Hybridisation of the underline atom changes in

Options

A. `underline(A)IH_(3) ` changes of `AlH_(4)^(-)`
B. `H_(2)underline(O)` changes to `H_(3)O^(4)`
C. `underline(N)H_(3)` changes to `NH_(4)^(+)`
D. in all cases

Answer

A

Explanation

Hybridisation`=(1)/(2)[("No of electrons in valance shell of atom")+("No of monovalent atoms around is")-("charge on cation")+("charge on anion")]`
For `AlH_(3)`,
Hybridisation of Al atom `=(1)/(2)[3+3-0+0]=3=sp^(2)`
For `AlF_(4)^(-)`,
Hybridisation of Al atom`=(1)/(2)[3+4-0+1]=4=sp^(3)`
(b) For `H_(2)O`.
Hybridisation of O atom `=(1)/(2)[6+2-0+0]=4=sp^(3)`
For `H_(3)O^(+)`, hybridisation of O atom`=(1)/(2)[6+3-1+0]=4=sp^(3)`
(c) For `NH_(3)`
Hybridisation of N atom `=(1)/(2)[5+3-0+0]=4=sp^(3)`
For `NH_(4)^(+)`, hybridisation of N atom`=(1)/(2)[5+4-1+0]`
`=4=sp^(3)` thus hybridisation only in option (a).