JEE Main 2024 — Coordination Compounds Question with Solution

The 'spin-only' magnetic moment for $${\mathrm{Mn}}^{7+}$$ is $$0$$ BM. During the titration of $${\mathrm{KMnO}}_4$$ with oxalic acid in an acidic medium, manganese is reduced to $${\mathrm{Mn}}^{2+}$$, which has a 'spin-only' magnetic moment of $$5.91$$ BM.

Thus, the difference in magnetic moment values is calculated as follows:

$$0\text{BM}$$ (for $${\mathrm{Mn}}^{7+}$$) subtracted from $$5.91\text{BM}$$ (for $${\mathrm{Mn}}^{2+}$$), which results in $$5.91\text{BM}$$.

Rounding this to the nearest integer, we get a difference of approximately $$6$$ BM.