Question

Oxidation number of $P$ in $P{O}_4^{3-}$, of S in $S{O}_4^{2-}$ and that of $Cr$ in $C{r}_2{O}_7^{2-}$ are respectively.

• A. $-3,+6$ and $+6$
• B. $+5,+6$ and $+6$
• C. $+3,+6$ and $+5$
• D. $+5,+3$ and $+6$

Answer 1

The correct option is B $+5,+6$ and $+6$

$\bullet$ Oxidation no. of $P$ in $P{O}_4^{3-}$ be $x$.

Oxidation state of $O$ in general $=-2$

Applying formula, $\text{Sum of total oxidation state of all atoms}=\text{Overall charge on the compound}.x+4(-2)=-3,x=+5$

$\therefore$ Oxidation state of $P=+5.$

Oxidation no. of $S$ in $S{O}_4^{2-}$ be $x$.

Applying above formula,

$x+4\times (-2)=-2x=+6$

$\therefore$ Oxidation state of $S=+6.$

Oxidation no. of $Cr$ in $C{r}_7^{2-}$ be $x$.

Applying above formula,

$2x+7\times (-2)=-2x=+6$

$\therefore$ Oxidation state of $Cr=+6.$