Question #18423

1 Answer
Jul 20, 2015

Answer:

Orbital hybridization actually tells you s-character.

Explanation:

Before deciding how much s-character a hybrid orbital has, you must first determine the type of hybrid orbital you're dealing with.

To do that, draw the Lewis structures of the molecules.

Let's start with xenon tetrafluoride, #XeF_4#. The molecule has a total of 36 valence electrons - 8 from xenon and 7 from each of the four fluorine atoms.

The xenon atom will be the central atom of the molecule and it will form a single bond with each of the four fluorine atoms, which will have 3 lone pairs of electrons attached.

The remaining 4 valence electrons will be placed as lone pairs on the xenon atom.

http://imglop.com/xef4-lewis-structure.asp

Now take a look at how many regions of electron density surround the central atom. This number, which is called steric number, will tell you what the hybridization of the central atom is.

The xenon atom is bonded to 4 fluorine atoms and has 2 lone pairs present, which means that it is surrounded by a total of 6 regions of electron density.

This means that it must use 6 hybrid orbitals

  • one s-orbital
  • three p-orbitals
  • two d-orbitals

The central atom is #sp^3d^2# hybridized. To get the s-character of these orbitals, simply divide the number of s-orbitals that were used to form the hybrids by the total number of orbitals used to form the hybrids.

In this case, you have one s-orbital and six total orbitals, which means that you get

#"1 s-orbital"/"6 orbitals in total" * 100 = color(green)("16.7% s-character")#

Now take a look at the carbonate ion, #CO_3^(2-)#. Its Lewis structure looks like this

http://chem-net.blogspot.ro/2012/01/simple-procedure-for-writing-lewis.html

You can draw three Lewis structures for the carbonate ion - these are called resonance structures.

Once again, the important thing to look for is the number of regions of electron density that surround the central atom.

Since the carbon atom is bonded to 3 oxygen atoms and has no lone pairs present, its steric number will be equal to 3.

This means that it uses 3 hybrid orbitals

  • one s-orbital
  • two p-orbitals

The central atom is #sp^2# hybridized. The s-character will now be

#"1 s-orbital"/"3 orbitals in total" * 100 = color(green)("33.3% s-character")#