Balance the following equation: XeF 6 + H 2 O → XeO 2 F 2 + HF

Given, equation is XeF 6 + H 2 O → XeO 2 F 2 + HF First balance the fluorine atoms on both sides of the equation followed by oxygen and hydrogen atoms. The balanced chemical equation is: XeF 6 + 2 H 2 O → XeO 2 F 2 + 4 HF Note: XeF 6 is a powerful fluorinating agent. They are readily hydrolysed even by traces of water. It also sublimes readily at 298 K .

How are XeF 2 and XeF 4 prepared? Give their structures.

Xenon difluoride is prepared by the combination of xenon and excess of fluorine at 673 K and 1 bar pressure. Xe + F 2 → 1 bar 673 K XeF 2 s Xenon tetrafluoride is prepared by the combination of xenon and excess of fluorine in the ratio of 1 : 5 at 873 K and 7 bar pressure as given below: Xe + 2 F 2 → 7 bar 873 K XeF 4 s XeF 2 has a linear structure with bond angle of 180 ° . Xenon atom has two bond pairs and three lone pairs of electrons. XeF 4 has square planar structure with bond angle of 90 ° . Xenon atom has four bond pairs and two lone pairs of electrons. The structures are as given below:

Why are halogens coloured.

Most of the halogens are coloured because they absorb radiations which are in the visible region which in turn excite the valence electrons to higher energy levels. As the amount of energy which is required to excite the electrons to a higher level is different for different halogens, each halogen has a different colour.