\( [CoF_6]^{3-} \)
The magnetic moment of a metal complex depends on the number of unpaired electrons in the d-orbitals of the central metal ion. The formula for magnetic moment \( \mu_{\text{eff}} \) is given by: \[ \mu_{\text{eff}} = \sqrt{n(n+2)} \] Where \( n \) is the number of unpaired electrons. The more unpaired electrons there are, the higher the magnetic moment. Let’s analyze the complexes:
- (1) \( [NiCl_4]^{2-} \): For Ni²⁺, which has an electronic configuration of \( 3d^8 \), in the presence of chloride ions (Cl⁻), a weak field ligand, it results in two unpaired electrons. The magnetic moment will be moderate.
- (2) \( [Mn(CN)_6]^{4-} \): \( \text{Mn}^{2+} \) has an electronic configuration of \( 3d^5 \), and \( \text{CN}^- \) is a strong field ligand, so it causes pairing of the electrons. In this complex, no unpaired electrons are present, making the magnetic moment zero (since \( [Mn(CN)_6]^{4-} \) is diamagnetic).
- (3) \( [Cr(NH_3)_6]^{3+} \): Cr³⁺ has an electronic configuration of \( 3d^3 \), and ammonia (NH₃) is a weak field ligand, so there will be three unpaired electrons, making this complex paramagnetic with a relatively high magnetic moment.
- (4) \( [CoF_6]^{3-} \): Co³⁺ has an electronic configuration of \( 3d^6 \), and fluoride ions (F⁻) are weak field ligands. This results in four unpaired electrons, so the complex will have a high magnetic moment.
Thus, the highest magnetic moment will be found in \( [CoF_6]^{3-} \) and \( [Cr(NH_3)_6]^{3+} \), with Cr³⁺ having a relatively higher number of unpaired electrons compared to the other complexes. The correct answer is Option (2).