# Question 9c01a

Apr 28, 2017

#### Answer:

No two will have used the same number of d-orbitals. However, $I C {l}_{2}^{+}$ and $I C {l}_{4}^{-}$ will each have two non-bonded electron pairs.

#### Explanation:

To apply the Valence Bond Theory, determine 1. the number of bonded pairs and 2. number of non-bonded pairs. Sum of BPrs + NBPrs => Number of Hybrids needed from valence structure. (Remember, Valence structure will be the highest principle quantum number of the electron configuration.)

Bonded Pairs (BPrs) = Number of elements attached to central element (we'll call these 'substrates'). Simply look at subscript after substrate element => number of bonded pairs.

Non-Bonded Pairs (NBPrs) = (Valence Number - Substrate Number) / 2
=> Valence Number is total number of valence electrons. If structure is cation, also subtract 1 ${e}^{-}$ to account for + charge or, if anion, add 1 ${e}^{-}$ to account for - charge.

=> Substrate Number is total electrons in valence of substrates when bonded. This will be '8' for all non-metals except hydrogen which will be '2'.

Number of Hybrids needed (all single bonds) = BPrs + NBPrs

$I C {l}_{2}^{+}$ => 2 Bonded Pr & 2 Non-Bonded Pr => 4 ${e}^{-}$pair => 4 hybrids needed.

I[Kr]$4 {d}^{10}$$5 {s}^{2}$$5 {p}_{x}^{2}$${p}_{y}^{2}$$5 {p}_{z}^{1}$ => I[Kr](4d^105s^2$5 {p}_{x}^{2}$${p}_{y}^{1}$5p_z^1)^+ + ${e}^{-}$
=> I[Kr][$4 {d}^{10}$$5 {\left(s {p}_{3}\right)}^{2}$${\left(s {p}_{3}\right)}^{2}$${\left(s {p}_{3}\right)}^{1}$(sp_3)^1]^+#

4 $\left(s {p}_{3}\right)$ Hybrid Orbitals => $A {X}_{4}$ Geometry => Tetrahedron with 2 bonded pair of electrons (Chloride substrates) and 2 non-bonded pair of electrons. (No d-orbitals used)

Apply same sequence to remaining formulas of interest.
$I C {l}_{2}^{-}$ => 2 Bonded Pr; 3 NBPr => Trigonal Bipyrimid Parent $\left(A {X}_{5}\right)$ => $\left(A {X}_{2} {E}_{3}\right)$ Geometry (Linear). (1 d-orbital used)

$I {F}_{7}$ => 7 BPrs; 0 NBPr => $A {X}_{7}$ Parent (Pentagonal Bipyrimid) which is also the structural geometry. (3 d-orbitals used)

$I C {l}_{4}^{-}$ => 4 BPrs; 2 NBPrs => $A {X}_{6}$ Parent (Octahedron) => $A {X}_{4} {E}_{2}$ Geometry (Square Planar). (2 d-orbitals used)