Question

The halogen that is most easily reduced is:
(A) \[{F_2}\]
(B) \[C{l_2}\]
(C) \[B{r_2}\]
(D) \[{I_2}\]

Answer 1

Hint: Reduction means gain of electron and oxidation is loss of electron. Halogens have a great tendency to reduce by accepting a single electron. By accepting an electron, halogens achieve noble gas configuration. The oxidising agent can be found out using the concept of bond dissociation energy and electron gain enthalpy.

Complete Step by Step Solution:
The bond dissociation energy is positive for both fluorine and chlorine. But the electron gain enthalpy is greater for chlorine hence in gaseous phase, chlorine is a better reducing agent than fluorine due to the summation of dissociation and electron gain enthalpy is greater for chlorine than fluorine. But in aqueous phase the hydration enthalpy for fluorine exceeds chlorine many times.

Hydration enthalpy is defined as ratio of charge by size. Fluorine and chlorine have the same charges that is $ - 1$ but the size of fluorine is less than chlorine so the hydration enthalpy is more.

Thus, to summarise\[{F_2}\] is the strongest oxidising agent because it has low enthalpy of dissociation and high enthalpy of hydration. The single bond between fluorine and fluorine is very weak because fluorine is small in size and the electrons present on fluorine cause repulsion and thus, the bond is easily dissociated.
Hence, the correct option is A.

Note: Oxidising agent is one which acts as an agent to oxidise others and thus, get reduced themselves. Whereas reducing agents are those which act as agents to reduce others and thus, get oxidised themselves. Non-metals are good oxidising agents and metals are good reducing agents.