Question

Atoms involved in bonds are usually more stable than they would be otherwise. This means that energy is released in the formation of the bond. For ionic bonds, Coulomb's law is used to calculate this energy:

E=14πϵ0q1q2d=kq1q2d

where E
is energy in joules (J
), k
is a constant equal to 1/(4πϵ0)=8.99×109 (J⋅m)/C2
, q1
and q2
are the charges of the two ions, and d
is the distance between the nuclei of the two ions in meters (m
). The charges on the ions (q1
and q2
) are expressed in units of electrical charge, coulombs (C
), and are calculated using the charge on the ion and the charge on an electron (1.60×10−19C
).

While the bond energy calculated by Coulomb's law pertains to a single ionic bond, lattice energy pertains to all the ionic bonding among a group of atoms arranged in a crystal lattice. Lattice energy is calculated by using the Coulomb's law equation, but with a different constant (unique to each substance) that takes into account the crystalline structure of the substance.

By convention, lattice energy is defined as the amount required to either break an ionic solid into individual gaseous ions, or to form an ionic solid from gaseous ions.

Part A
Calculate the amount of energy released in the formation of one mole of BeS
bonds (not lattice energy). The radius of the beryllium ion is 0.31 Å
, and the radius of the sulfide ion is 1.84 Å
. Note that 1Å=10−10m
.
Express your answer with the appropriate units to three significant figures.
View Available Hint(s)for Part A
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Energy released =

Part B
Arrange the following ionic compounds in order of decreasing amount of energy released in lattice formation: LiF
, CaO
, AlN
, and RbBr
.
Rank from greatest to least lattice energy. To rank items as equivalent, overlap them.
View Available Hint(s)for Part B

Answer 1

The energy released in the formation of one mole of BeS bonds is approximately 639.8 kJ/mol. Lattice energy is generally higher for substances with higher ionic charges and smaller ionic radii. AlN likely has the highest lattice energy among the given compounds, followed by CaO, LiF, and RbBr.

Step-by-step explanation:

To calculate the amount of energy released in the formation of one mole of BeS bonds, we use Coulomb's law, where E is the energy, k is Coulomb's constant (8.99×109 (J·m)/C2), q1 and q2 are the charges of the ions (in coulombs), and d is the distance between the nuclei of the ions (in meters). The formula becomes E = k × q1 × q2 / d. For Be2+ and S2-, the radii given convert to 3.1×10-11 m and 1.84×10-10 m respectively, and so d, the sum of the radii, is 2.15×10-10 m. The charges are twice the charge of an electron since both ions are divalent. Thus:

E = (8.99×109) × (2 × 1.60×10-19C)(2 × 1.60×10-19C) / 2.15×10-10 m

After calculating, the energy released for one BeS bond is approximately 10.63×10-19 J. To find the energy for one mole, multiply this value by Avogadro's number (6.022×1023 mol-1), which gives approximately 639.8 kJ/mol.

For Part B, estimating the lattice energy of ionic compounds requires considering the charges on the ions and the distance between them. Generally, compounds with higher ionic charges and smaller ionic radii will have the highest lattice energy. Without specific lattice energy values, a prediction can be made that AlN would have the highest, followed by CaO, then LiF, and RbBr would have the lowest lattice energy.