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popcornrox37

Answer:

Explanation:

So when we consider lattice energy we are looking at two different things:

1. The magnitude of the product of the cation and anion charges

2. The relative size of each cation and anion

However, the product of the charges takes priority over the size, so we will consider the charge first.

So lets look at A

For a we have NaCl, RbBr and CsBr

Na has a charge of +1 and Cl has a charge of -1 so the product of those two numbers is -1, taking the magnitude gives us 1.

Secondly, Rb has a charge of +1 and Br has a charge of -1 so the product of those two numbers is -1, taking the magnitude gives us 1.

Finally, Cs has a charge of +1 and Br has a charge of -1 so the product of those two numbers is -1, taking the magnitude gives us 1.

So as you can see there is no difference between our charge magnitudes which means we now have to consider size.

The smaller the size of the cations the stronger the lattice structure will be. Think of it like as we add more orbitals (n levels) the distance between the electrons and protons becomes larger and the force between them is weaker. However, in a smaller atom this is much stronger as we have less distance between the nucleus and the electrons.

So lets look at a periodic table: (here im using period number as "n")

Na and Cl are both in level n = 2 meaning they are pretty small (Cl is smaller due to effective nuclear charge)

Rb is in n = 4 and Br is in n = 3

Cs is in n = 5 and Br is in n = 3

So we can see that the smallest atoms are Na and Cl and therefore they have the highest lattice energy.

Both Rb and Cs have the anion Br- but Cs is bigger so can conclude that

NaCl > RbBr > CsBr

For b and c I will include the calculations and size comparisons if needed but the same process is repeated

b) Charge first:

BaO = +2 x -2 = -4 ---> 4 (magnitude)

KF = +1 x -1 = -1 ---> 1 (magnitude)

Here we dont need to consider size because BaO has a higher charge magnitude. Therefore, BaO > KF

c) Charge First

SrO = +2 x -2 = -4 ---> 4 (magnitude)

SrCl2 = +2 x -1 = -2 ----> 2 (magnitude)

Dont get tripped up here because it looks like Cl would have a total charge of -2 but we only consider the charge on the anion/cation itself not its subscript in a compound.

Since charge magnitude of SrO > SrCl2

The lattice energy of SrO > SrCl2

Let me know if you have any questions about this!

pstnonsonjoku

The lattice energy depends on the size of ions.

We define lattice energy as the energy released when a crystal lattice is formed from it's constituents understand standard conditions.

The lattice energy depends on the size of the ions present. The smaller the ions, the greater the lattice energy.

Based on this, we can see that in the first case, Na^+ is the smallest ion hence NaCl has largest lattice energy. In the second group, Ba^2+ and O^2- are smaller than K^+ and F^- hence BaO has a larger lattice energy. In the last group, the O^2- ion is smaller than the Cl^- hence SrO has a greater lattice energy.

Learn more:https://brainly.com/question/13694878

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