Question

Give reason all bonds in ${\mathrm{SF}}_4$ are not equivalent.

Answer 1

Structure of ${\mathrm{SF}}_4$:

• ${\mathrm{SF}}_4$ is known as sulfur tetrafluoride and it has ${\mathrm{sp}}^3\mathrm{d}$ hybridrization.
• In ${\mathrm{SF}}_4$, sulphur $\left(\mathrm{S}\right)$ is the central atom.
• The atomic number of sulphur is $16$, so the electronic configuration is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{px}}^{2}3{\mathrm{py}}^{1}3{\mathrm{pz}}^1$ .
• Fluorine has an atomic number $9$, so the electronic configuration is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^5$.
• In the ground state of sulphur, it has two lone pairs and two bond pairs of electrons.
• In an excited state, sulphur has one lone pair and four bond pairs of electrons to combine with four molecules of fluorine.
• Here $s{p}^{\mathit{3}}\mathrm{d}$ hybrid orbital of sulphur overlaps with four $2\mathrm{p}$ orbitals of fluorine from end to end to form Four sigma$\left(\mathrm{\sigma }\right)$ bonds and one lone pair.

Non-equivalent bonds in ${\mathrm{SF}}_4$:

• Its electron geometry is trigonal bipyramidal in which one of the positions is occupied by lone pair of electrons.
• The five electron pairs around Sulfur adopt trigonal bipyramidal geometry in which one position is occupied by a lone pair.
• The structure of ${\mathrm{SF}}_4$ is as follows:

• This lone pair finds a position that minimizes the number of 90^o repulsions it has with bonding electron pairs.
• It occupies an equatorial position with two 90^o repulsions and the bonded electrons occupy the axial and equatorial positions.
• The axial $\mathrm{S}-\mathrm{F}$ bonds are bent slightly away from the lone pair.
• There are two $\mathrm{S}-\mathrm{F}$ bonds which are positioned one at the equatorial position and the other two at the axial position.
• Hence, all the bonds in ${\mathrm{SF}}_4$ are not equivalent.

Therefore, the bonds in ${\mathrm{SF}}_4$ are not equivalent.