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

Aug 3, 2017

Apparently, ${\text{ICl}}_{4}^{-}$. It uses $s {p}^{3} {d}^{2}$ hybridization (i.e. octahedral electron geometry), for a square planar molecular geometry (i.e. two lone pairs).

Well, using the electron-counting method,

A)

${\text{ICl}}_{2}^{+}$ has...

• $7$ valence electrons from $\text{I}$
• $7 \times 2 = 14$ valence electrons total from $\text{Cl}$
• $- 1$ valence electrons due to the charge

$\implies 7 + 14 - 1 = 20$ valence electrons.

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

The $s {p}^{3}$ hybridization here requires zero $d$ orbitals, but there exist two lone pairs around $\text{I}$.

B)

${\text{ICl}}_{2}^{-}$ has...

• $7$ valence electrons from $\text{I}$
• $7 \times 2 = 14$ valence electrons total from $\text{Cl}$
• $+ 1$ valence electrons due to the charge

$\implies 7 + 14 + 1 = 22$ valence electrons.

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

The $s {p}^{3} d$ hybridization here (for $1 + 3 + 1 = \boldsymbol{\text{five}}$ electron groups) requires one $d$ orbital, but there exist three lone pairs around $\text{I}$.

C)

${\text{IF}}_{7}$ has...

• $7$ valence electrons from $\text{I}$
• $7 \times 7 = 49$ valence electrons total from $\text{F}$

$\implies 7 + 49 = 56$ valence electrons.

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

The hybridization here (for $\boldsymbol{\text{seven}}$ electron groups) will be $s {p}^{3} {d}^{3}$. Specifically, it will involve the linear combination of...

$s + {p}_{z} + \left({p}_{x} , {p}_{y}\right) + {d}_{{z}^{2}} + \left({d}_{{x}^{2} - {y}^{2}} , {d}_{x y}\right) \to s {p}^{3} {d}^{3}$

It requires three $d$ orbitals, but there exist zero lone pairs around $\text{I}$.

D)

${\text{ICl}}_{4}^{-}$ has...

• $7$ valence electrons from $\text{I}$
• $7 \times 4 = 28$ valence electrons total from $\text{Cl}$
• $1$ valence electron from the charge

$\implies 7 + 28 + 1 = 36$ valence electrons.

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

The hybridization here (for $\text{six}$ electron groups) will be $s {p}^{3} {d}^{2}$.

It requires two $d$ orbitals, and there exist two lone pairs around $\text{I}$.