Dichromate ion is treated with base, the oxidation number of Cr in the product formed is ………… .

Q: Why doesn't the oxidation state of Chromium change?

Because this is an acid-base equilibrium, not a redox reaction. The addition of $OH^-$ simply removes $H^+$ from the system, shifting the equilibrium without transferring electrons between Chromium and other species.

Q: What happens if we add acid instead of base?

Adding acid ($H^+$) lowers the $pH$ and shifts the equilibrium backward, converting the yellow chromate ions back into orange dichromate ions.

Q: How do we know oxygen is $-2$ here?

In most of its compounds (except peroxides, superoxides, and fluorides), oxygen exhibits an oxidation state of $-2$.

Q: Is $Cr_2O_7^{2-}$ a good oxidizing agent in basic medium?

Potassium dichromate is a strong oxidizing agent primarily in acidic medium. In basic medium, it converts to chromate, which is a much weaker oxidizing agent.

Q: Can we calculate the oxidation state directly from the dichromate ion?

Yes! In $Cr_2O_7^{2-}$, let $Cr$ be $x$. $2x + 7(-2) = -2 \implies 2x - 14 = -2 \implies 2x = 12 \implies x = +6$. The oxidation state is $+6$ in both ions!

Question

Dichromate ion is treated with base, the oxidation number of Cr in the product formed is ………… .

Sigma Percentile · Accepted Answer

The behavior of transition metal ions in aqueous solutions is one of the most visually stunning and conceptually rich areas of chemistry. In this problem, we explore the classic relationship between the dichromate and chromate ions—a transformation that is a favorite among JEE examiners.

### The Setup: A Shift in pH

Imagine a beaker filled with a vibrant orange solution of potassium dichromate ($K_2Cr_2O_7$). The dichromate ion, $Cr_2O_7^{2-}$, is stable in acidic conditions. However, the moment we introduce a base (like $NaOH$), increasing the $pH$ above 7, a remarkable color change occurs. The solution turns a bright, sunny yellow.

What is happening at the molecular level? The dichromate ions are reacting with the hydroxide ions ($OH^-$) to form chromate ions ($CrO_4^{2-}$). This is not a one-way street; it is a dynamic equilibrium governed by the $pH$ of the medium:

$$Cr_2O_7^{2-} (aq) + 2OH^- (aq) \rightleftharpoons 2CrO_4^{2-} (aq) + H_2O (l)$$

By Le Chatelier's Principle, adding a base consumes $H^+$ (or adds $OH^-$), driving the equilibrium to the right, favoring the formation of the yellow chromate ion.

### The Master Equation: Oxidation State Calculation

Now that we have identified our product as the chromate ion ($CrO_4^{2-}$), we need to determine the oxidation state of the central chromium atom.

Let the oxidation state of chromium be $x$. We know that oxygen, being highly electronegative, typically adopts an oxidation state of $-2$ in its oxides. The sum of the oxidation states of all atoms in a polyatomic ion must equal the net charge of the ion.

Setting up our algebraic equation for $CrO_4^{2-}$:

$$x + 4(-2) = -2$$

### Final Calculation

Let's solve this simple linear equation. Expanding the terms gives us:

$$x - 8 = -2$$

Moving the $-8$ to the other side:

$$x = +6$$

**The oxidation number of Chromium in the product is 6.**

### The Final Reveal: No Redox Here!

Here is the beautiful catch that often traps students: let's quickly calculate the oxidation state of chromium in the original dichromate ion ($Cr_2O_7^{2-}$).

$$2x + 7(-2) = -2 \implies 2x - 14 = -2 \implies 2x = 12 \implies x = +6$$

The oxidation state of chromium is $+6$ in *both* the reactant and the product! Despite the dramatic color change and the structural rearrangement, **no electrons were transferred**. This process is purely an acid-base equilibrium, not a redox reaction. Understanding this distinction is crucial for mastering inorganic chemistry.

Animated Solution for Chemistry - Redox Reactions: Dichromate ion is treated with base, the oxidation number of Cr in the product formed is ………… .

Enter Numerical Value:

Visualized Solution

Identifying the Reactant

The $p H$ Dependent Equilibrium

Product in Basic Medium

Setting up the Equation for $C r O_{4}^{2 -}$

Solving for $x$

Final Oxidation State

The Sigma Insight: Oxidation and Reduction

The Setup

The Master Equation

Final Calculation

The Final Reveal

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