Give reasons
(a) ${{H}_{2}}O$ is a liquid at room temperature while ${{H}_{2}}S$ is a gas
 (b) Ammonia is highly soluble in water
(C) The electron gain enthalpy of fluorine is exceptionally lower than chlorine.

Hint: To solve this question, we should have knowledge about the electronegativity of elements. Florine is the most electronegative element.
- In ammonia the central atom is nitrogen which is bonded to 3 hydrogen atoms. There is a lone pair on nitrogen.
- Electron gain enthalpy is the enthalpy which is released when an electron is added to an isolated gaseous atom.

Complete Solution :
(a)The oxygen atom present in water has a high electronegativity but the electronegativity of oxygen is less than that of fluorine as fluorine is the most electronegative element. So, the hydrogen atom present in other molecules of water forms a hydrogen bond with oxygen which results in intermolecular hydrogen bonding.
- While in the case of ${{H}_{2}}S$, the sulphur atom is less electronegative as compared to oxygen atom. So, the hydrogen bonding in case of ${{H}_{2}}S$ is negligible. Hence, ${{H}_{2}}O$ is a liquid at room temperature while ${{H}_{2}}S$ is a gas.
(b) Ammonia has the ability to form hydrogen bonds. The lone pair present on nitrogen form hydrogen bonding with the hydrogen atom present in water molecules. Hence, ammonia is highly soluble in water.
(c) fluorine has smaller size as compared to that of chlorine. As a result, there is strong interelectronic repulsion in the small 2p orbitals of fluorine. Hence, the electron gain enthalpy of fluorine is exceptionally lower than chlorine.

Note: (a) Each water molecule can form 4 hydrogen bonds. These four hydrogen bonds optimally arrange themselves tetrahedrally around each water molecule.
(b) Ammonia reacts with water to form ammonium hydroxide and hydroxyl ion.
(c) Do not get confused between electron gain enthalpy and electron affinity. The electron affinity of an element is positive.