Question #9c01a

1 Answer
Apr 28, 2017

Answer:

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^10##5s^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)