Reaction between halogens and concentrated nitric acid

Question

I have read that the reaction between $\ce{I2}$ and conc. $\ce{HNO3}$ gives $\ce{HIO3}$ and $\ce{NO2},$ but there is no mention of the other halogens' products. Do they all give halous acids and nitrogen dioxide? Specifically with chlorine, how does the reaction proceed?

Answer 1

This question can easily be addressed considering reduction potential of each compound. Concentrated nitric acid is a powerful oxidizing acid, which has a high reduction potential:

$$ \begin{align} \ce{NO3- + 4H+ + 3e- &<=> NO + 2H2O} &\quad E^\circ &= \pu{0.957 V} \tag1\\ \ce{NO3- + 3H+ + 2e- &<=> HNO2 + H2O} &\quad E^\circ &= \pu{0.934 V} \tag2\\ \ce{2NO3- + 4H+ + 2e- &<=> N2O4 + 2H2O} &\quad E^\circ &= \pu{0.803 V} \tag3 \end{align} $$

Now, let's look at reduction potential of halides:

$$ \begin{align} \ce{F2 + 2e- &<=> 2F-} &\quad E^\circ &= \pu{2.866 V} \tag4\\ \ce{Cl2 + 2e- &<=> 2Cl-} &\quad E^\circ &= \pu{1.3583 V} \tag5\\ \ce{Br2 + 2e- &<=> 2Br-} &\quad E^\circ &= \pu{1.066 V} \tag6\\ \ce{I2 + 2e- &<=> 2I-} &\quad E^\circ &= \pu{0.5355 V} \tag7\\ \end{align} $$ Accordingly, we can list oxidizing power at standard conditions in decreasing order:

$$\ce{F2} > \ce{Cl2} > \ce{Br2} > \ce{NO3-} > \ce{I2}$$

Therefore, at standard condition, concentrated $\ce{HNO3}$ acid (meaning $\ce{NO3-}$ in acidic medium) can easily oxidize only $\ce{I2}.$ One such example is documented in Ref.1. According to the authors, the oxidation of iodine by hyperazeotropic nitric acid ($\pu{16–21 M}$) was shown to proceed in at least two steps:

$$ \begin{align} \ce{I2 + 3HNO3 &<=>[$k_1$][$k_2$] 2I+ + 2NO3- + 2HNO2 + H2O}\\ \ce{I+ + 6HNO3 &<=>[$k_3$][$k_1$] I^5+ + 4NO3- + 2HNO2 + 2H2O} \end{align} $$

Since reduction potentials of $\ce{HNO3}$ and $\ce{Br2}$ are in such close range that each can oxidize the other. In fact, the kinetic studies of this phenomenon has been performed in literature (Ref.2).

In different aspect, there is a publication (Ref.3) claiming that the presence of bromine in fuming nitric acid makes it a brominating reagent (electrophilic aromatic substitution):

This bromination has worked well with deactivated aromatic nucleus. In absence of bromine, the reagent nitrate the ring. Thus, it is safe to suggest that $\ce{HNO3}$ might promote formation of '$\ce{Br+}$', meaning oxidation.

References:

1. J. C. Mailen, T. O. Tiffany, "The reaction of iodine with concentrated nitric acid," Journal of Inorganic and Nuclear Chemistry 1975, 37(1), 127-132 (https://doi.org/10.1016/0022-1902(75)80139-4).
2. Istvan Lengyel, Istvan Nagy, Gyorgy Bazsa, "Kinetic study of the autocatalytic nitric acid-bromide reaction and its reverse, the nitrous acid-bromine reaction," J. Phys. Chem. 1989, 93(7), 2801–2807 (https://doi.org/10.1021/j100344a021).
3. Alexander M. Andrievsky, Vera I. Lomzakova, Mikhail K. Grachev, Mikhail V. Gorelik, "Aromatic Bromination in Concentrated Nitric Acid," Open Journal of Synthesis Theory and Applications 2014, 3(2), Article ID:44941, 5 pages (doi: 10.4236/ojsta.2014.32003).

Answer 2

$\ce{HNO_3}$ does not react with $\ce{Cl_2}$, as $\ce{Cl_2}$ is produced in the famous "aqua regia", which is a mixture of nitric and chlorhydric acid, having reacted according to :$$\ce{HNO_3 + 3 HCl -> NOCl + Cl_2 + 2 H_2O}$$ This equation is given in Cotton and Wilkinson, Chapter 5-13. In Chapter 12-8, it is stated that NOCl is never pure. It is decomposed to $\ce{Cl_2 + NO}$ to the extent of $0.5$% at room temperature.

NOBr is produced in a mixture $\ce{NO + Br_2}$ and it is an equilibrium, where ~ $7$ % of NOBr is decomposed at room temperature and $1$ atm.

Ref.: F. Albert Cotton, Geoffrey Wilkinson, Advanced Inorganic Chemistry, A Comprehensive Text, 3rd Ed., 1972, Interscience Pub., John Wiley & Sons, Inc. New York, London.