Question

Identify the metal ions among $Co^{2+}, Ni^{2+}, Fe^{2+}, V^{3+}$ and $Ti^{2+}$ 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 ____________.

Hint:

For 3d series metal ions, the number of unpaired electrons in high spin octahedral complexes is: $d^1(1), d^2(2), d^3(3), d^4(4), d^5(5), d^6(4), d^7(3), d^8(2), d^9(1), d^{10}(0)$.

Answer 1

Correct Answer: 7

To identify the metal ions with a spin-only magnetic moment greater than 3.0 BM, we can use the formula for the magnetic moment: μ_so = √(n(n+2)), where n is the number of unpaired electrons.

First, we'll calculate μ_so for each metal ion: 

• $Co^{2+}$: Electronic configuration is [Ar] 3d⁷. In a high spin complex, Co²⁺ has 3 unpaired electrons, so μ_so = √(3(3+2)) = √15 ≈ 3.87 BM, which is > 3.0 BM.
• $Ni^{2+}$: Electronic configuration is [Ar] 3d⁸. In a high spin complex, Ni²⁺ has 2 unpaired electrons, so μ_so = √(2(2+2)) = √8 ≈ 2.83 BM, which is < 3.0 BM.
• $Fe^{2+}$: Electronic configuration is [Ar] 3d⁶. In a high spin complex, Fe²⁺ has 4 unpaired electrons, so μ_so = √(4(4+2)) = √24 ≈ 4.90 BM, which is > 3.0 BM.
• $V^{3+}$: Electronic configuration is [Ar] 3d². In a high spin complex, V³⁺ has 2 unpaired electrons, so μ_so = √(2(2+2)) = √8 ≈ 2.83 BM, which is < 3.0 BM.
• $Ti^{2+}$: Electronic configuration is [Ar] 3d². In a high spin complex, Ti²⁺ has 2 unpaired electrons, so μ_so = √(2(2+2)) = √8 ≈ 2.83 BM, which is < 3.0 BM.

Only $Co^{2+}$ and $Fe^{2+}$ have a spin-only magnetic moment greater than 3.0 BM. Their unpaired electrons are:

• $Co^{2+}$: 3 unpaired electrons
• $Fe^{2+}$: 4 unpaired electrons

Thus, the sum of unpaired electrons = 3 + 4 = 7.

 

This value of 7 is within the expected range of 7 to 7.