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Chapter 10: Problem 62

Predict the geometric shapes of (a) \(\mathrm{CO} ;\) (b) \(\mathrm{SiCl}_{4}\); (c) \(\mathrm{PH}_{3} ;\) (d) \(\mathrm{ICl}_{3} ;\) (e) \(\mathrm{SbCl}_{5} ;\) (f) \(\mathrm{SO}_{2} ;\) (g) \(\mathrm{AlF}_{6}^{3-}\).

Short Answer

Expert verified
The shapes of the molecules are: (a) CO, linear; (b) SiCl4, tetrahedral; (c) PH3, trigonal pyramidal; (d) ICl3, T-shaped; (e) SbCl5, trigonal bipyramidal; (f) SO2, bent or V-shaped; (g) AlF6^3-, octahedral.

Step by step solution

01

Determine the Central Atom

The first step is to identify the central atom in each molecule. The central atom is generally the one that is linked to all other atoms in the molecule. In our case, for (a) CO, the central atom is C; (b) SiCl4, the central atom is Si; (c) PH3, the central atom is P; (d) ICl3, the central atom is I; (e) SbCl5, the central atom is Sb; (f) SO2, the central atom is S; (g) AlF6^3-, the central atom is Al.
02

Count the Number of Electron Groups Around Central Atom

Count the number of bonding groups (single, double, or triple bonds) and non-bonding groups (lone pairs of electrons) around the central atom. For: (a) CO, 3 groups (two bonding, one non-bonding); (b) SiCl4, 4 groups (all bonding); (c) PH3, 4 groups (3 bonding, 1 non-bonding); (d) ICl3, 5 groups (3 bonding, 2 non-bonding); (e) SbCl5, 5 groups (all bonding); (f) SO2, 3 groups (two bonding, one non-bonding); (g) AlF6^3-, 6 groups (all bonding).
03

Determine the Molecular Geometry

Based on these number of groups and the VSEPR theory, the shapes are: (a) CO, linear; (b) SiCl4, tetrahedral; (c) PH3, trigonal pyramidal; (d) ICl3, T-shaped; (e) SbCl5, trigonal bipyramidal; (f) SO2, bent or V-shaped; (g) AlF6^3-, octahedral.

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