= Solution
Use $C=0.125g^2S(S+1)$ for \b[one] ion per formula unit, with high-spin $S=5/2,2,3/2,1$ for Mn, Fe, Co, Ni. Then
$$
\boxed{g\simeq(2.04,\ 2.20,\ 2.83,\ 2.47).}
$$
High-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.
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