Chapter 2, Problem 1P

1. The numbers of orbiting electrons in aluminum and silver are 13 and 47, respectively. Draw the electronic configuration for each, and discuss briefly why each is a good conductor.
2. Using the Internet, find the atomic structure of gold and explain why it is an excellent conductor of electricity.

To determine

(a)

To draw:

The electronic configuration of aluminum and silver. Also, state the reason due to which both aluminum and copper are considered a good conductor.

Explanation of Solution

Given:

The numbers of orbiting electrons in aluminum and silver are $13$ and $47$ respectively.

The arrangement of electron inside the atom and outside the nucleus follows a particular pattern. There is a shell structure in which all electron falls and the energy of the electrons decide which shell to fall. The number of electrons in each shell is always $2n^2$, where $n$ is the shell number.

In shell $K(n=1)$, the number of electron hold is only $2\times 1^2=2$.

In Shell $L(n=2)$, the number of electrons that can hold is $2\times 2^2=2\times 4=8$.

Inside shell, actually electron falls in sub-shell. The name of sub-shell are $s,p,d$ and $f$.

In a nutshell, the larger the value of $n$ the more number of electrons a shell can hold. In this case, some of the electrons may fall in the sub-shell, which are away from nucleus like $d\text{and}f$. In these sub-shell, electrons are loosely bound so they have the higher mobility which results in higher conductivity. In addition, if an electron is not in paired, then also it has higher level of mobility.

The electronic configuration of any element is based on below pattern:

For aluminum, we know that the numbers of orbital electrons are 13.

So, arrange the electrons will be as below

  $\begin{array}{l}{\text{1s}}^{\text{2}}\\ {\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}\\ {\text{3s}}^{\text{2}}{\text{3p}}^{\text{1}}\end{array}$

The electronic configuration of Aluminum is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{1}}$.

Here we can see, Aluminum has one free electron in its valence shell. Therefore, this free electron can move easily and hence, this causes the higher conductivity for Aluminum.

For silver, we know that the numbers of orbital electrons are 47.

So, arrange the electrons will be as follows:

  $\begin{array}{l}{\text{1s}}^{\text{2}}\\ {\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}\\ {\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}\\ {\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}\\ {\text{5s}}^{\text{1}}\end{array}$

The electronic configuration of silver is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{6}}{\text{5s}}^{\text{1}}{\text{4d}}^{\text{10}}$.

Silver is a good conductor because it has free electrons in its valence shell. It is clearly seen from the electronic configuration of silver ( ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{6}}{\text{5s}}^{\text{1}}{\text{4d}}^{\text{10}}$ ) that s orbit is holding only one electron and maximum limit is up to 2 electrons. This means that it has free electrons in its valence shell. In addition, the electrons in $d\text{and}f$ sub-shells are always loosely bound and it increases the conductivity. Hence, we can say silver is highly conductive.

To determine

(b)

The atomic structure of the gold. Also, state the reason due is an excellent conductor of electricity.

Explanation of Solution

Atomic structure of gold.

The electronic configuration of any element is based on below pattern:

For gold, we know that the numbers of orbital electrons are $79$.

So, arrange the electrons for gold will be as follows:

  $\begin{array}{l}{\text{1s}}^{\text{2}}\\ {\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}\\ {\text{3s}}^{\text{2}}{\text{3p}}^{\text{1}}{\text{3d}}^{\text{10}}\\ {\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{4f}}^{\text{14}}\\ {\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{5d}}^{\text{10}}\\ {\text{6s}}^{\text{1}}\end{array}$

The electronic configuration of gold is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{6}}{\text{5s}}^{\text{2}}{\text{4d}}^{\text{10}}{\text{5p}}^{\text{6}}{\text{6s}}^{\text{1}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}$.

Gold is a good conductor of electricity because it has free electrons in its valence shell.

It is clearly seen from the electronic configuration of gold that, s orbit is holding only one electron and maximum limit is up to $2$ electrons, which means it has free electrons in its valence shell. In addition, the electrons in $d\text{and}f$ sub-shells are always loosely bound and it increases the conductivity. Hence, we can say gold is highly conductive.