Question

Lowest oxidation state of phosphorous is in:

• A. $H_3{PO}_2$
• B. $H_3{PO}_4$
• C. $H_4{P_{2}O}_7$
• D. $H_3{PO}_3$

Answer 1

The correct option is D $H_3{PO}_3$

oxidation state of oxygen ;- $-2$

oxidation state of hydrogen ;-$1$

$H_{3}P{O}_2$ :-

$1\ast 3+x+(-2)\ast 2=0$

$x=1$

$H_{3}P{O}_4$:-

$1\ast 3+x+(-2)\ast 4=0$

$x=5$

$H_4{P}_2{O}_7$:-

$1\ast 4+2x+(-2)\ast 7=0$

$x=5$

$H_{3}P{O}_3$:-

$1\ast 3+x+(-2)\ast 3=0$

$x=3$

As we can see from the result:-

$H_{3}P{O}_3$ has the lowest oxidation state.