Problem 12 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Amongst the following identify the species with an atom in \(+6\) oxidation state [2000S] (a) \(\mathrm{MnO}_{4}^{-}\) (b) \(\mathrm{Cr}(\mathrm{CN})_{6}^{3-}\) (c) \(\mathrm{NiF}_{6}^{2-}\) (d) \(\mathrm{CrO}_{2} \mathrm{Cl}_{2}\)

Step-by-Step Solution

Answer

\(\mathrm{CrO}_{2} \text{Cl}_{2}\) has the atom in a \(+6\) oxidation state.

Step 1: Analyze \(\mathrm{MnO}_{4}^{-}\)

The permanganate ion, \(\mathrm{MnO}_{4}^{-}\), consists of manganese. Oxygen typically has an oxidation state of \(-2\). Using the formula: \([x + 4(-2) = -1]\), solve for \(x\), the oxidation state of manganese.\(x - 8 = -1 \Rightarrow x = +7\).

Step 2: Analyze \(\mathrm{Cr}( ext{CN})_{6}^{3-}\)

In the complex \(\mathrm{Cr}( ext{CN})_{6}^{3-}\), cyanide \((\text{CN})\) has an oxidation state of \(-1\). Let chromium's oxidation state be \(x\). Using the formula: \([x + 6(-1) = -3]\), solve for \(x\).\(x - 6 = -3 \Rightarrow x = +3\).

Step 3: Analyze \(\mathrm{NiF}_{6}^{2-}\)

In \(\mathrm{NiF}_{6}^{2-}\), each fluoride \((\text{F})\) has an oxidation state of \(-1\). Let the oxidation state of nickel be \(x\). Use the formula: \([x + 6(-1) = -2]\), and solve for \(x\).\(x - 6 = -2 \Rightarrow x = +4\).

Step 4: Analyze \(\mathrm{CrO}_{2} ext{Cl}_{2}\)

In \(\mathrm{CrO}_{2} ext{Cl}_{2}\), oxygen has an oxidation state of \(-2\), and chlorine has an oxidation state of \(-1\). Let chromium's oxidation state be \(x\). Using the formula: \([x + 2(-2) + 2(-1) = 0]\), solve for \(x\).\(x - 4 - 2 = 0 \Rightarrow x = +6\).

Key Concepts

Permanganate Ion

The permanganate ion, represented as \(\mathrm{MnO}_{4}^{-}\), features the metal manganese in various chemical reactions. The understanding of this ion's oxidation state is crucial because it plays a significant role in redox reactions. Typically, oxygen carries an oxidation state of \(-2\). For the permanganate ion, we use the formula \([x + 4(-2) = -1]\) to determine the oxidation state of manganese.Here's how it works:

• Each oxygen contributes \(-2\) to the overall charge.
• With four oxygen atoms, the total contribution is \(-8\).
• The ion has a net charge of \(-1\), allowing us to set up the equation: \(x - 8 = -1\).

Solving this yields \(x = +7\). Therefore, manganese in the permanganate ion holds an oxidation state of \(+7\). This high oxidation state enables its effectiveness as a strong oxidizing agent in various chemical processes.

Chromium Complexes

Chromium forms several important compounds, one of them being \(\mathrm{Cr}(\text{CN})_{6}^{3-}\), a complex with cyanide \((\text{CN})\) ions. Understanding the oxidation state within such complexes involves recognizing the charge contribution of cyanide, which is \(-1\). To identify the oxidation state of chromium, we utilize the formula \([x + 6(-1) = -3]\).Here’s the breakdown:

• Each cyanide ion contributes \(-1\).
• With six cyanide ions, the total contribution from cyanide is \(-6\).
• The complex carries a net charge of \(-3\), allowing for the equation: \(x - 6 = -3\).

Solving the equation, we find \(x = +3\). Therefore, the oxidation state of chromium in this complex is \(+3\). This knowledge is particularly useful in studying the chemistry and reactivity patterns of chromium complexes.

Nickel Fluorides

Nickel fluorides are part of an interesting group of nickel compounds, among which \(\mathrm{NiF}_{6}^{2-}\) is notable. Here, determining the oxidation state of nickel involves considering that each fluoride ion \((\text{F})\) has an oxidation state of \(-1\). We apply the formula \([x + 6(-1) = -2]\) to solve for nickel.Understanding the specifics:

• Each fluorine atom contributes \(-1\).
• Having six fluorines gives a total contribution of \(-6\).
• With a net charge of \(-2\), the relevant formula is \(x - 6 = -2\).

By solving the formula, we derive that \(x = +4\). Therefore, nickel in \(\mathrm{NiF}_{6}^{2-}\) has an oxidation state of \(+4\). This example demonstrates how the unique properties of fluorides impact the oxidation states of metals in chemical compounds.

Oxidation State Calculations

Oxidation state calculations are central to understanding the behavior of elements within compounds. An oxidation state, also known as oxidation number, represents the number of electrons lost or gained by an atom in a chemical substance. These calculations allow chemists to track electrons in redox reactions, assisting them in categorizing a substance as an oxidizing or reducing agent.When calculating oxidation states:

• Let the unknown oxidation state be \(x\).
• Consider the typical oxidation states: for example, oxygen is usually \(-2\), fluorine is \(-1\), and chlorine can vary.
• Set up an equation based on the compound's net charge, preserving the overall electric neutrality.

For instance, in \(\mathrm{CrO}_{2} \mathrm{Cl}_{2}\), we find chromium's oxidation state by setting up \([x + 2(-2) + 2(-1) = 0]\). Solving such problems requires substituting known values into the formula and solving for \(x\). Hence, accurately determining oxidation states is crucial for predicting reaction outcomes.