Q46 / 180 · COMEDK 2023 (Evening Shift) · Chemistry
Match the Coordination compounds given in Column I with their characteristic features listed in Column II.
S.No. Coordination compounds S.No. Characteristic features W $[Co(NH_3)_5Cl]Cl_2$ P Oxidation state $= +3$, Configuration $= d^5$, $\mu = 5.92$ BM X $K_4[Mn(CN)_6]$ Q Oxidation state $= +3$, Configuration $= d^3$, $\mu = 3.87$ BM Y $[CrCl_2(py)_3]$ R Oxidation state $= +3$, Configuration $= d^6$, $\mu = 0$ BM Z $Cs[FeCl_4]$ S Oxidation state $= +2$, Configuration $= d^5$, $\mu = 1.732$ BM
| S.No. | Coordination compounds | S.No. | Characteristic features |
|---|---|---|---|
| W | $[Co(NH_3)_5Cl]Cl_2$ | P | Oxidation state $= +3$, Configuration $= d^5$, $\mu = 5.92$ BM |
| X | $K_4[Mn(CN)_6]$ | Q | Oxidation state $= +3$, Configuration $= d^3$, $\mu = 3.87$ BM |
| Y | $[CrCl_2(py)_3]$ | R | Oxidation state $= +3$, Configuration $= d^6$, $\mu = 0$ BM |
| Z | $Cs[FeCl_4]$ | S | Oxidation state $= +2$, Configuration $= d^5$, $\mu = 1.732$ BM |
- A$\mathrm{W}=\mathrm{S} \quad \mathrm{X}=\mathrm{R} \quad \mathrm{Y}=\mathrm{Q} \quad \mathrm{Z}=\mathrm{P}$
- B$\mathrm{W}=\mathrm{S} \quad \mathrm{X}=\mathrm{P} \quad \mathrm{Y}=\mathrm{Q} \quad \mathrm{Z}=\mathrm{R}$
- C$\mathrm{W}=\mathrm{R} \quad \mathrm{X}=\mathrm{S} \quad \mathrm{Y}=\mathrm{P} \quad \mathrm{Z}=\mathrm{Q}$
- Correct answerD$\mathrm{W}=\mathrm{R} \quad \mathrm{X}=\mathrm{S} \quad \mathrm{Y}=\mathrm{Q} \quad \mathrm{Z}=\mathrm{P}$
Solution
For W: $\left[\mathrm{Co}(\mathrm{NH}_3)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$ : Co is in $+3$ oxidation state. $Co^{3+}$ is $d^6$. With strong field ligands like $\mathrm{NH}_3$, it forms a low spin complex with $\mu = 0$ BM. This corresponds to R.
For X: $\mathrm{K}_4\left[\mathrm{Mn}(\mathrm{CN})_6\right]$. Mn is in $+2$ oxidation state. $Mn^{2+}$ is $d^5$. $\mathrm{CN}^{-}$ is a strong field ligand, causing pairing, resulting in one unpaired electron ($t_{2g}^5 e_g^0$). $\mu = \sqrt{1(1+2)} = 1.732$ BM. This corresponds to S.
For Y: $\left[\mathrm{CrCl}_3(\mathrm{py})_3\right]$. Cr is in $+3$ oxidation state. $Cr^{3+}$ is $d^3$. In octahedral geometry, $t_{2g}^3 e_g^0$, three unpaired electrons. $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ BM. This corresponds to Q.
For Z: $\mathrm{Cs}\left[\mathrm{FeCl}_4\right]$. Fe is in $+3$ oxidation state. $Fe^{3+}$ is $d^5$. $\mathrm{Cl}^{-}$ is a weak field ligand, forming a high spin tetrahedral complex. Five unpaired electrons, $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM. This corresponds to P.
Answer: $\mathrm{W}=\mathrm{R} \quad \mathrm{X}=\mathrm{S} \quad \mathrm{Y}=\mathrm{Q} \quad \mathrm{Z}=\mathrm{P}$
For X: $\mathrm{K}_4\left[\mathrm{Mn}(\mathrm{CN})_6\right]$. Mn is in $+2$ oxidation state. $Mn^{2+}$ is $d^5$. $\mathrm{CN}^{-}$ is a strong field ligand, causing pairing, resulting in one unpaired electron ($t_{2g}^5 e_g^0$). $\mu = \sqrt{1(1+2)} = 1.732$ BM. This corresponds to S.
For Y: $\left[\mathrm{CrCl}_3(\mathrm{py})_3\right]$. Cr is in $+3$ oxidation state. $Cr^{3+}$ is $d^3$. In octahedral geometry, $t_{2g}^3 e_g^0$, three unpaired electrons. $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ BM. This corresponds to Q.
For Z: $\mathrm{Cs}\left[\mathrm{FeCl}_4\right]$. Fe is in $+3$ oxidation state. $Fe^{3+}$ is $d^5$. $\mathrm{Cl}^{-}$ is a weak field ligand, forming a high spin tetrahedral complex. Five unpaired electrons, $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM. This corresponds to P.
Answer: $\mathrm{W}=\mathrm{R} \quad \mathrm{X}=\mathrm{S} \quad \mathrm{Y}=\mathrm{Q} \quad \mathrm{Z}=\mathrm{P}$
Hence, the correct answer is option: 'D'.