Chapter U2.32, Problem 9E

Interpretation Introduction

Interpretation:

Between CH₃ and CH₄ which molecule is more likely exist in nature needs to be determined.

Concept introduction:

All noble gases have full octet valence shell electronic configurations. Therefore, they have the highest stability in nature. All the other atoms or molecule try to gain its neighboring noble gas like configuration.

Octet rule: A molecule will form a stable arrangement when it has 8 valence electrons by gaining its neighboring noble gas like configuration. This is called the Octet rule.

Answer to Problem 9E

CH₄ molecule is most likely to exist in nature.

Explanation of Solution

CH₃ molecule consists of 1 Carbon and 3 H atoms.

The valence shell electronic configuration of C is $\left[\text{He}\right]2s^{2}2p^2$ .

So, the number of valence electron $=4$

Again Hydrogen has only $1e^{-}$ so, the number of valence electron is also 1.

3 Hydrogen has a total valence electron $=3$

The total number of valence electrons in the CH₃ molecule is $=4+(1\times 3)=7$

So CH₃ molecule does not follow the Octet rule. Therefore, it is an unstable molecule and it does not exist naturally.

On the other hand, the CH₄ molecule consists of 1 Carbon and 4 H atoms

The valence shell electronic configuration of C is $\left[\text{He}\right]2s^{2}2p^2$

So, the number of valence electron $=4$

Again, hydrogen has only $1e^{-}$ . So, the number of valence electrons is also =1.

4 Hydrogen has a total valence electron $=4$ .

The total number of valence electrons in the CH₃ molecule is $=4+(1\times 4)=8$

So, the CH₄ molecule follows the Octet rule. Therefore, it is a very stable molecule and it exists naturally.