Question

17 sep use math calculate the electronegativity differences and determine the polarity for the bonds formed by the following pairs of atoms: k and f atoms and n and o atoms.

Explanation:

Step1: Recall Electronegativity Values

First, we need the electronegativity values of K, F, N, and O. From the Pauling scale: Electronegativity of K ($\chi_K$) is approximately 0.82, F ($\chi_F$) is 3.98, N ($\chi_N$) is 3.04, and O ($\chi_O$) is 3.44.

Step2: Calculate Electronegativity Difference for K - F

The formula for electronegativity difference ($\Delta\chi$) is $|\chi_{more\ electronegative} - \chi_{less\ electronegative}|$. For K and F: $\Delta\chi = |3.98 - 0.82| = 3.16$. A large difference (usually > 1.7) indicates an ionic bond (highly polar, essentially ionic).

Step3: Calculate Electronegativity Difference for N - O

For N and O: $\Delta\chi = |3.44 - 3.04| = 0.4$. A small difference (usually < 0.5) indicates a nonpolar covalent bond, but since it's between 0.4 - 1.7, it's a polar covalent bond (slightly polar).

Answer:

• For K - F: Electronegativity difference = 3.16, Bond type: Ionic (highly polar).
• For N - O: Electronegativity difference = 0.4, Bond type: Polar covalent (slightly polar).