A xenon compound 'A' upon partial hydrolysis gives XeO₂F₂. The number of lone pair of electrons present in compound A is _________. (Round off to the Nearest Integer)

Solution

Chemistry ma’am · 4 steps

Work backwards from XeO₂F₂ to XeF₆, then count lone pairs on Xe and on every F.

1. Identify A. Hydrolysis swaps Xe–F bonds for Xe=O bonds. Each O replaces two F atoms, so XeO₂F₂ comes from a compound with 2 + 4 = 6 fluorines. That is XeF₆.

   XeF₆ + 2H₂O → XeO₂F₂ + 4HF

2. Lone pairs on Xe. Xe has 8 valence electrons. Six of them go into the six Xe–F bonds, so 2 electrons are left. That is 1 lone pair on Xe, which is why XeF₆ is a distorted octahedron.
3. Lone pairs on F. Each F has 7 valence electrons and uses 1 in its bond, so it keeps 6, which is 3 lone pairs. Six fluorines give 6 × 3 = 18 lone pairs.
4. So. Total lone pairs in XeF₆ are 1 + 18 = 19.

The common slip: 18. Counting only the lone pairs on the fluorines and missing the one lone pair on Xe gives 18, not 19.

The idea to keep. When a question asks for lone pairs in a molecule, count them on every atom, not only on the central one. Each terminal F always carries 3 lone pairs.

Answer from the NTA final answer key. Solution by PrepWiser.