Question

In which of the following bond angles is maximum?
A. \[N{H_3}\]
B. \[N{H_4}^ + \]
C. $PC{l_3}$
D. $SC{l_2}$

Answer 1

Hint: A bond angle can be defined as the angle formed between three atoms across at least two bonds. Bond angles can be determined with the help of lone pair electron repulsion. Bond angle depends on hybridization of atoms, lone pair repulsion and electronegativity of the central atom.

Step-By-Step answer:
We have studied that the presence of lone pairs of electrons on the central metal atom always repels the bonded pair electrons due this repulsion bonds are displaced slightly and results in a decrease of bond angle. The electronegativity of the central metal atom in a molecule also affects bond angle as the electronegativity of the central metal atom increases, bond angle decreases. In the given molecules ammonia has a lone pair on the nitrogen atom, $PC{l_3}$ and $SC{l_2}$ also have a lone pair on the central metal atom. $PC{l_3}$ has three sigma bonds and one lone pair, the presence of lone pair exerts repulsion towards bond pair and bond angle reduces from actual tetrahedral bond angle. $SC{l_2}$ it has two bond pairs of electrons and two lone pairs of electrons at sulphur. \[N{H_4}^ + \] molecules do not have any lone pair on nitrogen. So option B is the correct answer to this problem that is \[N{H_4}^ + \].

Note: Linear molecules have bond angle of ${180^0}$. We know that smaller the bond angle means more repulsion so it tries to push the atoms bonded to the central metal atom away and makes the bond angle smaller than expected. The strength of repulsion is in the following order ; lone pair-lone pair $ > $ lone pair – bonding pair $ > $ bonding pair- bonding pair.