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The hybridizations of atomic orbitals of nitrogen in $NO_2^+, NO_3^-$ and $NH_4^+$ respectively are
$sp. sp^3$ and $sp^2$
$sp^2, sp^3$ and $sp$
$sp, sp^2$ and $sp^3$
$sp^2, sp$ and $sp^3$
Solution
Here's a nice short trick.
$1)$ Count and add all the electrons in valence shell
$2)$ Divide it by $8$
$3)$ On dividing by $8$ you will get a remainder and quotient, add quotient $+$ (remainder $/ 2$)
$4)$ Now get the hybridization corresponding to the number what you got
If $2$ its $sp$, if $3$ its $sp ^2$, if $4$ its $sp ^3$, if $5$ its $sp ^3 d$ and so on.
Ex: $NO ^{2+}$
$\rightarrow$ Total electrons in valence shell : $5+6 \times 2-1=16$.
$\rightarrow 16 / 8=2$. Quotient $=2$ and Remainder $=0$
$\rightarrow 2+0 / 2=2$ i.e $sp$
$N H ^{4+}$
$\rightarrow$ Total electrons in valence shell : $5+7 \times 4-1=32$.
$\rightarrow 32 / 8=4$. Quotient $=4$ and Remainder $=0$
$\rightarrow 4+0 / 2=4$ i.e $sp ^3$
$NO ^{3-}$
$\rightarrow$ Total electrons in valence shell : $5+6 \times 3+1=24$.
$\rightarrow 24 / 8=3$. Quotient $=3$ and Remainder $=0$
$\rightarrow 3+0 / 2=3$ i.e $sp ^2$