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Solutions 4: Coordination Complexes

These are solutions to select problems from homework set #4.

S4.0

\(V^{5+}\): Elementary Vanadium has electron configuration \([Ar] 3d^3 4s^2\). Hence \(V^{5+}\) ions have the same electron configuration as argon:

\[[V^{5+}] = [Ar] = 1s^2 2s^2 2p^6 3s^2 3p^6\]

\(Au^{3+}\): Elementary Gold has electron configuration \([Xe] 4f^{14} 5d^{10} 6s^1\).
Hence electron configuration of \(Au^{3+}\) ions is:

\[[Au^{3+}] = [Xe] 4f^{14} 5d^8 = 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^6 4f^{14} 5d^8\]

\(Fe^{2+}\): Elementary iron has electron configuration \([Ar] 4s^2 3d^6\).
Hence electron configuration of \(Fe^{2+}\) ions is:

\[[Fe^{2+}] = [Ar] 3d^6 = 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6\]

\(Co^{2+}\): Elementary cobalt has electron configuration \([Ar] 4s^2 3d^7\).
Hence electron configuration of \(Co^{2+}\) ions is:

\[[Co^{2+}] = [Ar] 3d^7 = 1s^2 2s^2 2p^6 3s^2 3p^6 3d^7\]

\(Ti^{4+}\): Elementary titanium has electron configuration \([Ar] 3d^2 4s^2\).
Hence electron configuration of \(Ti^{4+}\) ions is same as of argon:

\[[Ti^{4+}] = [Ar] = 1s^2 2s^2 2p^6 3s^2 3p^6\]

S4.1

Fac and mer isomerism can only occur in complex ions with structures that are octahedral in structure. These isomers involve the orientation of three ligands on the ion. Three ligands can be on the same side (or face) and be a fac isomer. Or they can be spread out across the meridian and be a mer isomer. Either way, only octahedral complexes have six ligands and therefore can sport fac and mer isomers. Answer is therefore  c) octahedral.

S4.2

To review isomerism, visit "Structural Isomers".

S4.3

the trans-dischlorobis(ethylenediamine)cobalt(III)ion

CHEM15.png

 cis-dichlorobis(ethylenediamine)cobalt(III)ion is a"Geometric Isomer" For review on ligands and naming of complexes visit "Ligands" and "Nomenclature of Coordination Complexes" and for review on isomers, visit "Geometric Isomers"

4.4

Not for exam 1

4.5

Not for exam 1

4.6

Not for exam 1

4.7

Not for exam 1

4.8
  1. Not for exam 1

S4.10

In period 4, transition elements have a variety of oxidation states. Manganese has the highest number of oxidation states, with Cr, Fe, and Co tied for second.

In period 4, the atomic radii trend shows that the radii decreases as you go from left to right within the period.

In period 4, most elements are paramagnetic, meaning they have unpaired electrons. Copper is diamagnetic when it is in the ion form of Cu+. Fe, Co, and Ni are able to become permanent magnets and display ferromagnetism.

http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/Periodic_Table_of_the_Elements/Atomic_Radii

S4.11

First, establish the oxidation and reduction elements

Oxidation: Fe2+ (aq) -> Fe3+(aq) + e-

Reduction: MnO4-(aq) -> Mn2+

Balance the overall reaction. Manganese is a transition element and goes from a positive 7 value to a +2. This demonstrates the vast range of oxidation numbers for the transition elements, one of their key attributes. To account for the shift of 5 electrons in reduction, we multiply the entire oxidation half reaction by 5.

Therefore, the balanced reaction is 5Fe2+ (aq) + MnO4-(aq) + 8H+ -> Fe3+(aq) + Mn2+ + 4H2O (l). Note the electrons do not appear in the final reaction.

S4.12

Initially, find the theoretical reduction potentials associating each of these in Table P1.

MnO4- / MnO2 = 1.70 V

MnO2/Mn2+ = 1.23 V

Mn2+/Mn (s) = -1.18 V

Next, we will use the additive nature of free energies along with the formula ΔG = -nFE°cell. We cannot simply add the E°cell together. We must calculate the ΔG using the ΔG formula and then add or subtract as seen necessary.

For reaction 1, the equation will be as follows. F is simply a constant. The trick in understanding this problem is to identify the number of electrons being moved around for each one (-n). You don’t have to actually compute the ΔG; simply remove the F (it is a constant and will be factored out) and plug in the E°cell and the number of electrons.

Thus, for MnO4- / Mn2+

E°cell = [3(1.70V) + (-2(1.23V))]/ 5 = 1.552 V

We divide by 5 because the manganese will change 5 in its oxidation states between MnO4- / Mn2+

(Reaction 2)

For MnO2/Mn(s)  E°cell = [2(1.23)V + 2(-1.18)V]/4 = 0.075 V

S4.13

Answer: Following noble gas formalism, the number of electrons from the metal plus the electrons from the carbonyls must be equal to the number of electrons in the nearest noble gas atom.

