An analysis of iron oxide shows that its formula is $Fe_{1.9}O_3$. What percentages of iron exists as $Fe^{3+}$ and $Fe^{4+}$ ions:

$Fe^{3+}=84.21\%$ and $Fe^{4+}=15.79\%$

The correct answer is Option 2. $Fe^{3+}=84.21\%$ and $Fe^{4+}=15.79\%$

The formula of the iron oxide is given as \(\text{Fe}_{1.9}\text{O}_3\), which suggests that the total number of iron atoms is 1.9. We are tasked with determining the percentages of iron that exist as \(\text{Fe}^{3+}\) and \(\text{Fe}^{4+}\).

Oxygen atoms (\(\text{O}^{2-}\)) contribute a total negative charge of \(3 \times (-2) = -6\).

The total positive charge from iron atoms must balance this negative charge, meaning the total charge from iron should be \(+6\).

Let \(x\) be the number of iron atoms in the \(\text{Fe}^{3+}\) oxidation state, and \(1.9 - x\) be the number of iron atoms in the \(\text{Fe}^{4+}\) oxidation state.

The total charge from \(\text{Fe}^{3+}\) ions is \(x \times (+3)\).

The total charge from \(\text{Fe}^{4+}\) ions is \((1.9 - x) \times (+4)\).

Since the total positive charge must be +6 to balance the -6 from oxygen, we can write the equation:

\(3x + 4(1.9 - x) = 6\)

Solving for \(x\):

\(3x + 7.6 - 4x = 6\)

\(-x + 7.6 = 6\)

\(x = 1.6\)

So, \(x = 1.6\) atoms of iron are in the \(\text{Fe}^{3+}\) oxidation state, and the remaining \(1.9 - 1.6 = 0.3\) atoms are in the \(\text{Fe}^{4+}\) oxidation state.

Percentage of \(\text{Fe}^{3+}\):

\(\left( \frac{1.6}{1.9} \right) \times 100 = 84.21\%\)

Percentage of \(\text{Fe}^{4+}\):

\(\left( \frac{0.3}{1.9} \right) \times 100 = 15.79\%\)

Thus, the correct answer is: Option 2: Fe\(^{3+}\) = 84.21\% and Fe\(^{4+}\) = 15.79\%