Question

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

Answer 1

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).

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Well, using the electron-counting method,

`A)`

`"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 `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"`.