Question

Determine the oxidation state of the underlined element in $K_4\underset{–}{P_2}{O}_7$.

• A. $+5$
• B. $+3$
• C. $+4$
• D. $+6$

Answer 1

Answer: (A)

$+5$

Let $x$ be the oxidation number of $P$ in $K_4{P}_2{O}_7$.
Since the overall charge on the complex is $0$, the sum of oxidation states of all elements in it should be equal to $0$.
Therefore, $4(+1)+2\left(x\right)+7(-2)=0$.
$4+2x-14=0$
$2x=10$
$x=+5$
Hence, the oxidation number of P in $K_4{P}_2{O}_7$ is $+5$.