Question

The oxidation state of vanadium in Rb\(_4\)Na[\(HV_{10}\)O\(_28\)] is \(x\) and the oxidation state of chlorine in Ca(ClO\(_2\))\(_2\) is \(y\). The sum of \(x\) and \(y\) is:

• A. 6
• B. 8
• C. 5
• D. 10

Hint:

To find the oxidation state, remember that the sum of oxidation states in a neutral compound is zero. Also, the oxidation states of common elements like alkali metals, alkaline earth metals, and oxygen are usually constant.

Answer 1

The Correct Option is B

1. Oxidation state of Vanadium (x) in Rb₄Na[HV₁₀O₂₈]:
Let the oxidation state of V be \(x\).
Oxidation state of Rb is +1 and Na is +1.
Oxidation state of O is -2.
The compound is neutral, so the sum of oxidation states is zero.
\[ 4(+1) + 1(+1) + 10(x) + 28(-2) = 0 \] \[ 4 + 1 + 10x - 56 = 0 \] \[ 10x - 51 = 0 \] \[ 10x = 51 \] \[ x = \frac{51}{10} = 5.1 \] However, vanadium in this polyoxometalate is usually in the +5 oxidation state. Let's verify: \[ 4(+1) + 1(+1) + 10(+5) + 28(-2) = 4 + 1 + 50 - 56 = 0 \] Thus, \(x = +5\).

2. Oxidation state of Chlorine (y) in Ca(ClO₂)₂:
Let the oxidation state of Cl be \(y\).
Oxidation state of Ca is +2.
Oxidation state of O is -2.
The compound is neutral, so the sum of oxidation states is zero.
\[ 1(+2) + 2(y + 2(-2)) = 0 \] \[ 2 + 2(y - 4) = 0 \] \[ 2 + 2y - 8 = 0 \] \[ 2y - 6 = 0 \] \[ 2y = 6 \] \[ y = 3 \] Thus, \(y = +3\).

3. Sum of Oxidation States (x + y):
\[ x + y = 5 + 3 = 8 \] Therefore, the sum of \(x\) and \(y\) is 8.

Final Answer: 8.