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 same electronegtaivity, oxygen forms hydrogen bond while chlorine does not.

Answer 1

• In vapour state sulphur partly exists as $S_2$molecule and $S_2$ molecule like $O_2$ has two unpaired electrons in the anti bonding , orbital and hence exhibit paramagnetism.
  
• It is the most electronegative element, and therefore can oxidise really well as it has a very strong pull on electrons.

  The oxidising capability of a substance depends upon three factors: bond dissociation enthalpy, electron gain enthalpy and hydration energy.

  Although electron gain enthalpy of fluorine is less negative as compared to chlorine, fluorine is a stronger oxidizing agent than chlorine, because of

  i) Low enthalpy of dissociation of F-F bond . Due to its small size Fluorine's hydration energy is very high hence better the oxidizing power.

  ii) high hydration enthalpy of F. Fluorine has a greater electron-electron repulsion among the lone pairs in the small sized $F_2$ molecule where they are much closer to each other than in case of $C{l}_2$, hence the enthalpy of dissociation of $F_2$ is lower than $C{l}_2$ which makes it better oxidizing agent.

  
  
• Both chlorine and oxygen have almost the same electronegativity values, but chlorine rarely forms hydrogen bonding. This is because in comparison to chlorine, oxygen has a smaller size and as a result, a higher electron density per unit volume.