Chapter 1, Problem 1EQ

Hydrogen is a Group I element and each hydrogen will contribute I valence electron. Carbon is a Group (Roman numeral) element and each carbon will contribute ___ (number) electrons. Every oxygen atom in a compound will contribute valence electrons.

Chloromethane has the molecular formula ${\text{CH}}_{\text{3}}\text{Cl}$. Its skeleton is

$\begin{array}{l}H\\ HCCl\\ H\end{array}$

and the number of valence electrons may be determined as follows. There are three hydrogen atoms, each of which contributes 1 valence electron; the single carbon contributes 4 electrons; and the single chlorine atom contributes 7 electrons, making a total of 14. A convenient tabular form of this calculation is

$\begin{array}{cccc}3& H& 1\times 3=& 3\\ 1& C& 4\times 1=& 4\\ 1& Cl& 7\times 1=& 7\\ & & & \overline{14}\end{array}$

Interpretation Introduction

Interpretation:

The group number and the valance electrons in the given elements C and O are to be identified.

Concept Introduction:

In an atom, the electrons revolve around the nucleus in orbits/shells.

The shells are numbered/named from the side of nucleus. The last shell from the side of nucleus is called as ultimate shell or valance shell.

Answer to Problem 1EQ

Carbon is a Group IV element and each carbon will contribute 4 electrons. Every oxygen atom in a compound will contribute 6 valence electrons.

Explanation of Solution

The of electrons present in the last shell (valance shell) are called as valance electrons. The number of valance electrons is equal to group number.

As Carbon is group IV element, it has 4 electrons in the last shell.

In the same manner, Oxygen belongs to group VI and has 6 electrons in the last shell.

Conclusion

The number of electrons present in the last shell of an atom are considered as valance electrons and they can be identified by group number of the element.