Chapter 5 Source Answers and Examples

This ion has the configuration of Aluminum, but without the 3 highest-energy electrons.

There are two ways of drawing this configuration.

Option 1:

Let’s start by identifying the electron configuration of Aluminum:

$\begin{aligned} \underline{\uparrow\downarrow} && \underline{\uparrow\downarrow}\ && \underline{\uparrow\downarrow}\ \underline{\uparrow\downarrow}\ \underline{\uparrow\downarrow} && \underline{\uparrow\downarrow}\ && \underline{\uparrow\phantom{\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}} \\ 1s^2 \phantom{1}&& 2s^2 \phantom{1}&& 2p^6 \phantom{\uparrow\downarrow1}&& 3s^2 \phantom{1}&& 3p^1 \phantom{\uparrow\downarrow1} \end{aligned}$

Now we take out the top three electrons, the two in the 3s orbital and the one in the 3p:

$\begin{aligned} \underline{\uparrow\downarrow} && \underline{\uparrow\downarrow}\ && \underline{\uparrow\downarrow}\ \underline{\uparrow\downarrow}\ \underline{\uparrow\downarrow} \\ 1s^2 \phantom{1}&& 2s^2 \phantom{1}&& 2p^6 \phantom{\uparrow\downarrow1} \end{aligned}$

This means our electron configuration is:

1s22s22p6!

Option 2:

Since we know that this Aluminum lost three electrons (as the charge is positive), let’s look at the periodic table, and go three steps left from Aluminum. We go Al -> Mg -> Na ->Ne. Therefore, we need to draw the electron configuration of Neon.

Neon has the electron configuration of:

$\begin{aligned} \underline{\uparrow\downarrow} && \underline{\uparrow\downarrow}\ && \underline{\uparrow\downarrow}\ \underline{\uparrow\downarrow}\ \underline{\uparrow\downarrow} \\ 1s^2 \phantom{1}&& 2s^2 \phantom{1}&& 2p^6 \phantom{\uparrow\downarrow1} \end{aligned}$

This is equivalent to the electron configuration of Al+3!

Either way, we achieve the same final configuration.

So, the first thing we need to figure out is how many bonds phosphorus has. We notice that it is attached to 5 chlorine. Since phosphorus would not have any electron pairs, our magic number is 5. This 5 will be the number of hybrid orbitals we need. Going through the list, we find that all the spn orbitals aren’t sufficient, but once we jump up into the d orbital, we find that sp3d gives us the five orbitals we need.

Therefore, our answer to the first part is sp3d.

When drawing the electron configuration for the hybridized Phosphorus we need to start with its base configuration:

$\begin{aligned} \underline{\uparrow\downarrow}&& \underline{\uparrow\phantom{\downarrow}}\ \underline{\uparrow\phantom{\downarrow}}\ \underline{\uparrow\phantom{\downarrow}} \\ 3s \phantom{1}&& 3p \phantom{37378^.} \end{aligned}$

Now, we need to pull one of the d orbitals down and combine it with the s orbital and all the p orbitals. As we took from the d orbitals, we need to show the rest, even though they are empty. Let’s just leave it empty for now:

$\begin{aligned} \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ && \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}} \\ 3sp^3d \phantom{25645^5375} && 3d \phantom{8472345^6} \end{aligned}$

Now, we have 5 electrons to distribute between these orbitals. They will fill up the lowest energy orbitals first, meaning the empty sp3d orbitals.

Putting the electrons in gives us our final answer:

$\begin{aligned} \underline{\uparrow\phantom{\downarrow}}\ \underline{\uparrow\phantom{\downarrow}}\ \underline{\uparrow\phantom{\downarrow}}\ \underline{\uparrow\phantom{\downarrow}}\ \underline{\uparrow\phantom{\downarrow}}\ && \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}}\ \underline{\phantom{\uparrow\downarrow}} \\ 3sp^3d \phantom{25645^5375} && 3d \phantom{8472345^6} \end{aligned}$