Question

Ratio of the number of lone pairs to the average oxidation state of $S$ in $S_4{O}_6^{2-}$ is:

• A. $7.2$
• B. $5.6$
• C. $4.4$
• D. $2.8$

Answer 1

The correct option is A $7.2$

Answer (a)
Here total number of lone pairs,
$=2\times 6+3\times 2=18$
let average oxidation state of $S$ be $y$.
Then, $4y+6\times (-2)=-2$
$\Rightarrow y=\frac{10}{4}=2.5$
So the ratio would be $=\frac{18}{2.5}=7.2$