Question

If an element X shows 2 oxidation states +2 and +3 and forms an oxide in such a way that the rate of the element +2 and +3 state is 1:3 in the compound. The formula of this oxide is

• A. $X_4{O}_{11}$
• B. $X_5{O}_{11}$
• C. $X_8{O}_{11}$
• D. $X_9{O}_{11}$

Answer 1

The correct option is C $X_8{O}_{11}$

Let the compound be $X_{n}Op$

since $X_{n}Op$ is neutral

$\frac{n}{4}x+2+\frac{3n}{4}x+3-2p=0$
$n[\frac{1}{2}+\frac{9}{4}]-2p=0$
$n\left[\frac{2+9}{4}\right]=2p$
$\frac{n}{p}=\frac{8}{11}$

$X_8{O}_{11}$