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 compound II has zero dipole moment then difference in number of lone pairs in both compounds is
________ .
Answer:
Solution:
Identify Element A and Compound I: Element A is Iodine (I). Compound I is IOF₃, which has a see-saw shape ( hybridization with 1 lone pair on I).
Total Lone Pairs in Compound I: I has 1 lone pair, O has 2 lone pairs, and 3 F atoms have 9 lone pairs (), giving a total of 12 lone pairs.
Identify Compound II: Compound II is IOF₅, which has hybridization, an octahedral geometry, and a non-zero dipole moment.
Total Lone Pairs in Compound II: I has 0 lone pairs, O has 2 lone pairs, and 5 F atoms have 15 lone pairs (), giving a total of 17 lone pairs.
Difference in Lone Pairs: The difference between the total lone pairs in both compounds is .
