# Question #06543

Nov 27, 2015

Here's what I got.

#### Explanation:

You are not wrong, the questions does indeed require you to use ionization energies.

As you know, ionization energy describes the energy needed to remove one mole of electrons from one mole of atoms in the gaseous state.

${\text{X " + color(blue)(" energy") color(white)(x) -> " X"^(+) + " e}}^{-}$

That represents the equation for the first ionization energy of a generic element $\text{X}$. The process can continue until all the electrons are removed from the atom.

The total energy needed to form an ${\text{Al}}^{2 +}$ cation will be the sum of the individual ionization steps required to get to that charge.

More specifically, you will have

${\text{Al" + color(blue)(E_1) -> "Al"^(+) + "e}}^{-}$

${\text{Al"^(+) + color(blue)(E_2) -> "Al"^(2+) + "e}}^{-}$

The first two ionization energies for aluminium are

This means that the total energy needed to form the ${\text{Al}}^{3 +}$ cation will be

${E}_{\text{total}} = \textcolor{b l u e}{{E}_{1}} + \textcolor{b l u e}{{E}_{2}}$

${E}_{\text{total" = 577.6 + 1816.6 = "2394 kJ/mol}}$

Now all you need to do is check the ionization energies for the transition metals given to you. Since the transition metal ion carries a $2 +$ charge, it follows that you need to add up the first two ionization energies.

http://www.theperiodicelements.com/elements/view/Co/ie.html

As it turns out, cobalt is indeed the answer, since it requires a total ionization energy of

${E}_{\text{total" = 760 + 1648 = "2408 kJ/mol}}$

to form the ${\text{Co}}^{2 +}$ cation. The values are not exactly the same, but they're the closest match I could find.

SIDE NOTE I suspect that the data given to you in the booklet will point to an exact match between the two energy values.