Question

Which of the following is a decreasing order of oxidation states of the central atoms?

• A. $PC{l}_5,HI{O}_4,C{l}_2{O}_7^{2-},C{l}_{2}O$
• B. $C{r}_2{O}_7^{2-},C{l}_{2}O,HI{O}_4,PC{l}_5$
• C. $HI{O}_4,C{r}_2{O}_7^{2-},PC{I}_5,C{l}_{2}O$
• D. $C{r}_2{O}_7^{2-},HI{O}_4,C{l}_{2}O,PC{l}_5$

Answer 1

The correct option is C $HI{O}_4,C{r}_2{O}_7^{2-},PC{I}_5,C{l}_{2}O$

Let the oxidation number of the central atom be $x$.

Oxidation state of side atoms be: $H=+1$, $Cl=I=-1$ and of $O=-2$.

$PC{l}_5$: $x+5\times (-1)=0$

$x=+5$

$HI{O}_4$: $1+x+4\times (-2)=0$

$x=+7$

$C{l}_2{O}_7^{-2}$: $2x+7\times (-2)=-2$

$x=+6$

$C{l}_{2}O$: $2x+(-2)=0$

$x=+1$

$C{r}_2{O}_7^{-2}$: $2x+7\times (-2)=-2$

$x=+6$

Thus the decreasing order of the oxidation state of central atom is as:

$HI{O}_4>C{r}_2{O}_7^{-2}=C{l}_2{O}_7^{-2}>PC{l}_5>C{l}_{2}O$