Question

In OF2, H2O and OCl2, there is sp3 hybridisation of central atom. The bond length and bond angle of OF2, H2O and OCl2 are given as:Which of the following is the correct statement about the differences in the bond angles of the compounds given above?

• A. (A) Hydrogen is the least electronegative, and electron density around the central atom increases.
• B. (B) Fluorine is the most electronegative, and electron density around the central atom increases.
• C. (C) Cl is the most electronegative, and electron density around the central atom increases.
• D. (D) Due to larger size of Cl, electron density around the central atom decreases.

Answer 1

Answer: (A)

(A) Hydrogen is the least electronegative, and electron density around the central atom increases.

Answer 2

Explanation:
The bond angle of OF2, H2O and OCl2 can be determined by the repulsion of the lone pairs and the bonding pairs. As the lone pair-bond pair repulsion increases, the bond angle increases.Since, H- is the least electronegative; the bonding pair of O-H will be closer to O, resulting in a higher electron density near O. This increases the repulsion with lone pairs in the outer shell of oxygen, and therefore bond angle increases. F is most electronegative, thus electron density of O-F bond near O is less than that in O-H. Thus, repulsion between lone pairs of O is lower, and bond angle decreases.Cl is less electronegative than O, so electron density around O increases and repulsion between the lone pairs increases. Thus, bond angle increases.