Question

Classify the type of bond for the following molecules. Show your work(calculation) using the electronegativity values:

H=2.10, Na= 0.93 Cl=3.16 O= 3.44 C=2.55

NaCl

CO

Answer 1

1) NaCl

first find electronegativity difference

so

dEN= 3.16 - 0.93

dEN = 2.23

we know that

If the ∆EN is between 1.6 more then its an ionic bond

also

if a metal is involved, then the bond is considered ionic

and

If only nonmetals are involved, the bond is considered polar covalent

Here Na is metal

so

the bond in NaCl is ionic

2) given

CO

find the electronegativity difference

dEN = 3.44 - 3.16

dEN = 0.28

we know that

if the Electron negativity difference between 0.2 - 0.5

then it is a non polar covalent bond

so

the bond in CO is a non polar covalent bond