Answer
a) In the compound $NO$, oxidation state of $N = +2$
b) In the compound $NO_{2}$, oxidation state of $N = +4$
c) In the compound $N_{2}O$, oxidation state of $ N = +1$
Work Step by Step
Algebraic sum of oxidation state for a compound is Zero. Now, let's find oxidation state of N in each atom.
a) In NO, oxidation state of O = -2. Let's say oxidation state of N = $x$
Hence, $x + (-2) = 0 $ By solving this we get $x = +2$
So, oxidation state of N = +2
b) In $NO_{2}$, oxidation state of O = -2. Let's say oxidation state of N = $x$
Hence, $x + (-2)\times2 = 0 $ By solving this we get $x = +4$
So, oxidation state of N = +4
c) In $N_{2}O$, oxidation state of O = -2. Let's say oxidation state of N = $x$
Hence, $x\times2 + (-2) = 0 $ By solving this we get $x = +1$
So, oxidation state of N = +1
