JEE Main 2026 — Jan 23, Shift 2 — Chemistry solution

JEE Main2026Jan 23, Shift 2Chemistry

Which statements are NOT TRUE about $X e O_{2} F_{2}$ ?

A. It has a see-saw shape.

B. Xe has 5 electron pairs in its valence shell in $X e O_{2} F_{2}$ .

C. The O-Xe-O bond angle is close to $18 0^{\circ}$ .

D. The F-Xe-F bond angle is close to $18 0^{\circ}$ .

E. Xe has 16 valence electrons in $X e O_{2} F_{2}$ .

Choose the correct answer from the options given below.

A
B, C and E only
B
B and D only
C
A and D only
D
B, D and E only

Solution

For $X e O_{2} F_{2}$ : Xe contributes 8 valence electrons, two F take 1 electron each in bonding, two O double-bonded. Steric number = 4 bond pairs + 1 lone pair = 5. Shape is see-saw (trigonal bipyramidal electron geometry with one equatorial lone pair).

(A) See-saw shape. True → not the answer.

(B) Xe has 5 electron pairs? Total electron pairs around Xe in $X e O_{2} F_{2}$ = 4 bonding regions + 1 lone pair = 5 regions, but the question writes 5 electron pairs; counting bonds, the two Xe=O are double bonds (each having 2 pairs), so total electron pairs around Xe is 7, not 5. Not True.

(C) Two O atoms occupy equatorial positions; O-Xe-O angle is around $105$ - $11 0^{\circ}$ , not $18 0^{\circ}$ . Not True.

(D) Two F atoms occupy axial positions; F-Xe-F angle is close to $18 0^{\circ}$ (slightly less due to lone pair). True → not the answer.

(E) Total valence electrons around Xe (bonding + lone pair) = $2 (F) \cdot 1 + 2 (O) \cdot 2 + lone pair = 2 + 4 + 2 = 14$ , not 16. Not True.

Hence B, C and E are not true.

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