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

Correct answer is (a) 0, +1, and -2.

Step 1: Determine Oxidation Number of Sulphur in S8

In elemental form as in \( \text{S}_8 \), the oxidation number of sulphur is 0 because the atoms are not combined with any other element and must, therefore, be balanced by themselves.

Step 2: Calculate Oxidation Number of Sulphur in S2F2

For \( \text{S}_2\text{F}_2 \), fluorine, which is more electronegative, has an oxidation number of -1. Let the oxidation number of sulphur be \( x \). The equation is: \[ 2x + 2(-1) = 0 \]Solving gives: \[ 2x - 2 = 0 \] \[ 2x = 2 \] \[ x = +1 \].Thus, each sulphur has an oxidation number of +1.

Step 3: Find Oxidation Number of Sulphur in H2S

In \( \text{H}_2\text{S} \), hydrogen has an oxidation number of +1. Let the oxidation number of sulphur be \( x \). The equation is: \[ 2(+1) + x = 0 \] Simplifying gives:\[ 2 + x = 0 \] \[ x = -2 \].Thus, the oxidation number of sulphur in \( \text{H}_2\text{S} \) is -2.

Key Concepts

Sulphur Compounds

Sulphur compounds play a significant role in chemistry, displaying diverse properties and occurring in various forms. The study of sulphur compounds involves understanding how sulphur bonds with other elements, such as fluorine or hydrogen. These relationships lead to variations in properties and functionalities.For example, in elemental form like in \(S_8\), sulphur exhibits an oxidation number of zero. This means each sulphur atom shares electrons equally with other sulphur atoms akin to how individuals might share a pizza where everyone gets an equal slice. Sulphur can also bond with fluorine to form \(S_2F_2\), where the presence of a more electronegative element like fluorine creates a different situation. Here, each sulphur atom steps up to a +1 oxidation state, showcasing an imbalance where sulphur atoms donate some electron density to fluorine atoms. Lastly, in compounds like hydrogen sulphide (\(H_2S\)), sulphur takes on an even more unique character. Hydrogen’s typical role as a donor of electron density leaves sulphur reduced to an oxidation number of -2, reflecting its increased electron density sharing.

Chemical Bonding

Chemical bonding explains how elements connect to each other to form compounds with new properties. Bonds form when electron sharing occurs between atoms, resulting in molecules with distinct behaviors. There are several types of bonds:

• Covalent bonds: Atoms share electron pairs, as seen in \(S_8\), where sulphur atoms equally share electrons.
• Ionic bonds: Electrons are transferred between atoms, creating charged ions. Although not explicitly ionic, the electronegativity differences in \(S_2F_2\) lead to partial ionic character due to electron pulling fluorine does.
• Polar covalent bonds: Unequal sharing of electrons between atoms occurs, leading to partial charges such as those in \(S_2F_2\).

Understanding these types of bonds is crucial for grasping the nature of chemical reactions and how compounds like sulphur compounds behave in different contexts. Each bond type introduces unique attributes to the compound, impacting its physical and chemical properties.

Elementary Chemistry

Elementary Chemistry is the introductory study of fundamental concepts in chemistry, essential for understanding more complex chemical behaviors. It provides the groundwork for insights into atomic structure, chemical reactions, and the periodic table.A core concept in elementary chemistry is the understanding of oxidation states or numbers. These numbers help chemists determine how atoms interact within a compound:

• An oxidation number indicates how many electrons an atom gains or loses.
• In \(S_8\), each sulphur atom retains an oxidation number of 0, showcasing no electron gain or loss among identical atoms.
• Contrast this with \(S_2F_2\), where the contrast in electronegativity between sulphur and fluorine leads each sulphur to hold a +1 oxidation number.
• For \(H_2S\), the hydrogen donation of electrons results in sulphur having an oxidation number of -2, indicating a gain of electron concentration.

These principles of oxidation numbers and elementary reactions aid in predicting and balancing chemical equations, thus forming the stepping stones toward deeper chemical comprehension.