Which of the following has an octahedral electron geometry? #(A)# #"ICl"_2^+#, #(B)# #"ICl"_2^-#, #(C)# #"IF"_7#, #(D)# #"ICl"_4^-#
1 Answer
Apparently,
Well, using the electron-counting method,
#"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 = 20# valence electrons.
And so, we form the skeletal structure and distribute valence electrons to get a bent molecular geometry:
The
#"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
#"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
#s + p_z + (p_x, p_y) + d_(z^2) + (d_(x^2-y^2), d_(xy)) -> sp^3d^3#
It requires three
#"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
It requires two