JEE Main 2026 — Coordination Compounds Question with Solution

The spin-only magnetic moment is given by the formula $\mu =\sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons.

A. $[Cr(H_{2}O{)}_6{]}^{2+}$: Chromium is in the $+2$ oxidation state. The electronic configuration of $C{r}^{2+}$ is $[Ar]3d^4$. Since $H_{2}O$ is a weak field ligand, no pairing occurs. The number of unpaired electrons is $n=4$.
$\mu =\sqrt{4(4+2)}=\sqrt{24}\approx 4.90$ BM. (A $\to$ III)

B. $[Co(H_{2}O{)}_6{]}^{2+}$: Cobalt is in the $+2$ oxidation state. The electronic configuration of $C{o}^{2+}$ is $[Ar]3d^7$. With $H_{2}O$ as a weak field ligand, the configuration is $t_{2g}^5{e}_g^2$. The number of unpaired electrons is $n=3$.
$\mu =\sqrt{3(3+2)}=\sqrt{15}\approx 3.87$ BM. (B $\to$ I)

C. $[Cu(H_{2}O{)}_6{]}^{2+}$: Copper is in the $+2$ oxidation state. The electronic configuration of $C{u}^{2+}$ is $[Ar]3d^9$. The number of unpaired electrons is $n=1$.
$\mu =\sqrt{1(1+2)}=\sqrt{3}\approx 1.73$ BM. (C $\to$ IV)

D. $[Mn(H_{2}O{)}_6{]}^{2+}$: Manganese is in the $+2$ oxidation state. The electronic configuration of $M{n}^{2+}$ is $[Ar]3d^5$. With $H_{2}O$ as a weak field ligand, the configuration is $t_{2g}^3{e}_g^2$. The number of unpaired electrons is $n=5$.
$\mu =\sqrt{5(5+2)}=\sqrt{35}\approx 5.92$ BM. (D $\to$ II)

The correct matching is A-III, B-I, C-IV, D-II.

Answer: A-III, B-I, C-IV, D-II