Answer
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Hint: Iodine is a heavy element and a non-metal. It is a solid halogen containing seven valence electrons. When iodine is treated with concentrated $HN{O_3}$, oxidation of iodine takes place. The products formed during the oxidation of ${I_2}$ by concentrated $HN{O_3}$ are iodic acid, nitrogen dioxide and water.
Complete step by step answer:
Here, we need to find out the number of moles produced of $N{O_2}$, when one mole of ${I_2}$ is oxidised by concentrated $HN{O_3}$.
Let us first have a look at the balanced chemical equation for the reaction of oxidation of one mole iodine by concentrated $HN{O_3}$.
${I_2} + 10HN{O_3} \to 2HI{O_3} + 10N{O_2} + 4{H_2}O$
From the above chemical equation, we can deduce that one mole of iodine is oxidised by 10 moles of concentrated $HN{O_3}$ to produce 2 moles of iodic acid, $HI{O_3}$, 10 moles of nitrogen dioxide, $N{O_2}$ and 4 moles of water.
Therefore, to oxidise one mole of ${I_2}$ we will require 10 moles of concentrated $HN{O_3}$ and it will produce 10 moles of $N{O_2}$.
So, the correct answer is D.
Additional information:
Nitric acid is an oxidising agent which can oxidise both metals and non-metals. It can oxidise metals to form soluble nitrate compounds. In concentrated form, it is a stronger oxidising agent than that of diluted form. Though most of the metals can get oxidised by $HN{O_3}$, gold and platinum are exceptions. The oxidising properties of $HN{O_3}$ can be enhanced by increase in pressure and temperature.
Note: The common misconception about oxidation of iodine by concentrated nitric acid is that iodine gets oxidised to form periodic acid, $HI{O_4}$. But, it is not true. The iodine gets oxidised to form iodic acid, $HI{O_3}$, though there are other non-metals getting oxidised to their highest oxidation state. The highest oxidation state of iodine is +7, but during oxidation by nitric acid, iodine gets oxidised to +5 state. The major reason for this is the behaviour of iodine atoms is that iodine acts as if it has only 5 valence electrons. Also, nitric acid is not strong enough to oxidise iodine to +7 state.
Complete step by step answer:
Here, we need to find out the number of moles produced of $N{O_2}$, when one mole of ${I_2}$ is oxidised by concentrated $HN{O_3}$.
Let us first have a look at the balanced chemical equation for the reaction of oxidation of one mole iodine by concentrated $HN{O_3}$.
${I_2} + 10HN{O_3} \to 2HI{O_3} + 10N{O_2} + 4{H_2}O$
From the above chemical equation, we can deduce that one mole of iodine is oxidised by 10 moles of concentrated $HN{O_3}$ to produce 2 moles of iodic acid, $HI{O_3}$, 10 moles of nitrogen dioxide, $N{O_2}$ and 4 moles of water.
Therefore, to oxidise one mole of ${I_2}$ we will require 10 moles of concentrated $HN{O_3}$ and it will produce 10 moles of $N{O_2}$.
So, the correct answer is D.
Additional information:
Nitric acid is an oxidising agent which can oxidise both metals and non-metals. It can oxidise metals to form soluble nitrate compounds. In concentrated form, it is a stronger oxidising agent than that of diluted form. Though most of the metals can get oxidised by $HN{O_3}$, gold and platinum are exceptions. The oxidising properties of $HN{O_3}$ can be enhanced by increase in pressure and temperature.
Note: The common misconception about oxidation of iodine by concentrated nitric acid is that iodine gets oxidised to form periodic acid, $HI{O_4}$. But, it is not true. The iodine gets oxidised to form iodic acid, $HI{O_3}$, though there are other non-metals getting oxidised to their highest oxidation state. The highest oxidation state of iodine is +7, but during oxidation by nitric acid, iodine gets oxidised to +5 state. The major reason for this is the behaviour of iodine atoms is that iodine acts as if it has only 5 valence electrons. Also, nitric acid is not strong enough to oxidise iodine to +7 state.
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