Match List I with List II : | List-I (molecule) | List-II (hybridization; shape) | |---|---| | A. $\mathrm{XeO}_{3}$ | I. $\mathrm{sp}^{3} \mathrm{~d}$; linear | | B. $\mathrm{XeF}_{2}$ | II. $\mathrm{sp}^{3}$; pyramidal | | C. $\mathrm{XeOF}_{4}$ | III. $\mathrm{sp}^{3} \mathrm{~d}^{3}$; distorted octahedral | | D. $\mathrm{XeF}_{6}$ | IV. $\mathrm{sp}^{3} \mathrm{~d}^{2}$; square pyramidal | Choose the correct answer from the options given below:

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · p-Block Elements

Match List I with List II :

List-I (molecule)	List-II (hybridization; shape)
A. $XeO_{3}$	I. $sp^{3} d$ ; linear
B. $XeF_{2}$	II. $sp^{3}$ ; pyramidal
C. $XeOF_{4}$	III. $sp^{3} d^{3}$ ; distorted octahedral
D. $XeF_{6}$	IV. $sp^{3} d^{2}$ ; square pyramidal

Choose the correct answer from the options given below:

Answer: (A)

For xenon compounds, determine hybridization by counting electron pairs (bonding + lone pairs):

**A.

XeO_{3}

**: 3 bonding pairs + 1 lone pair = 4 pairs →

sp^{3}

hybridization, pyramidal shape (II).

**B.

XeF_{2}

**: 2 bonding pairs + 3 lone pairs = 5 pairs →

sp^{3} d

hybridization, linear shape (I).

**C.

XeOF_{4}

**: 5 bonding pairs + 1 lone pair = 6 pairs →

sp^{3} d^{2}

hybridization, square pyramidal shape (IV).

**D.

XeF_{6}

**: 6 bonding pairs + 1 lone pair = 7 pairs →

sp^{3} d^{3}

hybridization, distorted octahedral shape (III).

Therefore, A-II, B-I, C-IV, D-III.

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option