Q24 / 180 · COMEDK 2024 (Morning Shift) · Chemistry
A d - block metal $\mathrm{X}(\mathrm{Z}=26)$ forms a compound $[\mathrm{X}(\mathrm{CN})_2(\mathrm{CO})_4]^{+}$. Calculate its spin magnetic moment value.
- A3.87 BM
- B2.83 BM
- C5.92 BM
- Correct answerD1.73 BM
Solution
The metal X is Iron (Fe) with atomic number $Z = 26$. The electronic configuration of Fe is $[Ar] 3d^{6} 4s^{2}$.
In the complex $[Fe(CN)_{2}(CO)_{4}]^{+}$, let the oxidation state of Fe be $x$. Since $CN^{-}$ is $-1$ and $CO$ is neutral, we have $x + 2(-1) + 4(0) = +1$, which gives $x = +3$.
The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^{5}$.
Both $CN^{-}$ and $CO$ are strong field ligands. According to crystal field theory, strong field ligands cause pairing of electrons in the $d$-orbitals. For a $d^{5}$ configuration with strong field ligands, the electrons pair up in the $t_{2g}$ orbitals: $(t_{2g})^{5} (e_{g})^{0}$.
The number of unpaired electrons $n = 1$.
The spin magnetic moment $\mu$ is calculated using the formula $\mu = \sqrt{n(n+2)}$ BM. Substituting $n = 1$, we get $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73$ BM.
Answer: 1.73 BM
In the complex $[Fe(CN)_{2}(CO)_{4}]^{+}$, let the oxidation state of Fe be $x$. Since $CN^{-}$ is $-1$ and $CO$ is neutral, we have $x + 2(-1) + 4(0) = +1$, which gives $x = +3$.
The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^{5}$.
Both $CN^{-}$ and $CO$ are strong field ligands. According to crystal field theory, strong field ligands cause pairing of electrons in the $d$-orbitals. For a $d^{5}$ configuration with strong field ligands, the electrons pair up in the $t_{2g}$ orbitals: $(t_{2g})^{5} (e_{g})^{0}$.
The number of unpaired electrons $n = 1$.
The spin magnetic moment $\mu$ is calculated using the formula $\mu = \sqrt{n(n+2)}$ BM. Substituting $n = 1$, we get $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73$ BM.
Answer: 1.73 BM
Hence, the correct answer is option: 'D'.