 Solution

Use $C=0.125g^{2}S(S+1)$ for one ion per formula unit, with high-spin $S=5/2,2,3/2,1$ for Mn, Fe, Co, Ni. ThenHigh-spin Mn(II), with an approximately orbitally nondegenerate $d^5$ ground state, stays closest to the free-electron value. Fe, Co and Ni have stronger orbital admixture, particularly Co in this effective Curie-law comparison. Spin-orbit coupling creates magnetic anisotropy, which can make reversal of the weak-ferromagnetic domains slower and enhance field-cooled/zero-field-cooled separation. A field-cooled sample selects domains as it orders; a zero-field-cooled sample initially has competing domains and can remain pinned when a small measuring field is applied.

This is a qualitative rationalization, not a strict proportionality between $g-2$ and the magnetization separation. The anisotropy tensor, domain-wall pinning and experimental timescale also matter; scalar $g$ values alone do not quantitatively fix the divergence.