Identify the process in which change in the oxidation state is five
(a) Cr2O7^2- - 2Cr^3+
(b) MnO4^

Q: Why do we always assume Oxygen is -2?

In the vast majority of compounds, Oxygen is highly electronegative and takes on a $-2$ oxidation state. The exceptions are peroxides (like $\text{H}_2\text{O}_2$) where it is $-1$, superoxides where it is $-1/2$, and compounds with Fluorine (like $\text{OF}_2$) where it can be positive. In standard oxoanions like chromate or permanganate, it is always $-2$.

Q: In option (a), there are two Chromium atoms. Shouldn't the total change be 6?

Yes, the total number of electrons transferred in the half-reaction is indeed $6$. However, the phrase 'change in the oxidation state' typically refers to the change experienced by a *single atom* of the element in question. The oxidation state of $\text{Cr}$ changes from $+6$ to $+3$, which is a change of $3$ per atom.

Q: What if the product is a neutral molecule like CO2? How do I find the charge?

For a neutral molecule, the sum of all the oxidation states of its constituent atoms must equal zero. So for $\text{CO}_2$, you set up the equation: $x + 2(-2) = 0$, which gives $x = +4$.

Q: Is the MnO4- to Mn2+ reaction always a change of 5?

It is a change of $5$ specifically in an **acidic medium**. In a neutral or faintly alkaline medium, $\text{MnO}_4^-$ reduces to $\text{MnO}_2$ (a change of $3$). In a strongly alkaline medium, it reduces to $\text{MnO}_4^{2-}$ (a change of $1$). This is a very important distinction to memorize!

Q: Why do we ignore the coefficients in front of the products, like the '2' in 2Cr3+?

The stoichiometric coefficients (like the $2$ in $2\text{Cr}^{3+}$) tell us how many ions are produced to balance the overall chemical equation. They do not affect the intrinsic oxidation state of the individual ion itself. A single $\text{Cr}^{3+}$ ion always has an oxidation state of $+3$, regardless of how many of them there are.

Question

Identify the process in which change in the oxidation state is five

Options:

(A) $	ext{Cr}_2	ext{O}_7^{2-} \longrightarrow 2	ext{Cr}^{3+}$

(B) $	ext{MnO}_4^- \longrightarrow 	ext{Mn}^{2+}$

(C) $	ext{CrO}_4^{2-} \longrightarrow 	ext{Cr}^{3+}$

(D) $	ext{C}_2	ext{O}_4^{2-} \longrightarrow 2	ext{CO}_2$

Sigma Percentile · Accepted Answer

### The Quest for the Magic Number Five

Welcome to a classic redox puzzle! In this problem, we are on a mission to find a specific chemical transformation. We need to identify the process where the central atom undergoes a change in its oxidation state of exactly **five**.

To crack this, we must become chemical detectives and calculate the oxidation states of the key elements in both the reactants and the products for every single option. Let's dive in and break them down one by one.

### Analyzing the Suspects

**Option (a): The Dichromate Reduction**
Let's look at the reaction: $	ext{Cr}_2	ext{O}_7^{2-} \longrightarrow 2	ext{Cr}^{3+}$
We need to find the oxidation state of Chromium ($	ext{Cr}$) in the dichromate ion. We know that Oxygen generally carries an oxidation state of $-2$. Let the oxidation state of $	ext{Cr}$ be $x$.
Setting up our algebraic equation based on the net charge of the ion:
$$2x + 7(-2) = -2$$
$$2x - 14 = -2$$
$$2x = 12 \implies x = +6$$
On the product side, we have the $	ext{Cr}^{3+}$ ion, which simply has an oxidation state of $+3$. 
The change in oxidation state per Chromium atom is $|6 - 3| = 3$. This is not five, so option (a) is out.

**Option (b): The Permanganate Reduction**
Next up is the famous permanganate reaction: $	ext{MnO}_4^- \longrightarrow 	ext{Mn}^{2+}$
Let's calculate the oxidation state of Manganese ($	ext{Mn}$) in the permanganate ion. Let it be $x$.
$$x + 4(-2) = -1$$
$$x - 8 = -1 \implies x = +7$$
On the product side, the Manganese ion is $	ext{Mn}^{2+}$, meaning its oxidation state is $+2$.
The change here is $|7 - 2| = 5$. 
Bingo! We found our match. The change is exactly five. But as good scientists, let's quickly verify the remaining options to be absolutely certain.

**Option (c): The Chromate Reduction**
Let's examine: $	ext{CrO}_4^{2-} \longrightarrow 	ext{Cr}^{3+}$
For the chromate ion, let the oxidation state of $	ext{Cr}$ be $x$.
$$x + 4(-2) = -2$$
$$x - 8 = -2 \implies x = +6$$
The product is again $	ext{Cr}^{3+}$ with an oxidation state of $+3$. 
The change is $|6 - 3| = 3$. Just like option (a), this is incorrect.

**Option (d): The Oxalate Oxidation**
Finally, let's check: $	ext{C}_2	ext{O}_4^{2-} \longrightarrow 2	ext{CO}_2$
For Carbon ($	ext{C}$) in the oxalate ion, let its oxidation state be $x$.
$$2x + 4(-2) = -2$$
$$2x - 8 = -2$$
$$2x = 6 \implies x = +3$$
In the product, Carbon Dioxide ($	ext{CO}_2$), the molecule is neutral. 
$$x + 2(-2) = 0 \implies x = +4$$
The change in oxidation state for Carbon is $|4 - 3| = 1$. Definitely not five.

### The Final Verdict

After a thorough investigation, it is crystal clear. The reduction of the permanganate ion ($	ext{MnO}_4^-$) to the manganese(II) ion ($	ext{Mn}^{2+}$) is the only process where the central atom undergoes an oxidation state change of exactly **five**.

This specific reaction is a cornerstone of redox titrations in acidic mediums, making it a high-yield concept for competitive exams. Therefore, the correct choice is **(b)**.

Animated Solution for Chemistry - Redox Reactions: Identify the process in which change in the oxidation state is five

Select Answer:

Visualized Solution

\text{Analyzing the Options}

\text{Option (a): } \text{Cr}_2\text{O}_7^{2-} \longrightarrow 2\text{Cr}^{3+}

\text{Option (b): } \text{MnO}_4^- \longrightarrow \text{Mn}^{2+}

\text{Option (c): } \text{CrO}_4^{2-} \longrightarrow \text{Cr}^{3+}

\text{Option (d): } \text{C}_2\text{O}_4^{2-} \longrightarrow 2\text{CO}_2

\text{Final Conclusion}

The Sigma Insight: Oxidation and Reduction

The Quest for the Magic Number Five

Analyzing the Suspects

The Final Verdict

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