Question

The correct sequence of the oxidation state of Fe, Ta, P and N elements in the given compounds are respectively :

$N{a}_2\left[Fe\right(CN{)}_{5}NO],K_{2}Ta{F}_7,M{g}_2{P}_2{O}_7,N{a}_2{S}_4{O}_6,N_{3}H$

• A. $+3,+5,+5.+2.5,-\frac{1}{3}$
• B. $+5,+3,+5,+3,+\frac{1}{3}$
• C. $+3,+3,+5,+5,-\frac{1}{3}$
• D. $+5,+5,+3,+2.5,+\frac{1}{3}$

Answer 1

The correct option is A $+3,+5,+5.+2.5,-\frac{1}{3}$

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

Oxidation number of $H=Na=+1,Mg=+2,CN=F=-1,$ and of $O=-2$.

A. $N{a}_2\left[Fe\right(CN{)}_{5}NO]$: $2\times 1+x+5\times (-1)+0=0$

$x+2-5=0$ "or" $x=+3$

B. $K_{2}Ta{F}_7$: $2\times 1+x+7\times (-1)=0$

$x+2-7=0$ "or" $x=+5$

C. $M{g}_2{P}_2{O}_7$: $2\times 2+2x+7\times (-2)=0$

$2x+4-14=0$ "or" $x=+5$

D. $N{a}_2{S}_4{O}_6$: $2\times 1+4x+6\times (-2)=0$

$4x+2-12=0$ "or" $x=+2.5$

E. $N_{3}H$: $3\times x+1=0$

"or" $x=-1/3$

Thus oxidation number of $Fe$ in $N{a}_2\left[Fe\right(CN{)}_{5}NO]$, $Ta$ in $K_{2}Ta{F}_7$, P in $M{g}_2{P}_2{O}_7$, S in $N{a}_2{S}_4{O}_6$ and $N$ in $N_{3}H$ is: +3,+5,+5,+2.5,-1/3, respectively.