# Which metal will form binary halide, "MX"_2 ?

## ELEMENT I II IE1 520 900 IE2 7300 1760

Jul 26, 2016

Element II.

#### Explanation:

The problem essentially wants you to figure out which of the two elements given are located in group 2 of the periodic table by examining the first and second ionization energies.

The problem provides you with the following data

$\textcolor{w h i t e}{a a a a a a a a \textcolor{b l a c k}{\boldsymbol{\text{Element I")aaaaaaaaaacolor(black)(bb"Element II}}} a a a a}$

color(purple)("IE"_1)color(white)(aaaaacolor(black)("520 kJ mol"^(-1))aaaaaaaacolor(black)("900 kJ mol"^(-1))aaa)

color(purple)("IE"_2)color(white)(aaaacolor(black)("7300 kJ mol"^(-1))aaaaaaacolor(black)("1760 kJ mol"^(-1))aaa)

As you know, ionization energy is a term used to denote the amount of energy needed in order to remove the outermost electron from one mole of atoms in the gaseous state to form one mole of cations.

The first ionization energy, ${\text{IE}}_{1}$, is defined as the energy needed to remove the outermost electron from one mole of atoms in the gaseous state to form one mole of $1 +$ cations

${\text{M"_ ((g)) + "IE"_ 1 -> "M"_ ((g))^(+) + "e}}^{-}$

The second ionization energy, ${\text{I}}_{2}$< is defined as the energy needed to remove the outermost electron from one mole of $1 +$ cations in the gaseous state to form one mole of $2 +$ cations

${\text{M"_ ((g))^(+) + "IE"_ 2 -> "M"_ ((g))^(2+) + "e}}^{-}$

Now, in order for an element to form ${\text{MX}}_{2}$ compounds, it must form $2 +$ cations. The idea here is that the element that will require the least amount of energy to go from neutral, $\text{M}$, to $2 +$ cation, ${\text{M}}^{2 +}$, will most likely form ${\text{MX}}_{2}$ compounds.

Notice that there's a significant jump between ${\text{IE}}_{1}$ and ${\text{IE}}_{2}$ for Element I. This tells you that the second electron that is being removed from the atom is located much closer to the nucleus than the first electron.

You can thus look at these values and say that Element I has $1$ valence electron, i.e. one electron in its outermost shell. This element will form $\text{MX}$ halides.

By comparison, Element II has a relatively small difference between ${\text{IE}}_{1}$ and ${\text{IE}}_{2}$, which tells you that both electrons are located on the same energy level, i.e. at the same distance from the nucleus.

The third ionization energy is not given, but you can presume that it will be significantly higher than ${\text{IE}}_{2}$.

This means that Element II is probably located in group 2 of the periodic table, which of course implies that it forms ${\text{MX}}_{2}$ halides.

As a conclusion, you can say that Element I will require a total of

${E}_{\text{total 1" = "IE"_1 + "IE}} _ 2$

${E}_{\text{total 1" = "520 kJ mol"^(-1) + "7300 kJ mol"^(-1) = "7820 kJ mol}}^{- 1}$

By comparison, Element II will require

${E}_{\text{total 2" = "900 kJ mol"^(-1) + "1760 kJ mol"^(-1) = "2660 kJ mol}}^{- 1}$

As you can see, you need significantly less energy to get Element II to form $2 +$ cations.