Question

practice: predicting bonds
fill in the reference box below to help you with this practice. then use a periodic table that shows the electronegativities of the elements to fill in the table. you will need to calculate the difference in electronegativity in order to determine the type of bond you predict the two elements would form.
reference:
δen > 2 =
δen between 0.5 - 2 =
δen < 0.5 =

element #1	element #2	δen	type of bond predicted
c	o
li	f
n	n
k	br
ba	n
cs	f
k	cl
c	cl
p	cl
li	s
n	o
p	o
pb	o
he	he

electrons unit
© it’s not rocket science™ 2023

Explanation:

Step1: Analyze K and Br

Find electronegativity of K (0.8) and Br (2.8). Calculate $\Delta\text{EN} = |2.8 - 0.8| = 2$.

Step2: Determine bond type

Check reference: $\Delta\text{EN}=2$ is between 0.5 - 2? Wait, no—wait, reference says $\Delta\text{EN}>2$ is ionic, 0.5 - 2 is polar covalent, <0.5 is covalent. Wait, 2 is equal to the upper bound? Wait, original reference: $\Delta\text{EN}>2$ (ionic), 0.5 - 2 (polar covalent, inclusive? Wait, the table for Na and Cl: $\Delta\text{EN}=2.1$ (ionic, since 2.1>2). Wait, K and Br: $\Delta\text{EN}=2$. Wait, maybe the reference is $\Delta\text{EN}\geq2$ for ionic? Wait, Na (0.9) and Cl (3.0): $\Delta\text{EN}=2.1$ (ionic). K (0.8) and Br (2.8): $\Delta\text{EN}=2.0$. Let's check the reference again. The reference box: $\Delta\text{EN}>2$ = Ionic, 0.5 - 2 = Polar Covalent, <0.5 = Covalent. So 2 is in the 0.5 - 2 range? Wait, no—maybe a typo, but let's follow the table. Wait, Na and Cl: $\Delta\text{EN}=2.1$ (ionic, since 2.1>2). K and Br: $\Delta\text{EN}=2.0$. Wait, maybe the reference is $\Delta\text{EN}\geq2$ for ionic? Alternatively, maybe the user's table has K and Br: $\Delta\text{EN}=2$. Let's see the reference: if $\Delta\text{EN}>2$ is ionic, then 2 is not >2, so polar covalent? Wait, no—maybe the reference is $\Delta\text{EN}\geq2$ for ionic. Let's check another row: Ba (0.9) and N (3.0): $\Delta\text{EN}=2.1$ (ionic). Cs (0.7) and F (4.0): $\Delta\text{EN}=3.3$ (ionic). K (0.8) and Cl (3.0): $\Delta\text{EN}=2.2$ (ionic). So maybe the reference is $\Delta\text{EN}\geq2$ for ionic. So K and Br: $\Delta\text{EN}=2.0$ (ionic? Wait, no—wait, electronegativity of Br is 2.8, K is 0.8: 2.8 - 0.8 = 2.0. So according to the reference, $\Delta\text{EN}>2$ is ionic, so 2.0 is not >2, so polar covalent? But Na and Cl: 3.0 - 0.9 = 2.1 (>2, ionic). So K and Br: 2.0, which is equal to 2, so maybe polar covalent? Wait, but let's check the table's existing entries. For example, C and O: $\Delta\text{EN}=1$ (polar covalent). N and N: 0 (covalent). So K and Br: $\Delta\text{EN}=2$—so according to the reference, 0.5 - 2 (inclusive) is polar covalent? Wait, the reference says "between 0.5 - 2"—so 0.5 ≤ ΔEN ≤ 2 is polar covalent, >2 is ionic, <0.5 is covalent. So K and Br: ΔEN=2, so polar covalent? Wait, no—wait, maybe the electronegativity values are different. Wait, actual electronegativity: K is 0.8, Br is 2.8, so ΔEN=2.0. So according to the reference, ΔEN between 0.5 - 2 (so 0.5 ≤ ΔEN ≤ 2) is polar covalent. So K and Br: polar covalent? Wait, but Na and Cl: ΔEN=2.1 (ionic, since 2.1>2). So yes, 2.0 is in 0.5 - 2, so polar covalent? Wait, but maybe the reference is a typo, and >2 is ionic, ≥2 is ionic. But according to the given reference, we have to follow. So K and Br: ΔEN=2, so polar covalent? Wait, no—let's check the table's K and Cl: K (0.8) and Cl (3.0): ΔEN=2.2 (ionic, since 2.2>2). So K and Br: ΔEN=2.0, which is not >2, so polar covalent.

Step3: Analyze Ba and N

Ba (0.9) and N (3.0): ΔEN=2.1 (ionic, since 2.1>2).

Step4: Analyze Cs and F

Cs (0.7) and F (4.0): ΔEN=3.3 (ionic, since 3.3>2).

Step5: Analyze K and Cl

K (0.8) and Cl (3.0): ΔEN=2.2 (ionic, since 2.2>2).

Step6: Analyze C and Cl

C (2.5) and Cl (3.0): ΔEN=0.5 (polar covalent, since 0.5 is in 0.5 - 2).

Step7: Analyze P and Cl

P (2.1) and Cl (3.0): ΔEN=0.9 (polar covalent, 0.9 in 0.5 - 2).

Step8: Analyze Li and S

Li (1.0) and S (2.5): ΔEN=1.5 (polar covalent, 1.5 in 0.5 - 2).

Step9: Analyze N and O

N (3.0) and O (3.5): ΔEN=0.5 (polar covalent, 0.5 in 0.5 - 2).

Step10: Analyze P and O

P (2.1) and O (3.5): ΔEN=1.4 (polar covalent, 1.4 in 0.5 - 2).

Step11: Analyze Pb and O

Pb (1.0) and O (3.5): ΔEN=2.5 (ionic, since 2.5>2).

Step12: Analyze He and He

He (0) and He (0): ΔEN=0 (covalent, since 0 < 0.5).

Let's focus on K and Br:

Step1: Electronegativity Values

Electronegativity of K: 0.8, Br: 2.8.

Step2: Calculate ΔEN

$\Delta\text{EN} = |2.8 - 0.8| = 2.0$.

Step3: Compare with Reference

Reference: $\Delta\text{EN} > 2$ = Ionic; $0.5 \leq \Delta\text{EN} \leq 2$ = Polar Covalent; $\Delta\text{EN} < 0.5$ = Covalent.
Since $0.5 \leq 2.0 \leq 2$, the bond type is Polar Covalent.

Answer:

For K and Br, the type of bond predicted is Polar Covalent.