Question #79d88

The total number of valence electrons in a molecule of acetaldehyde, #"CH"_3"CHO"#, can be calculated by adding the number of valence electrons of each atom that makes up the molecule.

A molecule of acetaldehyde contains

• two atoms of carbon, #2 xx "C"#
• four atoms of hydrogen, #4 xx "H"#
• one atom of oxygen, #1 xx "O"#

Now, you should know that you have

• #"For C: 4 valence electrons"#
• #"For H: 1 valence electron"#
• #"For O: 6 valence electrons"#

This means that the total number of valence electrons present in a molecule of acetaldehyde is equal to

#"no. of e"^(-) = overbrace(2 xx "4 e"^(-))^(color(blue)("from 2 atoms of C")) + overbrace(4 xx "1 e"^(-))^(color(darkorange)("from 4 atoms of H")) + overbrace(1 xx "6 e"^(-))^(color(purple)("from 1 atom of O"))#

#"no. of e"^(-) = "18 e"^(-)#

You can double-check the calculations by looking at the Lewis structure of acetaldehyde, or ethanal.

You know that each single bond is made up of #2# valence electrons and that a double bond is made up of #4# valence electrons.

As you can see, you have #5# single bonds and #1# double bond, which corresponds to a total of

#5 xx "2 e"^(-) + 1 xx "4 e"^(-) = "14 e"^(-)#

The two lone pairs of electrons present on the oxygen atom bring the total number of valence electrons to #18#.