Assume strong-field ligands CN^- and NH3 (with Co^3+, Fe^2+, Fe^3+) give low-spin octahedral complexes, H2O gives high-spin ones, and for tetrahedral…

Work out the $d$-configuration, spin state, number of unpaired electrons $n$ and CFSE for each species. $[\mathrm{Ni(H_2O)_6}]^{2+}$ is $t_{2g}^6e_g^2$: $n=2$, CFSE $=-1.2\Delta_o$. $[\mathrm{NiCl_4}]^{2-}$ is tetrahedral $e^4t_2^4$: $n=2$, CFSE $=(-2.4+1.6)\Delta_t=-0.8\times\tfrac49\Delta_o=-0.36\Delta_o$, so pair (A) has equal moments but unequal CFSE. $[\mathrm{Cr(H_2O)_6}]^{3+}$ is $t_{2g}^3$: $n=3$, CFSE $-1.2\Delta_o$, and paired with Ni(II) it has the same CFSE but $n=3$ versus $2$, so (B) fails. $[\mathrm{Fe(CN)_6}]^{4-}$ ($d^6$, low spin, $t_{2g}^6$) and $[\mathrm{Co(NH_3)_6}]^{3+}$ ($d^6$, low spin) are both diamagnetic with CFSE $-2.4\Delta_o$, so (C) works. $[\mathrm{Fe(H_2O)_6}]^{3+}$ has $n=5$, CFSE $0$, while $[\mathrm{Fe(CN)_6}]^{3-}$ has $n=1$, CFSE $-2.0\Delta_o$, so (D) fails.