JEE Main 2026 — Chemistry Inorganic Chemistry

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

A. $[Cr(H_2O)_6]^{2+}$[Cr(H2​O)6​]2+: Chromium is in the $+2$+2 oxidation state. The electronic configuration of $Cr^{2+}$Cr2+ is $[Ar] 3d^4$[Ar]3d4. Since $H_2O$H2​O is a weak field ligand, no pairing occurs. The number of unpaired electrons is $n = 4$n=4.

$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$μ=4(4+2)​=24​≈4.90 BM. (A $\rightarrow$→ III)

B. $[Co(H_2O)_6]^{2+}$[Co(H2​O)6​]2+: Cobalt is in the $+2$+2 oxidation state. The electronic configuration of $Co^{2+}$Co2+ is $[Ar] 3d^7$[Ar]3d7. With $H_2O$H2​O as a weak field ligand, the configuration is $t_{2g}^5 e_g^2$t2g5​eg2​. The number of unpaired electrons is $n = 3$n=3.

$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$μ=3(3+2)​=15​≈3.87 BM. (B $\rightarrow$→ I)

C. $[Cu(H_2O)_6]^{2+}$[Cu(H2​O)6​]2+: Copper is in the $+2$+2 oxidation state. The electronic configuration of $Cu^{2+}$Cu2+ is $[Ar] 3d^9$[Ar]3d9. The number of unpaired electrons is $n = 1$n=1.

$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73$μ=1(1+2)​=3​≈1.73 BM. (C $\rightarrow$→ IV)

D. $[Mn(H_2O)_6]^{2+}$[Mn(H2​O)6​]2+: Manganese is in the $+2$+2 oxidation state. The electronic configuration of $Mn^{2+}$Mn2+ is $[Ar] 3d^5$[Ar]3d5. With $H_2O$H2​O as a weak field ligand, the configuration is $t_{2g}^3 e_g^2$t2g3​eg2​. The number of unpaired electrons is $n = 5$n=5.

$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$μ=5(5+2)​=35​≈5.92 BM. (D $\rightarrow$→ II)

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

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