Hybridisation of the underline atom changes in:

Solution :
TIPS/Formulae :

$$\eqalign{ & \left( {\text{A}} \right)\,\,{\text{For}}\,\,Al{H_3}, \cr & {\text{Hybridisation of }}Al\,\,{\text{atom}} = \frac{1}{2}\left[ {3 + 3 - 0 + 0} \right] = 3 = s{p^2} \cr & {\text{For}}\,\,AlH_4^ - , \cr & {\text{Hybridisation of }}Al\,\,{\text{atom}} = \frac{1}{2}\left[ {3 + 4 - 0 + 1} \right] = 4 = s{p^3} \cr & \left( {\text{B}} \right)\,\,{\text{For}}\,{H_2}O, \cr & {\text{Hybridisation of}}\,\,O\,\,{\text{atom}}\, = \frac{1}{2}\left[ {6 + 2 - 0 + 0} \right] = 4 = s{p^3} \cr & {\text{For}}\,{H_3}{O^ + },\,\,{\text{Hybridisation of }}\,O\,\,\,{\text{atom}}\, = \frac{1}{2}\left[ {6 + 3 - 1 + 0} \right] = 4 = s{p^3} \cr & \left( {\text{C}} \right)\,\,{\text{For}}\,\,N{H_3} \cr & {\text{Hybridisation of }}N\,{\text{atom}}\, = \frac{1}{2}\left[ {5 + 3 - 0 + 0} \right] = 4 = s{p^3} \cr & {\text{For}}\,\,NH_4^ + ,\,\,{\text{Hybridisation of}}\,N\,{\text{atom}}\, = \frac{1}{2}\left[ {5 + 4 - 1 + 0} \right] = 4 = s{p^3} \cr} $$
$${\text{Thus hybridisation changes only in option (A)}}{\text{. }}$$