# Question #16995

May 5, 2015

All but $A s C {l}_{5}$.

Orbital hybridization is determined by the steric number, which represents the number of regions of electron density that surround an atom.

Regions of electron density can be bonds - simple, double, or triple bonds count as 1 region - or lone pairs of electrons.

Since the steric number determines orbital hybridization, you can work backwards and use orbital hybridization to determine the steric number.

In your case, $s {p}^{3} {d}^{2}$ hybridization implies the existance of 6 hybrid orbitals, formed from one s, three p, and two d-orbitals. In turn, 6 hybrid orbitals correspond to 6 regions of electron density lcoated around the central atom.

That being said, you could eyeball the given molecules and predict which could match this description, even without having to draw their respective Lewis structure.

• $S e {F}_{6}$, selenium hexafluoride

The selenium atom is bonded to 6 fluorine atoms, and since selenium has 6 valence electrons (it's part of the oxygen family), no lone pairs could possibly be present on it.

As a result, you get 6 bonds and 0 lone pairs $\to$ $s {p}^{3} {d}^{2}$ hybridization.

• $X e C {l}_{4}$, xenon tetrachloride

Xenon is a noble gas, so it has 8 valence electrons. Since it only forms 4 single bonds with the chlorine atoms, which account for 4 of the 8 valence electrons, it will also have 2 lone pairs present.

As a result, you get 4 bonds and 2 lone pairs $\to$ $s {p}^{3} {d}^{2}$ hybridization.

• $I {F}_{5}$, iodine pentrafluoride

Iodine is a halogen, which means it has 7 valence electrons. Since it forms 5 single bonds with the fluorine atoms, it will have 2 valence electrons present as a lone pair.

As a result, you get 5 bonds and 1 lone pair $\to$ $s {p}^{3} {d}^{2}$ hybridization.

• $A s C {l}_{5}$, arsenic pentachloride

Arsenic is part of the nitrogen family, which means it has 5 valence electrons. Since it forms 5 single bonds with the chlorine atoms, you won't have any lone pairs present.

As a result, you get 5 bonds and 0 lone pairs $\to$ $s {p}^{3} d$ hybridization.

SIDE NOTE As practice, you could actually draw the Lewis structures for all the molecules and double-check the answer.