Question

Given LiCl, NaCl, KCl, MgCl2, and SrCl2, arrange the compounds in order of increasing lattice energy.

Answer 1

The order of the given compounds in their increasing lattice energy is as follows:

$\boxed{{\text{KCl}} < {\text{NaCl}} < {\text{LiCl}} < {\text{SrC}}{{\text{l}}_2} < {\text{MgC}}{{\text{l}}_{\text{2}}}}$

Further explanation:

Lattice energy:

The amount of energy released when ions are combined to form an ionic compound or the energy required to break the ionic compound into its constituent gaseous ions is known as lattice energy. It can have positive and negative values. It cannot be measured directly and therefore is denoted by $\Delta {{\text{H}}_{{\text{lattice}}}}$.

Factors on which lattice energy depends:

1. Charge on ions: It is directly related to the lattice energy. High charges on the ions increase the electrostatic forces of attraction and therefore lattice energy also increases.

2. Size of ions: It is inversely related to the lattice energy. As the ionic size increases, the distance between the nuclei also increases and therefore lattice energy decreases.

LiCl, NaCl, and KCl have the same anion $\left( {{\text{C}}{{\text{l}}^ - }} \right)$. The charges on Li, Na, K are also same (+1). So the lattice energy is decided by the size of the ions. Li, Na, and K are present in the same group of the periodic table. Li is present above Na and Na is present above K. The size of Li-ion is the smallest and that of K ion is the largest. Therefore, the lattice energy of LiCl is the highest, followed by NaCl and KCl will have the least lattice energy among these three compounds.

${\text{MgC}}{{\text{l}}_2}$ and ${\text{SrC}}{{\text{l}}_2}$ also have the same anion $\left( {{\text{C}}{{\text{l}}^ - }} \right)$. But these cations have higher charges than those of LiCl, NaCl, and KCl so their lattice energies are higher. The size of Mg is smaller than that of Sr so the lattice energy of ${\text{MgC}}{{\text{l}}_2}$ will be more than that of ${\text{SrC}}{{\text{l}}_2}$.

The resultant order of the given compounds in their increasing lattice energy is as follows:

${\text{KCl}} < {\text{NaCl}} < {\text{LiCl}} < {\text{SrC}}{{\text{l}}_2} < {\text{MgC}}{{\text{l}}_{\text{2}}}$

Learn more:

1. Identify the exothermic change: brainly.com/question/1875234

2. Ranking of elements according to first ionization energy: brainly.com/question/1550767

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Solid state

Keywords: lattice energy, LiCl, NaCl, KCl, MgCl2, SrCl2, anion, Cl-, size of ions, charge on ions¸ increase, decrease.

Answer 2

Answer:-  $KCl$

Explanations:- Lattice energy depends on two factors, charge and size.

High charge and small size gives higher lattice energy where as low charge and bigger size gives lower lattice energy.

in LiCl, NaCl and KCl, the anion is same and also the charges for Li, Na and K are also same. The deciding factor here is the size of cations. Since the size increases as we move down a group, the order of size of these three atoms is Li<Na<K.

The order of lattice energy is exactly opposite as it's increases as the size decreases.

Now, if we look at magnesium chloride and strontium chloride then again the anion is common but the metals have higher charge as compared to the alkali metals(Li, Na and K). So, lattice energy values must be higher for these two compounds. If we compare Mg and Sr then size of Mg is smaller and so the lattice energy would be greater for this.

Hence, the increasing order of lattice energy is $KCl$ .