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: Assign known oxidation states.
In \( \text{K}_2\text{Cr}_2\text{O}_7 \), potassium is always +1 and oxygen is normally -2, except in peroxides.
Step 2: Set up the neutrality equation.
Let the oxidation state of Cr be x. The compound is neutral, and there are 2 K, 2 Cr, and 7 O atoms:
\[ 2(+1) + 2(x) + 7(-2) = 0 \]
Step 3: Solve for x.
\[ 2 + 2x - 14 = 0 \Rightarrow 2x = 12 \Rightarrow x = +6 \]
This confirms chromium is in its maximum common oxidation state in dichromate.
\[ \boxed{+6} \]