p-Block Elements (Group 15–18) · Question 3

JEE Advanced 2026 Paper 2Multiple CorrectMedium📈 Topic in 17% papers+4−1

Correct statement(s) about the compounds P, Q and R is(are) Xe(g)+F2(g) (1:5 ratio) 873 K, 7 barP P+O2F2 143 KQ+O2 Q+H2O complete hydrolysisR+HF

P has two lone pairs of electrons on the central atom.
Q has a perfect octahedral geometry.
Q can act as a fluorinating agent.
The molecular structure of R is trigonal pyramidal.

Answer

(1) P has two lone pairs of electrons on the central atom., (3) Q can act as a fluorinating agent., (4) The molecular structure of R is trigonal pyramidal.

Solution

Xe : F2 = 1 : 5 at 873 K, 7 bar gives P = XeF4 (square planar, 4 bp + 2 lp): (A) correct.

XeF4+O2F2 143 KXeF6+O2, so Q = XeF6.

XeF6 has 6 bp + 1 lp, so it is a distorted octahedron: (B) wrong; it is a strong fluorinating agent: (C) correct.

XeF6+3H2O XeO3+6HF; R = XeO3, 3 bp + 1 lp, trigonal pyramidal: (D) correct.

Answer: (A), (C), (D)

JEE Advanced 2026 Paper 2

Keep practising — p-Block Elements (Group 15–18)

The heating of NH4NO2 at 60−70 °C and NH4NO3 at 200−250 °C is associated with the formation of nitrogen containing compounds X and Y, respectively. X and Y, respectively, are

N2 and N2O
NH3 and NO2
NO and N2O
N2 and NH3

JEE Advanced 2025 Paper 1

The complete hydrolysis of ICl, ClF3 and BrF5, respectively, gives

IO-, ClO2- and BrO3-
IO3-, ClO2- and BrO3-
IO-, ClO- and BrO2-
IO3-, ClO4- and BrO2-

JEE Advanced 2025 Paper 2

The compound(s) with P–H bond(s) is(are)

H3PO4
H3PO3
H4P2O7
H3PO2

JEE Advanced 2025 Paper 2

At room temperature, disproportionation of an aqueous solution of in situ generated nitrous acid (HNO2) gives the species

H3O+, NO3- and NO
H3O+, NO3- and NO2
H3O+, NO- and NO2
H3O+, NO3- and N2O

JEE Advanced 2024 Paper 1

Based on VSEPR model, match the xenon compounds given in List-I with the corresponding geometries and the number of lone pairs on xenon given in List-II and choose the correct option.