Question

1. using electronegativities, find δen for n–h and identify the bond type. h has an en of 2.2 while n has an en of 3.0.

Explanation:

Step1: Recall the formula for ΔEN

The formula for the difference in electronegativity ($\Delta\text{EN}$) is the absolute value of the difference between the electronegativities of the two atoms. So, $\Delta\text{EN} = |\text{EN}_A - \text{EN}_B|$.

Step2: Calculate ΔEN for N - H

Given $\text{EN}_N = 3.0$ and $\text{EN}_H = 2.2$. Substitute into the formula: $\Delta\text{EN} = |3.0 - 2.2| = 0.8$.

Step3: Identify the bond type

• If $\Delta\text{EN} < 0.5$, the bond is non - polar covalent.
• If $0.5\leq\Delta\text{EN}< 2.0$, the bond is polar covalent.
• If $\Delta\text{EN}\geq2.0$, the bond is ionic.

Since $\Delta\text{EN} = 0.8$ which is in the range $0.5\leq\Delta\text{EN}< 2.0$, the bond type is polar covalent.

Answer:

The value of $\Delta\text{EN}$ for N - H is $0.8$, and the bond type is polar covalent.