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The hybridisation of orbitals of N atoms in $NO_3^-,NO_2^+$ and $NH_4^+$ are respectively

$\begin{array}{1 1}(a)\;sp,sp^2,sp^3&(b)\;sp^2,sp,sp^3 \\(c)\;sp,sp^3,sp^2&(d)\;sp^2,sp^3,sp\end{array}$

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$NO_3^-$ hybridisation is $sp^2$(Trigonal planar)
$NO_2^+$ is $sp$(Linear)
$NH_4^+$ is $sp^3$(Tetrahydral)
Hence (b) is the correct answer.
answered Mar 6, 2014 by sreemathi.v
 

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