Solution

For neutral molecules,
Number of electron pairs $$=$$ Number of atoms bonded to it $$+\dfrac12$$ [Gp. no of central atom - Valency of central atom]
$$\therefore$$ For $$CH_4$$, no. of e$$^-$$ pairs $$= 4 +$$ $$\displaystyle{\dfrac{1}{2}} [4 - 4]=4$$ (sp$$^3$$ hybridisation)
For $$SF_4$$, no. of e$$^-$$ pairs $$= 4 + \displaystyle{\frac{1}{2}}[6 - 4] =$$ $$5$$ ($$sp^3d$$ hybridisation)
For ions,
No. of electron pairs = No. of atoms bonded to it $$+ \dfrac12$$ [Gp. no. of central atom - Valency of central atom $$\pm$$ No. of electrons]
$$\therefore$$ For $$BF_4^-$$, no. e$$^-$$ pairs $$= 4 +$$ $$\displaystyle{\dfrac{1}{2}}[3 - 4 + 1] = 4$$ ($$sp^3$$ hybridisation)
For $$NH^+_4$$, no. of e$$^-$$ pairs $$= 4 +$$ $$\displaystyle{\frac{1}{2}}[5 - 4 - 1] = 4$$ ($$sp^3$$ hybridisation)
Thus, in $$SF_4$$, the central atom does not have $$sp^3$$ hybridization.

Therefore, the correct option is B.