Q
QuestionChemistry

Consider the molecule, XeF^4. The electron domain geometry (EDG) is Question Blank 1 of 2 The molecular geometry (MG) is Options: seesaw octahedral trigonal bipyramidal square planar square pyramidal tetrahedral
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Answer

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Step 1:
: Determine the electron configuration of Xe and F.

Its electron configuration is $$[He] 2s^{2} 2p^{5}$$.
Xenon (Xe) is in the 6th period and has an atomic number of 54. Fluorine (F) is in the 2nd period and has an atomic number of 9.

Step 2:
: Calculate the total valence electrons in the XeF^4 molecule.

Xenon contributes 8 valence electrons (2 from 6s and 6 from 6p). Each fluorine contributes 7 valence electrons (2 from 2s and 5 from 2p). There are 4 fluorine atoms in XeF^4, so they contribute a total of 28 valence electrons (7 electrons/fluorine * 4 fluorines). The total number of valence electrons in XeF^4 is 8 (from Xe) + 28 (from F) = 36 valence electrons.

Step 3:
: Determine the arrangement of electron domains around Xe.

Xenon has 8 valence electrons, so it needs to form 4 bonds to achieve an octet. In XeF^4, xenon forms 4 bonds with 4 fluorine atoms. The electron domain geometry (EDG) around Xe is trigonal bipyramidal, with 3 equatorial positions and 2 axial positions.

Step 4:
: Calculate the number of electron domains around each atom.

Each fluorine atom forms a single bond with xenon, so there are 1 electron domain around each fluorine atom. Xenon forms 4 bonds with fluorine atoms, so there are 4 electron domains around xenon.

Step 5:
: Determine the molecular geometry (MG) based on the EDG and the number of electron domains around each atom.

In the trigonal bipyramidal EDG, when there are 4 electron domains around the central atom (xenon), the molecular geometry is a seesaw.

Final Answer

The molecular geometry (MG) of XeF^4 is a seesaw.