Question

Formula of chromic acid is ${\text{H}}_2\text{Cr}{\text{O}}_4$. Formula of divalent metal chromate is:
A. $\text{MCr}{\text{O}}_4$
B. ${\text{M}}_2\text{Cr}{\text{O}}_4$
C. ${\text{M}}_2{\left(\text{Cr}{\text{O}}_4\right)}_3$
D. ${\text{M}}_3\text{Cr}{\text{O}}_4$

Answer 1

Hint: The formula of divalent metal chromate can be determined by checking the oxidation number of chromium metal in all compounds. The oxidation number of metal will be two and the oxidation number of chromium metal will remain the same in chromic acid and its chromate.

Complete step by step answer:
The chromate will form by substituting the hydrogens in chromic acid. The divalent metal will replace the two hydrogens, so the charge of metal will be balanced by the charge of two hydrogens. So, the oxidation number of chromium metal in chromate will remain the same as in chromic acid.
In neutral molecules, the sum of the oxidation number is equal to zero.
The oxidation number of chromium metal in ${\text{H}}_2\text{Cr}{\text{O}}_4$ is as follows:
The oxidation number of oxygen is - 2 and hydrogen is + 1.
$\left(+1\times 2\right)+\left(x\times 1\right)+\left(-2\times 4\right)=0$
$x=+6$
The oxidation number of chromium metal is + 6 in ${\text{H}}_2\text{Cr}{\text{O}}_4$.
Determine the oxidation number of chromium in all as follows:
The oxidation number of chromium metal in $\text{MCr}{\text{O}}_4$ is as follows:
The oxidation number of metal is + 2.
$\left(+2\times 1\right)+\left(x\times 1\right)+\left(-2\times 4\right)=0$
x = + 6
The oxidation number of chromium metal is + 6 in $\text{MCr}{\text{O}}_4$.
So, the oxidation number of chromium is the same in chromic acid and $\text{MCr}{\text{O}}_4$. So, option (A) is correct.
The oxidation number of chromium metal in ${\text{M}}_2\text{Cr}{\text{O}}_4$ is as follows:
$\left(+2\times 2\right)+\left(x\times 1\right)+\left(-2\times 4\right)=0$
x = + 4
The oxidation number of chromium metal is + 4 in ${\text{M}}_2\text{Cr}{\text{O}}_4$.
So, the oxidation number of chromium is not the same in chromic acid and ${\text{M}}_2\text{Cr}{\text{O}}_4$. So, option (B) is incorrect.
The oxidation number of chromium metal in ${\text{M}}_2{\left(\text{Cr}{\text{O}}_4\right)}_3$ is as follows:
$\left(+2\times 2\right)+\left(x\times 3\right)+\left(-2\times 12\right)=0$
3x = + 20
x = + 6.6
The oxidation number of chromium metal is + 6.6 in ${\text{M}}_2{\left(\text{Cr}{\text{O}}_4\right)}_3$.
The maximum oxidation that can be shown by chromium is + 6. The oxidation number cannot be fractional, because during oxidation or reduction, a complete electron is removed or accepted.
So, the oxidation number of chromium is not the same in chromic acid and ${\text{M}}_2{\left(\text{Cr}{\text{O}}_4\right)}_3$. So, option (C) is incorrect.
The oxidation number of chromium metal in ${\text{M}}_3\text{Cr}{\text{O}}_4$ is as follows:
$\left(+2\times 3\right)+\left(x\times 1\right)+\left(-2\times 4\right)=0$
x = + 2
The oxidation number of chromium metal is + 2 in ${\text{M}}_3\text{Cr}{\text{O}}_4$.
So, the oxidation number of chromium is not the same in chromic acid and ${\text{M}}_3\text{Cr}{\text{O}}_4$ so, option (D) is incorrect.

Therefore, option (A) $\text{MCr}{\text{O}}_4$ is correct.

Note: Oxidation number is the number of electrons gained or lost by the atoms. The valency shows the oxidation number. Divalent means the oxidation number of metal is + 2.