Q.74. Identify the metal ions among Co²⁺, Ni²⁺, Fe²⁺, V³⁺ and Ti²⁺ having a spin-only magnetic momen

Q: What is the correct answer for JEE Main Chemistry Q74?

The correct answer is 7. Sol. Magnetic moment $\mu = \sqrt{n(n+2)}$. If $\mu > 3.0$ BM, then $\sqrt{n(n+2)} > 3.0 \Rightarrow n(n+2) > 9$. For $n=3$, $3(3+2)=15 > 9$. For $n=2$, $2(2+2)=8 3.0$ BM). Co²⁺ = [Ar] 3d⁷ (high spin, n=3, $\mu >

Q: Q.74. Identify the metal ions among Co²⁺, Ni²⁺, Fe²⁺, V³⁺ and Ti²⁺ having a spin-only magnetic moment value more than 3.0 BM. The sum of unpaired electrons present in the high spin octahedral…

The correct answer is 7. Sol. Magnetic moment $\mu = \sqrt{n(n+2)}$. If $\mu > 3.0$ BM, then $\sqrt{n(n+2)} > 3.0 \Rightarrow n(n+2) > 9$. For $n=3$, $3(3+2)=15 > 9$. For $n=2$, $2(2+2)=8 3.0$ BM). Co²⁺ = [Ar] 3d⁷ (high spin, n=3, $\mu >

Question

Q.74. Identify the metal ions among Co²⁺, Ni²⁺, Fe²⁺, V³⁺ and Ti²⁺ having a spin-only magnetic moment value more than 3.0 BM. The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is .

FirstInTest · Accepted Answer

Sol. Magnetic moment $\mu = \sqrt{n(n+2)}$. If $\mu > 3.0$ BM, then $\sqrt{n(n+2)} > 3.0 \Rightarrow n(n+2) > 9$. For $n=3$, $3(3+2)=15 > 9$. For $n=2$, $2(2+2)=8 < 9$. So, $n$ must be $\ge 3$. V³⁺ = [Ar] 3d² (n=2, $\mu < 3.0$ BM). Ti²⁺ = [Ar] 3d² (n=2, $\mu < 3.0$ BM). Ni²⁺ = [Ar] 3d⁸ (n=2, $\mu < 3.0$ BM). Fe²⁺ = [Ar] 3d⁶ (high spin, n=4, $\mu > 3.0$ BM). Co²⁺ = [Ar] 3d⁷ (high spin, n=3, $\mu > 3.0$ BM). Only for Fe²⁺ and Co²⁺ $\mu$ is more than 3.0 B.M. Number of unpaired electrons = 4 (for Fe²⁺) + 3 (for Co²⁺) = 7.

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