Question

What is the correct mode of hybridization of the central atom in the following compounds? \[N{O_2}^ + \],\[S{F_4}\], and \[P{F_6}^ - \].
(a)\[s{p^2},{\rm{ }}s{p^3},{d^2}s{p^3}\]
(b)\[s{p^3},{\rm{ }}s{p^3}{d^2},{\rm{ }}s{p^3}{d^2}\]
(c)\[sp,{\rm{ }}s{p^3}d,{\rm{ }}s{p^3}{d^2}\]
(d) \[sp,{\rm{ }}s{p^2},{\rm{ }}s{p^3}\]

Answer 1

Hint: The mixing of atomic orbitals of elements, to produce the new orbital of different energy, shapes and size is called hybridized orbitals, and this phenomenon is called hybridization. The hybridization can be classified as\[sp,{\rm{ }}s{p^2},{\rm{ }}s{p^3},s{p^3}d,s{p^3}{d^2}\].

Complete step by step solution:By employing the hybridization method, we can easily determine the correct hybridization of any given molecule.
To determine the hybridization of a given molecule, first, we have to count the total number of sigma and lone pair of electrons present around the atom for which we will determine the hybridization.
Now one by one we will check the hybridization for the given molecules.

 (a) The \[N{O_2}^ + \] molecule contains 2 sigma bonds. The value of 2 sigma bonds will be equal to \[sp\] hybridization. Because all the five valence electrons of the nitrogen atom are engaged in bonding with two oxygen atoms, therefore, \[N{O_2}^ + \]does not have any lone pair of electrons i.e. it has linear geometry and shape.
(b) The \[S{F_6}\] molecule contains 4 sigma bonds and one lone pair of electrons. The value of 5 will be equal to \[s{p^3}d\] hybridization and it has trigonal bi-pyramidal geometry and a see-saw shape.
(c) The \[P{F_6}^ - \]molecule contains 6 sigma bonds. The value of 6 sigma bonds will be equal to \[s{p^3}{d^2}\]hybridization. Because all the five valence electrons of the phosphorus atom are engaged in bonding with fluorine atoms, therefore, \[P{F_6}^ - \]does not have any lone pair of electrons i.e. it has octahedral geometry and shape.

Therefore from the above explanation we can say option (c) will be the correct option:

Note: The\[N{O_2}^ + \]is named a nitronium ion.
The\[P{F_6}^ - \]is considered isoelectronic with sulphur hexafluoride (\[S{F_6}\]).