Question

Question #9c01a

Answer 1

No two will have used the same number of d-orbitals. However, `ICl_2^+` and `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^-` 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

`ICl_2^+` => 2 Bonded Pr & 2 Non-Bonded Pr => 4 `e^-`pair => 4 hybrids needed.

I[Kr]`4d^10` `5s^2` `5p_x^2` `p_y^2` `5p_z^1` => I[Kr](#4d^105s^2# `5p_x^2` `p_y^1` `5p_z^1)^+` + `e^-`
=> I[Kr][`4d^10` `5(sp_3)^2` `(sp_3)^2` `(sp_3)^1` `(sp_3)^1]^+`

4 `(sp_3)` Hybrid Orbitals => `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.
`ICl_2^-` => 2 Bonded Pr; 3 NBPr => Trigonal Bipyrimid Parent `(AX_5)` => `(AX_2E_3)` Geometry (Linear). (1 d-orbital used)

`IF_7` => 7 BPrs; 0 NBPr => `AX_7` Parent (Pentagonal Bipyrimid) which is also the structural geometry. (3 d-orbitals used)

`ICl_4^-` => 4 BPrs; 2 NBPrs => `AX_6` Parent (Octahedron) => `AX_4E_2` Geometry (Square Planar). (2 d-orbitals used)