Consider the following element combinations. Classify the bonds formed between each pair as ionic, polar covalent, or nonpolar covalent qualitatively based solely on each element's position on the periodic table? Do not conduct calculations.

1 Answer
Nathan L.
Jun 24, 2017

#color(red)("Rb" + "I"#: ionic
#color(blue)("Br" + "Br"#: nonpolar covalent
#color(green)("S" + "I"#: polar covalent
#color(blue)("N" + "N"#: nonpolar covalent
#color(green)("O" + "F"#: polar covalent
#color(red)("Cs" + "Cl"#: ionic
#color(red)("Ca" + "O"#: ionic
#color(green)("P" + "I"#: polar covalent

Explanation:

Something worth knowing about deciding between nonpolar and polar is that only two of the same element can form an ideally nonpolar bond.

With this in mind, the #"Br"-"Br"# and #"N"-"N"# are both nonpolar covalent bonds.

Bonds between a metal and a nonmetal are almost always considered "ionic", so the #"Rb"-"I"#, #"Cs"-"Cl"#, and #"Ca"-"O"# are all ionic bonds.

And therefore, the remaining binary compounds (#"S"-"I"#, #"O"-"F"#, and #"P"-"I"#), which are those with two nonmetals, form polar covalent bonds, those in which electrons are dispersed unevenly throughout the bond.

In summary,

#color(red)("Rb" + "I"#: ionic
#color(blue)("Br" + "Br"#: nonpolar covalent
#color(green)("S" + "I"#: polar covalent
#color(blue)("N" + "N"#: nonpolar covalent
#color(green)("O" + "F"#: polar covalent
#color(red)("Cs" + "Cl"#: ionic
#color(red)("Ca" + "O"#: ionic
#color(green)("P" + "I"#: polar covalent

Related questions
See all questions in Ionic vs. Molecular Bonds
Impact of this question
27095 views around the world
You can reuse this answer
Creative Commons License