The hybridisation of the central atom can be calculated as

H = $${1 \over 2}\left[ \matrix{ \left( \matrix{ No.\,of\,electrons \hfill \cr in\,valence\,shell \hfill \cr of\,atom \hfill \cr} \right) + \left( \matrix{ No.\,of\,monovalent \hfill \cr atoms\,around \hfill \cr central\,atom \hfill \cr} \right) \hfill \cr - \left( \matrix{ Ch{\mathop{\rm arge}\nolimits} \,on \hfill \cr cation \hfill \cr} \right) + \left( \matrix{ Ch{\mathop{\rm arge}\nolimits} \,on \hfill \cr anion \hfill \cr} \right) \hfill \cr} \right]$$

$$ \therefore $$ For BF₃, H = $${1 \over 2}\left[ {(3) + (3) - (0) + (0)} \right] \Rightarrow 3 \Rightarrow $$ sp² hybridisation

$$ \therefore $$ For NO₂^-, H = $${1 \over 2}\left[ {(5) + (0) - (0) + (1)} \right] \Rightarrow 3 \Rightarrow $$ sp² hybridisation

According to MOT, the molecular orbital electronic configuration of

$${N_2}:{\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( {\pi 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {\pi 2{p_z}} \right)^2}$$

$$ \therefore $$ B.O = $${{10 - 4} \over 2} = 3$$

$${N_2}^ - :{\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( {\pi 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {\pi 2{p_z}} \right)^2}{\left( {{\pi ^*}2{p_x}} \right)^1}$$

$$ \therefore $$ B.O = $${{10 - 5} \over 2} = 2.5$$

$${N_2}^{2 - }:{\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( {\pi 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {\pi 2{p_z}} \right)^2}{\left( {{\pi ^*}2{p_x}} \right)^1}{\left( {{\pi ^*}2{p_y}} \right)^1}$$

$$ \therefore $$ B.O = $${{10 - 6} \over 2} = 2$$

Hence the order : $$N_2^{2-}$$ < $$N_2^-$$ < $$N_2$$

BF₃ $$ \to $$ sp²

NO₂^- $$ \to $$ sp²

NH₂^- $$ \to $$ sp³

H₂O $$ \to $$ sp³

Methanol can undergo intermolecular association through H-bonding as the - OH group in alcoholos is highly polarised.

$$ - - - \mathop O\limits^{\mathop |\limits^{C{H_3}} } \hbox{---}H - - - \mathop O\limits^{\mathop |\limits^{C{H_3}} } \hbox{---}H - - - \mathop O\limits^{\mathop |\limits^{C{H_3}} } \hbox{---}H - - - $$

As a result the order to convert liquid CH₃OH to gaseous state, the strong hydrogen bonds must be broken.