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JEE Advanced Chemistry Chemical Bonding and Molecular Structure 2026 JEE Advanced 2026 (Paper 1)

JEE Advanced Chemistry Question (2026) — Solution

Question

Consider the following species: SOCl _2, XeOF _4, ClF _3, ClF _5, XeF _5^+, SO _3^ 2- , XeF _3^+, SF _4 List-I contains different molecular shapes and List-II contains total number of species with the same molecular shapes from the given species. Match each entry in List-I with the appropriate entry in List-II and choose the correct option. List-I List-II (P) See-saw (1) one (Q) T-Shaped (2) two (R) Trigonal Planar (3) three (S) Square Pyramidal (4) four (5) zero

Options

  1. A. P 1; Q 2; R 5; S 3
  2. B. P 5; Q 4; R 2; S 3
  3. C. P 3; Q 2; R 1; S 4
  4. D. P 1; Q 3; R 5; S 4

Answer

A. P 1; Q 2; R 5; S 3

Step-by-step solution

Let us determine the hybridization and shape of each given species by finding the number of bond pairs (bp) and lone pairs (lp) on the central atom: SOCl _2: Central atom S has 6 valence electrons. It forms 3 bonds (one with O, two with Cl) and has 1 lone pair. Steric number = 4 (sp^3). Shape is Trigonal Pyramidal. XeOF _4: Central atom Xe has 8 valence electrons. It forms 5 bonds (one with O, four with F) and has 1 lone pair. Steric number = 6 (sp^3d^2). Shape is Square Pyramidal. ClF _3: Central atom Cl has 7 valence electrons. It forms 3 bonds with F and has 2 lone pairs. Steric number = 5 (sp^3d). Shape is T-shaped. ClF _5: Central atom Cl has 7 valence electrons. It forms 5 bonds with F and has 1 lone pair. Steric number = 6 (sp^3d^2). Shape is Square Pyramidal. XeF _5^+: Central atom Xe has 8 valence electrons. The positive charge leaves 7 valence electrons. It forms 5 bonds with F and has 1 lone pair. Steric number = 6 (sp^3d^2). Shape is Square Pyramidal. SO _3^ 2- : Central atom S has 6 valence electrons. The -2 charge gives 8 valence electrons. It forms 3 bonds with O and has 1 lone pair. Steric number = 4 (sp^3). Shape is Trigonal Pyramidal. XeF _3^+: Central atom Xe has 8 valence electrons. The positive charge leaves 7 valence electrons. It forms 3 bonds with F and has 2 lone pairs. Steric number = 5 (sp^3d). Shape is T-shaped. SF _4: Central atom S has 6 valence electrons. It forms 4 bonds with F and has 1 lone pair. Steric number = 5 (sp^3d). Shape is See-saw. Counting the species for each shape: (P) See-saw: 1 ( SF _4) (Q) T-Shaped: 2 ( ClF _3, XeF _3^+) (R) Trigonal Planar: 0 (None) (S) Square Pyramidal: 3 ( XeOF _4, ClF _5, XeF _5^+) Matching with List-II: P 1 Q 2 R 5 S 3 Answer: P 1; Q 2; R 5; S 3

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