Problem 13 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The oxidation number of sulphur in \(\mathrm{S}_{8}, \mathrm{~S}_{2} \mathrm{~F}_{2}, \mathrm{H}_{2} \mathrm{~S}\) respectively, are [1999-2 Marks] (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

(a) \(0, +1\) and \(-2\).

Step 1: Determine Oxidation Number in Elemental Sulphur

The oxidation number of sulphur in its elemental form \(\mathrm{S}_8\) is 0. This is because each sulfur atom shares two electrons equally with its neighboring sulfur atoms in the ring.

Step 2: Determine Oxidation Number of Sulphur in \(\mathrm{S}_2\mathrm{F}_2\)

Fluorine has an oxidation number of \(-1\). In \(\mathrm{S}_2\mathrm{F}_2\), the total oxidation number for two fluorine atoms is \(-2\). Let the oxidation number of each sulfur(atom be \(x\). The equation is: \[2x + 2(-1) = 0\] Solving, \[2x - 2 = 0\] \[2x = 2\] \[x = +1\]. So, the oxidation number of sulfur in \(\mathrm{S}_2\mathrm{F}_2\) is \(+1\).

Step 3: Determine Oxidation Number of Sulphur in \(\mathrm{H}_2\mathrm{S}\)

Hydrogen has an oxidation number of \(+1\). In \(\mathrm{H}_2\mathrm{S}\), the total oxidation number for two hydrogen atoms is \(+2\). Let the oxidation number of sulfur be \(y\). The equation is: \[2(+1) + y = 0\] Solving, \[2 + y = 0\] \[y = -2\]. Thus, the oxidation number of sulfur in \(\mathrm{H}_2\mathrm{S}\) is \(-2\).

Step 4: Match with Options

The calculated oxidation numbers for sulfur are \(0\) in \(\mathrm{S}_8\), \(+1\) in \(\mathrm{S}_2\mathrm{F}_2\), and \(-2\) in \(\mathrm{H}_2\mathrm{S}\). Identifying the matching option from the given choices, the correct answer is (a) \(0,+1\) and \(-2\).

Key Concepts

Sulfur Chemistry

Sulfur is a fascinating element due to its diverse chemical behavior. It is often found in nature in its elemental form and also as a component of numerous chemical compounds. This versatility is due to its ability to form various bonds and oxidation states, making sulfur chemistry a complex but intriguing field.

Elemental sulfur, like in \(\mathrm{S}_8\), forms a ring structure where each sulfur atom is bonded to two others. Because these bonds involve equal sharing of electrons, each sulfur atom has an oxidation number of zero. This is common for elements in their pure form. Being able to form such stable rings is part of what makes sulfur unique among the elements and critical in various natural processes.

Oxidation States

Oxidation states, also known as oxidation numbers, are important in understanding the electron transfer in chemical reactions. They tell us how many electrons an atom can gain, lose, or share when it forms a chemical compound.

In sulfur compounds, like \(\mathrm{S}_2\mathrm{F}_2\) and \(\mathrm{H}_2\mathrm{S}\), sulfur can exhibit different oxidation states. For instance, in \(\mathrm{S}_2\mathrm{F}_2\), we find sulfur in a "+1" state due to its combination with fluorine, an element with a strong tendency to attract electrons, leaving sulfur slightly electron-deficient.

In the case of \(\mathrm{H}_2\mathrm{S}\), sulfur has an oxidation state of \(-2\) because hydrogen, although slightly electropositive, lets sulfur hold onto more electrons compared to its oxidized state. This results in the sulfur atom gaining electrons, thus obtaining a negative oxidation state.

Chemical Compounds Analysis

Analyzing chemical compounds involves understanding the relationships between the elements that compose them, especially through their oxidation states. When examining a compound, one should identify the typical oxidation states of its components to analyze their interactions.

For example, in the compound \(\mathrm{S}_2\mathrm{F}_2\), recognizing fluorine's standard \(-1\) oxidation state helps to deduce sulfur's state. Knowing that the net charge of the compound is zero, you can set up an equation balancing the total charges, which aids in determining sulfur's role in the compound.

Similarly, for \(\mathrm{H}_2\mathrm{S}\), understanding that hydrogen usually adopts a "+1" state, assists in calculating sulfur's state as \(-2\) due to the overall neutral nature of the compound. Thus, practicing such analyses sharpens chemistry problem-solving skills and deepens understanding of chemical properties and behaviors.