Problem 39 - Chemistry An Atoms First Approach

Question

Predict the type of bond (ionic, covalent, or polar covalent) one would expect to form between the following pairs of elements. a. \(\mathrm{Rb}\) and \(\mathrm{Cl}\) b. \(S\) and \(S\) c. \(C\) and \(F\) d. Ba and S e. \(\mathrm{N}\) and \(\mathrm{P}\) f. \(B\) and \(H\)

Step-by-Step Solution

Answer

a. Rb and Cl form an ionic bond. b. S and S form a covalent bond. c. C and F form a polar covalent bond. d. Ba and S form an ionic bond. e. N and P form a polar covalent bond. f. B and H form a covalent bond.

Step 1: Finding Electronegativity Values

Find the electronegativity values of Rb and Cl using a periodic table or other resources. For Rb, the value is 0.82, and for Cl, the value is 3.16.

Step 2: Calculate the Electronegativity Difference

Subtract the lower value from the higher value. In this case, 3.16 - 0.82 = 2.34.

Step 3: Determine the Bond Type

As the electronegativity difference is greater than 1.7, the bond between Rb and Cl is ionic. b. S and S

Step 1: Finding Electronegativity Values

Find the electronegativity values of S, which is 2.58.

Step 2: Calculate the Electronegativity Difference

As both elements are the same, the difference is zero (2.58 - 2.58 = 0).

Step 3: Determine the Bond Type

Since the electronegativity difference is less than 0.4, the bond between S and S is covalent. c. C and F

Step 1: Finding Electronegativity Values

Find the electronegativity values of C and F. For C, the value is 2.55, and for F, the value is 3.98.

Step 2: Calculate the Electronegativity Difference

Subtract the lower value from the higher value. In this case, 3.98 - 2.55 = 1.43.

Step 3: Determine the Bond Type

As the electronegativity difference is between 0.4 and 1.7, the bond between C and F is polar covalent. d. Ba and S

Step 1: Finding Electronegativity Values

Find the electronegativity values of Ba and S. For Ba, the value is 0.89, and for S, the value is 2.58.

Step 2: Calculate the Electronegativity Difference

Subtract the lower value from the higher value. In this case, 2.58 - 0.89 = 1.69.

Step 3: Determine the Bond Type

Since the electronegativity difference is greater than 1.7, the bond between Ba and S is ionic. e. N and P

Step 1: Finding Electronegativity Values

Find the electronegativity values of N and P. For N, the value is 3.04, and for P, the value is 2.19.

Step 2: Calculate the Electronegativity Difference

Subtract the lower value from the higher value. In this case, 3.04 - 2.19 = 0.85.

Step 3: Determine the Bond Type

As the electronegativity difference is between 0.4 and 1.7, the bond between N and P is polar covalent. f. B and H

Step 1: Finding Electronegativity Values

Find the electronegativity values of B and H. For B, the value is 2.04, and for H, the value is 2.20.

Step 2: Calculate the Electronegativity Difference

Subtract the lower value from the higher value. In this case, 2.20 - 2.04 = 0.16.

Step 3: Determine the Bond Type

Since the electronegativity difference is less than 0.4, the bond between B and H is covalent.

Key Concepts

Electronegativity

Electronegativity is the ability of an atom to attract electrons towards itself in a chemical bond. It acts like a tug-of-war between atoms over shared electrons.
Elements vary in their electronegativities, and these values are available on the periodic table.

• High electronegativity means an element strongly attracts electrons.
• Low electronegativity means it weakly attracts electrons.

The differences in electronegativity values between two bonding atoms greatly influence the type of chemical bond they form.
This is why understanding electronegativity is crucial for predicting the nature of a chemical bond between different elements.

Ionic Bond

An ionic bond occurs when the electronegativity difference between two atoms exceeds a certain threshold, typically more than 1.7. This substantial difference in electronegativity values indicates that one atom can dominate and transfer electrons from another.
This transfer leads to the formation of electronic charges, known as ions, where:

• The atom gaining electrons becomes a negatively charged ion (anion).
• The atom losing electrons becomes a positively charged ion (cation).

For instance, consider Rb and Cl. With an electronegativity difference of 2.34, Rb donates an electron to Cl, thus creating an ionic bond.

Covalent Bond

Covalent bonds arise when two atoms share electrons to achieve a more stable electronic configuration.
Such bonds typically form between atoms with similar electronegativities, resulting in a difference of less than 0.4.
This sharing ensures that neither atom has a strong inclination to completely take the electron from the other, leading to a balanced distribution.
A classic example would be S and S, as they both possess the same electronegativity. Their shared electrons create a pure covalent bond.

Polar Covalent Bond

A polar covalent bond is a situation where the difference in electronegativity between two atoms is between 0.4 and 1.7.
This difference creates a bond where the electron distribution is unequal.

• The more electronegative atom attracts the shared electrons more strongly, gaining a slight negative charge.
• The less electronegative atom becomes slightly positive as a result of an unequal electron cloud distribution.

Examples include C and F, where a difference of 1.43 leads to electron sharing that is skewed towards F, making the bond polar covalent. Polar covalent bonds are essential in the creation of molecules with unique properties, like water.