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Question

Geometry of the Xenon oxyfluoride (XeOF4) is________


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Solution

The geometry of Xenon oxyfluoride (XeOF4)

  1. For determining the geometry we first need to draw the lewis dot structure of XeOF4
  2. First knowing the valence electrons present in the chemical is necessary in order to draw the Lewis dot structure.
  3. In XeOF4 Valence electrons=Xe+O+4×F
  4. Electronic configuration of Xe with atomic number=54 is1s22s22p63s23p63d104s24p64d105s25p6, valence electron=8
  5. Electronic configuration of oxygen with atomic number=16 is 1s22s22p4, valence electron=6
  6. Electronic configuration of chlorine with atomic number=17 is 1s22s22p63s23p5, valence electron=7
  7. Total valence electron=Xe+O+4×F8+6+4×742
  8. Therefore, these 42 valence electrons must be distributed among the elements such that each one can attain its octet. In the given compound, Xenon is an exception because it can expand its octet.
  9. The steric number can be used to predict hybridization. Steric number = sigma bond number + lone pair number
  10. There are five sigma bonds and one lone pair present in XeOF4, therefore the steric number is5+1=6 , and the hybridization will besp3d2.
  11. Geometry of xenon oxyfluoride left XeOF4 right is class 11 chemistry CBSE
  12. The hybrid orbitals will be created by the center atom using a combination of its s, p, and as many of its d orbitals as are required. In order to formXeOF4, Xenon must have its s orbital, all three of its p-orbitals, and two of its d-orbitals. As a result, Xenon will be in the hybridization state sp3d2 and its shape will be square pyramidal.

Therefore, Xenon oxyfluoride ((XeOF4) molecular geometry is square pyramidal.


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