Chapter 7, Problem 49A

Interpretation Introduction

Interpretation:

The number of valence electrons of elements labelled from A to G in the periodic table needs to be determined.

Concept introduction:

Periodic table is tabular display of elements which have their atomic number in increasing order. Moreover, the elements are grouped in such a way that those elements have similar electronic configuration and same properties.

Answer to Problem 49A

The elements A is Aluminum which will become ${\text{Al}}^{\text{3+}}$, B is Barium which will become ${\text{Ba}}^{\text{2+}}$, C is Rubidium which will become ${\text{Rb}}^{\text{+}}$, D is Nitrogen which will become ${\text{N}}^{\text{3-}}$, E is Iodine which will become ${\text{I}}^{\text{-}}$, F is Argon which will not form any ion, and G is Selenium which will become ${\text{Se}}^{\text{2-}}$.

Explanation of Solution

The given periodic table of elements are given below.

Comparing with the periodic table

1. Element A belongs to Group 13 and is identified as Aluminum. It has electronic configuration as ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{1}}$. This element will tend to lose 3 electrons to get octet configuration. Hence, $\text{Al}$ will lose 3 electrons to form ${\text{Al}}^{\text{3+}}$
2. Element B belongs to Group 2 and is identified as Barium. It has electronic configuration as $\left[\text{Xe}\right]{\text{ 6s}}^2$. It will lose two electrons to become ${\text{Ba}}^{\text{2+}}$
3. Element C belongs to Group 1 and is identified as Rubedium. It has electronic configuration as $\left[\text{Kr}\right]{\text{ 5s}}^1$. It will lose one electron to become ${\text{Rb}}^{\text{+}}$
4. Element D belongs to Group 15 and is identified as Nitrogen. It has electronic configuration as ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{3}}$. It will gain 3 electrons to become ${\text{N}}^{\text{3-}}$ .
5. Element E belongs to Group 17 and is identified as Iodine. It has electronic configuration as $\left[\text{Kr}\right]{\text{ 4d}}^{10}{\text{5s}}^2{\text{5p}}^5$. It will gain 1 electron to become ${\text{I}}^{\text{-}}$ .
6. Element Fbelongs to Group 18 and is identified as Argon. It has electronic configuration as $\left[\text{Ne}\right]3{\text{s}}^2{\text{3p}}^6$. It will not lose or gain electrons as it has octet.
7. Element G belongs to Group 16 and is identified as Selenium. It has electronic configuration as $\left[\text{Ar}\right]{\text{ 3d}}^{10}4{\text{s}}^2{\text{4p}}^4$. It will gain 2 electrons to become ${\text{Se}}^{\text{2-}}$ .

Conclusion

The elements A is Aluminum which will become ${\text{Al}}^{\text{3+}}$, B is Barium which will become ${\text{Ba}}^{\text{2+}}$, C is Rubidium which will become ${\text{Rb}}^{\text{+}}$, D is Nitrogen which will become ${\text{N}}^{\text{3-}}$, E is Iodine which will become ${\text{I}}^{\text{-}}$, F is Argon which will not form any ion, and G is Selenium which will become ${\text{Se}}^{\text{2-}}$.