Problem 26 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Arrange the following in increasing oxidation number of iodine. [1986-1 Mark] \(\mathrm{I}_{2}, \mathrm{HI}, \mathrm{HIO}_{4}, \mathrm{ICl}\)

Step-by-Step Solution

Answer

\( \mathrm{HI} < \mathrm{I}_2 < \mathrm{ICl} < \mathrm{HIO_4} \)

Step 1: Determine Oxidation State of Iodine in I2

In the molecule \( \mathrm{I}_2 \), which is a diatomic molecule made of two identical iodine atoms, the oxidation state of iodine is zero. This is because there is no electronegativity difference between the two atoms, resulting in no charge separation.

Step 2: Determine Oxidation State of Iodine in HI

In \( \mathrm{HI} \), hydrogen has an oxidation state of \(+1\). Since the compound is neutral, iodine must have an oxidation state of \(-1\) to balance the charge.

Step 3: Determine Oxidation State of Iodine in HIO4

In \( \mathrm{HIO_4} \), oxygen has an oxidation state of \(-2\). There are four oxygens, so the total contribution is \(-8\). Hydrogen contributes \(+1\). To balance this in a neutral compound, iodine must have an oxidation state of \(+7\).

Step 4: Determine Oxidation State of Iodine in ICl

In \( \mathrm{ICl} \), chlorine is more electronegative than iodine and has an oxidation state of \(-1\). Therefore, iodine must have an oxidation state of \(+1\) to balance the charge.

Step 5: Arrange the Compounds

Based on the calculated oxidation numbers, arrange the compounds in order of increasing oxidation states of iodine. The compounds and their iodine oxidation states are: \( \mathrm{HI} (-1) \), \( \mathrm{I}_2 (0) \), \( \mathrm{ICl} (+1) \), and \( \mathrm{HIO_4} (+7) \). Thus, the order is: \( \mathrm{HI} < \mathrm{I}_2 < \mathrm{ICl} < \mathrm{HIO_4} \).

Key Concepts

Iodine Compounds

Iodine is a versatile element that forms a variety of compounds with different elements. Whether it's forming simple diatomic molecules or more complex combinations, iodine's ability to form bonds is impressive.
Some common iodine compounds include:

• \( \mathrm{I_2} \) (iodine molecule): This is a simple diatomic molecule, where two iodine atoms share electrons equally. As the element is in its pure form, it serves as a baseline for oxidation number calculations.
• \( \mathrm{HI} \) (hydrogen iodide): A compound where iodine forms a bond with hydrogen. Here, the iodine is reduced, giving it a negative oxidation number.
• \( \mathrm{HIO_4} \) (periodic acid): In this compound, iodine is bonded to hydrogen and oxygen in a highly oxidized state. It is interesting to explore because iodine reaches its highest common oxidation state here.
• \( \mathrm{ICl} \) (iodine monochloride): A mixed halide compound showcasing a bond between iodine and a more electronegative halogen, leading to iodine being in a positive oxidation state.

Understanding these compounds helps us appreciate the variability and bonding characteristics iodine exhibits across different chemical environments.

Oxidation State Determination

Determining the oxidation state, also known as oxidation number, involves figuring out how electrons are distributed in a compound. It plays a crucial part in understanding the behavior of atoms in reactions. Here’s how to go about it:
Steps for Determining Oxidation States:

• Initially, recognize the standard oxidation states of common elements: hydrogen is usually \(+1\), oxygen \(-2\), unless in special cases.
• For a neutral molecule, the sum of the oxidation states must equal zero. This means that the positive charges must balance the negative charges.
• When analyzing charged species or ions, their net charge must equal the sum of the oxidation states of all their atoms.

Applying these rules to iodine compounds:

• In \( \mathrm{I_2} \), iodine is in elemental form, thus its oxidation state is 0.
• In \( \mathrm{HI} \), hydrogen has an oxidation state of \(+1\), making iodine \(-1\) to balance it to zero.
• In \( \mathrm{HIO_4} \), the oxidation state of iodine is computed by ensuring the sum \(+1 + x + 4(-2) = 0\) becomes balanced, resulting in \(x = +7\).
• Finally, in \( \mathrm{ICl} \), with chlorine at \(-1\), iodine takes on \(+1\) to neutralize.

Knowing these methods will equip you with the tools required to solve complex chemical problems effortlessly.

Chemical Bonding Concepts

Chemical bonding is the backbone of molecular structure and behavior, defining how atoms link together to form compounds. Let's delve into the different bonding aspects particularly relevant to iodine and its compounds.
Ionic vs. Covalent Bonds:

• Covalent Bonding: This occurs when two atoms share electrons. In the case of \( \mathrm{I_2} \), this is quite straightforward since both atoms have identical electronegativity. Thus, electrons are shared equally leading to a nonpolar covalent bond.
• Ionic Bonding: Features a transfer of electrons and is prevalent when there is a significant difference in electronegativity between the atoms. Although iodine compounds like \( \mathrm{HI} \) might look ionic, they actually possess covalent characteristics due to the closeness in electronegativity.

The type of chemical bonding affects the oxidation states we assign:

• In \( \mathrm{HIO_4} \), iodine forms bonds with oxygen which is highly electronegative, pulling the shared electrons towards oxygen, leading to a higher oxidation state in iodine.
• For \( \mathrm{ICl} \), chlorine is slightly more electronegative than iodine, slightly pulling the shared electrons towards it, and attributing a positive oxidation state to iodine.

Understanding these bonding differences aids in predicting and rationalizing the properties and reactivities of iodine in various chemical settings. This fundamental insight is key for explaining numerous chemical phenomena.