Q10.57P - Chemistry: Molecular Nature Of Matter And Change

Question

Which molecule in each pair has the greater dipole moment? Give the reason for your choice.

1. ${\mathbf{SO}}_2\mathbf{or}{\mathbf{SO}}_3$
   
2.  $\mathbf{ICl}\mathbf{ }\mathbf{or}\mathbf{ }\mathbf{IF}$
3. ${\mathbf{SiF}}_4\mathbf{or}{\mathbf{SF}}_4$
   
4. ${\mathbf{H}}_2\mathbf{O}\mathbf{or}{\mathbf{H}}_2\mathbf{S}$
   

Step-by-Step Solution

Answer

1. ${\mathrm{SO}}_2$ has a greater dipole moment.
2.  IF has a greater dipole moment.
3. ${\mathrm{SF}}_4$ has a greater dipole moment.
4. ${\mathrm{H}}_2\mathrm{O}$ has a greater dipole moment

Step 1: Definition of Molecular Polarity, Dipole Moment

Molecules that share a net imbalance of charge show molecular polarity. The product of the partial charges and the distance between them is called the dipole moment $\left(\mathbf{\mu }\right)$.

• A bond is said to be polar when it joins the atoms with different electronegativity. 
• In diatomic molecules with only two atoms with one bond, the molecule is polar due to bond polarity. 
• In molecules with more than two atoms, shape and bond polarity will give the molecular polarity. 
• Polar molecules have partial charges around the atoms showing dipole moment.
• The polarity of a bond is determined by the atom’s electronegativity values. 
• We also know that electronegativity values of elements increase across the period and decrease down the group. 
• Calculating the electronegativity difference $\left(\mathrm{\Delta EN}\right)$ of the molecules will help to find out which molecule is more polar. 
• Molecules with a high value of  $\left(\mathrm{\Delta EN}\right)$ has the most polar bond with a greater dipole moment.

 

Step 2: Molecules with a Greater Dipole Moment

The shape of the molecule plays an important role in determining the dipole moment. Therefore, not only the polar bonds but the shape of the molecule also determines the molecular polarity. Let us consider the shapes of the given molecules to find whether the molecules have a dipole moment or not. 

(a) The molecule ${\mathrm{SO}}_2$  has a bent shape. As O (EN = 3.4) has more electronegativity value than S (EN = 2.5) with an electronegativity difference of 0.9, making the S-O bond more polar as each bond dipole points towards O. The lone pair also contributes to the net dipole moment. As the bond polarities are not counterbalanced, the molecule shows a significant dipole moment making the molecule polar.

While in ${\mathrm{SO}}_3$ the shape of the molecule is trigonal pyramidal. Even though the S-O bond is more polar as each bond dipole points towards O, the bond polarities are counterbalanced and the molecule shows no significant dipole moment making the molecule nonpolar. Therefore, ${\mathrm{SO}}_2$ has a greater dipole moment than ${\mathrm{SO}}_3$ .

(b) The molecules ICl and IF both have a linear shape. In molecule ICl, as Cl (EN = 3.16) has more electronegativity value than I (EN = 2.66) with an electronegativity difference of 0.5, making the I-Cl bond more polar the bond dipole points toward Cl. Hence, the molecule shows a significant dipole moment making the molecule polar.

In molecules IF, as F (EN = 4.00) has more electronegativity value than I (EN = 2.66) with an electronegativity difference of 1.34, making the I-F bond more polar than I-Cl with the bond dipole pointing towards F. As F is more electronegative than Cl, the molecule IF shows more dipole moment than ICl.

 

(c) The molecule ${\mathrm{SF}}_4$  has a see-saw shape. As F (EN = 4.0) has more electronegativity value than S (EN = 2.5) with an electronegativity difference of 1.5, making the S-F bond more polar as each bond dipole points towards F. The lone pair also contribute for net dipole moment. As the bond polarities are not counterbalanced, the molecule shows a significant dipole moment making the molecule polar.

The molecule  ${\mathrm{SiF}}_4$ has a tetrahedral shape. Fluorine, F (EN = 4.0) has more electronegativity value than Si (EN = 1.9) with an electronegativity difference of 2.1, making the Si-F bond more polar as each bond dipole points towards F. As the bond polarities are counterbalanced, the molecule shows no significant dipole moment making it a nonpolar molecule. Therefore, the molecule  data-custom-editor="chemistry" ${\mathrm{SF}}_4$ shows more dipole moments than ${\mathrm{SiF}}_4$.

(d) The molecule ${\mathrm{H}}_2\mathrm{O}$  has a V shape. As O (EN = 3.4) has more electronegativity value than H (EN = 1.0) with an electronegativity difference of 2.4, making the O-H bond more polar as each bond dipole points towards O. The two lone pairs also contribute to the net dipole moment. As the bond polarities are not counterbalanced, the molecule shows a significant dipole moment making the molecule polar.

Similarly, the molecule ${\mathrm{H}}_2\mathrm{S}$ also has a V shape as  ${\mathrm{H}}_2\mathrm{O}$. As S (EN = 2.5) has more electronegativity value than H (EN = 1.0) with an electronegativity difference of 1.5, making the H-S bond more polar as each bond dipole points towards S. The two lone pairs also contribute to the net dipole moment.  As the bond polarities are not counterbalanced, the molecule shows a significant dipole moment making the molecule polar.  Even though both ${\mathrm{H}}_2\mathrm{O}$ and data-custom-editor="chemistry" ${\mathrm{H}}_2\mathrm{S}$  have the same shape, O is more electronegative than S,  ${\mathrm{H}}_2\mathrm{O}$ shows more dipole moment than data-custom-editor="chemistry" ${\mathrm{H}}_2\mathrm{S}$ .

Therefore, ${\mathrm{SO}}_2$ has a greater dipole moment than ${\mathrm{SO}}_3$ , IF ${\mathrm{SF}}_4$ has a greater dipole moment than ICl, has a greater dipole moment than  ${\mathrm{SiF}}_4$,  ${\mathrm{H}}_2\mathrm{O}$has a greater dipole moment than  data-custom-editor="chemistry" ${\mathrm{H}}_2\mathrm{S}$.