Hybridization Analysis for Given Species
To determine the hybridization, we calculate the steric number (SN) for the central atom in each species. The steric number is the sum of the number of atoms bonded to the central atom and the number of lone pairs on the central atom. It can be calculated using the formula: SN = (Valence electrons of central atom + electrons from ligands - charge) / 2.
The hybridization corresponds to the steric number as follows:
- SN = 2: $sp$
- SN = 3: $sp^2$
- SN = 4: $sp^3$
- SN = 5: $sp^3d$
- SN = 6: $sp^3d^2$
- SN = 7: $sp^3d^3$
| Species | Central Atom | Valence e- | Bonds with F | Charge | SN Calculation (VE + Ligands - Charge) / 2 | Steric Number (SN) | Lone Pairs (SN - Bonds) | Hybridization |
| $XeF_6$ | Xe | 8 | 6 | 0 | (8 + 6 - 0) / 2 | 7 | 1 | $sp^3d^3$ |
| $BrF_6^+$ | Br | 7 | 6 | +1 | (7 + 6 - 1) / 2 | 6 | 0 | $sp^3d^2$ |
| $IF_5$ | I | 7 | 5 | 0 | (7 + 5 - 0) / 2 | 6 | 1 | $sp^3d^2$ |
| $XeF_4$ | Xe | 8 | 4 | 0 | (8 + 4 - 0) / 2 | 6 | 2 | $sp^3d^2$ |
Identifying Species without $sp^3d^2$ Hybridization
Based on the calculations:
- $BrF_6^+$ has a steric number of 6, resulting in $sp^3d^2$ hybridization.
- $IF_5$ has a steric number of 6, resulting in $sp^3d^2$ hybridization.
- $XeF_4$ has a steric number of 6, resulting in $sp^3d^2$ hybridization.
- $XeF_6$ has a steric number of 7, resulting in $sp^3d^3$ hybridization.
Therefore, $XeF_6$ is the species where $sp^3d^2$ hybridization is not associated.
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