Question

Write shape, hybridization and no of lone pair of electrons for following: ${\mathrm{SF}}_4$

Answer 1

Lone pair of ${\mathbf{SF}}_{\mathbf{4}}$ molecule

• The atomic number of Sulfur $\left(\mathrm{S}\right)$ is $16$.
• The electronic configuration of the Sulfur atom is as follows:

$\mathrm{S}=1s^{2}2s^{2}2p^{6}3s^{2}3p^4$

• The number of valence electrons of sulfur is $6$.
• Sulfur form four sigma bonds with four Florine $\left(\mathrm{F}\right)$ atoms to form a Sulfur tetrafluoride $\left({\mathrm{SF}}_4\right)$ molecule.
• ${\mathrm{SF}}_4$ molecule has one lone pair on Sulphur atom.

Hybridization of ${\mathbf{SF}}_{\mathbf{4}}$ molecule

• In ${\mathrm{SO}}_2$, Sulfur has four bond pairs and one lone pair, so it requires five hybrid orbitals.
• It will form an $s{p}^{3}d$ hybridized orbital.

Shape of ${\mathbf{SF}}_{\mathbf{4}}$ molecule

• According to VSEPR theory, the order of repulsion between electron pairs is as follows:
• lone pair-lone pair $>$ lone pair-bond pair $>$ bond pair-bond pair repulsions
• It has trigonal bipyramidal geometry, but due to bond pair-lone pair repulsion, ${\mathrm{SF}}_4$ will form a see-saw shape.
• Structure of ${\mathrm{SF}}_4$: