Problem 54 - A Complete Resource Book in Chemistry for JEE Main

Question

Among the following, identify the species with an atom in \(+6\) oxidation state (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

(d) \( \mathrm{CrO}_{2} \mathrm{Cl}_{2} \) is the species with Cr in the +6 oxidation state.

Step 1: Identify Oxidation States in \( \mathrm{MnO}_{4} \)

The permanganate ion \( \mathrm{MnO}_{4} \) has oxygen atoms with an oxidation state of \(-2\). Set the equation for the oxidation number of Mn: let \( x \) be the oxidation state of Mn. Then the equation is \( x + 4(-2) = -1 \). Solve for \( x \): \( x - 8 = -1 \). Thus, \( x = +7 \). Mn is in the +7 oxidation state.

Step 2: Calculate Oxidation States in \( \mathrm{Cr} (\mathrm{CN})_{6}^{3-} \)

The ligand cyanide \( \mathrm{CN}^- \), contributes \(-1\) for each of the 6 CN groups, totaling \(-6\). Let \( x \) be the oxidation state of Cr. The overall charge is \(-3\), so \( x + (-6) = -3 \). Solving gives \( x = +3 \). Cr is in the +3 oxidation state.

Step 3: Determine Oxidation States in \( \mathrm{NiF}_{6}^{2-} \)

Each fluorine atom has an oxidation state of \(-1\). Let \( x \) be the oxidation state of Ni. The equation is \( x + 6(-1) = -2 \). Therefore, \( x - 6 = -2 \), solving gives \( x = +4 \). Ni is in the +4 oxidation state.

Step 4: Compute Oxidation States in \( \mathrm{CrO}_{2} \mathrm{Cl}_{2} \)

In \( \mathrm{CrO}_{2} \mathrm{Cl}_{2} \), the oxygen atoms have an oxidation state of \(-2\), while chlorine atoms have \(-1\). Let the oxidation state of Cr be \( x \). The equation becomes: \( x + 2(-2) + 2(-1) = 0 \). Simplifying gives \( x - 4 - 2 = 0 \), so \( x = +6 \). Cr is in the +6 oxidation state.

Key Concepts

Permanganate Ion

The permanganate ion, represented as \( \mathrm{MnO}_{4}^- \), is a polyatomic ion featuring manganese bonded to four oxygen atoms. In chemical compounds, the oxidation states help us understand the electron distribution within the molecule. This is especially important for reactions involving redox processes. In the permanganate ion, each oxygen atom typically has an oxidation state of \(-2\). To find the oxidation state of manganese \( \text{Mn} \) in \( \mathrm{MnO}_4^- \), set \( x \) as the oxidation state of manganese, allowing us to form the equation:

• \( x + 4(-2) = -1 \)

By solving this, \( x - 8 = -1 \) simplifies to \( x = +7 \). This tells us that manganese has an oxidation state of \(+7\) in permanganate. Understanding this allows chemists to predict the behavior of the permanganate ion in various reactions, particularly its role as an oxidizing agent.

Cyanide Ligand

Cyanide ligands consist of a carbon atom triple-bonded to a nitrogen atom, and collectively they carry a negative charge, denoted as \( \mathrm{CN}^- \). As a ligand, cyanide is known for its strong field due to the ability to donate electron pairs to a central metal ion. In coordination chemistry, this donation can significantly alter the oxidation state of the metal. For example, in \( \mathrm{Cr} (\mathrm{CN})_{6}^{3-} \), chromium \( \mathrm{Cr} \) coordinates with six cyanide ligands. Given the oxidation state contribution of each cyanide is \(-1\), the equation:

• \( x + 6(-1) = -3 \)

where \( x \) is the oxidation state of chromium, allows us to find that \( x = +3 \). Knowing the oxidation state of metals in complexes help chemists predict the compound's properties and reactivity.

Fluorine Oxidation State

Fluorine, the most electronegative element, nearly always possesses an oxidation state of \(-1\) in compounds. This makes it a powerful oxidizing agent in reactions with less electronegative elements. When calculating the oxidation state of a metal in a compound with multiple fluorine atoms, each contributes \(-1\). As seen in \( \mathrm{NiF}_{6}^{2-} \), let \( x \) be the oxidation state of nickel (Ni):

• \( x + 6(-1) = -2 \)

Solving gives \( x = +4 \). This showcases fluorine's role in determining the electron count around the metal ion, which is crucial in predicting the compound's chemical behavior.

Chromium Compound Oxidation States

Chromium is an element known for exhibiting a multitude of oxidation states, typically ranging from \( +2 \) to \( +6 \). This versatility is due to its unique electronic configuration, which allows it to donate or share a variety of electron numbers in chemical processes. In particular, the compound \( \mathrm{CrO}_{2}\mathrm{Cl}_{2} \) presents an example where chromium adopts the \(+6\) oxidation state.The equation accounting for all atoms in the compound:

• \( x + 2(-2) + 2(-1) = 0 \)

Simplifies to \( x = +6 \) for chromium. This high oxidation state of chromium is often associated with strong oxidizing behavior, influencing its role in industrial and environmental chemistry processes.