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7. a) Classify the following bonds as ionic, polar covalent, or nonpolar qualita

ID: 504920 • Letter: 7

Question

7. a) Classify the following bonds as ionic, polar covalent, or nonpolar qualitatively based solely on each element's position on the periodic table. S-S Na-N O-I S-Cl Cu-Cl

b) Classify the bonds (ionic, polar covalent, nonpolar) formed between each pair of elements based on electronegativity values in the following table: Element 1.) Sc 2.) Be 3.) H 4.) P 5.) S 6.) I 7.) Br 8.) Cl 9) O 10) F

Electronegativity 1) 1.3 2) 1.5 3) 2.1 4) 2.1 5) 2.5 6) 2.5 7) 2.8 8) 3.0 9) 3.5 10) 4.0

H-Br H-P S-O Be-F Be-Cl

Explanation / Answer

Guide line for ionic, covalent or nonpolar bonds:
Electronegativity difference between bonded atoms = 0 then its nonpolar.
= more than 0 and less than 1.7 are polar covalent.
= greater than 1.7 are ionic.

7. a) S-S (difference between two atoms would be zero due to same atom hence, nonpolar qualitatively)

Na-N (difference between two atoms would be more due to more difference in there periodic position hence, ionic bond)

O-I (difference between two atoms would be less due to less difference in their periodic position hence, polar covalent bond)

S-Cl (difference between two atoms would be less due to less difference in their periodic position hence, polar covalent bond)

Cu-Cl (difference between two atoms would be less due to less difference in there periodic position hence, polar covalent bond)

b) H-Br (difference of electronegativity between two atoms = 2.8-2.1 = 0.7 hence, polar covalent bond)

H-P (difference of electronegativity between two atoms = 2.1-2.1 = 0.0 hence, non polar bond)

S-O (difference of electronegativity between two atoms = 3.5-2.5 = 1.0 hence, polar covalent bond)

Be-F (difference of electronegativity between two atoms = 4.0-1.5 = 2.5 hence, ionic bond)

Be-Cl (difference of electronegativity between two atoms = 3.0-2.1 = 0.9 hence, polar covalent bond)

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