Question

xenon fluoride compound has xenon 53.5 %.what is the oxiation state of xenon.?

Answer 1

xenon fluoride compound has xenon 53.5 %. what is the oxidation state of xenon?

**Solution:**

If we permit we have 100 g of xenon fluoride, we will have 53.5 g of xenon and 46.5 g (100-53.5) of fluorine. The equivalent of fluorine is:

$$E(F) = \frac{Ar(F)}{valency} = \frac{19.0}{1} = 19.0 \, \text{g/mol}.$$

We use the equivalent's law, to determine the equivalent of xenon:

$$\frac{m(Xe)}{m(F)} = \frac{E(Xe)}{E(F)}, \text{ hence } E(Xe) = \frac{m(Xe) \cdot E(F)}{m(F)} = \frac{53.5 \cdot 19.0}{46.5} = 21.9 \, \text{g/mol}$$

The atom mass of xenon is 131.3 g/mol, so the valency of xenon in this compound is

$$E(Xe) = \frac{131.3}{21.9} = 6.$$

The compound is XeF₆.

**Answer:** in compound XeF₆ the oxidation state (valency) of xenon is 6.