Question #96fc8

Jan 24, 2017

It is easy to remember

Explanation:

The main idea you have to remember is that the elements in the left and right most columns of the periodic table have fixed valency. Hence they have more or less constant charge. Some elements do possess more than one ionic state like N (+3, +5).

So to calculate the ionic state of iron in $F {e}_{2} {\left(S {O}_{4}\right)}_{3}$ let us start with known elements.

S has oxidation state of +6, if in doubt go for the parent acid and calculate from that state. e.g. ${H}_{2} S {O}_{4}$.

H has charge of +1, O has charge of -2.
So we have 4 Oxygen molecules of each -2 charge. Hence the balance equation would be

$2 + x + 4 \left(- 2\right) = 0$
$x - 6 = 0$
$x = + 6$

Now move on to $F {e}_{2} {\left(S {O}_{4}\right)}_{3}$.

We have 2 Fe atoms, 3 S atoms and 12 O atoms. Hence balance equation is

$2 y + 3 \left(+ 6\right) + 12 \left(- 2\right) = 0$
$2 y - 6 = 0$
$y = + 3$

Thus ionic state of Fe in $F {e}_{2} {\left(S {O}_{4}\right)}_{3}$ is +3.

Another way to do this is by remembering how molecular formula is written.

$F {e}^{x} S {O}_{4}^{- 2} \setminus \rightarrow F {e}_{2} {\left(S {O}_{4}\right)}_{x}$ Now you compare with the formula and you get +3.