Which of the following has the highest bond angle?
A. ${H_2}O$
B. ${H_2}S$
C. $N{H_3}$
D. $P{H_3}$

Hint: A bond angle is an angle which is formed between three atoms across at least two bonds. We can compare the bond angles in different molecules on the basis of their lone pairs and bond pairs. This is explained by valence shell electron pair repulsion (VSEPR) and Non- bonding electron pair repulsion (NBEPR). Hence, in order to compare the bond angles in the above compounds, we will first calculate the number of bond pairs and lone pairs in it.

Complete answer:
Total electron pairs are the sum of total lone pairs and bond pairs. Electron pairs are calculated by adding the total number of valence electrons in the molecule and then dividing it by the total number of atoms in the molecule.
We should remember that the greater the lone pair of electrons the more will be the repulsion between them and the angle between the bonds reduces.
Now, we will calculate the number of bond pairs and lone pairs in the above molecules as follows:
We will denote electron pairs by E, valence electrons by V, bond pairs by B and Lone pairs by L.
A. ${H_2}O$
$E = \dfrac{{2 + 6}}{2}$
$E = 4$
Out of these electron pairs, we can see from the formula that there are two bond pairs as two hydrogen atoms are bonded to oxygen.
$B = 2$
$L = 2$
B. ${H_2}S$
$E = \dfrac{{2 + 6}}{2}$
$E = 4$
Out of these electron pairs, we can see from the formula that there are two bond pairs as two hydrogen atoms are bonded to sulphur.
$B = 2$
$L = 2$
In this case the repulsion between the lone pairs will be less as compared to ${H_2}O$ due to the larger size of sulphur and hence, bond pair – bond pair repulsion is much faster than lone pair- bond pair repulsion. Therefore, due to the larger size of sulphur, no hybridization takes place in this molecule and its bond angle will be least among these given compounds.
C. $N{H_3}$
$E = \dfrac{{3 + 5}}{2}$
$E = 4$
Out of these electron pairs, we can see from the formula that there are three bond pairs as three hydrogen atoms are bonded to nitrogen atoms.
$B = 3$
$L = 1$
D. $P{H_3}$
$E = \dfrac{{3 + 5}}{2}$
$E = 4$
Out of these electron pairs, we can see from the formula that there are three bond pairs as three hydrogen atoms are bonded to nitrogen atoms.
$B = 3$
$L = 1$
Bond angle of $P{H_3}$ will be less than that of ammonia because of the greater size of the nitrogen.
Therefore, based on the above explanation we can conclude that $N{H_3}$ has the highest bond angle among the given compounds.
Hence, the correct option is C. $N{H_3}$ .

Note:
Some compounds do not undergo hybridization because the size of the central atom is large and there are no repulsions between the lone pairs and bond pairs. These types of molecules are known as drago molecules and the bond angle in these molecules is approximately equal to the ${90^o}$ .