Q10.56P - Chemistry: Molecular Nature Of Matter And Change

Question

Consider the molecules  and ${\mathbf{BF}}_3,{\mathbf{PF}}_3,{\mathbf{BrF}}_3,{\mathbf{SF}}_4$.

a.Which has bonds that are the most polar?

b.Which have a molecular dipole moment?

Step-by-Step Solution

Answer

a.The molecule $B{F}_3$ has bonds that are the most polar.

b.Molecules $P{F}_3,Br{F}_3$ and $S{F}_4$ have a molecular 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.$\left(\mu \right)$

• 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

Step 2: Molecules with Bonds that are Most Polar

Let us consider the given molecules with their type of bond for determining the most polar bond.

Molecule	Bond
$B{\mathbf{F}}_3$	B-F
${\mathbf{PF}}_3$	P-F
${\mathbf{BrF}}_3$	Br-F
${\mathbf{SF}}_4$	S-F
$S{F}_6$	S-F

• 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. 
• Nonmetals are more electronegative than metals. 

Fluorine is the most electronegative element within the tabular array

• Calculating the electronegativity difference$\left(\Delta EN\right)$ of the molecules will help to find out which molecule is more polar.

Molecules with a high value of $\left(\Delta EN\right)$ has the most polar bond

For,

B-F bond, the EN of  $B=2.0$ and EN of $F=4.0$ so the .$\left(\Delta EN\right)=4.0-2.0=2.0$

P-F bond, the EN of  $P=2.1$ and EN of $F=4.0$ so the .$\left(\Delta EN\right)=4.0-2.1=1.9$

Br-F bond, the EN of $Br=2.8$  and EN of$F=4.0$ , so the$\left(\Delta EN\right)=4.0-2.8=1.2$ 

.S-F bond, the EN of  $S=2.5$ and EN of $F=4.0$, so the $\left(\Delta EN\right)=4.0-2.5=1.5$.

The B-F bond has a high value of $\left(\Delta EN\right)$ , thus has the most polar bond.

Step 3: Molecules with a Molecular 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 havea dipole moment or not.  

The molecule $B{F}_3$  has a trigonal planar shape. As  $F(EN=4.0)$ has more electronegativity value than $B(EN=2.0)$ with an electronegativity difference of 2.0, making the B-F bond more polar as each bond dipole points towards F. The bond polarities are counterbalanced and the molecule showsno significant dipole moment making the molecule nonpolar.

The molecule $P{F}_3$, has a trigonal pyramidal shape. As $F(EN=4.0)$  has more electronegativity value than $P(EN=2.1)$ with an electronegativity difference of 1.9, making the P-F bond more polar as each bond dipole points towards F. The lone pair also contribute for the net dipole moment. As the bond polarities are not counterbalanced, the molecule shows a significant dipole moment making the molecule polar.

The molecule $Br{F}_3$  has a trigonal bipyramidal or T shape. As  $F(EN=4.0)$ has more electronegativity value than $Br(EN=2.8)$ with an electronegativity difference of 1.2, making the Br-F bond more polar as each bond dipole points towards F. The lone pairs also contribute for the net dipole moment. As the bond polarities are not counterbalanced, the molecule showsa significant dipole moment making the molecule polar

The molecule $S{F}_4$ 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 the net dipole moment. As the bond polarities are not counterbalanced, the molecule shows a significant dipole moment making the molecule polar.

The molecule  $S{F}_6$ has an octahedral 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. As the bond polarities are counterbalanced, the molecule shows no significant dipole moment making it a nonpolar molecule.

Thus, $B{F}_3$ has bonds that are the most polar and molecules $P{F}_3,Br{F}_3$ and $S{F}_4$ have a molecular dipole moment