Which of the following pairs will show the same magnetic moment ("spin only')?

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$[Cr (H_{2} O)_{6}]$ and $[Fe {(H_{2} O)}_{5} {NO}^{2 +}$

${[Mn (CN)_{6}]}^{4 -}$ and ${[Fe (CN)_{6}]}^{3 -}$

$[Ni (CO)_{4}]$ and ${[Zn {({NH}_{3})}_{4}]}^{2 +}$

All of these

answer is D.

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Detailed Solution

Since they have the same number of electrons in their respective pair they have the same magnetic moment.

(1) Hybridization of $C r^{3 +}, 3 d^{3}, s p^{3} d^{2}$ with three unpaired electrons (weak field as well as strong field ligand). hybridization of three unpaired electrons by $F e^{+}, 3 d^{7}, s p^{3} d^{2}$ (one 4s electron jumps to 3d orbitals for the pairing)

(2) After pairing, $M n^{2 +}, 3 d^{5}, d^{2} s p^{3}$ hybridises with one unpaired electron ( ${CN}^{-}$ is a strong field ligand). After pairing, $F e^{3 +}, 3 d^{5}, d^{2} s p^{5}$ hybridises with one unpaired electron ( ${CN}^{-}$ is a strong field ligand).

(3) $Z n^{2 +}, 3 d^{10}, s p^{3}$ hybridization and all electrons are paired. $N i, 3 d^{8} 4 s^{2}, s p^{3}$ hybridization and all electrons are paired since CO is a strong field ligand (4s- electron hops to 3d-orbitals for the pairing).