Q41 / 180 · COMEDK 2024 (Evening Shift) · Chemistry
Based on Valence Bond Theory, match the complexes listed in Column I with the number of unpaired electrons on the central metal ion, given in Column II
No. Complex ions No. Number of unpaired electrons A $[FeF_6]^{3-}$ P 0 B $[Fe(CN)_6]^{4-}$ Q 1 C $[Fe(H_2O)_6]^{2+}$ R 5 D $[Fe(CN)_6]^{3-}$ S 4
| No. | Complex ions | No. | Number of unpaired electrons |
|---|---|---|---|
| A | $[FeF_6]^{3-}$ | P | 0 |
| B | $[Fe(CN)_6]^{4-}$ | Q | 1 |
| C | $[Fe(H_2O)_6]^{2+}$ | R | 5 |
| D | $[Fe(CN)_6]^{3-}$ | S | 4 |
- Correct answerA$\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{Q}$
- B$\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}$
- C$\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{Q} \quad \mathrm{C}=\mathrm{R} \quad \mathrm{D}=\mathrm{P}$
- D$\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{Q} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{R}$
Solution
For $\left[\mathrm{Fe} \mathrm{F}_6\right]^{3-}$, the oxidation state of Fe is $+3$. The electronic configuration of $\mathrm{Fe}^{3+}$ is $[\mathrm{Ar}] 3d^5$. Since $\mathrm{F}^{-}$ is a weak field ligand, it does not cause pairing. Thus, there are $5$ unpaired electrons. This matches (R).
For $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$, the oxidation state of Fe is $+2$. The electronic configuration of $\mathrm{Fe}^{2+}$ is $[\mathrm{Ar}] 3d^6$. Since $\mathrm{CN}^{-}$ is a strong field ligand, it causes pairing of electrons in the $3d$ orbitals. All $6$ electrons pair up, resulting in $0$ unpaired electrons. This matches (P).
For $\left[\mathrm{Fe}(\mathrm{H}_2 \mathrm{O})_6\right]^{2+}$, the oxidation state of Fe is $+2$. The electronic configuration of $\mathrm{Fe}^{2+}$ is $[\mathrm{Ar}] 3d^6$. Since $\mathrm{H}_2 \mathrm{O}$ is a weak field ligand, it does not cause pairing. The $3d^6$ configuration has $4$ unpaired electrons. This matches (S).
For $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$, the oxidation state of Fe is $+3$. The electronic configuration of $\mathrm{Fe}^{3+}$ is $[\mathrm{Ar}] 3d^5$. Since $\mathrm{CN}^{-}$ is a strong field ligand, it causes pairing. Four electrons pair up, leaving $1$ unpaired electron. This matches (Q).
Comparing these results: A=R, B=P, C=S, D=Q.
Answer: $\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{Q}$
For $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$, the oxidation state of Fe is $+2$. The electronic configuration of $\mathrm{Fe}^{2+}$ is $[\mathrm{Ar}] 3d^6$. Since $\mathrm{CN}^{-}$ is a strong field ligand, it causes pairing of electrons in the $3d$ orbitals. All $6$ electrons pair up, resulting in $0$ unpaired electrons. This matches (P).
For $\left[\mathrm{Fe}(\mathrm{H}_2 \mathrm{O})_6\right]^{2+}$, the oxidation state of Fe is $+2$. The electronic configuration of $\mathrm{Fe}^{2+}$ is $[\mathrm{Ar}] 3d^6$. Since $\mathrm{H}_2 \mathrm{O}$ is a weak field ligand, it does not cause pairing. The $3d^6$ configuration has $4$ unpaired electrons. This matches (S).
For $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$, the oxidation state of Fe is $+3$. The electronic configuration of $\mathrm{Fe}^{3+}$ is $[\mathrm{Ar}] 3d^5$. Since $\mathrm{CN}^{-}$ is a strong field ligand, it causes pairing. Four electrons pair up, leaving $1$ unpaired electron. This matches (Q).
Comparing these results: A=R, B=P, C=S, D=Q.
Answer: $\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{Q}$
Hence, the correct answer is option: 'A'.