1.73 BM
5.90 BM
Option 2: 1.73 BM
To determine the spin-only magnetic moment of the Hexacyanomanganate(II) ion, [Mn(CN)6]4-, we need to find the number of unpaired electrons in the Mn2+ ion.
1. Electronic configuration of Mn:
Mn (Z = 25): [Ar] 3d5 4s2
2. Electronic configuration of Mn2+:
Mn2+: [Ar] 3d5
3. Effect of CN- ligand:
CN- is a strong field ligand. In the presence of strong field ligands, the d-electrons pair up. However, since there are 5 d-electrons, there will be one unpaired electron in the t2g orbital. The configuration will be t2g5eg0
4. Number of unpaired electrons (n):
n = 1
5. Spin-only magnetic moment formula:
μs = √[n(n+2)] BM
6. Calculation:
μs = √[1(1+2)] BM
μs = √[1(3)] BM
μs = √3 BM
μs ≈ 1.73 BM
Therefore, the spin-only magnetic moment of [Mn(CN)6]4- is approximately 1.73 BM.
The correct answer is:
Option 2: 1.73 BM
The spin-only magnetic moment (\(\mu\)) value (B.M.) of the compound with the strongest oxidising power among \(Mn_2O_3\), \(TiO\), and \(VO\) is ……. B.M. (Nearest integer).
List-I | List-II |
---|---|
(A) Mn2+ | (I) Pyrolusite ore |
(B) Spin only Magnetic Moment | (II) An alloy of 4f metal, iron and traces of S, C, Al and Ca |
(C) MnO2 | (III) μs = √n(n + 2) BM |
(D) Misch metal | (IV) Highest oxidation states |