Question

Assign reason for the following : (i) Sulphur in vapour state exhibits paramagnetism. (ii)Fluorine is the strongest oxidising agent among halogens. (iii) In spite of having the same electronegativity, oxygen forms hydrogen bonds while chlorine does not.

Answer 1

Hint: The atomic number of sulphur is 16 and its symbol is S. The atomic number of fluorine is 9 and its symbol is F. The atomic number of oxygen is 8 and its symbol is O. The atomic number of chlorine is 17 and its symbol is Cl. The number of valence shell and oxidation state determine the electronegativity and oxidising ability of an element.

Complete Step by step solution:
Under ordinary conditions, sulphur exists as $ {S_8} $ in solid state but in vapour state sulphur exists as $ {S_2} $ molecule. The structure of $ {S_2} $ molecules is similar to $ {O_2} $ molecules. According to molecular orbital theory, in the structure of oxygen molecules there are two unpaired electrons remaining in the antibonding orbital. Hence, oxygen is paramagnetic. Also, in sulphur molecules, there are two electrons in antibonding orbitals which are unpaired. Hence, due to the unpaired electrons, sulphur in vapour state exhibits paramagnetism.
The oxidizing capability of any substance depends upon the three factors. The three factors are bond dissociation enthalpy, hydration energy and electron gain enthalpy. Electron gain enthalpy among halogens is highest for chlorine yet fluorine is the strongest oxidizing agent because of the following reasons,
In fluorine, the bond dissociation enthalpy is very low. This is because in fluorine, the electron-electron repulsion among the lone pairs is greater due to its small size.
Hydration enthalpy of fluoride ions is very high.
Due to the above reasons, the oxidising ability of fluorine is highest among the halogens.
The size of oxygen is smaller than that of chlorine. The smaller the size, the more is its ability to form hydrogen bonds. This is because oxygen has a small size and hence the electron density per unit volume is high. Due to high electron density, hydrogen bond formation is more favoured. Hence, in spite of having the same electronegativity, oxygen forms hydrogen bonds while chlorine does not.

Note:
Along with oxygen, nitrogen and fluorine can also form hydrogen bonds. Fluorine is also the most electronegative element followed by oxygen. Even chlorine, bromine and iodine form bonds with hydrogen such as HCl, HBr, HI respectively.