Question

The hydrogen atom is not actually electronegative enough to form bonds to xenon. were the xenon-hydrogen bond to exist, what would be the structure of xeh4? double-click any atom and type xe to change the label. draw the molecule by placing atoms on the grid and connecting them with bonds. show all lone pairs of electrons.

Answer 1

Answer : The structure of $XeH_4$ will be square-planar.

Explanation :

In the given molecule $XeH_4$, 'Xe' is the central atom and 'H' is the terminal atom.

Xenon has 8 valence electrons and hydrogen has 1 valence electron. Therefore, the total number of valence electrons are 8 + 4(1) = 12 electrons.

The number of electrons used in Xe-H bonding = 8 electrons

The remaining electrons which are used as lone pair on central atom (Xe) = 12 - 8 = 4 electrons

There are 4 bonding pairs and 2 lone pairs of electrons, they will be arranged in the octahedral arrangement around the central atom with 2 lone pairs of electrons on central atom. The lone pairs are arranged linearly across the central atom. The resulting structure will be square-planar.

The structure of $XeH_4$ is shown below.