Question

Among the following bonds, which has the most polar character?
(A) $C-O$
(B) $C-Br$
(C) $C-F$
(D) $C-S$

Answer 1

Hint: If the difference in electronegativity for the atoms in a bond is greater than 0.4, then consider that the bond is polar. If the difference in electronegativity is less than 0.4, then the bond is nonpolar.

Complete step by step answer:
Bonds that are partially ionic and polar covalent bonds. Nonpolar covalent bonds, with equal sharing of bond electrons, when electronegativities of the two atoms are equal.
To find the polar character bond between two atoms which are two different electronegativity atoms can be accumulated by a formula.
One way of estimating the ionic character of a bond that is, a magnitude of the charge separation in a polar covalent bond is to calculate the difference in electronegativity between the two atoms,

     $$\mathrm{\Delta }X=X_B-X_A--(1)$$
By using the above equation (1), let us find the most polar character from the given options.

 (A) from C-O bond, the electronegativity of carbon ($X_C$ ) = 2.55
           The electronegativity of oxygen ($X_O$ ) = 3.44
     $$\mathrm{\Delta }{X}_{CO}=X_O-X_C$$

     $$\mathrm{\Delta }{X}_{CO}=3.44-2.55=0.89$$ ---(2)

(B) from C-Br bond, the electronegativity of carbon ($X_C$ ) = 2.55
The electronegativity of bromine ($X_{Br}$ ) = 2.96
     $$\mathrm{\Delta }{X}_{CBr}=X_{Br}-X_C$$
     $$\mathrm{\Delta }{X}_{CBr}=2.96-2.55=0.41$$ ----(3)

(C ) from C-F bond, the electronegativity of carbon ($X_C$ ) = 2.55
The electronegativity of fluorine ($X_F$ ) = 3.98
     $$\mathrm{\Delta }{X}_{CF}=X_F-X_C$$
     $$\mathrm{\Delta }{X}_{CF}=3.98-2.55=1.43$$ ----(4)

(D) from C-S bond, the electronegativity of carbon ($X_C$ ) = 2.55
The electronegativity of sulfur ($X_S$ )= 2.58
     $$\mathrm{\Delta }{X}_{CS}=X_S-X_C$$
     $$\mathrm{\Delta }{X}_{CS}=2.58-2.55=0.03$$ ---(5)
From the above equations, the following order of polar character of given bonds C-O, C-Br, C-F, C-S respectively,

     $$\mathrm{\Delta }{X}_{CF}>\mathrm{\Delta }{X}_{CO}>\mathrm{\Delta }{X}_{CBr}>\mathrm{\Delta }{X}_{CS}$$
From the above order, hence C-F bond is the most polar character. Because electronegativity of F is more than other elements. So the C-F bond is more polar character than C-Br, C-S, and C-O
The correct answer is C-F bond.
So, the correct answer is “Option C”.

Note: When there are no polar bonds in a molecule, there is no permanent change difference between one part of the molecule and another, the molecule is nonpolar. A molecule can possess polar bonds that are symmetrically distributed.