Chemistry Q53 – JEE-MAIN 2026

Home Chemistry Chemical BondingQuestion #217

_{4}

is isostructural with :

Clue 1 of 3: VSEPR Theory

To determine if molecules are isostructural, compare their molecular geometries, which are derived from the VSEPR theory.

🧠 Full Solution Path

Step 1: Determine the molecular geometry of SF

_{4}

.✦ Active

The central atom in $S F_{4}$ is Sulfur (S), which has 6 valence electrons. It forms 4 single bonds with Fluorine (F) atoms, using 4 electrons. Remaining electrons = $6 - 4 = 2$ electrons, which form 1 lone pair. Steric number = (number of bond pairs) + (number of lone pairs) = $4 + 1 = 5$ . Hybridization is $s p^{3} d$ . Electron domain geometry is trigonal bipyramidal. Molecular geometry (shape) is **see-saw** (due to one lone pair occupying an equatorial position).

💡 Teacher's Secret Hint

Remember that lone pairs influence the molecular geometry by occupying space and causing repulsion.

Step 2: Determine the molecular geometry of each given option.○ Expand

A) $B r F_{4}^{⊖}$ : Br has 7 valence electrons + 4 from F + 1 for negative charge = 12 electrons. 4 bond pairs, 2 lone pairs. Steric number = $4 + 2 = 6$ . Hybridization $s p^{3} d^{2}$ . Shape: Square planar. (Not see-saw) B) $C H_{4}$ : C has 4 valence electrons + 4 from H = 8 electrons. 4 bond pairs, 0 lone pairs. Steric number = $4 + 0 = 4$ . Hybridization $s p^{3}$ . Shape: Tetrahedral. (Not see-saw) C) $I F_{4}^{\oplus}$ : I has 7 valence electrons + 4 from F - 1 for positive charge = 10 electrons. 4 bond pairs, 1 lone pair. Steric number = $4 + 1 = 5$ . Hybridization $s p^{3} d$ . Shape: **See-saw**. (Isostructural with $S F_{4}$ ) D) $X e F_{4}$ : Xe has 8 valence electrons + 4 from F = 12 electrons. 4 bond pairs, 2 lone pairs. Steric number = $4 + 2 = 6$ . Hybridization $s p^{3} d^{2}$ . Shape: Square planar. (Not see-saw) E) $X e O_{2} F_{2}$ : Xe has 8 valence electrons. 2 F atoms form single bonds (2 electrons). 2 O atoms form double bonds (4 electrons). Total bonding electrons = $2 + 4 = 6$ . Remaining electrons = $8 - 6 = 2$ , forming 1 lone pair. 4 bond domains (2 single, 2 double), 1 lone pair. Steric number = $4 + 1 = 5$ . Hybridization $s p^{3} d$ . Shape: **See-saw**. (Isostructural with $S F_{4}$ )

💡 Teacher's Secret Hint

For molecules with multiple bonds (like $X e O_{2} F_{2}$ ), count each multiple bond as one electron domain for steric number calculation.

Step 3: Identify the options with the same molecular geometry as SF

_{4}

.○ Expand

Both $I F_{4}^{\oplus}$ (Option C) and $X e O_{2} F_{2}$ (Option E) have a see-saw molecular geometry, which is the same as $S F_{4}$ . Therefore, $S F_{4}$ is isostructural with $I F_{4}^{\oplus}$ and $X e O_{2} F_{2}$ . Comparing this with the given choices, "C and E Only" is the correct option.

💡 Teacher's Secret Hint

Isostructural means having the same molecular geometry and hybridization, which implies the same number of bond pairs and lone pairs around the central atom.

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Chemistry Question 53 – JEE-MAIN 2026

Clue 1 of 3: VSEPR Theory

🧠 Full Solution Path