Question

Account for the following:
Bond angle in $$N{H}_4^{+}$$ is higher than $$N{H}_3.$$

Answer 1

Hint: We know that for the formation of bond angles there must be at least three atoms and two bonds. Bond angle is the angle which is formed between three atoms across at least two bonds. This angle is different for different compounds.

Complete answer:
Ammonium ions have tetrahedral shape, and ammonia have a lone pair to donate to form ammonium ion, also ammonium ions have no lone pair present to distort the normal tetrahedral angle. We know bond angle is the angle which is formed between two bonds. It depends on many factors which are stated below:
Forces of attraction or repulsion between atoms of molecules may increase or decrease the bond angle. Hybridization also affects the bond angle. As the character of hybrid orbital increases bond angle increases. Electronegativity and lone pairs also affect the bond angle. Increasing the lone pair and decreasing the electronegativity decreases the bond angle. As stated above, the number of lone pairs less is the bond angle due to lone pair-lone pair repulsions.
In $$N{H}_4^{+},$$ all are bond pairs whereas in ammonia the lone pair of electrons on nitrogen repels the bond pairs and reduces the bond angle.

Note:
Remember that ammonia acts as base; it is accepting hydrogen ions from the water. Ammonium ion is its conjugate acid; it can release hydrogen ions to reform back to ammonia. Ammonia is a weak base and ammonium ion is a weak acid.