Problem 154 - A Complete Resource Book in Chemistry for JEE Main

Question

Identify the compounds in which the sulphur atoms are in different oxidation states? (a) \(\mathrm{K}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}\) (b) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) (c) \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\) (d) \(\mathrm{K}_{2} \mathrm{~S}_{2} \overline{\mathrm{O}}_{\mathrm{s}}\)

Step-by-Step Solution

Answer

Compound (c) has sulphur atoms in different oxidation states: \( \mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6} \).

Step 1: Understand the Question

We need to determine in which of the given compounds the sulphur atoms exist in different oxidation states.

Step 2: Determine the Oxidation State Formula

To find the oxidation states of sulphur, use the fact that the sum of oxidation states in a neutral compound is zero. For a compound, calculate the total charge using known oxidation states of other elements and solve for sulphur.

Step 3: Analyze (a) \( \mathrm{K}_{2} \mathrm{~S}_{2} \mathrm{O}_{7} \)

In \( \mathrm{K}_{2} \mathrm{~S}_{2} \mathrm{O}_{7} \), potassium has an oxidation state of +1 and oxygen is -2. Let the oxidation state of sulphur be \( x \). The equation is:\[ 2(+1) + 2x + 7(-2) = 0 \]Solving this gives \( x = +6 \). Both sulphur atoms are in the +6 state.

Step 4: Analyze (b) \( \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \)

In \( \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \), sodium is +1 and oxygen is -2. The equation is:\[ 2(+1) + 2x + 3(-2) = 0 \]Solving yields \( x = +2 \). Both sulphur atoms are in the same oxidation state.

Step 5: Analyze (c) \( \mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6} \)

In \( \mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6} \), sodium is +1 and oxygen is -2. Let's consider two sulphur atoms are in \(+5\) state and the other two are in \(-1\):\[ 2(+1) + 4(+5\text{ or }-1) + 6(-2) = 0 \]Using different \( x \) values show some are +5, some are -1, fulfilling mixed states.

Step 6: Analyze (d) \( \mathrm{K}_{2} \mathrm{~S}_{2} \overline{\mathrm{O}}_{8} \)

In \( \mathrm{K}_{2} \mathrm{~S}_{2} \mathrm{O}_{8} \), potassium is +1 and oxygen is -2. The equation is:\[ 2(+1) + 2x + 8(-2) = 0 \]Solving for \( x \) gives \( x = +7 \). Both sulphur atoms are in the +7 state.

Key Concepts

Sulphur Compounds

Sulphur compounds are an integral part of chemistry due to the diverse oxidation states that sulfur can exhibit. These compounds are found across various chemical environments and often play crucial roles in biological and industrial processes.
Sulphur, a chalcogen like oxygen, is commonly associated with other nonmetals such as oxygen, forming sulfur oxides and thiosulfates. The ability of sulfur to assume varying oxidation states allows these compounds to participate in a wide range of chemical reactions. These reactions can involve oxidation or reduction, dramatically changing their chemical nature and functionality. This variability in oxidation states makes sulfur a versatile element.

• Sulfur can bond with a simple element like oxygen to form sulfates and sulfites, which are commonly found in nature.
• In industrial applications, sulfur compounds are involved in the synthesis of chemicals like sulfuric acid, an important industrial chemical.
• Biologically, these compounds range from vital amino acids like cysteine and methionine to sulfate ions critical to metabolism.

Understanding these properties of sulfur compounds helps us appreciate their role in sustaining life and their importance in various technological applications.

Chemical Reactions

Chemical reactions are transformative processes where substances, known as reactants, are converted into different substances, called products. During chemical reactions, the breaking and forming of bonds between atoms occur, which is accompanied by energy changes. This process underpins a vast range of applications from rusting, combustion, digestion, to synthesis of new materials.
Every chemical reaction adheres to the Law of Conservation of Mass, implying that the number of each type of atom remains constant before and after the reaction. Chemical equations succinctly describe these reactions, identifying the reactants, products, and their stoichiometric measures. Chemical reactions can be categorized into several types depending on the nature of the reactants and products:

• **Synthesis reactions**, where two or more simple substances combine to form a more complex compound.
• **Decomposition reactions**, which involve breaking down a compound into simpler substances.
• **Single displacement reactions**, where one element replaces another in a compound.
• **Double displacement reactions**, involving the exchange of components between two compounds.

These reactions are essential in the study of chemistry, as understanding them can explain how substances interact and transform, enabling innovations and solutions in science and industry.

Oxidation Number Calculation

Calculating oxidation numbers is crucial for understanding the behavior and transformation of substances in chemical reactions. The oxidation number is an indicator of the degree of oxidation (loss of electrons) or reduction (gain of electrons) that an atom undergoes within a compound. This concept is particularly important in redox reactions, where transfer of electrons between species is involved.
To calculate the oxidation number, follow these simple rules:

• The oxidation number for an atom in its elemental form is always zero. For example, O extsubscript{2} and S extsubscript{8} have an oxidation number of zero.
• In monoatomic ions, the oxidation number equals the charge of the ion. For instance, Na extsuperscript{+} has an oxidation number of +1.
• Oxygen usually has an oxidation number of -2, and hydrogen is typically +1 when bonded to nonmetals.
• The sum of oxidation numbers in a neutral compound is zero, while in a polyatomic ion, it equals the ion's charge.

By applying these rules, you can accurately determine the oxidation states of elements within compounds, as seen in our exercise with various sulfur compounds. This method not only helps in balancing chemical reactions but also in predicting the reactivity and properties of substances.