JEE Chemistry 2025 Question with Solution

Given: The low-spin complexes are $K_3[Co(NO_3)_6], \, K_4[Fe(CN)_6], \, K_3[Fe(CN)_6], \, Cu_2[Fe(CN)_6], \, Zn_2[Fe(CN)_6]$.

Find: The sum of the spin-only magnetic moment values of the complexes having yellow colour.

From the extracted solution, the yellow complexes identified are $K_3[Co(NO_3)_6]$ and $K_4[Fe(CN)_6]$.

For $K_3[Co(NO_3)_6]$:

$$\mu = \sqrt{n(n+2)} = 0 \, \text{B.M.}$$

For $K_4[Fe(CN)_6]$:

$$\mu = \sqrt{n(n+2)} = 0 \, \text{B.M.}$$

Therefore, the required sum is

$$0 + 0 = 0$$

So, the nearest integer answer is $0$.

Given: We need the spin-only magnetic moment of the yellow low-spin complexes.

Find: Sum of their magnetic moments.

The solution states that in low-spin complexes, electrons pair in lower-energy orbitals, giving no unpaired electrons for the relevant yellow complexes.

For a spin-only magnetic moment,

$$\mu = \sqrt{n(n+2)}$$

where $n$ is the number of unpaired electrons.

For the yellow complexes considered in the solution:

1. $K_3[Co(NO_3)_6]$ has $n = 0$
2. $K_4[Fe(CN)_6]$ has $n = 0$

Hence,

$$\mu = \sqrt{0(0+2)} = 0 \, \text{B.M.}$$

for each.

Therefore,

$$\text{Total magnetic moment} = 0 + 0 = 0 \, \text{B.M.}$$

Thus, the answer is $0$.