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

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

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

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

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

Hence, the correct option is D.