Question

Consider the following four molecules. Which of these satisfy the octet rule and which do not? Drag the appropriate items to their respective bins.

Answer 1

Concept and reason

The concept used is to identify the molecules that obey octet rule and the molecules that do not obey.

Fundamentals

A Lewis structure is the structure that shows the distribution of electrons around the atoms in a molecule.

Octet rule says that each atom of a molecule tend to bond so that they have eight electrons in its valence shell like the noble gas configuration.

B.

Consider ${\rm{P}}{{\rm{F}}_5}$ .

The central P atom contributes 5 valence electrons. Each F atom contributes 7 valence electrons.

$\begin{array}{l}\\{\rm{Total valence electrons}} = 5 + 5\left( 7 \right)\\\\{\rm{ = 40}}\\\end{array}$

P can form five bonds with five F atoms. So, 10 electrons are used as bonding electrons.

The remaining 30 electrons are assigned as three lone pairs on each F atom.

The Lewis structure is as follows:

${\rm{P}}{{\rm{F}}_5}$ does not obey octet rule.

Consider ${\rm{BB}}{{\rm{r}}_3}$ .

The central B atom contributes 3 valence electrons. Each ${\rm{Br}}$ atom contributes 7 valence electrons.

$\begin{array}{l}\\{\rm{Total valence electrons}} = 3 + 3\left( 7 \right)\\\\{\rm{ = 24}}\\\end{array}$

B forms three bonds with three ${\rm{Br}}$ atoms. So, 6 electrons are used as bonding electrons.

The remaining 18 electrons are assigned as three lone pairs on each ${\rm{Br}}$ atom.

The Lewis structure is as follows:

${\rm{BB}}{{\rm{r}}_3}$ does not obey octet rule.

Consider ${\rm{C}}{{\rm{S}}_2}$ .

The central C atom contributes 4 valence electrons. Each S atom contributes 6 valence electrons.

$\begin{array}{l}\\{\rm{Total valence electrons}} = 4 + 2\left( 6 \right)\\\\{\rm{ = 16}}\\\end{array}$

C forms two bonds with two S atoms. So, 4 electrons are used as bonding electrons.

The remaining 10 electrons are assigned as two lone pairs on each S atom and two double bonds between C and S atoms.

The Lewis structure is as follows:

${\rm{C}}{{\rm{S}}_2}$ obeys octet rule.

Consider ${\rm{CO}}_3^{2 - }$ .

The central C atom contributes 4 valence electrons. Each O atom contributes 6 valence electrons.

$\begin{array}{l}\\{\rm{Total valence electrons}} = 4 + 3\left( 6 \right)\\\\{\rm{ = 22}}\\\end{array}$

Due to the presence of two negative charges, to electrons are added. So, total valence electrons is ${\rm{24}}$ .

C forms three bonds with three O atoms. So, 6 electrons are used as bonding electrons.

The remaining 18 electrons are assigned as two lone pairs on each O atom and one double bond between one of the C and O atoms. Since there are two negative charges, the remaining two O atoms have two electrons more.

The Lewis structure is as follows:

${\rm{CO}}_3^{2 - }$ obeys octet rule.

Ans: Part B

Therefore, the molecules that obey the octet rule and that do not obey the octet rule are as follows: