# Question 79d88

Jun 8, 2017

$\text{18 valence electrons}$

#### Explanation:

The total number of valence electrons in a molecule of acetaldehyde, $\text{CH"_3"CHO}$, 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, $2 \times \text{C}$
• four atoms of hydrogen, $4 \times \text{H}$
• one atom of oxygen, $1 \times \text{O}$

Now, you should know that you have

• $\text{For C: 4 valence electrons}$
• $\text{For H: 1 valence electron}$
• $\text{For O: 6 valence electrons}$

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"))#

${\text{no. of e"^(-) = "18 e}}^{-}$

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

$5 \times {\text{2 e"^(-) + 1 xx "4 e"^(-) = "14 e}}^{-}$

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