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Chapter 23: Problem 41

A four-coordinate complex \(\mathrm{MA}_{2} \mathrm{~B}_{2}\) is prepared and found to have two different isomers. Is it possible to determine from this information whether the complex is square planar or tetrahedral? If so, which is it?

Short Answer

Expert verified
The complex MA₂B₂ can be either square planar or tetrahedral. However, considering the number of isomers, a square planar complex has two isomers (cis- and trans-), while a tetrahedral complex has no isomers. Since the given information states that the complex has two different isomers, it can be concluded that the complex MA₂B₂ is square planar.

Step by step solution

01

Identify possible isomers for square planar geometry

In a square planar complex, the molecule is arranged in a planar fashion with the 4 ligands at the vertices of a square. The possible isomers for a square planar complex are cis- and trans-isomers. Cis-isomers have the two A ligands and the two B ligands adjacent to each other, whereas trans-isomers have one A and one B ligand adjacent to each other. Therefore, there are two different isomers for a square planar complex.
02

Identify possible isomers for tetrahedral geometry

In a tetrahedral complex, the molecule is arranged in a three-dimensional tetrahedral configuration with 4 ligands at the vertices of a tetrahedron. In the case of the MA₂B₂ complex with a tetrahedral geometry, there is no possibility of having different isomers because all vertices of a tetrahedron are adjacent. Regardless of the positioning of the A and B ligands, they always possess a single arrangement.
03

Comparing isomer numbers

Now, we have determined that a square planar MA₂B₂ complex has two isomers (cis- and trans-) while a tetrahedral MA₂B₂ complex doesn't have any isomers. The given information implies that the complex has two different isomers, which matches the isomer count for a square planar MA₂B₂ complex.
04

Conclusion

Based on the given information and the analysis of the isomers for square planar and tetrahedral geometries, we can conclude that the complex MA₂B₂ is square planar.

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Most popular questions from this chapter

The molecule methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\) can act as a monodentate ligand. The following are equilibrium reactions and the thermochemical data at \(298 \mathrm{~K}\) for reactions of methylamine and en with \(\mathrm{Cd}^{2+}(a q):\) $$ \begin{array}{c} \mathrm{Cd}^{2+}(a q)+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) \rightleftharpoons\left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q) \\ \Delta H^{\circ}=-57.3 \mathrm{~kJ} ; \quad \Delta S^{\circ}=-67.3 \mathrm{~J} / \mathrm{K} ; \quad \Delta G^{\circ}=-37.2 \mathrm{~kJ} \\\ \mathrm{Cd}^{2+}(a q)+2 \mathrm{en}(a q) \rightleftharpoons\left[\mathrm{Cd}(\mathrm{en})_{2}\right]^{2+}(a q) \\ \Delta H^{\circ}=-56.5 \mathrm{~kJ} ; \quad \Delta S^{\circ}=+14.1 \mathrm{~J} / \mathrm{K} ; \quad \Delta G^{\circ}=-60.7 \mathrm{~kJ} \end{array} $$ (a) Calculate \(\Delta G^{\circ}\) and the equilibrium constant \(K\) for the following ligand exchange reaction: \(\left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q)+2 \operatorname{en}(a q) \rightleftharpoons\) $$ \left[\mathrm{Cd}(\mathrm{en})_{2}\right]^{2+}(a q)+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) $$ Based on the value of \(K\) in part (a), what would you conclude about this reaction? What concept is demonstrated? (b) Determine the magnitudes of the enthalpic \(\left(\Delta H^{\circ}\right)\) and the entropic \(\left(-T \Delta S^{\circ}\right)\) contributions to \(\Delta G^{\circ}\) for the ligand exchange reaction. Explain the relative magnitudes. (c) Based on information in this exercise and in the "A Closer Look" box on the chelate effect, predict the sign of \(\Delta H^{\circ}\) for the following hypothetical reaction: $$ \begin{aligned} \left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q) &+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \\ \left[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}(a q)+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) \end{aligned} $$

Pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\), abbreviated py, is the molecule (a) Why is pyridine referred to as a monodentate ligand? (b) For the equilibrium reaction $$ \left[\mathrm{Ru}(\mathrm{py})_{4}(\mathrm{bipy})\right]^{2+}+2 \mathrm{py} \rightleftharpoons\left[\mathrm{Ru}(\mathrm{py})_{6}\right]^{2+}+\text { bipy } $$ what would you predict for the magnitude of the equilibrium constant? Explain your answer.

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: $$ \begin{array}{ll} \mathrm{Hb}+\mathrm{O}_{2} \longrightarrow \mathrm{HbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\\ \mathrm{Hb}+\mathrm{CO} \longrightarrow \mathrm{HbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{array} $$ Using these data, estimate the equilibrium constant at \(298 \mathrm{~K}\) for the equilibrium $$ \mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2} $$

Write the formula for each of the following compounds, being sure to use brackets to indicate the coordination sphere: (a) tetraaquadibromomanganese(III) perchlorate (b) bis(bipyridyl) cadmium(II) chloride (c) potassium tetrabromo( ortho-phenanthroline)cobaltate (III) (d) cesium diamminetetracyanochromate(III) (e) tris(ethylenediammine)rhodium(III) tris(oxalato)cobaltate(III)

For each of the following polydentate ligands, determine (i) the maximum number of coordination sites that the ligand can occupy on a single metal ion and (ii) the number and type of donor atoms in the ligand: (a) ethylenediamine (en), (b) bipyridine (bipy), (c) the oxalate anion \(\left(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right),(\mathrm{d})\) the \(2-\) ion of the porphine molecule (Figure 23.13 ); (e) [EDTA] \(]\) -
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