Question

The pair of compounds having the same hybridization for the central atom is:

• A. \( \text{Ni(CO)}_4 \) and \( [\text{PtCl}_4]^{2-} \)
• B. \( \text{Ni(CO)}_4 \) and \( \text{XeO}_2\text{F}_2 \)
• C. \( \text{XeF}_4 \) and \( [\text{SF}_4]^{2-} \)
• D. \( [\text{Co(NH}_3)_6]^{3+} \) and \( [\text{Co(H}_2\text{O})_6]^{3+} \)

Hint:

When determining hybridization, focus on the number of bonds and lone pairs around the central atom. The geometry and hybridization are determined by these factors.

Answer 1

The Correct Option is D

Step 1: Hybridization of the central atom.
- In \( \text{Ni(CO)}_4 \), the nickel atom undergoes sp\(^3\) hybridization. - In \( [\text{PtCl}_4]^{2-} \), the platinum atom also undergoes sp\(^3\) hybridization. - In \( \text{XeO}_2\text{F}_2 \), xenon undergoes sp\(^3\) hybridization. - In \( \text{XeF}_4 \), xenon undergoes sp\(^3\)d\(^2\) hybridization. - In \( [\text{SF}_4]^{2-} \), sulfur undergoes sp\(^3\)d hybridization. - In \( [\text{Co(NH}_3)_6]^{3+} \) and \( [\text{Co(H}_2\text{O})_6]^{3+} \), cobalt undergoes d\(^2\)sp\(^3\) hybridization in both cases.
Step 2: Conclusion.
The pair of compounds with the same hybridization for the central atom is (4) \( [\text{Co(NH}_3)_6]^{3+} \) and \( [\text{Co(H}_2\text{O})_6]^{3+} \) as both involve the central cobalt atom with d\(^2\)sp\(^3\) hybridization.

Answer 2

This question asks which pair of compounds share the same hybridization at their central atom, so each option needs its hybridization worked out independently and then compared within the pair.

1. \( \text{Ni(CO)}_4 \) and \( [\text{PtCl}_4]^{2-} \): In \( \text{Ni(CO)}_4 \), nickel is in the zero oxidation state with a \( d^{10} \) configuration, and with strong-field CO ligands it adopts \( sp^3 \) hybridization giving a tetrahedral shape. In \( [\text{PtCl}_4]^{2-} \), platinum is in the +2 state with a \( d^8 \) configuration, and this ion is square planar, corresponding to \( dsp^2 \) hybridization, not \( sp^3 \). Since the hybridizations differ, this pair does not match.
2. \( \text{Ni(CO)}_4 \) and \( \text{XeO}_2\text{F}_2 \): \( \text{Ni(CO)}_4 \) is \( sp^3 \) hybridized as established above. In \( \text{XeO}_2\text{F}_2 \), xenon has one lone pair along with two Xe=O and two Xe-F bonding regions, five electron domains overall, giving \( sp^3d \) hybridization. Since \( sp^3 \) and \( sp^3d \) differ, this pair does not match.
3. \( \text{XeF}_4 \) and \( [\text{SF}_4]^{2-} \): \( \text{XeF}_4 \) has four bonding pairs and two lone pairs around xenon, six electron domains, giving \( sp^3d^2 \) hybridization. The sulfur centre in \( [\text{SF}_4]^{2-} \) does not reach this same six-domain arrangement, so the two central atoms have different hybridizations, and this pair does not match.
4. \( [\text{Co(NH}_3)_6]^{3+} \) and \( [\text{Co(H}_2\text{O})_6]^{3+} \): In both complexes, cobalt is in the +3 oxidation state with a \( d^6 \) configuration, and in both cases six ligands are arranged octahedrally around the metal, using two \( d \)-orbitals along with one \( s \)- and three \( p \)-orbitals, giving \( d^2sp^3 \) hybridization in each. This pair matches.

Working through the electron-domain count and resulting hybridization for each central atom shows that only the cobalt pair shares an identical hybridization scheme.

Therefore, the correct answer is \( [\text{Co(NH}_3)_6]^{3+} \) and \( [\text{Co(H}_2\text{O})_6]^{3+} \).