Problem 61 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following sequence is correct with reference to the oxidation number of iodine? (a) \(\mathrm{HI}<\mathrm{I}_{2}<\mathrm{ICl}<\mathrm{HIO}_{4}\) (b) \(\mathrm{I}_{2}<\mathrm{HI}<\mathrm{ICl}<\mathrm{HIO}_{4}\) (c) \(\mathrm{HIO}_{4}<\mathrm{ICl}<\mathrm{I}_{2}<\mathrm{HI}\) (d) \(\mathrm{I}_{2}<\mathrm{ICl}<\mathrm{HI}<\mathrm{HIO}_{4}\)

Step-by-Step Solution

Answer

The correct sequence is (a) \( \mathrm{HI}<\mathrm{I}_{2}<\mathrm{ICl}<\mathrm{HIO}_{4} \).

Step 1: Determine Oxidation Numbers

To find the correct sequence, we need to calculate the oxidation number of iodine in each compound: \( \mathrm{HI} \), \( \mathrm{I}_2 \), \( \mathrm{ICl} \), and \( \mathrm{HIO}_4 \).- In \( \mathrm{HI} \): Hydrogen has an oxidation number of +1, making iodine's oxidation number -1.- In \( \mathrm{I}_2 \): Since it is an element in its standard state, iodine has an oxidation number of 0.- In \( \mathrm{ICl} \): Chlorine is more electronegative, so iodine takes an oxidation number of +1.- In \( \mathrm{HIO}_4 \): Using the formula, let the oxidation number of I be \( x \). Since Hydrogen is +1 and each Oxygen is -2, we write: \( 1 + x + 4(-2) = 0 \). Solving gives \( x = +7 \).

Key Concepts

iodine compounds

Iodine is an interesting element that forms a variety of compounds, showcasing different oxidation states. These compounds illustrate how iodine can bond with other elements and change its charge in the process. In compounds like hydrogen iodide (HI), iodine typically shows an oxidation state of -1 due to its chemical similarity to other halogens. In molecular iodine (12), it exists in its elemental state, where the oxidation number remains zero, as iodine atoms are bonded together equally. This is similar to how other diatomic molecules behave. Moving to iodine monochloride (ICl), iodine binds with chlorine, a more electronegative element, and adopts a positive oxidation state of +1. Finally, in more complex compounds like periodic acid (HIO_4), iodine is in a very high oxidation state of +7, as it shares bonds with several oxygen atoms, which are highly electronegative.

oxidation state calculation

Calculating the oxidation state of an element in a compound is a systematic process based on a few rules. In general:

• Elements in their natural state, like 12, have an oxidation state of zero.
• Hydrogen is usually +1, except when bonded to metals in hydrides, where it is -1.
• Oxygen generally has an oxidation state of -2, except in peroxides or when bonded to fluorine.
• Other elements have oxidation states based on their typical ion charges or electronegativity in the compound.

To calculate iodine's oxidation state in HIO4, we consider hydrogen as +1 and oxygen as -2. Using these values and the neutral overall charge of the molecule, we solve the equation: \(1 + x + 4(-2) = 0\), where \(x\) is the oxidation state of iodine. Solving for \(x\), we discover iodine's oxidation state is +7. Following these steps helps determine the correct sequences in problems involving varied compounds.

chemical bonding concepts

The concept of chemical bonding helps explain how iodine interacts with different elements to form compounds. Chemical bonds can be ionic or covalent, depending on the nature of the elements involved. Iodine predominantly forms covalent bonds, where electrons are shared between atoms.
For instance, iodine forms a polar covalent bond with hydrogen in HI due to the electronegativity difference, resulting in an oxidation state of -1 for iodine. In ICl, the bond is also covalent, but iodine shares electrons with a more electronegative chlorine atom, leading to a positive +1 state for iodine.
In complex polyatomic molecules like HIO4, electron sharing with multiple oxygen atoms highlights iodine’s versatility. Such bonding scenarios underscore how variations in electronegativity and molecule structure influence iodine's oxidation state in its compounds.