Question

Which one of the following has (have) octahedral geometry?
i) $SbC{l_6}^ -$
ii) $SnC{l_6}^{2 - }$
iii) $Xe{F_6}$
iv) $IO_6^{5 - }$
A) i), ii) and iv)
B) i), iii) and iv)
C) i), ii) and iii)
D) All of these

Answer 1

The geometry of the molecule can be described using its hybridisation. Hybridisation is the process of combining atomic orbitals to form new hybrid atomic orbitals. These hybrid orbitals should be appropriate for electron pairing and forming of new bonds according to the Valence Bond Theory.

Complete answer:
We are given 4 compounds, all of which are polyatomic molecules/ ions. The hybridisation of polyatomic compounds can be given by the formula:
$Hybridisation = \dfrac{{no.of{\text{ }}valence{\text{ }}electron\operatorname{s} {\text{ }}in{\text{ }}the{\text{ }}central{\text{ }}atom + no.ofHydrogen{\text{ }}Atoms + no.of{\text{ }}Halide{\text{ }}Atoms \pm Formal{\text{ }}Ch\arg e}}{2}$
If the answer obtained is:

Answer Obtained/Steric Number	Hybridisation	Geometry
2	sp	Linear
3	$s{p^2}$	Trigonal Planar
4	$s{p^3}$	Tetrahedral
5	$s{p^3}d$	Trigonal Bipyramidal
6	$s{p^3}{d^2}$	Octahedral

Let us consider the given compounds one by one.
i) $SbC{l_6}^ -$ : The central atom is Sb which belongs to group 15 of the periodic table. The no. of valence electrons are 5. The hybridisation as by the formula will be $= \dfrac{{5 + 6 + 1}}{2} = \dfrac{{12}}{2} = 6 \to s{p^3}{d^2}$ . Therefore, the geometry would be Octahedral with zero lone pairs. The structure of the molecule will be:

ii) $SnC{l_6}^{2 - }$ : The central atom is Sn which belongs to group 14 of the periodic table. The no. of valence electrons is 4. The hybridisation as by the formula will be $= \dfrac{{4 + 6 + 2}}{2} = \dfrac{{12}}{2} = 6 \to s{p^3}{d^2}$ . Therefore, the geometry would be Octahedral with zero lone pairs. The structure of the molecule will be:

iii) $Xe{F_6}$ : The central atom is Xe which belongs to group 18 of the periodic table. The no. of valence electrons is 8. The hybridisation as by the formula will be $= \dfrac{{8 + 6}}{2} = \dfrac{{14}}{2} = 7 \to s{p^3}{d^3}$ . Therefore, the geometry would be a distorted octahedral with one lone pair present.
iv) $IO_6^{5 - }$ : The central atom is Iodine which belongs to group 17 of the periodic table. The no. of valence electrons is 7. The hybridisation as by the formula will be $= \dfrac{{7 + 5}}{2} = \dfrac{{12}}{2} = 6 \to s{p^3}{d^2}$ . Therefore, the geometry would be octahedral with zero lone pairs present.

i), ii) and iv) have octahedral geometry.
The correct option is (A).

Note:
The bond order and other properties of Diatomic species can be found out with the Help of Molecular Orbital Diagrams (MODs). MODs for polyatomic species are found to be complicated to understand. If there are lone pairs present on the central atom, the Hybridisation and geometry of the molecule will remain the same, their shape will change accordingly. For example, the shape of a molecule having Tetrahedral geometry with one L.P will be Pyramidal shaped. Similarly, molecules with Trigonal Planar geometry and one L.P will have angular shape.