Question

In a compound XY, if the electronegativity difference between X and Y is greater than 1.7, then the compound XY is soluble in:
A. Benzene
B. Carbon tetrachloride
C. Water
D. Carbon disulfide

Answer 1

We know that Electronegativity is a measure of an attraction of an atom for electrons in a bond. Electronegativity tells us how much electrons are needed for a particular atom. We know that the electronegativity of an atom takes the value from 0 to 4; the greater the value, the atom would be more electronegative, and the more would be attraction of the electrons in a bond.

Complete step by step answer:
Given data contains,
The electronegative difference between X and Y is greater than 1.7.
If the electronegativity difference between the two atoms is less than ${\text{0}}{\text{.5}}$ units the type of bond is nonpolar.
If the electronegativity difference between the two atoms is ${\text{0}}{\text{.5 - 1}}{\text{.9}}$ units then the type of bond is polar covalent.
If the electronegativity difference between two atoms is greater than ${\text{1}}{\text{.9}}$ units then the type of bond is ionic.
Now, coming back to the question, we have to know that if the electronegativity value is greater than 1.7 units, then the compound XY should contain ionic bonds and the compound should be an ionic compound.
We know that ionic compounds are soluble in polar solvent and in the given options, the solvent that is polar is water. So, the compound XY will be water soluble and Therefore, the option (C) is correct.
The solvents benzene, carbon tetrachloride, and carbon disulfide are non-polar in nature and they dissolve only nonpolar compounds. So, option (A), (B) and (D) are incorrect.

So, the correct answer is Option C.

Note: Let us see how the type of bonding is determined based on electronegativity difference.
Consider the example below of hydrogen fluoride. Let us find the type of bonding that is found in hydrogen fluoride.
The electronegative value of hydrogen is $2.1$
The electronegative value of fluorine is ${\text{4}}$
The electronegativity difference is calculated as,
$E.D = 4.0\left( F \right) - 2.1\left( H \right)$
$E.D = 1.9$
The electronegative difference between hydrogen and fluorine is ${\text{1}}{\text{.9}}$ units. The type of bond in $HF$ is polar covalent because the value lies in the range of ${\text{0}}{\text{.5 - 1}}{\text{.9}}$ units.