Question

Which of the following is/are correct?

$\square$ [CrCl$_2$(CN)$_2$(NH$_3$)$_2$] and [CrCl$_3$(NH$_3$)$_3$], both have d$^2$sp$^3$ hybridisation

$\square$ Magnetic moment of [PdCl$_4$]$^{2-}$ is zero

$\square$ [Cu(NH$_3$)$_4$]$^{2+}$, [Pt(NH$_3$)$_4$]$^{2+}$ and [Ni(CN)$_4$]$^{2-}$, all have dsp$^2$ hybridization of central metal

$\square$ It is difficult to substitute chelating ligands compared to similar monodentate ligands

• A. [CrCl$_2$(CN)$_2$(NH$_3$)$_2$] and [CrCl$_3$(NH$_3$)$_3$], both have d$^2$sp$^3$ hybridisation
• B. Magnetic moment of [PdCl$_4$]$^{2-}$ is zero
• C. [Cu(NH$_3$)$_4$]$^{2+}$, [Pt(NH$_3$)$_4$]$^{2+}$ and [Ni(CN)$_4$]$^{2-}$, all have dsp$^2$ hybridization of central metal
• D. It is difficult to substitute chelating ligands compared to similar monodentate ligands

Answer 1

Answer: (B), (C), (D)

Options 2, 3, and 4 are correct.

Answer 2

1. Option 1:

   • For [CrCl$_2$(CN)$_2$(NH$_3$)$_2$]: Cr is in +4 oxidation state (d$^2$) and can form a d$^2$sp$^3$ hybridized octahedral complex.

   • For [CrCl$_3$(NH$_3$)$_3$]: Cr is +3 (d$^3$); with weak-field Cl$^⁻$ present, it generally forms a high‐spin complex using outer (4d) orbitals giving sp$^3$d$^2$ hybridization rather than d$^2$sp$^3$.

   Thus, Option 1 is incorrect.

2. Option 2:

   [PdCl$_4$]$^{2-}$: Pd(II) is a d$^8$ ion and in a square‐planar geometry (typical for Pd(II)) all electrons are paired.

   Thus, its magnetic moment is zero.

   Option 2 is correct.

3. Option 3:

   • [Cu(NH$_3$)$_4$]$^{2+}$ (d$^9$), [Pt(NH$_3$)$_4$]$^{2+}$ (d$^8$) and [Ni(CN)$_4$]$^{2-}$ (d$^8$) all adopt square‐planar geometries.

   • In square‐planar complexes, the central metal uses dsp$^2$ hybridization.

   Option 3 is correct.

4. Option 4:

   Chelate complexes are more stable (kinetically inert) than analogous monodentate ones due to the chelate effect, making ligand substitution more difficult.

   Option 4 is correct.