Element A has an electronegativity of 0.8 and element B has an electronegativity of 3.0. Which statement best describes the bonding in A3B? A) The AB bond is largely covalent with a on A. B) The AB bond is largely covalent with a δ+ on A. C) The compound is largely ionic with A as the cation. D) The compound is largely ionic with A as the anion.

Pulings proposed the method to determine if the compound is ionic in nature or covalent in nature , by finding the difference between the electronegativity of the respective cation and anion .

The ion with higher electronegativity is the anion and the ion with lower electronegativity is the cation.

The electronegativity difference above 1.7 make the compound ionic in nature.

Hence, from the question ,

A is the cation and B is the anion.

And the electronegativity difference above 1.7 so the compound is ionic in nature.

Histrionicus Histrionicus · Answer 2 · 2021-11-23T15:28:57+01:00

Element A has an electronegativity of 0.8 and element B has an electronegativity of 3.0. The statement best describes the bonding in A3B - C) The compound is largely ionic with A as the cation.

It is given that element A has an electronegativity 0.8 and element B has an electronegativity of 3.

The type of bond is based on the electronegativity difference.

If the electronegativity difference = equal, the bond is non-polar covalent.
If the electronegativity difference = greater than 0 and less than 2, the bond is polar covalent
If the electronegativity difference = greater than 2, the bond is ionic.

here the electronegativity difference

= 3 - 0.8 = 2.2

Thus, the compound will be ionic and we know that an element with less electronegativity is a cation, therefore, element A is a cation.

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