Question

Polar covalent bond is present in:

• A. $NaCl$
• B. $C{a}_3{N}_2$
• C. $K_{3}N$
• D. $F_2$
• E. $N{H}_3$

Answer 1

The correct option is E $N{H}_3$

Electronegativity Difference	Type of Bond Formed
0.0 to 0.2	nonpolar covalent
0.3 to 1.4	polar covalent
> 1.5	ionic

The electronegativity difference provides a way of predicting the kind of bond that will form between two elements.

$NaCl$ Electron negativity difference $=0.9-3.0=2.1$(ionic bond)

$C{a}_3{N}_2$ Electronegativity difference $=1.0-3.0=2$ (ionic bond)

$K_{3}N$ Electronegativity difference $=0.8-3.0=2.2$ (ionic bond)

$F_2$ Electronegativity difference $=4.0-4.0=0$ Bond $=$ Non-polar covalent

$N{H}_3$ Electronegativity difference $=3.0-2.1=0.9$ Bond $=$ Polar covalent