Question

Draw the Lewis structure for the I₃^-. Include all lone pairs of electrons.

Answer 1

Concepts and reason

The concept used to solve this question is to draw the Lewis structure of ${{\rm{I}}_{\rm{3}}}^ -$ ion including all the lone pairs of electrons. The Lewis structure of any molecule or ion can be drawn by the counting the total number of valence electrons in the molecule or ion.

Fundamentals

Lewis structure: It is a graphical representation of the electron distribution around atoms. It is also known as electron dot structures. These structures show the bonding between atoms of a molecule and the lone pairs of electrons if exist in each atom in the molecule.

Lewis structure of any molecule or ion can be drawn by the following steps.

1) Count the total number of valence electrons in the molecule or ion, which can be calculated by adding the valence electrons of each atom in the molecule.

2) After calculating the total number of valence electrons, the central atom is identified and number of atoms bonded to the central atom is calculated.

3) The total number of atoms bonded to the central atom is indicated as ‘n’.

4) The number of lone pairs (LPs) present on the central atom can be calculated by the following formula.

${\rm{Number of lone pairs (LPs) on the central atom = }}\frac{{{\rm{Total number of valence electrons - 8n}}}}{2}$

${\rm{Here, n = total number of atoms attached to the central atom}}{\rm{.}}$

5) After putting the lone pairs (LPs) on the central atom, put lone pairs on the atoms bonded central atom according to octet rule.

Octet rule: It states the atoms of main group elements in the molecule forms bonds in such a way that it has eight electrons in its valence shell.

But there is an exception that in hypervalent compounds, the central atom has more than 8 electrons in its valence shell. This is possible by utilizing its d-orbitals found in the third principal energy level and beyond. This concept is also known as expanded octet. For example, phosphorous, sulfur, chlorine, bromine and iodine form an expanded octet.

The given ion is ${{\rm{I}}_{\rm{3}}}^ -$ ion.

There are three iodine atoms present in ${{\rm{I}}_{\rm{3}}}^ -$ ion, each iodine atom has 7 valence electrons and there is a negative charge on the ion.

$\begin{array}{c}\\{\rm{Hence, the total number of valence electrons in }}{{\rm{I}}_{\rm{3}}}^{\rm{ - }}{\rm{ ion = 3 (7) + 1}}\\\\{\rm{ = 22}}\\\end{array}$

Among the three iodine atoms, one must be the central atom and other two iodine atoms are bonded to the central iodine atom.

Thus, ${\rm{the total number of atoms bonded to the central atom, n = 2}}{\rm{.}}$

The number of lone pairs (LPs) present on the central atom can be calculated by the following formula.

$\begin{array}{c}\\{\rm{Number of lone pairs (LPs) on the central atom = }}\frac{{{\rm{Total number of valence electrons - 8n}}}}{2}\\\\ = {\rm{ }}\frac{{22 - 8(2)}}{2}\\\\ = \frac{6}{2}\\\\ = 3\\\end{array}$

Therefore, the number of lone pairs of electrons present on central iodine atom is 3.

The number of lone pairs present on the central iodine atom is 3.

So, first write three iodine atoms and put three lone pairs on central iodine atom and single bonds between central iodine atom and other two iodine atoms.

At last, put the lone pairs on the terminal iodine atoms according to octet rule and a negative charge on the whole molecule.

Therefore, the Lewis structure of ${{\rm{I}}_{\rm{3}}}^ -$ ion is as follows:

Ans:

The Lewis structure of ${{\rm{I}}_{\rm{3}}}^ -$ ion is as follows: