# Question #64b32

Jul 18, 2015

I'd say the answer is (1) one.

#### Explanation:

Ionization energy is defined as the energy needed to remove one or more electrons from a mole of isolated atoms in the gaseous state.

This process leads to the formation of a mole positive ions, or cations. Ionization energies are given in kJ per mole.

So, you know that your metal's first four ionization energies are

• 212 kJ/mole $\to$ the first electron;
• 1326 kJ/mole $\to$ the second electron;
• 1537 kJ/mole $\to$ the third electron;
• 1886 kJ/mole $\to$ the fourth electron.

The most important thing to notice here is that there's a noticeable difference between the first ionization energy, which is the energy needed to remove the most loosely held electron from a mole of atoms in the gaseous state, and the second, third, and fourth ionization energies.

In fact, the ${2}^{\text{nd}}$, ${3}^{\text{rd}}$, and ${4}^{\text{th}}$ ionization energies are quite similar in magnitude. This tells you that these electrons probably come from the same energy shell.

If you consider the fact that you need only 212 kJ/mole to remove the first electron, then it's easy to conclude that the first electron is located in a higher energy shell.

The facts that this electron is located further away from the nucleus when compared with the other three, and that it benefits from better screening from the nucleus, justify the relatively small energy needed to remove it from the atom.

By comparison, the other three electrons are closer to the nucleus, since they occupy a lower energy level, and benefit from less screening.

In simple terms, the first electron is more loosely held than the following three electrons, which is why it can be removed with less energy.

As a result, the electron located in the outermost shell of the atom, which in your case is of course the first one, acts as a valence electron.

The metal thus has one valence electron.