Question

Select the correct statement (s)
A.$C{H_3} - C - C{H_3}$ angle in ${(C{H_3})_2}C = C{H_2}$ is smaller and the $C{H_3} - C = C{H_2}$angle is larger than the trigonal $120^\circ$
B.$O - I - F$angle is less than $90^\circ (89^\circ )$ in $IOF_4^ -$
C.In $SeOC{l_2},C - Se - Cl$ angle is less than the $Cl - Se - O$ angle
D.$POC{l_3}$ is tetrahedral with a double bond between P and O. There is no lone pair on the central atom.

Answer 1

We must know that the bond angle is the angle present between two atoms which is occupied in a covalent species. Hence, there should be at least two bonds in that covalent species. The lone pair repulsion mainly affects the bond angle. Therefore, if the central atom contains the lone pair of electrons, the angle of the compound will be decrease.

Complete answer:
We must need to remember that the bond angle of $C{H_3} - C - C{H_3}$ present in isobutene is smaller and $C{H_3} - C = C{H_2}$ is angle is larger than the trigonal $120^\circ$. And this is due to the steric hindrance which is present in the methyl group. Hence, option (A) is correct.
The angle of $O - I - F$ present in $IOF_4^ -$ is less than $90^\circ$ and it is equal to$({89^o})$. This is because of the repulsion between the electrons of iodine – oxygen double bond and the electrons of iodine – fluorine single bond which is present in the $IOF_4^ -$. Hence, option (B) is incorrect.
In selenium oxychloride, the angle of $C - Se - Cl$ is less than the angle of $Cl - Se - O$. Because, there is repulsion between the electrons selenium and oxygen double bond and the electrons of selenium and chlorine double bond. Hence, option (C) is incorrect.
The shape of phosphorus oxide chloride is tetrahedral and it has a double bond between phosphorus and oxygen. And it does not have any lone pair on the central atom. And it undergoes $s{p^2}$ hybridization with analogous structure of $C{H_3}Cl$. Hence, the option (D) is correct.

Hence, the option (A) is correct.

Note:
We have to know that the bond angle mainly depends on the shape and hybridization of a compound. If the compounds have the same hybridization, then the sum of lone pair and bond pair is equal to the same. Hence, the bond angle becomes equal. And the electronegativity and shape of molecules also affect the bond angle. The bond angle decreases with decreasing electronegativity.