  1. Zr(CO); Zr has 40 electrons, 14 electrons are contributed from 7 CO to total 54, which is the number of electrons in Xe.
  2. W(CO); W has 74 electrons, 12 electrons are contributed from 6 CO to contribute to a total of 86, which is the number of electrons in Rn.
  3. Pt(CO)4 ; Pt has 78 electrons, 4 CO contribute another 8 electrons for a total of 86, which is the number of electrons in Rn.
  4. The states of various metal carbonyl structures can be attributed to their ability to function in a lattice structure. Generally, compounds that do not properly fit into crystalline structures due to issues involving repulsion or symmetry will be liquids. Conversely, structures that fit well in the crystal lattice, such asTungsten hexacarbonyl, will be solid. Also, we must take the intermolecular forces into consideration. Molecules with weak mass tend to be liquid while higher molecular mass structures have an affinity to be solid.
  5. This structure would most likely be salt-like and therefore ionic, due to the potassium ion. The structure would consist of Kand Mo(CO)6- ions. 

S4.14

Write the electron configuration for each of the following transition elements and ions.

  1.  \(V: [Ar] 4s^2 3d^3\)
  2. \(Mn^{3+}: [Ar] 3d^4\)
  3. \(Fe^{2+}: [Ar] 3d^6\)
  4. \(Cr: [Ar] 4s^1 3d^5\)
  5. \(Co^{2+}: [Ar] 3d^7\)

S4.15

A main group element usually only has one oxidation state that corresponds to its periodic family number although there are some exceptions. (Such as Tl (+1,+3) , Pb (+2, +4) , and Sn (+2, +4) where the lower number represents an inert pair, a pair of electrons not being ionized). However it is common for transition elements exhibit many oxidation states. Most main group elements are colorless, while most transition elements are colored due to d-orbital interactions. Main group elements are also primarily diamagnetic (having no unpaired electrons), while transition elements are mainly paramagnetic (having one or more unpaired electrons)

S4.16

Complete and balance the following reaction:

\[Mo_2O_3 (s) + Al (s) \rightarrow 2Mo(s) + Al_2O_3(s) \]

S4.17

Create potential half-reaction equations for the following in acidic solutions.

a) \(Cr^{3+} (aq)\) as a reducing agent.

Oxidation:

\[Cr^{3+}(aq) \rightarrow Cr^{4+} (aq) + e^-\]

b) \(Sc^{2+} (aq)\) as an oxidizing agent.

Reduction:

\[Sc^{2+}(aq) + 2H^+(aq) +e^- \rightarrow Sc^{3+} (aq) + H_2O (l)\]

S4.18

Use the given standard reduction potentials to construct a standard electrode potential diagram relation to the following species in acidic solution.

 

VO2+ (aq) + 2H+(aq) + e- --> V3+ (aq) + H2O (l)    E0= 0.337V

V3+ (aq) + e- ---> V2+ (aq)       E0= -0.255V

V2+(aq) + 2e- --> V(s)      E0= -1.13

Answer:

a) VO2+/V2+:

E0: (0.337V - 0.255V)/2 = .041V

b) V3+/V

E0: (-0.255V - 1.13V)/3 = -.462V

S4.19

What formulas would you expect for the metal carbonyls of:

  1. Tungsten (W): \(W(CO)_6\); W has 74 electrons and 6 CO contributes another 12 for a total number of 86, The number of electrons that are in Radon (Rn)
  2. Ruthenium (Ru): \(Ru(CO)_5\); Ru has 44 electrons and 5 CO contributes another 10 for a total number of 54, The number of electrons that are in Xenon (Xe)
  3. Technetium (Tc): \(Tc(CO)_5^-\) ; The Tc anion has 44 electrons and 5 CO contributes another 10 for a total number of 54, The number of electrons that are in Xenon (Xe)

S4.20

Write the electron configurations for the following transition elements: a) Sc b) Cr3+ c) Mn7+ d) Fe2+ e) Cu f) Cu2+

a) [Ar]4s23d1 b) [Ar]3d3 c)[Ar] d) [Ar]3d6 e) [Ar]4s13d10 f) [Ar]3d9

Electronic Configurations

S4.21

Compare the transition metals with the main group metals.  You may wish to consider oxidation states, formation of complexes, colors of compounds, and magnetic properties.

Transition Metals Main Group Metals
multiple oxidation states 1 or 2 oxidation states
form complex ions Don’t form complex ions
Typically have color No colors
Many are paramagnetic and certain metals exhibit ferromagnetism (Fe, Co, Ni) Main group metals in group I are usually paramagnetic due to their s1 orbital while group II metals are usually diamagnetic due to their s2 orbitals.

Oxidation States of Transition Metals

S4.21

Why does copper react with HCl, but not zinc? 

The reason this happens is because Zn is more reactive than H. This allows the zinc to form a compound with the Cl atoms.

The equation would look like this: Zn (s) + HCl (aq) --> ZnCl2 + H2

Cu is LESS reactive than H so it makes the reaction not possible. 

The Cell Potential

S4.22

 Write plausible half-equations in acidic solution: 

  1. MnO4- as a reducing agent 
  2. Ni2+ as an oxidizing agent

We can do this by referring to the standard electrode (reduction) potentials. 

a) MnO4- (aq) + 4H+ (aq) + 2e- --> MnO2(s) + H2O (l) 

b) Oxidizing agent = reduced, so you want to flip the equation so the electrons are on the right hand side of the arrows

Ni (s) --> Ni2+ (aq) + 2e

The Cell Potential

S4.23

Construct a cell diagram from the given reaction couple: \(Cr^{3+} / Cr^{2+}: E'_{cell} = -0.9 V ; H+/H2 E'cell = 0 V

Anode (oxidation) ll cathode (reduction) 

\[Pt (s) | Cr^{3+}(aq), Cr^{2+} (aq) || H^+ (aq) | H_2 (g) | Pt (s) \]

see The Cell Potential