Chemistry - CHEMICAL BONDING Question with Solution | TestHub
An element "A" forms a compound I comprising of A, F and O having see-saw shape. Another compound II hybridised central atom) of A, F & O having non-zero dipole moment has adjacent bond angle less than as compared to in see-saw shape compound.
If "A" is Xe, then difference in number of lone pairs in both compounds is .
Answer:
Solution:
Identify Compound I: For XeO₂F₂, Xe has 4 bonding pairs and 1 lone pair (Steric Number = 5), giving it a see-saw shape with 11 total lone pairs (Xe = 1, O = 2 × 2 = 4, F = 3 × 2 = 6).
Identify Compound II: For XeOF₄, Xe has 5 bonding pairs and 1 lone pair (Steric Number = 6, sp³d² hybridized), giving it a square pyramidal shape with a non-zero dipole moment and 15 total lone pairs (Xe = 1, O = 2, F = 4 × 3 = 12).
Calculate the difference: The difference in the total number of lone pairs between both compounds is:
