Question

Consider the following low-spin complexes \[ \text{K}_3[\text{Co(NO}_3)_6], \, \text{K}_4[\text{Fe(CN)}_6], \, \text{K}_3[\text{Fe(CN)}_6], \, \text{Cu}_2[\text{Fe(CN)}_6], \, \text{Zn}_2[\text{Fe(CN)}_6] \] The sum of the spin-only magnetic moment values of complexes having yellow colour is: \[ \text{B.M. (answer is nearest integer)} \]

Hint:

For low-spin complexes, the metal ion undergoes hybridization that results in paired electrons in the lower energy orbitals, leading to no unpaired electrons and hence a magnetic moment of 0 B.M.

Answer 1

Correct Answer: 0

To determine the magnetic moment, we first need to consider the electronic configuration of the metal ions and their respective spin states. We are looking for the complexes with yellow colour, which are low-spin complexes.
- (I) \( \text{K}_3[\text{Co(NO}_3)_6] \):
- In the presence of a ligand field, Co\(^{3+}\) ( \(d^6\) configuration) undergoes \( \text{d}^2\text{sp}^3 \) hybridization, and it is a low-spin complex. - Magnetic moment:
\[ \mu = \sqrt{n(n+2)} = 0 \, \text{B.M.} \]
- (II) \( \text{K}_4[\text{Fe(CN)}_6] \):
- Fe\(^{2+}\) ( \(d^6\) configuration) undergoes \( \text{d}^2\text{sp}^3 \) hybridization, and it is also a low-spin complex. - Magnetic moment:
\[ \mu = \sqrt{n(n+2)} = 0 \, \text{B.M.} \]
Thus, the sum of the spin-only magnetic moment values for the complexes with yellow colour is \( 0 \, \text{B.M.} \). Therefore, the correct answer is \( \boxed{(0)} \).