Among the following, the CORRECT combinations are: A. IF₃ arrow T-shaped (sp³d)$\mathrm{IF}_3 \rightarrow \text{T-shaped } (sp^3d)$ B. IF₅ arrow Square pyramidal (sp³d²)$\mathrm{IF}_5 \rightarrow \text{Square pyramidal } (sp^3d^2)$ C. IF₇ arrow Pentagonal bipyramidal (sp³d³)$\mathrm{IF}_7 \rightarrow \text{Pentagonal bipyramidal } (sp^3d^3)$ D. ClO₄^- arrow Square planar (sp²d)$\mathrm{ClO}_4^- \rightarrow \text{Square planar } (sp^2d)$ Choose the correct answer from the options given below :

Core Logic

(A) IF₃$\mathrm{IF}_3$: Iodine has 7 valence electrons. 3$3$ are shared with F, leaving 2$2$ lone pairs. Total electron pairs = 3 (bond) + 2 (lone) = 5$3 \text{ (bond)} + 2 \text{ (lone)} = 5$. Hybridization is sp³d$sp^3d$. Geometry is trigonal bipyramidal, shape is T-shaped. (Correct)

(B) IF₅$\mathrm{IF}_5$: Iodine has 7 valence electrons. 5$5$ are shared with F, leaving 1$1$ lone pair. Total electron pairs = 5 (bond) + 1 (lone) = 6$5 \text{ (bond)} + 1 \text{ (lone)} = 6$. Hybridization is sp³d²$sp^3d^2$. Shape is Square pyramidal. (Correct)

(C) IF₇$\mathrm{IF}_7$: Iodine has 7 valence electrons, all 7$7$ shared with F. Zero lone pairs. Total pairs = 7. Hybridization is sp³d³$sp^3d^3$. Shape is Pentagonal bipyramidal. (Correct)

(D) ClO₄^-$\mathrm{ClO}_4^-$: Chlorine forms 4 bonds (mostly double) with Oxygen, plus one extra electron overall. 4 bond regions + 0 lone pairs = 4$4 \text{ bond regions} + 0 \text{ lone pairs} = 4$. Hybridization is sp³$sp^3$. Shape is Tetrahedral. (Incorrect)

Step 1: Final Conclusion

Only combinations A, B, and C are correct.

Pattern Recognition

Steric Number (SN) = (Valence e^-$e^-$ on central atom + Monovalent atoms - Cation charge + Anion charge) / 2. Use SN to quickly deduce hybridization: 4 arrow sp³$\rightarrow sp^3$, 5 arrow sp³d$\rightarrow sp^3d$, 6 arrow sp³d²$\rightarrow sp^3d^2$, 7 arrow sp³d³$\rightarrow sp^3d^3$.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: The p-Block Elements