Question

The formation of CsCl from Cs(s) and Cl2 (g) involves the following steps:
Cs(s) -> Cs(g)
1/2Cl2(g) -> Cl(g)
Cs(g) -> Cs + (g) + e –
Cl(g) + e - -> Cl - (g)
Cs+(g) + Cl - (g) -> CsCl(s)
Which of these steps absorb energy and which release energy?

Answer 1

Explanation:

1) $Cs(s)\rightarrow Cs(g)$

Energy will be absorbed by the Cs(s) to get converted into Cs(g). In order to break the strong association of particles present in solid to convert it into gaseous state energy will be needed. So, energy will be absorbed by the Cs solid in this reaction.

2) $\frac{1}{2}Cl_2(g)\rightarrow Cl(g)$

Energy will be absorbed by $Cl_2$ gas in order to break bond between two Cl atoms to form isolated(alone) single chlorine atom.So, the energy will be absorbed by the chlorine gas molecule in this reaction.

3) $Cs(g)\rightarrow Cs^+(g)+e^-$

Energy will be absorbed by the Cs(g) to get converted into $Cs^+(g)$ cation. In order to remove an electron from the outer most shell of Cs atom energy will be required by Cs atom, So, energy will be absorbed in this reaction.

4) $Cl(g)+e^-\rightarrow Cl^-(g)$

Energy will get release in formation of $Cl^-$ anion from Cl atom in gaseous state. Chlorine atom need one electron to attain noble gas configuration. So, when an electron is added to the outer most shell of chlorine it attains stability of fully filled outermost shell by which it releases energy on addition of an electron.

5) $Cs^+(g)+Cl^-(g)\rightarrow CsCl(s)$

Energy will get release on formation of CsCl(s) from$Cs^+$ cation and $Cl^-$ anion. since both are oppositely charged ion and due to strong electrostatic interaction will get converted into stable molecule of CsCl (s) with release in an energy.