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

1 Answer
Truong-Son N.
Jul 5, 2016

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

#"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, #"CH"_4# utilizes 8 valence electrons; 2 for each single bond.

So, #"CH"_4# has four electron groups and no nonbonding electrons, giving it a tetrahedral geometry.

#"XeF"_4# consists of xenon and fluorine, and:

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

Therefore, #"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#xx#2 + 4#xx#6 + 4 = 36, so the structure is:

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