Question

The transition metal complex(es) with zero magnetic moment, zero dipole moment and CFSE of –2.4 \( \Delta_o \) is(are)

• A. \([ \text{Mn(CO)}_5 ] \)
• B. \([ \text{trans}-\text{Ni}(\text{ethylene diamine})_2\text{Cl}_2 ] \)
• C. \([ \text{trans}-\text{Co(CN)}_4(\text{H}_2\text{O})]^{2-} \)
• D. \([ \text{trans-Fe(CN)}_4\text{Cl}_4]^{3-} \)

Hint:

For zero magnetic moment and dipole moment, look for low-spin, symmetric complexes with appropriate electron configurations.

Answer 1

The Correct Option is C, D

Step 1: Analysis of Requirements

The complex(es) must satisfy three criteria:

Zero Magnetic Moment ($\mu=0$): Requires zero unpaired electrons ($n=0$), meaning the complex is diamagnetic. This typically implies a low-spin configuration for $\text{d}^4$, $\text{d}^6$, $\text{d}^8$, or a $\text{d}^{10}$ configuration.

Zero Dipole Moment ($\mu_d=0$): Requires the complex to have a symmetrical geometry so that all bond dipoles cancel out. Common examples are $\text{square planar}$ (e.g., $\text{cis}$ isomers may have a dipole, $\text{trans}$ usually do not) and $\text{octahedral}$ ($\text{trans}$ isomers often have $\mu_d=0$).

CFSE $\mathbf{= -2.4\Delta_0}$: Crystal Field Stabilization Energy of $\mathbf{-2.4\Delta_0}$ is calculated for a $\mathbf{\text{d}^6}$ low-spin octahedral complex.

$$\text{CFSE}(\text{low-spin } \text{d}^6) = (4 \times -0.4\Delta_0) + (2 \times 0.6\Delta_0) = -1.6\Delta_0 + 1.2\Delta_0 = -0.4\Delta_0$$

Wait, this CFSE is incorrect. The low-spin $\text{d}^6$ configuration is $\mathbf{(t_{2g})^6(e_g)^0}$.

$$\text{CFSE}(\text{low-spin } \text{d}^6) = (6 \times -0.4\Delta_0) + (0 \times 0.6\Delta_0) = \mathbf{-2.4\Delta_0}$$

This confirms the complex must be $\mathbf{\text{d}^6}$ low-spin and octahedral.

Step 2: Analysis of Options

(A) $\mathbf{[\text{Mn}(\text{CO})_5(\text{CH}_3)]}$

• Oxidation State & $\mathbf{d}$-count: $\text{CO}$ and $\text{CH}_3$ are 2-electron donors. This is a 18-electron complex. $\text{Mn}$ is in the $+1$ oxidation state (formal charge on $\text{CH}_3$ is $-1$). $\text{Mn}(I) = \mathbf{\text{d}^6}$.
• Geometry: Octahedral ($\text{Mn}$ is bonded to $5\text{CO}$ and $1\text{CH}_3$).
• Magnetism/CFSE: $\text{CO}$ is a very strong field ligand; $\text{d}^6$ will be low-spin ($\mu=0$).
• $\text{CFSE} = -2.4\Delta_0$. Matches.
• Dipole Moment: The complex is $\text{Mn}(\text{CO})_5(\text{CH}_3)$. This is a mono-substituted octahedral complex, and substitution will be non-symmetrical, giving it a non-zero dipole moment. Does not match.

(B) $\mathbf{[\text{trans}-\text{Ni}(\text{ethylenediamine})_2\text{Cl}_2]}$

• Oxidation State & $\mathbf{d}$-count: $\text{en}$ (ethylenediamine) is neutral, $\text{Cl}$ is $-1$. $\text{Ni}$ is in the $+2$ oxidation state. $\text{Ni}(II) = \mathbf{\text{d}^8}$.
• Magnetism/CFSE: $\text{d}^8$ complexes in octahedral geometry are typically high spin.
• $\text{CFSE} = (6 \times -0.4\Delta_0) + (2 \times 0.6\Delta_0) = -2.4\Delta_0 + 1.2\Delta_0 = -1.2\Delta_0$. Does not match CFSE.
• $\text{d}^8$ has two unpaired electrons in $t_{2g}^6 e_g^2$, so $\mu \neq 0$. Does not match $\mu=0$.
• Dipole Moment: $\text{trans-M}(\text{AA})_2\text{X}_2$ (where AA is a symmetrical bidentate ligand) is symmetrical and has a zero dipole moment.

(C) $\mathbf{[\text{trans}-\text{Co}(\text{CN})_4(\text{H}_2\text{O})_2]^{-}}$

• Oxidation State & $\mathbf{d}$-count: $\text{CN}$ is $-1$, $\text{H}_2\text{O}$ is neutral. $\text{Co}$ is in the $+3$ oxidation state. $\text{Co}(III) = \mathbf{\text{d}^6}$.
• Geometry: Octahedral (6 ligands).
• Magnetism/CFSE: $\text{CN}$ is a very strong field ligand, forcing $\text{d}^6$ to be low-spin ($\mu=0$).
• $\text{CFSE} = -2.4\Delta_0$. Matches.
• Dipole Moment: The complex is $\text{trans}-\text{Co}(\text{CN})_4(\text{H}_2\text{O})_2^{-}$. Since $\text{CN}$ and $\text{H}_2\text{O}$ are different ligands, and the $\text{CN}$ ligands are in the equatorial plane, the two $\text{H}_2\text{O}$ dipoles cancel, but the $\text{CN}$ dipoles do not perfectly cancel the $\text{H}_2\text{O}$ dipoles. However, for $\text{trans-M}(\text{A})_4\text{B}_2$ geometry, the net dipole moment is often zero if $\text{A}$ and $\text{B}$ are reasonably symmetric, but $\text{CN}$ is strong $\sigma/\pi$ acceptor and $\text{H}_2\text{O}$ is a weak $\sigma$ donor. Due to the high symmetry of the $\text{trans}$ form, the net dipole moment is considered zero. Matches.

(D) $\mathbf{[\text{trans}-\text{Fe}(\text{CN})_4\text{Cl}_2]^{4-}}$

• Oxidation State & $\mathbf{d}$-count: $\text{CN}$ is $-1$, $\text{Cl}$ is $-1$. $\text{Fe}$ is in the $+2$ oxidation state. $\text{Fe}(II) = \mathbf{\text{d}^6}$.
• Geometry: Octahedral (6 ligands).
• Magnetism/CFSE: $\text{CN}$ is a very strong field ligand, forcing $\text{d}^6$ to be low-spin ($\mu=0$).
• $\text{CFSE} = -2.4\Delta_0$. Matches.
• Dipole Moment: The complex is $\text{trans}-\text{Fe}(\text{CN})_4\text{Cl}_2^{4-}$. This has the general form $\text{trans-M}(\text{A})_4\text{B}_2$, which is highly symmetric. The $\text{CN}$ dipoles cancel each other in the equatorial plane, and the two $\text{Cl}$ dipoles cancel each other along the axial axis, resulting in a zero dipole moment. Matches.

Step 3: Conclusion

Both complexes (C) and (D) satisfy all three criteria:

$\text{d}^6$ Low-Spin (gives $\mu=0$ and $\text{CFSE} = -2.4\Delta_0$).

$\text{trans-M}(\text{A})_4\text{B}_2$ Geometry (gives $\mu_d=0$).

Therefore, the correct options are (C) and (D).