Question

Formula of chromic acid is $H_{2}Cr{O}_4$. Formula of divalent metal chromate is:

• A. $MCr{O}_4$
• B. $M_{2}Cr{O}_4$
• C. $M_2(Cr{O}_4{)}_3$
• D. $M_{3}Cr{O}_4$

Answer 1

The correct option is A $MCr{O}_4$

Chromic acid is $H_{2}Cr{O}_4$, therefore valency of $Cr\left(x\right)$ is calculated as:
Its a neutral molecule therefore all valency should add upto zero
$2\times 1+x+4\times (-2)=0$
or, $x=+6$ for $Cr$ in chromic acid or chromate ion.
For divalent metal chromate, let the formula is: $M_a(Cr{O}_4{)}_b$
Since molecule is neutral and given that valency of $M$ is $+2$
$2\times a+[6+4\times (-2\left)\right]\times b=0$
or $a=b=1$
Hence molecular formula of the compound is $M_1(Cr{O}_4{)}_1$ or $M\left(Cr{O}_4\right)$
Option A is correct.