Question
Molecular shapes of $$S{F_4},C{F_4}$$ and $$Xe{F_4}$$ are
A.
the same, with 2, 0 and 1 lone pairs of electrons respectively
B.
the same, with 1, 1 and 1 lone pairs of electrons respectively
C.
different, with 0, 1 and 2 lone pairs of electrons respectively
D.
different, with 1, 0 and 2 lone pairs of electrons respectively
Answer :
different, with 1, 0 and 2 lone pairs of electrons respectively
Solution :
Thestructure of species can be predicted on the basis of hybridisation which in turn can be known by knowing the number of hybrid orbitals $$(H)$$ in that species
$$=\frac{1}{2}\left( {6 + 4 + 0 - 0} \right) = 5$$
For $$S{F_4}$$ : $$S$$ is $$s{p^3}d$$ hybridised in $$S{F_4}.$$ Thus $$S{F_4}$$ has 5 hybrid orbitals of which only four are used by $$F,$$ leaving one lone pair of electrons on sulphur.
For $$C{F_4}:H = \frac{1}{2}\left[ {4 + 4 + 0 - 0} \right]$$ $$ = 4$$ ∴ $$s{p^3}$$ hybridisaion
Since all the four orbitals of carbon are involved in bond formation, no lone pair is present on $$C$$ having four valence electrons
For $$Xe{F_4}$$ : $$H = \frac{1}{2}\left( {8 + 4 + 0 - 0} \right) = 6,$$ ∴ $$s{p^3}{d^2}$$ hybridization of the six hybrid orbitals, four form bond with $$F,$$ leaving behind two lone pairs of electrons on $$Xe$$ .
Thestructure of species can be predicted on the basis of hybridisation which in turn can be known by knowing the number of hybrid orbitals $$(H)$$ in that species
$$=\frac{1}{2}\left( {6 + 4 + 0 - 0} \right) = 5$$
For $$S{F_4}$$ : $$S$$ is $$s{p^3}d$$ hybridised in $$S{F_4}.$$ Thus $$S{F_4}$$ has 5 hybrid orbitals of which only four are used by $$F,$$ leaving one lone pair of electrons on sulphur.
For $$C{F_4}:H = \frac{1}{2}\left[ {4 + 4 + 0 - 0} \right]$$ $$ = 4$$ ∴ $$s{p^3}$$ hybridisaion
Since all the four orbitals of carbon are involved in bond formation, no lone pair is present on $$C$$ having four valence electrons
For $$Xe{F_4}$$ : $$H = \frac{1}{2}\left( {8 + 4 + 0 - 0} \right) = 6,$$ ∴ $$s{p^3}{d^2}$$ hybridization of the six hybrid orbitals, four form bond with $$F,$$ leaving behind two lone pairs of electrons on $$Xe$$ .