Question

The oxidation state of chromium in ($K_2Cr_2O_7$) is:

• A. +4
• B. +5
• C. +6
• D. +7

Hint:

Chromium in the $+6$ state is a very strong oxidizing agent. This is why $K_2Cr_2O_7$ is commonly used in laboratories for titration and organic oxidation reactions.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The sum of the oxidation states of all atoms in a neutral molecule is equal to zero.
Step 2: Key Formula or Approach:
Assign standard oxidation states: \(K = +1\) (Group 1 metal) and \(O = -2\) (standard for oxides).
Let the oxidation state of Chromium be \(x\).
Step 3: Detailed Explanation:
For \(K_2Cr_2O_7\):
\[ 2(\text{O.S. of } K) + 2(\text{O.S. of } Cr) + 7(\text{O.S. of } O) = 0 \]
\[ 2(+1) + 2(x) + 7(-2) = 0 \]
\[ 2 + 2x - 14 = 0 \]
\[ 2x - 12 = 0 \]
\[ 2x = 12 \]
\[ x = +6 \]
Step 4: Final Answer:
The oxidation state of chromium in \(K_2Cr_2O_7\) is +6.