UPSC Civil Services (Mains), Chemistry optional · 2018 · Paper I

Q1(d)(i). BeF₂ linear, SF₂ angular

Question

Explain the following on the basis of VBT and hybridization: (i) BeF₂ is linear whereas SF₂ is angular.

Be uses two sp hybrids with no lone pair: linear, 180°. S uses four sp³ hybrids, two of them holding lone pairs: the two S–F bonds are bent (98°).

one 2s electron is promoted to 2p (2s¹2p¹). The 2s and one 2p orbital mix to give two sp hybrids at 180°. Each overlaps a half-filled F 2p orbital.
BeF₂ has only two bond pairs and no lone pair, so the molecule is linear, F–Be–F 180° (gas-phase monomer).
four orbitals (3s and three 3p) mix to sp³ hybrids pointing to the corners of a tetrahedron. Two hold lone pairs; two hold one electron each and overlap F 2p orbitals.
two bonds out of a tetrahedral set, so SF₂ is angular (V-shaped). The lone pairs repel more strongly than bond pairs (lp–lp > lp–bp > bp–bp), so F–S–F closes from 109.5° to 98°.

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

Be (2s²): one 2s electron is promoted to 2p (2s¹2p¹). The 2s and one 2p orbital mix to give two sp hybrids at 180°. Each overlaps a half-filled F 2p orbital.
BeF₂ has only two bond pairs and no lone pair, so the molecule is linear, F–Be–F 180° (gas-phase monomer).
S (3s²3p⁴): four orbitals (3s and three 3p) mix to sp³ hybrids pointing to the corners of a tetrahedron. Two hold lone pairs; two hold one electron each and overlap F 2p orbitals.
Only the atoms define the shape: two bonds out of a tetrahedral set, so SF₂ is angular (V-shaped). The lone pairs repel more strongly than bond pairs (lp–lp > lp–bp > bp–bp), so F–S–F closes from 109.5° to 98°.

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