Question #9c01a

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Question #9c01a

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Answer 1 · 2017-04-28T23:29:05

No two will have used the same number of d-orbitals. However, $I C l_{2}^{+}$ $ICl_2^+$ and $I C l_{4}^{-}$ $ICl_4^-$ 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 $e^{-}$ $e^-$ to account for + charge or, if anion, add 1 $e^{-}$ $e^-$ 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

$I C l_{2}^{+}$ $ICl_2^+$ => 2 Bonded Pr & 2 Non-Bonded Pr => 4 $e^{-}$ $e^-$ pair => 4 hybrids needed.

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

4 $(s p_{3})$ $(sp_3)$ Hybrid Orbitals => $A X_{4}$ $AX_4$ 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.
$I C l_{2}^{-}$ $ICl_2^-$ => 2 Bonded Pr; 3 NBPr => Trigonal Bipyrimid Parent $(A X_{5})$ $(AX_5)$ => $(A X_{2} E_{3})$ $(AX_2E_3)$ Geometry (Linear). (1 d-orbital used)

$I F_{7}$ $IF_7$ => 7 BPrs; 0 NBPr => $A X_{7}$ $AX_7$ Parent (Pentagonal Bipyrimid) which is also the structural geometry. (3 d-orbitals used)

$I C l_{4}^{-}$ $ICl_4^-$ => 4 BPrs; 2 NBPrs => $A X_{6}$ $AX_6$ Parent (Octahedron) => $A X_{4} E_{2}$ $AX_4E_2$ Geometry (Square Planar). (2 d-orbitals used)

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Question #9c01a

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