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Standard XII

Chemistry

Oxidation States

A metal oxide...

Question

A metal oxide $A B$ has formula $A_{0.90} O$ in which $A$ is present in $A^{+ 2}$ and $A^{+ 3}$ form. 77.77 $%$ of $A^{+ 2}$ present in

True

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False

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Solution

The correct option is A True
Let the number of $A^{2 +}$ ion $= x$

Then the number of cations $=$ ( 0.90 - x)
Total no. of cations $= 2 x + 3 (0.90 - x)$
$= 2 x + 2.7 - 3 x$
$= - x + 2.88$

no. of anions $O_{2} = - 2 \times 1 = - 2$

No. of cations $=$ No. of anions
$- x + 2.7 = 2$
$- x = - 2.7 + 2$
$x = 0.7$

% of $A^{2 +}$ ion $= 100 \times \frac{0.7}{0.90} = 77.77$ %

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A metal oxide AB has formula A0.90O in which A is present in A+2 and A+3 form. 77.77 % of A+2 present in

The correct option is A TrueLet the number of A2+ ion =x Then the number of cations = ( 0.90 - x) Total no. of cations =2x+3(0.90−x)=2x+2.7−3x =−x+2.88no. of anions O2=−2×1=−2No. of cations = No. of anions −x+2.7=2−x=−2.7+2 x=0.7 % of A2+ ion =100×0.70.90=77.77%

A metal oxide $A B$ has formula $A_{0.90} O$ in which $A$ is present in $A^{+ 2}$ and $A^{+ 3}$ form. 77.77 $%$ of $A^{+ 2}$ present in