Calculate the difference in electronegativity (ΔEN) for the bonds that would form between the
groups of atoms listed below. State whether the bond would be ionic, polar covalent or non-polar
covalent.
a. N and Cl
b. B and Br
c. B and F
d. Al and O
e. N and F
f. C and H
g. Mg and I
h. Li and S
i. C and F
j. Na and N
To calculate the difference in electronegativity (ΔEN) for the bonds, we can use the Pauling electronegativity values. The larger the difference in electronegativity, the more polar the bond will be.
Here are the Pauling electronegativity values for the atoms involved:
N: 3.04
Cl: 3.16
B: 2.04
Br: 2.96
F: 3.98
Al: 1.61
O: 3.44
C: 2.55
H: 2.20
Mg: 1.31
I: 2.66
Li: 0.98
S: 2.58
Na: 0.93
a. ΔEN = |EN(Cl) - EN(N)| = |3.16 - 3.04| = 0.12
The difference in electronegativity is small, so the bond between Cl and N would be non-polar covalent.
b. ΔEN = |EN(Br) - EN(B)| = |2.96 - 2.04| = 0.92
The difference in electronegativity is moderate, so the bond between Br and B would be polar covalent.
c. ΔEN = |EN(F) - EN(B)| = |3.98 - 2.04| = 1.94
The difference in electronegativity is large, so the bond between F and B would be ionic.
d. ΔEN = |EN(O) - EN(Al)| = |3.44 - 1.61| = 1.83
The difference in electronegativity is large, so the bond between O and Al would be ionic.
e. ΔEN = |EN(F) - EN(N)| = |3.98 - 3.04| = 0.94
The difference in electronegativity is moderate, so the bond between F and N would be polar covalent.
f. ΔEN = |EN(H) - EN(C)| = |2.20 - 2.55| = 0.35
The difference in electronegativity is small, so the bond between H and C would be non-polar covalent.
g. ΔEN = |EN(I) - EN(Mg)| = |2.66 - 1.31| = 1.35
The difference in electronegativity is large, so the bond between I and Mg would be ionic.
h. ΔEN = |EN(S) - EN(Li)| = |2.58 - 0.98| = 1.60
The difference in electronegativity is large, so the bond between S and Li would be ionic.
i. ΔEN = |EN(F) - EN(C)| = |3.98 - 2.55| = 1.43
The difference in electronegativity is large, so the bond between F and C would be polar covalent.
j. ΔEN = |EN(N) - EN(Na)| = |3.04 - 0.93| = 2.11
The difference in electronegativity is large, so the bond between N and Na would be ionic.
To calculate the difference in electronegativity (ΔEN), we can use the Pauling Electronegativity scale. The scale ranges from 0.7 (the electronegativity of cesium) to 4.0 (the electronegativity of fluorine). The larger the difference in electronegativity, the more likely the bond will be ionic or polar covalent.
Here are the answers for each group of atoms:
a. N (electronegativity: 3.04) and Cl (electronegativity: 3.16)
ΔEN = |3.16 - 3.04| = 0.12
The difference in electronegativity is small, so this bond is considered polar covalent.
b. B (electronegativity: 2.04) and Br (electronegativity: 2.96)
ΔEN = |2.96 - 2.04| = 0.92
The difference in electronegativity is moderate, so this bond is considered polar covalent.
c. B (electronegativity: 2.04) and F (electronegativity: 3.98)
ΔEN = |3.98 - 2.04| = 1.94
The difference in electronegativity is large, so this bond is considered ionic.
d. Al (electronegativity: 1.61) and O (electronegativity: 3.44)
ΔEN = |3.44 - 1.61| = 1.83
The difference in electronegativity is large, so this bond is considered ionic.
e. N (electronegativity: 3.04) and F (electronegativity: 3.98)
ΔEN = |3.98 - 3.04| = 0.94
The difference in electronegativity is moderate, so this bond is considered polar covalent.
f. C (electronegativity: 2.55) and H (electronegativity: 2.20)
ΔEN = |2.55 - 2.20| = 0.35
The difference in electronegativity is small, so this bond is considered non-polar covalent.
g. Mg (electronegativity: 1.31) and I (electronegativity: 2.66)
ΔEN = |2.66 - 1.31| = 1.35
The difference in electronegativity is large, so this bond is considered ionic.
h. Li (electronegativity: 0.98) and S (electronegativity: 2.58)
ΔEN = |2.58 - 0.98| = 1.60
The difference in electronegativity is large, so this bond is considered ionic.
i. C (electronegativity: 2.55) and F (electronegativity: 3.98)
ΔEN = |3.98 - 2.55| = 1.43
The difference in electronegativity is large, so this bond is considered polar covalent.
j. Na (electronegativity: 0.93) and N (electronegativity: 3.04)
ΔEN = |3.04 - 0.93| = 2.11
The difference in electronegativity is large, so this bond is considered ionic.
To calculate the difference in electronegativity (ΔEN) and determine the type of bond formed between two atoms, you need to understand electronegativity values and the range of values that correspond to different types of bonds.
1. Determine the electronegativity values for the atoms involved in each bond:
You can find a periodic table that includes electronegativity values or search for a table online. Some commonly used values are:
N (Nitrogen): 3.04
Cl (Chlorine): 3.16
B (Boron): 2.04
Br (Bromine): 2.96
F (Fluorine): 3.98
Al (Aluminum): 1.61
O (Oxygen): 3.44
Mg (Magnesium): 1.31
I (Iodine): 2.66
Li (Lithium): 0.98
S (Sulfur): 2.58
C (Carbon): 2.55
Na (Sodium): 0.93
2. Calculate the difference in electronegativity (ΔEN):
ΔEN = |Electronegativity of atom 1 - Electronegativity of atom 2|
3. Determine the type of bond:
- If ΔEN is less than 0.5, the bond is non-polar covalent.
- If ΔEN is between 0.5 and 1.7, the bond is polar covalent.
- If ΔEN is greater than 1.7, the bond is ionic.
Now, let's calculate the ΔEN for each bond:
a. N and Cl:
ΔEN = |3.04 - 3.16| ≈ 0.12
The bond is non-polar covalent.
b. B and Br:
ΔEN = |2.04 - 2.96| ≈ 0.92
The bond is polar covalent.
c. B and F:
ΔEN = |2.04 - 3.98| ≈ 1.94
The bond is ionic.
d. Al and O:
ΔEN = |1.61 - 3.44| ≈ 1.83
The bond is ionic.
e. N and F:
ΔEN = |3.04 - 3.98| ≈ 0.94
The bond is polar covalent.
f. C and H:
ΔEN = |2.55 - 2.20| ≈ 0.35
The bond is non-polar covalent.
g. Mg and I:
ΔEN = |1.31 - 2.66| ≈ 1.35
The bond is polar covalent.
h. Li and S:
ΔEN = |0.98 - 2.58| ≈ 1.60
The bond is polar covalent.
i. C and F:
ΔEN = |2.55 - 3.98| ≈ 1.43
The bond is polar covalent.
j. Na and N:
ΔEN = |0.93 - 3.04| ≈ 2.11
The bond is ionic.
In summary, the types of bonds are as follows:
a. N and Cl: non-polar covalent
b. B and Br: polar covalent
c. B and F: ionic
d. Al and O: ionic
e. N and F: polar covalent
f. C and H: non-polar covalent
g. Mg and I: polar covalent
h. Li and S: polar covalent
i. C and F: polar covalent
j. Na and N: ionic