Question

Determine whether each of the following molecules is polar or nonpolar.

SiCl2F2

CO2

XeO4

XeF2

Answer 1

Concepts and reason

This problem is based on polarity of a molecule.

A molecule that has negative charge density on one end and positive charge density on the other end is called polar molecule. From the shape of a molecule polarity of the molecule can be determined. A polar molecule has an asymmetric shape with one or more polar bonds due to difference in electronegativity of the atoms.

Fundamentals

First determine the Lewis structure to tell that if a molecule is polar or nonpolar.

Then determine the shape of molecule. Linear, trigonal pyramidal and tetrahedral shapes are symmetric, while bent and trigonal pyramidal are asymmetric shapes.

A polar molecule is not symmetric in shape.

A nonpolar molecule has symmetric shape. Also, it’s all atoms surrounding the central atom are same. If the atoms attached to central atom are different, the molecule may be polar. In this case identify whether the bonds are polar or nonpolar based on the difference in electronegativities. The molecule is nonpolar, if there are no polar bonds.

If there are polar bonds, draw arrows pointing toward the more electronegative element.

(a)

Number of valence electrons in Si, Cl and F are 4, 7 and 7 respectively.

Total number of valence electrons is calculated as follows:

$n = {n_{{\rm{Si}}}} + 2{n_{{\rm{Cl}}}} + 2{n_{\rm{F}}}$

Substitute 4 for ${n_{{\rm{Si}}}}$ , 7 for ${n_{{\rm{Cl}}}}$ and 7 for ${n_{\rm{F}}}$ .

$\begin{array}{c}\\n = {\rm{4}} + {\rm{2}}\left( {\rm{7}} \right) + 2\left( 7 \right)\\\\ = 4 + 14 + 14\\\\ = 32\\\end{array}$

(a)

Total number of valence electrons is 32. These electrons are arranged in such a way that the valency of each atom is completed.

The Lewis structure for ${\rm{SiC}}{{\rm{l}}_2}{{\rm{F}}_2}$ is as follows:

(a)

Si atom has four atoms attached to it. There are two ${\rm{Si}} - {\rm{Cl}}$ and ${\rm{Si}} - {\rm{F}}$ bonds. Cl and F are more electronegative than Si. F is more electronegative than Cl.

Draw arrows to represent polar bonds.

The arrows do not balance each other.

Thus, the molecule is polar.

(b)

Number of valence electrons in C and O are 4 and 6 respectively.

Total number of valence electrons is calculated as follows:

$n = {n_{\rm{C}}} + 2{n_{\rm{O}}}$

Substitute 4 for ${n_{\rm{C}}}$ , and 6 for ${n_{\rm{O}}}$ .

$\begin{array}{c}\\n = {\rm{4}} + {\rm{2}}\left( 6 \right)\\\\ = 4 + 12\\\\ = 16\\\end{array}$

(b)

Total number of valence electrons are 16.

The Lewis structure is as follows:

(b)

The shape of ${\rm{C}}{{\rm{O}}_2}$ is linear as follows:

The two oxygen atoms are more electronegative than carbon atom.

Draw arrows pointing towards oxygen.

The two arrows for ${\rm{C}} - {\rm{O}}$ bond dipole exactly balances each other.

Thus, the molecule ${\rm{C}}{{\rm{O}}_2}$ is nonpolar.

(c)

Number of valence electrons in Xe and O are 8 and 6 respectively.

Total number of valence electrons is calculated as follows:

$n = {n_{{\rm{Xe}}}} + 4{n_{\rm{O}}}$

Substitute 8 for ${n_{{\rm{Xe}}}}$ , and 6 for ${n_{\rm{O}}}$ .

$\begin{array}{c}\\n = 8 + 4\left( 6 \right)\\\\ = 8 + 24\\\\ = 32\\\end{array}$

(c)

Total number of valence electrons is 32.

Thus, The Lewis structure is as follows:

(c)

The geometric shape of ${\rm{Xe}}{{\rm{O}}_4}$ is tetrahedral:

And O is more electronegative than ${\rm{Xe}}$ .

Draw the arrows pointing towards O, to show the bond dipole.

The bond dipole of four ${\rm{Xe}} - {\rm{O}}$ dipoles are equal. They cancel each other.

Thus, the molecule ${\rm{Xe}}{{\rm{O}}_4}$ is non-polar.

(d)

Number of valence electrons in Xe and F are 8 and 7 respectively.

Total number of valence electrons is calculated as follows:

$n = {n_{{\rm{Xe}}}} + 2{n_{\rm{F}}}$

Substitute 8 for ${n_{{\rm{Xe}}}}$ , and 7 for ${n_{\rm{F}}}$ .

$\begin{array}{c}\\n = 8 + 2\left( 7 \right)\\\\ = 8 + 14\\\\ = 22\\\end{array}$

(d)

Total number of valence electrons is 22.

Thus placing as central atom, arrange fluorine atoms around it, and Lewis structure is drawn as follows:

(d)

The geometric shape of ${\rm{Xe}}{{\rm{F}}_2}$ is linear.

And F is more electronegative than Xe.

Draw the arrows pointing towards F, to show the bond dipole.

The bond dipole of two ${\rm{Xe}} - {\rm{F}}$ bonds are equal.

Thus, the molecule ${\rm{Xe}}{{\rm{F}}_2}$ is nonpolar.

Ans: Part a

The molecule ${\rm{SiC}}{{\rm{l}}_2}{{\rm{F}}_2}$ is polar.

Part b

The molecule ${\rm{C}}{{\rm{O}}_2}$ is nonpolar.

Part c

The molecule ${\rm{Xe}}{{\rm{O}}_4}$ is nonpolar.

Part d

The molecule ${\rm{Xe}}{{\rm{F}}_2}$ is nonpolar.