Step 1: Understand the formula for spin-only magnetic moment.

The spin-only magnetic moment (μ) is calculated using the formula: $\mu = \sqrt{n(n+2)} \text{ B.M.}$ where 'n' is the number of unpaired electrons.

Step 2: Determine the number of unpaired electrons (n) for the given magnetic moment.

We are given that μ = 2.82 B.M. Let's solve for 'n': $2.82 = \sqrt{n(n+2)}$ Squaring both sides: $(2.82)^2 = n(n+2)$ $7.9524 = n^2 + 2n$ This value is very close to 8. Let's test integer values for 'n':

• If n = 1, μ = $\sqrt{1(1+2)} = \sqrt{3} \approx 1.73$ B.M.
• If n = 2, μ = $\sqrt{2(2+2)} = \sqrt{8} \approx 2.83$ B.M.
• If n = 3, μ = $\sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ B.M.

The given value of 2.82 B.M. corresponds to n = 2 unpaired electrons. So, we need to find the complex among the options that has 2 unpaired electrons.

Step 3: Analyze each complex to find the number of unpaired electrons.

A: $\mathrm{Ni}(\mathrm{CO})_4$ (Nickel tetracarbonyl)

• Oxidation state of Ni: CO is a neutral ligand, so the oxidation state of Ni is 0.
• Electronic configuration of Ni(0): The atomic number of Ni is 28. Its configuration is $[\mathrm{Ar}] 3d^8 4s^2$.
• Ligand and Hybridization: CO is a strong field ligand. It causes the pairing of electrons. The 4s electrons are pushed into the 3d orbitals.
• The configuration becomes $[\mathrm{Ar}] 3d^{10}$.
• All 3d orbitals are completely filled. The electronic arrangement is: 3d: [↑↓][↑↓][↑↓][↑↓][↑↓].
• Number of unpaired electrons (n): n = 0.
• Magnetic moment: μ = 0 B.M.

B: $\left[\mathrm{NiCl}_4\right]^{2-}$ (Tetrachloronickelate(II))

• Oxidation state of Ni: Let the oxidation state of Ni be 'x'. $x + 4(-1) = -2 \implies x = +2$.
• Electronic configuration of Ni$^{2+}$: From Ni ($[\mathrm{Ar}] 3d^8 4s^2$), we remove two 4s electrons to get Ni$^{2+}$: $[\mathrm{Ar}] 3d^8$.
• Ligand and Hybridization: Cl$^{-}$ is a weak field ligand. It does not cause electron pairing. The complex has a coordination number of 4 and will be tetrahedral ($sp^3$ hybridization).
• The electronic arrangement in the 3d orbitals is based on Hund's rule: 3d: [↑↓][↑↓][↑↓][↑][↑].
• Number of unpaired electrons (n): There are 2 unpaired electrons. n = 2.
• Magnetic moment: μ = $\sqrt{2(2+2)} = \sqrt{8} \approx 2.83$ B.M. This matches the given value.

C: $\mathrm{Ni}\left(\mathrm{PPh}_3\right)_4$ (Tetrakis(triphenylphosphine)nickel(0))

• Oxidation state of Ni: PPh$_3$ is a neutral ligand, so the oxidation state of Ni is 0.
• Electronic configuration of Ni(0): $[\mathrm{Ar}] 3d^8 4s^2$.
• Ligand and Hybridization: PPh$_3$ is a strong field ligand. Similar to CO, it causes the 4s electrons to pair up in the 3d orbitals, resulting in a $[\mathrm{Ar}] 3d^{10}$ configuration.
• Number of unpaired electrons (n): n = 0.
• Magnetic moment: μ = 0 B.M.

D: $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ (Tetracyanidonickelate(II))

• Oxidation state of Ni: Let the oxidation state of Ni be 'x'. $x + 4(-1) = -2 \implies x = +2$.
• Electronic configuration of Ni$^{2+}$: $[\mathrm{Ar}] 3d^8$.
• Ligand and Hybridization: CN$^{-}$ is a strong field ligand. It will force the pairing of the two unpaired electrons in the 3d orbitals.
• The electronic arrangement becomes: 3d: [↑↓][↑↓][↑↓][↑↓][ ]. One 3d orbital is now empty.
• The complex will be square planar with $dsp^2$ hybridization.
• Number of unpaired electrons (n): n = 0.
• Magnetic moment: μ = 0 B.M.

Step 4: Conclusion.

Comparing the calculated magnetic moments:

• $\mathrm{Ni}(\mathrm{CO})_4$: μ = 0 B.M.
• $\left[\mathrm{NiCl}_4\right]^{2-}$: μ ≈ 2.83 B.M.
• $\mathrm{Ni}\left(\mathrm{PPh}_3\right)_4$: μ = 0 B.M.
• $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$: μ = 0 B.M.

The complex with a spin-only magnetic moment of 2.82 B.M. is $\left[\mathrm{NiCl}_4\right]^{2-}$.