## Which pair of atoms forms a nonpolar covalent bond? which pair of atoms forms a nonpolar covalent bond? na and cl c and o n and cl b and o?

Question

Which pair of atoms forms a nonpolar covalent bond? which pair of atoms forms a nonpolar covalent bond? na and cl c and o n and cl b and o?

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1. [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

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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

[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.n

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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

[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 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

[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 Cln

The electronegativity of nitrogen and chlorine is nearly equal so N and Cl will form a non-polar covalent bond.n

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1. Identification of ionic bonding: https://brainly.com/question/1603987n

2. Which is a covalent compound? https://brainly.com/question/2083444n

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Subject: Chemistryn

Chapter: Ionic and covalent compoundsn

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Keywords: electronegativity difference, electropositive, electronegative, polar, non-polar, ionic, 0.4, 1.7, N, Cl, O, C, B.n

2. Types of Bonds can be predicted by calculating thendifference in electronegativity.

nn

If, Electronegativity difference is,

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Lessnthan 0.4 then it is Non Polar Covalent

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n

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nBetween 0.4 and 1.7 then it is Polar Covalent

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nGreater than 1.7 then it is Ionic

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For Na and Cl,

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n E.N of Chlorine = 3.16

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n E.N of Sodium = 0.93

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n n ________

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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)