Which of the following has an octahedral electron geometry? $(A)$ $(A)$ ${ICl}_{2}^{+}$ $"ICl"_2^+$ , $(B)$ $(B)$ ${ICl}_{2}^{-}$ $"ICl"_2^-$ , $(C)$ $(C)$ ${IF}_{7}$ $"IF"_7$ , $(D)$ $(D)$ ${ICl}_{4}^{-}$ $"ICl"_4^-$

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Which of the following has an octahedral electron geometry? $(A)$ $(A)$ ${ICl}_{2}^{+}$ $"ICl"_2^+$ , $(B)$ $(B)$ ${ICl}_{2}^{-}$ $"ICl"_2^-$ , $(C)$ $(C)$ ${IF}_{7}$ $"IF"_7$ , $(D)$ $(D)$ ${ICl}_{4}^{-}$ $"ICl"_4^-$

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Answer 1 · 2017-08-03T00:18:04

Apparently, ${ICl}_{4}^{-}$ $"ICl"_4^(-)$ . It uses $s p^{3} d^{2}$ $sp^3d^2$ hybridization (i.e. octahedral electron geometry), for a square planar molecular geometry (i.e. two lone pairs).

Well, using the electron-counting method,

$A)$ $A)$

${ICl}_{2}^{+}$ $"ICl"_2^(+)$ has...

$7$ $7$ valence electrons from $I$ $"I"$
$7 \times 2 = 14$ $7 xx 2 = 14$ valence electrons total from $Cl$ $"Cl"$
$- 1$ $-1$ valence electrons due to the charge

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

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

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

$B)$ $B)$

${ICl}_{2}^{-}$ $"ICl"_2^(-)$ has...

$7$ $7$ valence electrons from $I$ $"I"$
$7 \times 2 = 14$ $7 xx 2 = 14$ valence electrons total from $Cl$ $"Cl"$
$+ 1$ $+1$ valence electrons due to the charge

$\Rightarrow 7 + 14 + 1 = 22$ $=> 7 + 14 + 1 = 22$ valence electrons.

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

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

$C)$ $C)$

${IF}_{7}$ $"IF"_7$ has...

$7$ $7$ valence electrons from $I$ $"I"$
$7 \times 7 = 49$ $7 xx 7 = 49$ valence electrons total from $F$ $"F"$

$\Rightarrow 7 + 49 = 56$ $=> 7 + 49 = 56$ valence electrons.

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

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

$s + p_{z} + (p_{x}, p_{y}) + d_{z^{2}}^{2} + (d_{x^{2} - y^{2}}^{2}, d_{x y}) \to s p^{3} d^{3}$ $s + p_z + (p_x, p_y) + d_(z^2) + (d_(x^2-y^2), d_(xy)) -> sp^3d^3$

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

$D)$ $D)$

${ICl}_{4}^{-}$ $"ICl"_4^(-)$ has...

$7$ $7$ valence electrons from $I$ $"I"$
$7 \times 4 = 28$ $7 xx 4 = 28$ valence electrons total from $Cl$ $"Cl"$
$1$ $1$ valence electron from the charge

$\Rightarrow 7 + 28 + 1 = 36$ $=> 7 + 28 + 1 = 36$ valence electrons.

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

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

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

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Which of the following has an octahedral electron geometry? $(A)$ $(A)$ ${ICl}_{2}^{+}$ $"ICl"_2^+$ , $(B)$ $(B)$ ${ICl}_{2}^{-}$ $"ICl"_2^-$ , $(C)$ $(C)$ ${IF}_{7}$ $"IF"_7$ , $(D)$ $(D)$ ${ICl}_{4}^{-}$ $"ICl"_4^-$

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Which of the following has an octahedral electron geometry? (A)(A) ICl+2"ICl"_2^+, (B)(B) ICl−2"ICl"_2^-, (C)(C) IF7"IF"_7, (D)(D) ICl−4"ICl"_4^-

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Which of the following has an octahedral electron geometry? $(A)$ $(A)$ ${ICl}_{2}^{+}$ $"ICl"_2^+$ , $(B)$ $(B)$ ${ICl}_{2}^{-}$ $"ICl"_2^-$ , $(C)$ $(C)$ ${IF}_{7}$ $"IF"_7$ , $(D)$ $(D)$ ${ICl}_{4}^{-}$ $"ICl"_4^-$