[tex]boxed{{text{N and Cl}}}[/tex]will form a non-polar covalent compound.

Further Explanation:

Electronegativity:n

It is the tendency of an atom to attract the shared electrons in the bond towards itself. More the electronegativity of the atom, higher will be its tendency to attract the bonding electrons towards itself. The electronegative atom develops a partial negative charge and the electropositive atom develops a partial positive charge.n

n

The polarity of the bond can be determined by the electronegativity difference [tex]left({Delta {text{EN}}}right)[/tex]. [tex]Delta{text{EN}}[/tex] is the electronegativity difference between the two atoms that are bonded to each other.n

The formula to calculate [tex]Delta{text{EN}}[/tex] in XY bond is as follows:

[tex]{{Delta {text{EN}}=left( {{text{electronegativity of Y}}} right) – left( {{text{electronegativity of X}}} right)[/tex] …….(1)

Here,n

X is the electropositive atom.n

Y is the electronegative atom.n

Higher the electronegativity difference between the two atoms, more will be the polarity of the bond and vice-versa.n

n1.tIf [tex]Deltatext{EN}}[/tex] is less than 0.4, the bond formed will be non-polar covalent bond.n

2.tIf [tex]Delta{text{EN}}[/tex] is between 0.4 and 1.7, the bond formed will be polar covalent bond.n

3.tIf [tex]Delta{text{EN}}[/tex] is more than 1.7, the bond formed will be an ionic bond.n

Na and Cln

Chlorine is more electronegative than sodium. Substitute 3 for the electronegativity of Y and 0.9 for the electronegativity of X in equation (1) to calculate the electronegativity difference between Na and Cl.n

The electronegativity difference between sodium and chlorine comes out to be greater than 1.7. So Na and Cl will form an ionic bond.n

n

C and On

Oxygen is more electronegative than carbon. Substitute 3.5 for the electronegativity of Y and 2.5 for the electronegativity of X in equation (1) to calculate the electronegativity difference between C and O.n

The electronegativity difference between carbon and chlorine is between 0.4 and 1.7. So C and O will form a polar covalent bond.

B and On

Oxygen is more electronegative than boron. Substitute 3.5 for the electronegativity of Y and 2 for the electronegativity of X in equation (1) to calculate the electronegativity difference between B and O.n

## Answers ( 2 )

[tex]boxed{{text{N and Cl}}}[/tex]will form a non-polar covalent compound.

Further Explanation:Electronegativity:nIt is the tendency of an atom to

attracttheshared electronsin the bond towards itself. More the electronegativity of the atom, higher will be its tendency to attract the bonding electrons towards itself. The electronegative atom develops a partial negative charge and the electropositive atom develops a partial positive charge.nn

The polarity of the bond can be determined by the

electronegativity difference [tex]left({Delta {text{EN}}}right)[/tex]. [tex]Delta{text{EN}}[/tex] is the electronegativity difference between the two atoms that are bonded to each other.nThe formula to calculate [tex]Delta{text{EN}}[/tex] in XY bond is as follows:

[tex]{{Delta {text{EN}}=left( {{text{electronegativity of Y}}} right) – left( {{text{electronegativity of X}}} right)[/tex] …….(1)

Here,n

X is the electropositive atom.n

Y is the electronegative atom.n

Higher the electronegativity difference between the two atoms, more will be the polarity of the bond and vice-versa.n

n1.tIf [tex]Deltatext{EN}}[/tex] is less than 0.4, the bond formed will be non-polar covalent bond.n2.tIf [tex]Delta{text{EN}}[/tex] is between 0.4 and 1.7, the bond formed will be polar covalent bond.n3.tIf [tex]Delta{text{EN}}[/tex] is more than 1.7, the bond formed will be an ionic bond.nNa and ClnChlorine is more electronegative than sodium. Substitute 3 for the electronegativity of Y and 0.9 for the electronegativity of X in equation (1) to calculate the electronegativity difference between Na and Cl.n

[tex]begin{aligned}Delta text{EN}_{text{Na-Cl}}&=3-0.9&=2.1end{aligned}[/tex]

The electronegativity difference between sodium and chlorine comes out to be greater than 1.7. So Na and Cl will form an

ionic bond.nnC and OnOxygen is more electronegative than carbon. Substitute 3.5 for the electronegativity of Y and 2.5 for the electronegativity of X in equation (1) to calculate the electronegativity difference between C and O.n

[tex]begin{aligned}Delta text{EN}_{text{C-O}}&=3.5-2.5&=1end{aligned}[/tex]

The electronegativity difference between carbon and chlorine is between 0.4 and 1.7. So C and O will form a

polar covalent bond.B and OnOxygen is more electronegative than boron. Substitute 3.5 for the electronegativity of Y and 2 for the electronegativity of X in equation (1) to calculate the electronegativity difference between B and O.n

[tex]begin{aligned}Delta text{EN}_{text{B-O}}&=3.5-2&=1.5end{aligned}[/tex]

The electronegativity difference between boron and oxygen is between 0.4 and 1.7. So B and O will form a

polar covalent bond.N and ClnThe electronegativity of nitrogen and chlorine is nearly equal so N and Cl will form a

non-polar covalent bond.nnLearn more:n1.Identification of ionic bonding: https://brainly.com/question/1603987n2.Which is a covalent compound? https://brainly.com/question/2083444nn

Answer details:nGrade:High SchoolnSubject:ChemistrynChapter:Ionic and covalent compoundsnn

Keywords:electronegativity difference, electropositive, electronegative, polar, non-polar, ionic, 0.4, 1.7, N, Cl, O, C, B.nTypes of Bonds can be predicted by calculating thendifference in electronegativity.

nn

If, Electronegativity difference is,

nn

nn

Lessnthan

0.4then it isNon Polar Covalentnn

n

nn

nBetween

0.4and1.7then it isPolar Covalentnn

nn

nGreater than

1.7then it isIonicnn

nn

For Na and Cl,nn

n E.N of Chlorine = 3.16

nn

n E.N of Sodium = 0.93

nn

n n ________

nn

n E.N Difference 2.23 (

Ionic Bond)For C and O,E.N of Oxygen = 3.44

E.N of Carbon = 2.55

________

E.N Difference 0.89 (

Polar CovalentBond)For N and Cl,E.N of Chlorine = 3.16

E.N of Notrogen = 3.04

________

E.N Difference 0.12 (

Non-Polar CovalentBond)For B and O,E.N of Oxygen = 3.44

E.N of Boron = 2.04

________

E.N Difference 1.40 (

Polar CovalentBond)