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Chapter 23: Problem 50

Identify each of the following coordination complexes as either diamagnetic or paramagnetic: (a) \(\left.\left[\mathrm{ZnBr}_{4}\right)\right]^{2-}\) (b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (c) \(\mathrm{OsO}_{4}\) (d) \(\left[\mathrm{PtCl}_{4}\right]^{2-}\)

Short Answer

Expert verified
The following coordination complexes can be classified as: (a) \(\left[\mathrm{ZnBr}_{4}\right]^{2-}\) is diamagnetic. (b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) is paramagnetic. (c) \(\mathrm{OsO}_{4}\) is diamagnetic. (d) \(\left[\mathrm{PtCl}_{4}\right]^{2-}\) is diamagnetic.

Step by step solution

01

(a) Analyzing \(\left[\mathrm{ZnBr}_{4}\right]^{2-}\)

First, we'll determine the oxidation state of Zn in the complex. Since the complex has a -2 charge, and each Br ion has a -1 charge, the oxidation state of Zn must be +2. The electron configuration of Zn in its +2 oxidation state (Zn²⁺) is: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10}\] As all the d-orbitals are completely filled, there are no unpaired electrons, and therefore, the complex is diamagnetic.
02

(b) Analyzing \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\)

Next, we'll find the oxidation state of Mn in this complex. As each H₂O ligand has a neutral charge, the oxidation state of Mn must be +3. The electron configuration of Mn in its +3 oxidation state (Mn³⁺) is: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^4\] There are four unpaired electrons in the 3d orbitals. Therefore, the complex is paramagnetic.
03

(c) Analyzing \(\mathrm{OsO}_{4}\)

To find the oxidation state of Os in OsO₄, note that each O atom has a charge of -2 and the complex has a neutral charge. Thus, the oxidation state of Os must be +8. The electron configuration of Os in its +8 oxidation state (Os⁸⁺) is: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6\] As all the d-orbitals are completely filled, there are no unpaired electrons. Hence, the complex is diamagnetic.
04

(d) Analyzing \(\left[\mathrm{PtCl}_{4}\right]^{2-}\)

Finally, let's determine the oxidation state of Pt in this complex. Since each Cl ion has a -1 charge and the complex has a -2 charge, the oxidation state of Pt must be +4. The electron configuration of Pt in its +4 oxidation state (Pt⁴⁺) is: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2\] As all the d-orbitals are completely filled, there are no unpaired electrons. Thus, the complex is diamagnetic. In summary: (a) \(\left[\mathrm{ZnBr}_{4}\right]^{2-}\) is diamagnetic. (b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) is paramagnetic. (c) \(\mathrm{OsO}_{4}\) is diamagnetic. (d) \(\left[\mathrm{PtCl}_{4}\right]^{2-}\) is diamagnetic.

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Most popular questions from this chapter

When Alfred Werner was developing the field of coordination chemistry, it was argued by some that the optical activity he observed in the chiral complexes he had prepared was due to the presence of carbon atoms in the molecule. To disprove this argument, Werner synthesized a chiral complex of cobalt that had no carbon atoms in it, and he was able to resolve it into its enantiomers. Design a cobalt(III) complex that would be chiral if it could be synthesized and that contains no carbon atoms. (It may not be possible to synthesize the complex you design, but we will not worry about that for now.)

(a) Using Werner's definition of valence, which property is the same as oxidation number, primary valence or secondary valence? (b) What term do we normally use for the other type of valence? (c) Why can \(\mathrm{NH}_{3}\) serve as a ligand but \(\mathrm{BH}_{3}\) cannot?

If the lobes of a given \(d\) -orbital point directly at the ligands, will an electron in that orbital have a higher or lower energy than an electron in a \(d\) -orbital whose lobes do not point directly at the ligands?

Indicate the coordination number and the oxidation number of the metal for each of the following complexes: (a) \(\mathrm{Na}_{2}[\mathrm{Co}(\mathrm{EDTA})]\) (b) \(\mathrm{KMnO}_{4}\) (c) \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}\) (d) \(\mathrm{K}_{3} \mathrm{Fe}(\mathrm{CN})_{6}\) (e) \(\mathrm{Rh}\left(\mathrm{PPh}_{3}\right)_{3} \mathrm{Cl}\) (f) $\mathrm{Zn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\left(\mathrm{NH}_{3}\right)_{2}$

Carbon monoxide is toxic because it binds more strongly to the iron in hemoglobin (Hb) than does \(\mathrm{O}_{2}\), as indicated by these approximate standard free-energy changes in blood: $$ \begin{aligned} \mathrm{Hb}+\mathrm{O}_{2} & \longrightarrow \mathrm{HbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\ \mathrm{Hb}+\mathrm{CO} & \longrightarrow \mathrm{HbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{aligned} $$ Using these data, estimate the equilibrium constant at 298 K for the equilibrium $$ \mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2} $$
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