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq) → 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)

    Ecell = E(cathode/reduction) - E(anode/oxidation) = 1.33V - (0.771V) = 0.559V

    (b) ΔG० = -nFEcell = -(6)(96485C/mol)(0.559V) = 323610.69J/mol = 323.61kJ/mol

    (c) ΔG० = -RTln(K)

    323610.69J/mol = -(8.3145J/molK)(298K)ln(K)

    ln(K) = -130.61

    K = 1.89 x 10-57

    (d) Since K is very small, this reaction will not go to completion substantially.

    Please review Aluminum reaction within "Chemistry of Aluminum"


    Predict the following reactions and balance it:

    \[Zn_{(s)}+K_2Cr_2O_{7 (s)} \rightarrow ? \]


    '\[Zn_{(s)}+K_2Cr_2O_{7 (s)} \rightarrow ZnCr_2O_{7(s)}+ 2 K_{(s)}\]


    Draw mer and Fac isomers for the ZA3B3 complex. Assume A and B and different ligands and Z is a random metal


    Describe Chelation and how it relates to polydentates. Give an example of a polydentate.


    Chelation is when a ligand binds to a metal more than one time. This relates to poly dentates because these ligands often chelate to metals. The example in lecture is EDTA, which Kraken due to the fact that it has binging sites 6 times to the metal.


    1. 8H+(aq)+MnO4-(aq)+5e- Mn2+(aq)+4H2O(l)
    2. 4Cr2+(aq)+O2(g)+4H+(aq) 4Cr3+(aq)+2H2O(l)


    half reaction: Cr2+(aq) Cr3+(aq)+e-

    Use half reactions:

    MnO4-(aq) + e- → MnO42-(aq) Eo = +.56

    MnO4-(aq) + 4H+(aq) + 3e- → MnO2(s) + 2H2O(l) Eo = +1.70V

    to construct

    MnO4- ----------MnO42-----------MnO2




    Now, we just need to find the reaction from MnO42- to MnO2,

    by going from MnO4- to MnO2, then from MnO2 to MnO42- we can find the route through

    MnO4-(aq) + e- → MnO42-(aq) Eo = +0.56 ΔG = -nFEo = -F(0.56)V

    MnO2(s) + 2H2O(l) → MnO4-(aq) + 4H+(aq) + 3e- Eocell = -1.70V ΔG = -3F(-1.70)V

    then combining the equations and adding ΔG together

    MnO4-(aq) + MnO2(s) + 2H2O(l) → MnO42-(aq) + MnO4-(aq) + 4H+(aq) + 2e-

    cancelling out spectators

    MnO2(s) + 2H2O(l) → MnO42-(aq) + 4H+(aq) + 2e- ΔG = -F(0.56 + 3(-1.70)) V

    This reaction therefore has a ΔG = -F(.56 + 3(-1.70)) V which is equal to ΔG = -2FEo since it transfers two electrons, and therefore

    Eo = -ΔG/(-2F) = -(-F(.56 + 3(-1.70))) / (-2F) = -(.56 + 3(-1.70))/2 = 2.27V

    So, Electrode potential diagram looks like

    MnO4- ----------MnO42-----------MnO2

    +.56V +2.27V




    Write half-equation to represent each of the following in an acidic solution.

    1. MnO4-(aq) as an oxidizing agent.
    2. Cr2+(aq) as a reducing agent.


    Write the possible half equations to represent each of the following

    1. Cu2+ as an oxidizing agent
    2. Mn2+ as a reducing agent


    1. Cu2+ + 2e- --> Cu
    2. Mn2+ --> Mn4+ + 2e-

    More Review


    Write an equation using \(Mn^{2+}\) ion as an oxidizing agent in basic solution. Write an equation using Fe(s) as a reducing agent.


    • Reduction: 2e-+ MnO2(s)+2H2O(l)->Mn2+ (aq)
    • Oxidation: Ag(s)->Ag+(aq)+e-


    Considering the following chromium species in acidic solution, assemble a standard electrode potential diagram.


    : chapter_27.jpg

    More Review


    Create a standard electrode potential diagram for the reduction of \(Fe^{3+}\) to \(Fe^{2+}\) in acidic solution.


    MnO4-/ MnO42- standard reduction potential is 0.56V. MnO42-/MnO2 standard reduction potential is 2.27V.

    So, the overall Eo=[2(2.27V)-0.56V]/3=1.33V

    Mn3+/Mn2+ standard reduction potential is 1.49V. Mn2+/Mn standard reduction potential: -1.18V



    Predict the molecular formulas for the following metal carbonyls and explain how many electrons are contributed by the metal and how many are contributed by the carbon monoxide and which noble gas electron configuration does the molecule exhibit.

    1. Titanium (T)
    2. Manganese (Mn)
    3. Tungsten (W)


    1. Ti(CO)7 Ti has 22 electrons and 7 CO contribute 14 electrons for a total of 36 electrons which the same number of electron as Kr.
    2. Pt(CO)4; Pt has 78 electrons and 4 CO contribute 8 electrons for a total of 86 electrons which is the element Rn.
    3. W(CO)6; W has 74 electrons and 6 CO contribute 12 electrons for a total of 86 electrons which is the element Rn.


    How many electrons would be in the metals carbonyl

    1. Chromium
    2. Nickel


    1. Chromium Cr(CO)6 36
    2. Nickel Ni(CO)4 36


    What chemical formulas would you expect for the metal carbonyls of

    1. Chromium hexacarbonyl
    2. Iron Pentacarbonyl
    3. Nickel Tetracarbonyl


    1. Cr(CO)6
    2. Fe(CO)5
    3. Ni(CO)4

    More Review


    What would the formulas be for the metal carbonyls of: iron, chromium, and vanadium?

    1. Fe(CO)5
    2. Cr(CO)6
    3. V(CO)6

    To review ligand bonding, please visit the page "Ligands".

    20: Transition Metals is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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