Question

The species having bond angle of \[120^\circ \]is:
A. \[BC{l_3}\]
B. \[P{H_3}\]
C. \[Cl{F_3}\]
D. \[NC{l_3}\]

Answer 1

Hint: We know that the bond angle is the angle between the two atoms bonded to the central atom. It is dependent on the non-bonded pairs and the bonded pair.

Complete solution
As per the VSEPR theory, the strength of the repulsion between a lone pair and a bond pair of electrons lies in between the repulsion between two lone pairs and between two bond pairs. The order of repulsion between electron pairs as follows:
Lone Pair- lone pair \[ > \] Lone Pair- bond- pair \[ > \] Bond Pair- bond pair.
Now, let us consider all the species one by one.
a. In \[BC{l_3}\], the number of valence electrons of boron is three and chlorine is one, hence the bonding pairs are three and non-bonding pairs is zero. Hence, the total bond formed will be three and the hybridization is \[s{p^2}\]. Hence, the structure of the \[BC{l_3}\] molecule is trigonal planar, and bond angle will remain \[{120^ \circ }\], as there are no non-bonded pair of electrons.
b. In \[P{H_3}\], the number of valence electron of phosphorus is five and hydrogen is one hence the bonding pairs are three and non-bonding pair is one as three electrons out of five from phosphorus valence shell are used up to make bond with hydrogen. Hence, the total bond formed will be three and the hybridization is \[s{p^3}\]. Hence, the structure of the \[P{H_3}\] molecule is pyramidal, and bond angle will remain \[{93^ \circ }\], as there is one non-bonded pair of electrons.
c. In \[Cl{F_3}\], the number of valence electron of chlorine is seven and fluorine is one, hence the bonding pairs are two and non-bonding pair are two as three electrons out of seven from chlorine valence shell are used up to make bond with chlorine. Hence, the total bond formed will be three and the hybridization is \[s{p^3}d\]. Hence, the structure of the \[Cl{F_3}\] molecule is trigonal bipyramidal, and bond angle will remain \[{90^ \circ }\], as the there are two non-bonded pairs of electrons which are in the plane. And the shape will be T-shape.
d. In \[NC{l_3}\], the number of valence electron of nitrogen is five and chlorine is one, hence the bonding pairs are three and non-bonding pair is one as three electrons out of five from nitrogen valence shell are used up to make bond with chlorine. Hence, the total bond formed will be three and the hybridization is \[s{p^3}\]. Hence, the structure of the \[NC{l_3}\] molecule is trigonal bipyramidal, and bond angle will remain \[{107.8^ \circ }\], as there is one non-bonded pair of electrons.
The structure of all species with bond angles are shown below.

Hence, we can say that the correct option is (A).

Note:
The value of angle and shape will be distorted as the non-bonding pair is present. Here, the number of valence electrons is very important to be known in the correct way. Here, the central metal will be the one which is less in number and the other will form the covalent bond by sharing of the electron.