# Are CH_4 and XeF_4 the same shape? Provide lewis structures to prove your argument.

Jul 5, 2016

No. The number of electron groups differ, hence they are different electron geometries (and molecular geometries).

${\text{CH}}_{4}$ is the most common example of a tetrahedral molecule.

• Carbon atom contributes 4 valence electrons
• Each hydrogen atom contributes 1 valence electron

Therefore, ${\text{CH}}_{4}$ utilizes 8 valence electrons; 2 for each single bond.

So, ${\text{CH}}_{4}$ has four electron groups and no nonbonding electrons, giving it a tetrahedral geometry.

${\text{XeF}}_{4}$ consists of xenon and fluorine, and:

• Xenon atom contributes 8 valence electrons
• Each fluorine atom contributes 7 valence electron

Therefore, ${\text{XeF}}_{4}$ utilizes 36 valence electrons; 2 for each single bond, 6 nonbonding electrons on each fluorine, and 4 nonbonding electrons on xenon.

Indeed, 4$\times$2 + 4$\times$6 + 4 = 36, so the structure is: