Problem 112 - A Complete Resource Book in Chemistry for JEE Main

Question

The oxidation number of sulphur in \(\mathrm{S}_{8}, \mathrm{~S}_{2} \mathrm{~F}_{2}, \mathrm{H}_{2} \mathrm{~S}\) respectively, are (a) \(0,+1\) and \(-2\) (b) \(+2,+1\) and \(-2\) (c) \(0,+1\) and \(+2\) (d) \(-2,+1\) and \(-2\).

Step-by-Step Solution

Answer

The oxidation numbers are (a) \(0,+1\) and \(-2\).

Step 1: Understanding Oxidation Number

The oxidation number of an element in a molecule refers to the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic. For example, the oxidation number of oxygen is usually -2, and that of hydrogen is usually +1, in compounds.

Step 2: Determine Oxidation Number of Sulfur in S8

In \(\mathrm{S}_{8}\), sulfur exists in its natural elemental form, hence the oxidation number in this form is \(0\).

Step 3: Determine Oxidation Number of Sulfur in S2F2

In \(\mathrm{S}_{2} \mathrm{~F}_{2}\), the fluorine atom is more electronegative and typically has an oxidation number of \(-1\). Thus, for two fluorine atoms: \2 \times (-1) = -2\. Given the compound should be neutral, the sum for the two sulfur atoms must be \(+2\). Hence, the oxidation number for sulfur is \(\frac{+2}{2} = +1\).

Step 4: Determine Oxidation Number of Sulfur in H2S

In \(\mathrm{H}_{2} \mathrm{S}\), the hydrogen atoms each have an oxidation number of \(+1\). So for two hydrogen atoms: \( 2 \times (+1) = +2\). The sulfur, in keeping with the neutral charge of the compound, must have an oxidation number of \(-2\).

Key Concepts

Oxidation State of Sulfur

In chemistry, the oxidation state, or oxidation number, is crucial to understand how elements interact within compounds. Particularly for sulfur, its oxidation state can vary, which is essential to comprehend reactions and bonding in chemical compounds. The oxidation state is an assigned value indicating an element's loss, gain, or sharing of electrons when forming compounds.

When sulfur is present in its elemental form, such as \(\mathrm{S}_8\), it has an oxidation state of \(0\). This neutrality means that sulfur atoms are neither gaining nor losing electrons in this naturally occurring state. Understanding these values helps predict the behavior and reactivity of sulfur in different chemical contexts.

Chemical Bonding

Chemical bonding involves the attraction between atoms that leads to molecule and compound formation. This concept is essential in recognizing how elements like sulfur interact with others, such as fluorine and hydrogen.

• Elements bond to achieve a stable electronic structure, often through sharing, donating, or accepting electrons.
• The type of bond, such as ionic or covalent, influences the oxidation states of the elements involved.

For example, in \(\mathrm{S}_2\mathrm{F}_2\), sulfur forms bonds with fluorine, which is highly electronegative. Fluorine, being powerful in attracting electrons, largely determines the oxidation state of sulfur by influencing electron distribution in the \(\mathrm{S}_2\mathrm{F}_2\) compound.

Elementary Sulfur Oxidation State

Elementary sulfur is unique compared to when sulfur is part of a compound.

• In its elemental state in nature, sulfur usually exists as \(\mathrm{S}_8\), where it forms a ring of directly bonded sulfur atoms.
• This natural state means all sulfur bonds are of the same type, and thus, each sulfur atom in the ring bears an oxidation number of \(0\).

The importance of identifying sulfur's oxidation state in such elementary form lies in establishing a baseline for comparison as sulfur transitions into various compounds. This benchmark helps in calculating the overall charge and valency interactions when sulfur participates in reactions, providing a clear insight into the behaviors and properties of sulfur in complex chemical processes.