Which of the following has an octahedral electron geometry? #(A)# #"ICl"_2^+#, #(B)# #"ICl"_2^-#, #(C)# #"IF"_7#, #(D)# #"ICl"_4^-#

1 Answer
Truong-Son N.
Aug 3, 2017

Apparently, #"ICl"_4^(-)#. It uses #sp^3d^2# hybridization (i.e. octahedral electron geometry), for a square planar molecular geometry (i.e. two lone pairs).

Well, using the electron-counting method,

#A)#

#"ICl"_2^(+)# has...

#=> 7 + 14 - 1 = 20# valence electrons.

And so, we form the skeletal structure and distribute valence electrons to get a bent molecular geometry:

The #sp^3# hybridization here requires zero #d# orbitals, but there exist two lone pairs around #"I"#.

#B)#

#"ICl"_2^(-)# has...

  • #7# valence electrons from #"I"#
  • #7 xx 2 = 14# valence electrons total from #"Cl"#
  • #+1# valence electrons due to the charge

#=> 7 + 14 + 1 = 22# valence electrons.

And so, we form the skeletal structure and distribute valence electrons to get a linear molecular geometry:

The #sp^3d# hybridization here (for #1 + 3 + 1 = bb"five"# electron groups) requires one #d# orbital, but there exist three lone pairs around #"I"#.

#C)#

#"IF"_7# has...

  • #7# valence electrons from #"I"#
  • #7 xx 7 = 49# valence electrons total from #"F"#

#=> 7 + 49 = 56# valence electrons.

And so, we form the skeletal structure and distribute valence electrons to get a pentagonal bipyramidal molecular geometry:

The hybridization here (for #bb"seven"# electron groups) will be #sp^3d^3#. Specifically, it will involve the linear combination of...

#s + p_z + (p_x, p_y) + d_(z^2) + (d_(x^2-y^2), d_(xy)) -> sp^3d^3#

It requires three #d# orbitals, but there exist zero lone pairs around #"I"#.

#D)#

#"ICl"_4^(-)# has...

  • #7# valence electrons from #"I"#
  • #7 xx 4 = 28# valence electrons total from #"Cl"#
  • #1# valence electron from the charge

#=> 7 + 28 + 1 = 36# valence electrons.

And so, we form the skeletal structure and distribute valence electrons to get a square planar molecular geometry:

The hybridization here (for #"six"# electron groups) will be #sp^3d^2#.

It requires two #d# orbitals, and there exist two lone pairs around #"I"#.

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