Question

Add lone pairs to these Lewis structures of polyhalide ions.

Answer 1

Concepts and reason

The problem is based on the concept of Lewis structure.

Lewis structure is the electron dot structure and represents the electron distribution in the atom. Lewis structure generally follows the octet rule.

Fundamentals

The electron distribution in the atom is represented by Lewis structure. The electrons in the outer shell are known as valence electrons. To complete the octet of the atom, valence shell should have 8 electrons. For the Lewis structure, the total number of valence electrons and number of electrons needed to complete the octet of the atom are counted. In the molecule, the total number of bonds in the molecule is counted and the skeletal structure of the compound is drawn. The total electrons are counted in the bond and the remaining electrons are added to the atom such that the octet of each atom is complete. The electrons which are not bonded to the bond are the non-bonding electrons and are termed as the lone pairs.

Part 1

Count the total number of valence electrons $\left( {{n_{{{\left[ {{\rm{Cl}}{{\rm{F}}_2}} \right]}^ - }}}} \right)$ in ${\left[ {{\rm{Cl}}{{\rm{F}}_2}} \right]^ - }$ as follows:

$\begin{array}{c}\\{n_{{{\left[ {{\rm{Cl}}{{\rm{F}}_2}} \right]}^ - }}} = \left( {1 \times 7} \right) + \left( {2 \times 7} \right) + 1\\\\ = 22\\\end{array}$

The total number of valence electrons $\left( {{n_{{{\left[ {{\rm{Cl}}{{\rm{F}}_2}} \right]}^ - }}}} \right)$ in ${\left[ {{\rm{Cl}}{{\rm{F}}_2}} \right]^ - }$ are 22 electrons.

Consider the skeletal structure of ${\left[ {{\rm{Cl}}{{\rm{F}}_2}} \right]^ - }$ as follows:

The skeletal structure has 4 electrons. 18 valence electrons are remaining. Give 6 electrons to each atom of ${\rm{Cl}}$ and ${\rm{F}}$ . Draw the Lewis structure of ${\left[ {{\rm{Cl}}{{\rm{F}}_2}} \right]^ - }$ as follows:

Part 2

Count the total number of valence electrons $\left( {{n_{{{\left[ {{\rm{Cl}}{{\rm{F}}_2}} \right]}^ + }}}} \right)$ in ${\left[ {{\rm{Cl}}{{\rm{F}}_2}} \right]^ + }$ as follows:

$\begin{array}{c}\\{n_{{{\left[ {{\rm{Cl}}{{\rm{F}}_2}} \right]}^ + }}} = \left( {1 \times 7} \right) + \left( {2 \times 7} \right) - 1\\\\ = 20\\\end{array}$

The total number of valence electrons $\left( {{n_{{{\left[ {{\rm{Cl}}{{\rm{F}}_2}} \right]}^ + }}}} \right)$ in ${\left[ {{\rm{Cl}}{{\rm{F}}_2}} \right]^ + }$ are 20 electrons.

Consider the skeletal structure of ${\left[ {{\rm{Cl}}{{\rm{F}}_2}} \right]^ + }$ as follows:

The skeletal structure has 4 electrons. 16 valence electrons are remaining. Give 6 electrons to each atom of ${\rm{Cl}}$ and ${\rm{F}}$ . Therefore, the octet of ${\rm{Cl}}$ and ${\rm{F}}$ is complete. Draw the Lewis structure of ${\left[ {{\rm{Cl}}{{\rm{F}}_2}} \right]^ + }$ as follows:

Part 3

Count the total number of valence electrons $\left( {{n_{{{\left[ {{\rm{Cl}}{{\rm{F}}_4}} \right]}^ - }}}} \right)$ in ${\left[ {{\rm{Cl}}{{\rm{F}}_4}} \right]^ - }$ as follows:

$\begin{array}{c}\\{n_{{{\left[ {{\rm{Cl}}{{\rm{F}}_4}} \right]}^ - }}} = \left( {1 \times 7} \right) + \left( {4 \times 7} \right) + 1\\\\ = 36\\\end{array}$

The total number of valence electrons $\left( {{n_{{{\left[ {{\rm{Cl}}{{\rm{F}}_4}} \right]}^ - }}}} \right)$ in ${\left[ {{\rm{Cl}}{{\rm{F}}_4}} \right]^ - }$ are 36 electrons.

Consider the skeletal structure of ${\left[ {{\rm{Cl}}{{\rm{F}}_4}} \right]^ - }$ as follows:

The skeletal structure has 8 electrons. 28 valence electrons are remaining. Give 4 electrons to ${\rm{Cl}}$ and 6 electrons to each atom of ${\rm{F}}$ . Draw the Lewis structure of ${\left[ {{\rm{Cl}}{{\rm{F}}_4}} \right]^ - }$ as follows:

Ans: Part 1

The Lewis structure for ${\left[ {{\rm{Cl}}{{\rm{F}}_2}} \right]^ - }$ with lone pairs is as follows: