Question

The oxidation state of Cr in $\text{K}_2\text{Cr}_2\text{O}_7$ is:

• A. +4
• B. +5
• C. +6
• D. +7

Hint:

Chromium in its $+6$ oxidation state is in its highest possible oxidation state, which explains why Potassium Dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$) is a very powerful oxidizing agent in chemical reactions.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the oxidation state of Chromium (Cr) in Potassium Dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$).
This is a redox chemistry question requiring algebraic determination of oxidation states.

Step 2: Key Formula or Approach:
The sum of oxidation states of all atoms in a neutral chemical compound is equal to zero.
We use the established oxidation numbers of known elements:
- Potassium (K), an alkali metal (Group 1), always has an oxidation state of $+1$ in its compounds.
- Oxygen (O), a chalcogen, typically has an oxidation state of $-2$ in non-peroxide compounds.

Step 3: Detailed Explanation:

• Let the oxidation state of Chromium (Cr) in $\text{K}_2\text{Cr}_2\text{O}_7$ be represented by $x$.

• The compound contains:
- 2 Potassium (K) atoms
- 2 Chromium (Cr) atoms
- 7 Oxygen (O) atoms

• We set up the algebraic equation for the neutral molecule:
\[ 2 \times (\text{oxidation state of K}) + 2 \times (\text{oxidation state of Cr}) + 7 \times (\text{oxidation state of O}) = 0 \]

• Substitute the known values into the equation:
\[ 2(+1) + 2(x) + 7(-2) = 0 \]

• Simplify the equation step-by-step:
\[ 2 + 2x - 14 = 0 \]
\[ 2x - 12 = 0 \]
\[ 2x = 12 \]
\[ x = +6 \]

• Therefore, the oxidation state of each Chromium atom is $+6$.


Step 4: Final Answer:
The calculated oxidation state of Chromium in Potassium Dichromate is $+6$.