If the difference in electronegativity of elements P and Q is 3.0. The bond that will be formed between them is

1. Analyze the question

The question asks for the type of chemical bond formed between two elements, P and Q, given that the difference in their electronegativity values is 3.0.

2. Explain the concept

Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. The difference in electronegativity ($\Delta EN$) between two bonded atoms determines the nature of the bond:

• Non-polar Covalent Bond: $\Delta EN$ is usually between 0 and 0.4. Electrons are shared equally.
• Polar Covalent Bond: $\Delta EN$ is between 0.5 and 1.7. Electrons are shared unequally.
• Ionic Bond: $\Delta EN$ is generally greater than 1.7 to 2.0. In this case, the difference is so large that the more electronegative atom completely "steals" the electron from the less electronegative atom, resulting in the formation of ions.

3. Show the solution

• Given: Difference in electronegativity ($\Delta EN$) = 3.0.
• Threshold: In chemistry, a $\Delta EN$ value greater than 1.7 is widely accepted as the threshold for an ionic bond.
• Reasoning: Since 3.0 is significantly higher than 1.7, the bond between elements P and Q is strongly ionic. The element with the higher electronegativity will pull the bonding electrons completely toward itself, creating a strong electrostatic attraction between the resulting positive and negative ions.

4. Identify the answer

Comparing our result to the given options: A. Metallic B. Covalent C. Co-ordinate D. Ionic

The correct option is D.

5. Tip

Memory Trick: Think of the "1.7 Rule." If the difference is less than 1.7, it's Covalent (sharing). If the difference is greater than 1.7, it's Ionic (transferring). Since 3.0 is much larger than 1.7, it is definitely ionic!