Chapter 7, Problem 68A

Interpretation Introduction

Interpretation:

The compound formed between elements C (Rb) and D (N) needs to be determined.

Concept introduction:

Molecules are bound together by different types of bonds namely ionic and covalent. Ionic bonds are formed when there is complete transfer of electrons from one atom to another. These atoms either lose or gain electrons to become negatively or positively charged ions. The forces of attraction between these ions causes the ionic bond formation.

Answer to Problem 68A

The compound formed between elements C and D is ${\text{Rb}}_{\text{3}}\text{N}$.

Explanation of Solution

The given periodic table wherein elements C and D are marked.

From the table, it can be seen that element C belongs to group 1 while element D belongs to group 15.

Element C is Rubidium and element D is Nitrogen.

Rubidium has atomic number of 37 with electronic configuration as $\left[\text{Kr}\right]5{\text{s}}^1$. There is one outermost electron which Rubidium will lose to form a Rubidium ion, ${\text{Rb}}^{+}$.

Nitrogen has atomic number of 7 with electronic configuration as ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^3$. It has 3 valence electrons and requires 3 more electron to get octet structure.

Hence, the compound formed between Rubidium and Nitrogen is ${\text{Rb}}_{\text{3}}\text{N}$

Conclusion

The compound between element C and element D is ${\text{Rb}}_{\text{3}}\text{N}$ as element C is Rubidium and element D is Nitrogen.