The "spin-only" magnetic moment [in units of Bohr magneton, (BM)] of Ni 2+ in aqueous solution would be (atomic number of Ni =28 )

Question

The "spin-only" magnetic moment [in units of Bohr magneton, (BM)] of Ni 2+ in aqueous solution would be (atomic number of Ni =28 ) (A) 2.83 (B) 4.90 (C) 0 (D) 1.73

Tardigrade Admin · Accepted Answer

28 Ni arrow[ Ar ] 3 d8 4 s2 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/46986b6a86670eb462a91443f3c1d98c-.png" /> Number of unpaired electrons (n)=2 μ=√n(n+2)=√2(2+2)=√8 ≈ 2.83

Q. The "spin-only" magnetic moment [in units of Bohr magneton, (BM)] of $N i^{2 +}$ in aqueous solution would be (atomic number of $N i = 28$ )

2.83

4.90

0

1.73

$2 8^{N i} \to [A r] 3 d^{8} 4 s^{2}$

Number of unpaired electrons $(n) = 2$

$μ = n (n + 2) = 2 (2 + 2) = 8 \approx 2.83$

Q. The "spin-only" magnetic moment [in units of Bohr magneton, (BM)] of Ni2+ in aqueous solution would be (atomic number of Ni=28 )

2.83

4.90

0

1.73

28Ni→[Ar]3d84s2 Number of unpaired electrons (n)=2 μ=n(n+2)​=2(2+2)​=8​≈2.83

Q. The "spin-only" magnetic moment [in units of Bohr magneton, (BM)] of $N i^{2 +}$ in aqueous solution would be (atomic number of $N i = 28$ )

$2 8^{N i} \to [A r] 3 d^{8} 4 s^{2}$

Number of unpaired electrons $(n) = 2$

$μ = n (n + 2) = 2 (2 + 2) = 8 \approx 2.83$