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

The fastest way to spot the matching pair is to notice that hybridization depends mainly on the coordination number and geometry around the central atom, not on which particular ligand is attached.
Both \( [\text{Co(NH}_3)_6]^{3+} \) and \( [\text{Co(H}_2\text{O})_6]^{3+} \) have cobalt in the same +3 oxidation state, surrounded by six donor atoms in an octahedral arrangement, so both must use the same \( d^2sp^3 \) hybridization scheme regardless of whether the ligand is ammonia or water.
None of the other pairs share this kind of matching geometry: \( \text{Ni(CO)}_4 \) is tetrahedral \( (sp^3) \) while \( [\text{PtCl}_4]^{2-} \) is square planar \( (dsp^2) \), \( \text{XeO}_2\text{F}_2 \) is a different five-domain geometry \( (sp^3d) \), and \( \text{XeF}_4 \) (six domains, \( sp^3d^2 \)) does not match the four-domain \( [\text{SF}_4]^{2-} \).
Therefore, the correct answer is the cobalt pair, \( [\text{Co(NH}_3)_6]^{3+} \) and \( [\text{Co(H}_2\text{O})_6]^{3+} \).

Answer 2

Another way to approach this is to first count the number of electron domains around each central atom, since hybridization is a direct consequence of that count, and then check which two compounds in a pair share the same domain count.

1. \( \text{Ni(CO)}_4 \) and \( [\text{PtCl}_4]^{2-} \): Nickel here has four bonding domains and no lone pairs, giving \( sp^3 \). Platinum in \( [\text{PtCl}_4]^{2-} \) also has four bonded chlorides, but because it is a \( d^8 \) metal ion, it retains a filled non-bonding \( d \) orbital that forces a square-planar arrangement, \( dsp^2 \), instead. The matching domain count does not translate into the same hybridization here.
2. \( \text{Ni(CO)}_4 \) and \( \text{XeO}_2\text{F}_2 \): \( \text{Ni(CO)}_4 \) has four domains (\( sp^3 \)), while xenon in \( \text{XeO}_2\text{F}_2 \) carries four bonding regions plus one lone pair, five domains total (\( sp^3d \)). The domain counts differ, so the hybridizations differ.
3. \( \text{XeF}_4 \) and \( [\text{SF}_4]^{2-} \): Xenon in \( \text{XeF}_4 \) has four bonds plus two lone pairs, six domains (\( sp^3d^2 \)). Sulfur in \( [\text{SF}_4]^{2-} \) does not reach six domains under the same reasoning, so the domain counts and hence the hybridizations do not align.
4. \( [\text{Co(NH}_3)_6]^{3+} \) and \( [\text{Co(H}_2\text{O})_6]^{3+} \): Both have cobalt bonded to exactly six donor atoms with no lone pairs on the metal itself, six domains in each case, and in both, two inner \( d \)-orbitals combine with one \( s \) and three \( p \) orbitals to give \( d^2sp^3 \). The domain count matches, and so does the actual orbital combination used.

Counting electron domains around each central atom and confirming that the resulting orbital combination is identical shows that only the two cobalt(III) octahedral complexes share the exact same hybridization.

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