InterviewSolution
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InterviewSolution
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(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. |
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Answer» (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. |
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