(a) What happens when :
(i) HCl reacts with finely powdered iron.
(ii) `Cl_(2)` reacts with hot concentrated solution of NaOH.
(b) Give appropriate reason for each of the following :
(i) Sulphur vapour exhibits some paramagnetic character.
(ii) `NH_(3)` more basic than `PH_(3)`.
(iii) Dioxygen is a gas but sulphur a solid.

Explanation

(i) When HCl reacts with finely powdered iron, it forms ferrous chloride.
`Fe+2HClrarrFeCl_(2)+H_(2)`
(ii) `underset("(Conc.)")(3Cl_(2)+6NaOH)overset("Hot")rarrunderset("Sodium Chlorate")(5NaCl+NaClO_(3)+3H_(2)O)`
(b) (i) In vapour state, sulphur partly exists as `S_(2)` molecules which has two unpaired electrons in the antibonding `pi^(**)`-molecular orbitals like `O_(2)` and hence, `S_(2)` exhibits paramagnetism.
(ii) Due to the presence of a long pair of electrons on N and P, both `NH_(3)` and `PH_(3)` act as lewis bases and accept a proton to form an addition `N-H` and `P-H` bonds.
However, due to smaller size of N over, P, N-H bond thus formed is much stronger than `P-H` bond. Therefore, `NH_(3)` has higher proton affinity than `PH_(3)`.
(iii) Due to small size and high electronegativity, oxygen forms `ppi-ppi` multiple bonds. As a results, oxygen exists as `O_(2)` molecules which are held together by weak Van der Waals forces which can be easily overcome by collisions of molecules at room temperature. Hence, `O_(2)` is a gas at room temperature where as, sulphur has bigger size and lower electronegativity, does not form `ppi-ppi` multiple bond. Instead it prefers to form `S-S` single bond and has greater tendency for catenation and exist as `S_(8)` molecule. These molecules are held together by much stonger bond at room temperatrue. Hence, sulphur is a solid at room temperature.