Spin Only Magnetic Moment of Nickel Complexes

To determine which complex exhibits a spin-only magnetic moment ($\mu_s$) of $2.82\text{ B.M.}$, we use the spin-only formula:

$\mu_s = \sqrt{n(n + 2)}\text{ B.M.}$

where $n$ is the number of unpaired electrons.

Given $\mu_s = 2.82\text{ B.M.}$:
$2.82 = \sqrt{n(n + 2)}$
$(2.82)^2 \approx 8 = n(n + 2)$
$n^2 + 2n - 8 = 0$
$(n + 4)(n - 2) = 0 \implies n = 2$

Thus, the complex must contain $2$ unpaired electrons.

Now, let us analyze the oxidation state, electronic configuration, and magnetic properties of nickel in each given complex:

1. $\text{Ni(CO)}_4$:

   • Oxidation state of $\text{Ni} = 0$.
   • Electronic configuration of $\text{Ni}^0$: $[\text{Ar}] 3d^8 4s^2$.
   • $\text{CO}$ is a strong field ligand and causes complete pairing of electrons, reorganizing the configuration to $3d^{10} 4s^0$.
   • Number of unpaired electrons ($n$) = $0$.
   • Magnetic moment ($\mu_s$) = $0\text{ B.M.}$

2. $[\text{NiCl}_4]^{2-}$:

   • Oxidation state of $\text{Ni} = +2$.
   • Electronic configuration of $\text{Ni}^{2+}$: $[\text{Ar}] 3d^8$.
   • $\text{Cl}^-$ is a weak field ligand and does not cause pairing of electrons.
   • It forms a tetrahedral geometry ($sp^3$ hybridization) where the $d$-orbitals split into $e$ and $t_2$ sets: $e^4 t_2^4$.
   • Number of unpaired electrons ($n$) = $2$.
   • Magnetic moment ($\mu_s$) = $\sqrt{2(2 + 2)} = \sqrt{8} \approx 2.83\text{ B.M.} \approx 2.82\text{ B.M.}$

3. $\text{Ni(PPh}_3)_4$:

   • Oxidation state of $\text{Ni} = 0$.
   • Electronic configuration of $\text{Ni}^0$: $[\text{Ar}] 3d^8 4s^2$.
   • $\text{PPh}_3$ acts as a strong field ligand causing pairing of electrons to give a $3d^{10}$ configuration.
   • Number of unpaired electrons ($n$) = $0$.
   • Magnetic moment ($\mu_s$) = $0\text{ B.M.}$

4. $[\text{Ni(CN)}_4]^{2-}$:

   • Oxidation state of $\text{Ni} = +2$.
   • Electronic configuration of $\text{Ni}^{2+}$: $[\text{Ar}] 3d^8$.
   • $\text{CN}^-$ is a strong field ligand causing pairing of all $8$ $d$-electrons into four $d$-orbitals, resulting in a square planar geometry ($dsp^2$ hybridization).
   • Number of unpaired electrons ($n$) = $0$.
   • Magnetic moment ($\mu_s$) = $0\text{ B.M.}$

Therefore, the complex showing a magnetic moment of $2.82\text{ B.M.}$ is $[\text{NiCl}_4]^{2-}$.

Correct Option: B