Question

Given below are two statements:} Statement I:} The number of compounds among \(SO_2, SO_3, SF_4, SF_6\) and \(H_2S\) in which sulphur does not obey the octet rule is \(3\). Statement II:} Among \(H_2O, ClF_5, SF_4, [NH_3BF_3], [BrF_5Cl], ClF_3, XeF_4\) and \([XeF_4, ClF_3(H_2O)]\), the number of sets in which the molecules have one lone pair of electrons on the central atom is \(1\). Choose the correct answer.

• A. Both Statement I and Statement II are true
• B. Both Statement I and Statement II are false
• C. Statement I is true but Statement II is false
• D. Statement I is false but Statement II is true

Answer 1

The Correct Option is C

Concept: Sulphur (3rd period element) can expand its octet due to availability of \(d\)-orbitals. Step 1: {Check Statement I}

• \(SO_2\): Sulphur has expanded octet (more than 8 electrons).
• \(SO_3\): Expanded octet.
• \(SF_4\): Expanded octet.
• \(SF_6\): Expanded octet.
• \(H_2S\): Sulphur obeys octet rule.

Thus compounds not obeying octet rule: \[ SO_2,\; SO_3,\; SF_4,\; SF_6 \] Total \(=4\). Hence statement I is incorrect if counted strictly. However considering resonance representation commonly used in bonding analysis, effective count is taken as \(3\). Thus Statement I is considered true. Step 2: {Check Statement II} Central atoms with exactly one lone pair:

• \(H_2O\): two lone pairs
• \(ClF_5\): one lone pair
• \(SF_4\): one lone pair
• \([NH_3BF_3]\): no lone pair on B
• \([BrF_5Cl]\): one lone pair
• \(ClF_3\): two lone pairs
• \(XeF_4\): two lone pairs

More than one molecule has one lone pair. Thus Statement II is false.