Give the hybridization and no. of lone pairs over the central atom (Molecule - Hybridization - No. of lone pairs)<br>i) $$\text{XeF}_4$$ ________ _________<br>ii) $$\text{ICl}_3$$ ________ _________<br>iii) $$\text{NH}_3$$ ________ _________

PLACEHOLDER ANSWER - the answer is a set of entries. In $$\text{XeF}_4$$, Xe has 4 bond pairs and 2 lone pairs, so it is $$sp^3d^2$$ with 2 lone pairs. In $$\text{ICl}_3$$, I has 3 bond pairs and 2 lone pairs, so it is $$sp^3d$$ with 2 lone pairs. In $$\text{NH}_3$$, N has 3 bond pairs and 1 lone pair, so it is $$sp^3$$ with 1 lone pair.