Question

Identify the correct pair of species with the correct description.

$\square$ CO$_2$ $\longrightarrow$ CO$_3^{2-}$ - Change in both hybridization and shape

$\square$ XeF$_4$ $\longrightarrow$ XeF$_5^-$ - Change in hybridization and number of lone pair on central atom

$\square$ SO$_3$ $\longrightarrow$ SO$_3^{2-}$ - Change in both hybridization and number of d-p$\pi$ bonds

$\square$ H$_2$O $\longrightarrow$ H$_3$O$^+$ - No change in hybridization as well as shape

• A. CO$_2$ $\longrightarrow$ CO$_3^{2-}$ - Change in both hybridization and shape
• B. XeF$_4$ $\longrightarrow$ XeF$_5^-$ - Change in hybridization and number of lone pair on central atom
• C. SO$_3$ $\longrightarrow$ SO$_3^{2-}$ - Change in both hybridization and number of d-p$\pi$ bonds
• D. H$_2$O $\longrightarrow$ H$_3$O$^+$ - No change in hybridization as well as shape

Answer 1

Answer: (A), (C)

Options 1 and 3

Answer 2

1. CO₂ → CO₃²⁻

   • CO₂: Linear shape, carbon is sp hybridized.
   • CO₃²⁻: Trigonal planar shape, carbon is sp² hybridized.
   • Conclusion: Both the hybridization and the shape change.

2. XeF₄ → XeF₅⁻

   • XeF₄: Has 4 bonded F atoms and 2 lone pairs (electron geometry: octahedral, molecular shape: square planar).
   • XeF₅⁻: Has 5 F atoms and 1 lone pair (electron geometry: octahedral, molecular shape: square pyramidal).
   • Conclusion: Only the number of lone pairs changes, the overall electron geometry (and hence the hybridization; often taken as sp³d² in both) does not change.

3. SO₃ → SO₃²⁻

   • SO₃: Trigonal planar with sulfur sp² hybridized; typically drawn with three S=O double bonds involving d–p π interactions.
   • SO₃²⁻ (sulfite ion): Trigonal pyramidal shape due to the presence of a lone pair on sulfur and is considered to have sp³ hybridization with a reduced number of effective d–p π bonds.
   • Conclusion: Both the hybridization and the number of d–p π bonds change.

4. H₂O → H₃O⁺

   • H₂O: Bent shape, oxygen is sp³ hybridized with 2 lone pairs.
   • H₃O⁺: Trigonal pyramidal shape, oxygen is still sp³ hybridized (one lone pair remains).
   • Conclusion: While the hybridization remains sp³, the molecular shape changes from bent to trigonal pyramidal. (Thus, “no change” is incorrect.)