JEE Chemistry 2024 Question with Solution

Given: The complex is $[Ni(NH_3)_6]^{2+}$ and the atomic number of Ni is $28$.

Find: The value of $x$ in the magnetic moment written as $x \times 10^{-1} \, \text{BM}$.

NH$_3$ acts as a weak field ligand with $Ni^{2+}$, so the complex is outer orbital with hybridization $sp^3d^2$.

For $Ni^{2+}$:

$$Ni^{2+} = 3d^8$$

So there are $2$ unpaired electrons.

Using the spin-only formula:

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

Substituting $n = 2$,

$$\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.82 \, \text{BM}$$

This can be written as $28.2 \times 10^{-1} \, \text{BM}$.

Therefore, the value of $x$ is $28$, so the correct option is A.

Given: $Ni^{2+}$ in the complex $[Ni(NH_3)_6]^{2+}$.

Find: The spin-only magnetic moment in the form $x \times 10^{-1} \, \text{BM}$.

The electronic configuration is:

$$Ni^{2+} = 3d^8$$

In the $3d$ orbitals, the arrangement leaves $2$ electrons unpaired.

Hence,

$$\text{No. of unpaired electrons} = 2$$

Now apply the spin-only magnetic moment formula:

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

Therefore,

$$\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.82 \, \text{BM}$$

Rewriting,

$$2.82 \, \text{BM} = 28.2 \times 10^{-1} \, \text{BM}$$

So the required value is $x = 28$. Therefore, the correct option is A.