Question

The correct order of the calculated spin-only magnetic moments of complexes A to D is:

(A) $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$
(B) $[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6]{\mathrm{Cl}}_2$
(C) ${\mathrm{Na}}_2\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]$
(D) ${\mathrm{PdCl}}_2{\left({\mathrm{PPh}}_3\right)}_2$

• A. (C) < (D) < (B) < (A)
• B. (A) ≈ (C) ≈ (D) < (B)
• C. (A) ≈ (C)<(B) ≈ (D)
• D. (C) ≈ (D) < (B) < (A)

Answer 1

The correct option is B

(A) ≈ (C) ≈ (D) < (B)

Explanation for the correct option:

Option B: (A) ≈ (C) ≈ (D) < (B)

1. $\mathrm{Pd}{\left({\mathrm{PPh}}_3\right)}_2{\mathrm{Cl}}_2$ : Here Pd is in +2 oxidation state and the configuration of ${\mathrm{Pd}}^{2+}$ is $\left[\mathrm{Kr}\right]4{\mathrm{d}}^8$. As the CFSE value for Pd is very high so all the electrons will be paired and hence magnetic moment for this complex will be zero.
2. $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$: Here Ni is in a 0 oxidation state and the configuration of Ni is $\left[\mathrm{Ar}\right]3{\mathrm{d}}^{8}4{\mathrm{s}}^2$. As here the ligand is carbonyl which is a strong field ligand, all the electrons will be paired and hence magnetic moment for this complex will be zero.
3. ${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$: Here Ni is in a +2 oxidation state and the configuration of ${\mathrm{Ni}}^{2+}=$$\left[\mathrm{Ar}\right]3{\mathrm{d}}^8$. As here the ligand is cyanide which is a strong field ligand, all the electrons will be paired and hence the magnetic moment for this complex will be zero.
4. ${\left[\mathrm{Ni}{\left(\mathrm{H}2\mathrm{O}\right)}_6\right]}^{2+}$: Here Ni is in a +2 oxidation state and the configuration of ${\mathrm{Ni}}^{2+}=$$\left[\mathrm{Ar}\right]3{\mathrm{d}}^8$.. As here the ligand is water which is a weak field ligand, the electrons will not be paired and there are two unpaired electrons in this complex hence magnetic moment for this complex will be $\sqrt{8}\mathrm{BM}.$
5. So the order of magnetic moment is (A) ≈ (C) ≈ (D) < (B).
6. Hence correct option.

Explanation for the incorrect options:

Option A: (C) < (D) < (B) < (A):

1. $\mathrm{Ni}{\left(\mathrm{CO}\right)}_4$ Ni=$\left[\mathrm{Ar}\right]3{\mathrm{d}}^{8}4{\mathrm{s}}^2$ (strong field ligand) μ=0B.M;diamagnetic.
2. $[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6]{\mathrm{Cl}}_2$​ : ${\left[\mathrm{Ni}{\left(\mathrm{H}2\mathrm{O}\right)}_6\right]}^{2+}$, ${\mathrm{Ni}}^{2+}=$$\left[\mathrm{Ar}\right]3{\mathrm{d}}^8$ (weak field ligand) ∴ Number of unpaired ${\mathrm{e}}^{-}=2\mathrm{\mu }=\mathrm{n}(\mathrm{n}+2)=2(2+2)=22\mathrm{B}.\mathrm{M}.$
3. ${\mathrm{Na}}_2\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]$ ${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$ ${\mathrm{Ni}}^{2+}=$$\left[\mathrm{Ar}\right]3{\mathrm{d}}^8$ (strong field ligand) Number of unpaired ${\mathrm{e}}^{-}=0\therefore \mathrm{\mu }=0\mathrm{B}.\mathrm{M}$
4. _​ ${\mathrm{PdCl}}_2{\left(\mathrm{PP}{\mathrm{h}}_3\right)}_2$${\mathrm{Pd}}^{2+}=4{\mathrm{d}}^8$ (diamagnetic) μ=0B.M
5. Hence incorrect option.

Option C: (A) ≈ (C)<(B) ≈ (D)

1. The magnetic spin of A and C can't be the same.
2. Similarly, the magnetic spin of B and D cannot be the same.
3. Hence incorrect option.

Option D: (C) ≈ (D) < (B) < (A)

1. The magnetic spin of D and C can't be the same.
2. Their magnetic field cannot be lesser than A and B.

Therefore, The correct order of the calculated spin-only magnetic moments of complexes A to D is: (A) ≈ (C) ≈ (D) < (B).