Question

If an element A shows two oxidation states $+2$ and $+3$. A forms oxides in such a way that they have a ratio 1 : 3 in a compound, the formula of the compound will be

• A. $A_8{O}_{11}$
• B. $A_4{O}_{11}$
• C. $A_9{O}_{11}$
• D. $A_5{O}_{11}$

Answer 1

The correct option is A $A_8{O}_{11}$

As we can see from the options, the number of oxygen atoms in the oxide of element A is 11.

Let us consider that the number of atoms of element A in $+2$ oxidation state to be $x$ and the number of atoms of element A in $+3$ oxidation state to be $3x$

For a neutral molecule, the number of $+ve$ and $-ve$ charges should be equal and their sum will be zero.
$\therefore (+2\times x)+(+3\times 3x)+(-2\times 11)=0$
On solving we get, $11x-22=0,x=2$

So, the number of atoms of element A in $+2$ oxidation state, $x=2$
And the number of atoms of element A in $+3$ oxidation state, $3x=3\times 2=6$

$\therefore$ The total number of atoms of element A in the compound $=2+6=8$

The Molecular formula of the compound is $A_8{O}_{11}$