Question

Oxidation state of nitrogen is incorrectly given for:

• A. $\left[Co\right(N{H}_3{)}_{5}Cl]C{l}_2=-3$
• B. $N{H}_{2}OH=-1$
• C. $(N_2{H}_5{)}_{2}S{O}_4=+2$
• D. $M{g}_3{N}_2=-3$

Answer 1

Answer: (C)

$(N_2{H}_5{)}_{2}S{O}_4=+2$

For the given compounds, we have to find the oxidation state of nitrogen

for option A

Let x be the

oxidation state of N in $\left[Co\right(N{H}_3{)}_{5}Cl]C{l}_2$

the sum of oxidation states of all elements in it should be equal to 0.

Therefore, consider the NH3,

x+3=0 or, x=−3

the oxidation state of N=−3.

for option B

$N{H}_2^{+}$ $O{H}^{-}$

let oxidation sate of N be x so x + 2 = 1

$\Rightarrow x=-1$

for option C

The oxidation number for sulphate ion, $SO4$ is equal to - 2.

So, the oxidation number of two ion $N2H5$ will be 2

Or, the oxidation number of the ion $N2H5$ will be 2/2 = 1.

The oxidation number of hydrogen is equal to +1.

Let us consider the oxidation number of nitrogen to be x.

2x + (5x1) = 1

⇒2x= -4

⇒x= -4/2= -2

Hence oxidation number of nitrogen in the given compound is equal to-2

for option D

3 x$M{g}^{2+}$ +2x =0

$\Rightarrow$ x $=\frac{-6}{2}$

x $=-3$

Hence the correct OPT: C