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

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

Step 1: Determine the oxidation state of Mn in \( \mathrm{MnO}_{4}^{-} \)

To find the oxidation state of Mn in \( \mathrm{MnO}_{4}^{-} \), set up an equation. Oxygen is normally \(-2\) in compounds. The sum of the oxidation states in the molecule should equal the charge of the ion, which is \(-1\).Letting \( x \) be the oxidation state of Mn: \[ \begin{align*} x + 4(-2) &= -1 \ x - 8 &= -1 \ x &= +7 \end{align*} \]Thus, Mn is in the \(+7\) oxidation state.

Step 2: Determine the oxidation state of Cr in \( \mathrm{Cr} (\mathrm{CN})_{6}^{3-} \)

In \( \mathrm{Cr} (\mathrm{CN})_{6}^{3-} \), \( \mathrm{CN}^{-} \) has a charge of \(-1\). Since there are six cyano ligands, their total charge is \(-6\). The charge of the entire complex is \(-3\).Letting \( y \) be the oxidation state of Cr:\[ \begin{align*} y + 6(-1) &= -3 \ y - 6 &= -3 \ y &= +3 \end{align*} \]Thus, Cr is in the \(+3\) oxidation state.

Step 3: Determine the oxidation state of Ni in \( \mathrm{NiF}_{6}^{2-} \)

In \( \mathrm{NiF}_{6}^{2-} \), each fluorine atom has an oxidation state of \(-1\). With six fluorine atoms, their total contribution is \(-6\). The charge on the whole ion is \(-2\).Letting \( z \) be the oxidation state of Ni:\[ \begin{align*} z + 6(-1) &= -2 \ z - 6 &= -2 \ z &= +4 \end{align*} \]Thus, Ni is in the \(+4\) oxidation state.

Step 4: Determine the oxidation state of \( \mathrm{Cr} \) in \( \mathrm{CrO}_{2} \mathrm{Cl}_{2} \)

In \( \mathrm{CrO}_{2} \mathrm{Cl}_{2} \), we need to find the oxidation state of Cr. Oxygen generally has an oxidation state of \(-2\), totaling \(-4\) for \( \mathrm{CrO}_{2} \). Chlorine generally is \(-1\), totaling \(-2\) for \( \mathrm{Cl}_{2} \). The compound is neutral.Letting \( w \) be the oxidation state of Cr:\[ \begin{align*} w + 2(-2) + 2(-1) &= 0 \ w - 4 - 2 &= 0 \ w &= +6 \end{align*} \]Thus, Cr is in the \(+6\) oxidation state.

Key Concepts

MnO4- oxidation state

The molecule permanganate, represented as \( \mathrm{MnO}_4^- \), contains manganese (Mn) and oxygen (O). Oxygen typically has an oxidation state of \(-2\). In a compound like this, the sum of oxidation states must equal the overall charge of the ion, which is \(-1\).
The formula to set for manganese in this scenario looks like this: \( x + 4(-2) = -1 \), where \( x \) is defined as the oxidation state of manganese. Solving this equation:

• The total contribution from oxygen is \( 4 \times (-2) = -8 \).
• Thus, \( x - 8 = -1 \).
• Solving for \( x \) gives \( x = +7 \).

Therefore, manganese in \( \mathrm{MnO}_4^- \) is in the \(+7\) oxidation state. This means that manganese atoms are highly oxidized in this compound, lending the ion significant oxidizing properties.

Cr(CN)6 3- oxidation state

Chromium hexacyanide, or \( \mathrm{Cr(CN)}_6^{3-} \), consists of chromium (Cr) and cyanide ions (CN). Each cyanide ion, \( \mathrm{CN}^- \), carries a charge of \(-1\). Since there are six cyanide ions, their total charge amounts to \(-6\).
For the entire complex, which has a charge of \(-3\), the equation looks like this: \( y + 6(-1) = -3 \), with \( y \) representing the oxidation state of chromium. Solving the equation involves:

• The contribution from cyanide is \( 6 \times (-1) = -6 \).
• Thus, \( y - 6 = -3 \).
• Solving for \( y \) gives \( y = +3 \).

So, chromium is in a \(+3\) oxidation state in this complex. This intermediate oxidation state allows chromium in \( \mathrm{Cr(CN)}_6^{3-} \) to be both a reducing and an oxidizing agent under different conditions.

NiF6 2- oxidation state

Nickel hexafluoride, denoted as \( \mathrm{NiF}_6^{2-} \), comprises nickel (Ni) and fluorine (F). Fluorine generally has an oxidation state of \(-1\). With six fluorine atoms, their collective contribution to the charge adds up to \(-6\).
For the whole ion, which has a \(-2\) charge, the equation can be developed as follows: \( z + 6(-1) = -2 \), where \( z \) is the oxidation state of nickel. Solving this equation involves:

• The total contribution from fluorine is \( 6 \times (-1) = -6 \).
• Thus, \( z - 6 = -2 \).
• Solving for \( z \) results in \( z = +4 \).

Therefore, nickel is in a \(+4\) oxidation state within this complex. In this state, nickel can engage in various chemical reactions, exhibiting different catalytic properties due to its unusual oxidation state.

CrO2Cl2 oxidation state

The compound \( \mathrm{CrO}_2\mathrm{Cl}_2 \) includes chromium (Cr), oxygen (O), and chlorine (Cl). Here, oxygen typically possesses an oxidation state of \(-2\), resulting in a total of \(-4\) for the two oxygen atoms. Chlorine generally has an oxidation state of \(-1\), contributing a total of \(-2\) for the two chlorine atoms.
For a neutral compound, the equation becomes \( w + 2(-2) + 2(-1) = 0 \), with \( w \) as the oxidation state of chromium. To solve:

• The contribution from oxygen is \( 2 \times (-2) = -4 \).
• The contribution from chlorine is \( 2 \times (-1) = -2 \).
• Thus, the equation is \( w - 4 - 2 = 0 \).
• Solving for \( w \) results in \( w = +6 \).

As a result, chromium in \( \mathrm{CrO}_2\mathrm{Cl}_2 \) is in the \(+6\) oxidation state, showcasing the versatile oxidation chemistry of chromium. This high state indicates strong oxidizing behavior and significant reactivity in various chemical processes.