# In oxide X2O7 element X has the highest oxidation state. This element in a compound with hydrogen makes up 99.219%. What is this element?

Jul 21, 2016

Here's what I got.

#### Explanation:

The idea here is that for two atoms that are covalently bonded, oxidation numbers are assigned by assuming that the more electronegative of the two atoms "takes" all the bonding electrons shared by two atoms.

For a given element, the highest oxidation state occurs when all of its valence electrons are "taken" by the atoms to which it is covalently bonded.

In this case, element $\text{X}$ has a $\textcolor{b l u e}{+ 7}$ oxidation state in ${\text{X"_2"O}}_{7}$, since you have

stackrel(color(blue)(?))("X")_ 2 stackrel(color(blue)(-2))("O")_ 7

which implies

2 xx color(blue)(?) + 7 xx (-2) = 0

color(blue)(?) = 14/2 = color(blue)(+7)

Now, what this means is that your unknown element has $7$ valence electrons, so right from the start you know that you're looking for either a transition metal or a halogen.

The first thing to do here is check the halogens by using the fact that element $\text{X}$ forms a compound with hydrogen in which it holds a 99.219% percent composition.

Since halogens and hydrogen form compounds that have the general form $\text{HX}$, you know that one mole of this compound will contain one mole of hydrogen and one mole of element $\text{X}$.

If you take ${M}_{M}$ to be the molar mass of element $\text{X}$, you can write

(M_Mcolor(red)(cancel(color(black)("g mol"^(-1)))))/((1 xx M_M + 1 xx 1.00794) color(red)(cancel(color(black)("g mol"^(-1))))) xx 100 = 99.219%

This will get you

$100 \cdot {M}_{M} = 99.219 \cdot {M}_{M} + 99.219 \cdot 1.00794$

Rearrange to find

$0.781 \cdot {M}_{M} = 100.0068 \implies {M}_{M} = \frac{100.0068}{0.781} = 128.05$

This means that you element has a molar mass of ${\text{128.05 g mol}}^{- 1}$. The closest match to this value is the molar mass of iodine, $\text{I}$, which is listed as

M_("M I") = "126.9045 g mol"^(-1)

So my guess would be that your unknown element is iodine, which would make your oxide

${\text{I"_2"O}}_{7} \to$ diiodine heptoxide

The problem here is that this is a hypothetical compound, meaning that it doesn't really exist, at least as far as I know.

For example, dehydrating periodic acid, ${\text{H"_5"IO}}_{6}$, which is the highest oxoacid of iodine, i.e. here iodine has a $\textcolor{b l u e}{+ 7}$ oxidation state, produces diiodine pentoxide, ${\text{I"_2"O}}_{5}$, as opposed to ${\text{I"_2"O}}_{7}$.