The atomic radius of halogens increases as we go down the group due to the addition of new shells. As a result, the bond length of halogen $\ce{X-X}$ increases down the group. So, less energy is required to break the bond (bond dissociation enthalpy decreases). So, as per this assumption, iodine should be the most reactive halogen. But this is not true. Rather, fluorine is the most reactive halogen. It reacts violently with almost all chemicals. Bond cleavage of fluorine is much easier that that of other halogens, which may be due to the repulsive force of the lone pairs of electrons.

$$\ce{2F2(g) + 2H2O(g) → O2(g) + 4HF(g) + heat}$$

So, what about iodine? This source gives the values of bond dissociation enthalpy. It shows that the B.D.E value of iodine ($\pu{151 kJ/mol}$) is less than that of fluorine ($\pu{158 kJ/mol}$). So, iodine should be the most reactive halogen and not halogen. But this does not happen. In most of the reactions of iodine, equilibrium is maintained. To push the reaction forward, a catalyst is used.

$$\ce{I2(l) + H2O(l) <=> OI-(aq) + 2H+(aq) + I-(aq)}$$

Another similar phenomenon is that acidity of hydrohalic acid increases on going the group. $\ce{H-X}$ bond length increases on going down the group. Bond dissociation enthalpy decreases. Bond breaks into corresponding ions. So, $\ce{HI}$ is able to furnish $\ce{H+}$ ion faster than $\ce{HF}$. So, $\ce{HI}$ is more acidic than $\ce{HF}$. But, the reverse is observed. $\ce{HF}$ is so reactive that it reacts with glass and has to be stored in wax bottles.

$$\ce{SiO2 + 4 HF -> SiF4(g) + 2 H2O}$$

$$\ce{SiO2 + 6 HF -> H2SiF6 + 2 H2O}$$

In most of the reactions of hydroiodic acid, equilibrium is maintained. To push the reaction forward, a catalyst is used.

$$\ce{H2(g) + I2(g) <=> 2 HI(g)}$$

My question: Why is it so?

  • 1
    $\begingroup$ A catalyst does NOT affect the position of equilibrium! $\endgroup$
    – jimchmst
    Apr 14, 2022 at 6:26

1 Answer 1


In comparing the reactivity of fluorine or iodine, there are a couple of factors to consider. Part of it is the strength of the $\ce{X–X}$ bond, which is a barrier that tends to discourage them from reacting. Obviously, the higher this barrier, the less reactive the halogen is (in general).

However, unless you're trying to consider the relative reactivity towards homolytic cleavage of the bond, it isn't sufficient to look at the bond strength. The other bit is the driving force for the formation of the products – in other words, something that encourages them to react.

In this case, the formation of fluorine-containing products is generally much more thermodynamically favourable than that of the corresponding iodine-containing products. Some factors influencing this are:

  1. The electron affinity of fluorine is much larger, which makes reduction to fluoride much easier
  2. $\ce{E–F}$ bonds are stronger than $\ce{E–I}$ bonds, in both an ionic and covalent sense.
  3. Fluoride has a larger solvation enthalpy than iodide

For example, factors 1 and 3 lead to the large differences in the reduction potentials: $E^\circ(\ce{F2}/\ce{F-}) = \pu{+2.87 V}$ versus $E^\circ(\ce{I2}/\ce{I-}) = \pu{+0.54 V}$.

Silica is readily attacked by $\ce{HF}$ not because it is a strong acid, but rather because of the strength of $\ce{Si-F}$ bond thus formed.

That deals with the thermodynamic parts of "reactivity", in the sense that it explains why some reactions with iodine require elevated temperatures for any significant conversion, whereas reactions with fluorine proceed readily at room temperature.

The same factors, though, apply to the kinetic aspects of "reactivity". The activation energy is largely dictated by bonds that one has to break or form, for example, and the rate of electron transfer processes usually increases with $E^\circ$. Exceptions to this arise when – for example – the reactants have solid state structures which must be broken apart before reaction can occur. However, that is not the case here.


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