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Chapter 55: Problem 3

Using VSEPR theory, predict the shapes of the following molecules: (a) \(\mathrm{AlCl}_{3}\) (b) \(\mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{BeF}_{2}\) (d) \(\mathrm{SF}_{6}\) (e) \(\mathrm{IF}_{5}\)

Short Answer

Expert verified
Shaping of molecules as per VSEPR Theory are as follows: (a) AlCl3 is Trigonal Planar, (b) H2O is Bent, (c) BeF2 is Linear, (d) SF6 is Octahedral, (e) IF5 is Square Pyramidal.

Step by step solution

01

AlCl3

This compound is made up of one aluminum (Al) and three chlorine (Cl) atoms. First, determine the number of valence electrons Al has (3) and Cl has (7), totalling to 24. Since Al is the center atom, draw Al surrounded by Cl atoms. The electron pairs will form a bond with each of the Cl atoms, forming a planar structure. Hence, the shape of this molecule using VSEPR theory is \(Trigonal Planar\).
02

H2O

This compound is made up of 2 Hydrogen (H) and one Oxygen (O) atoms. Oxygen has 6 valence electrons and Hydrogen has 1, totalling to 8. Here, the center atom is O. Draw O surrounded by H atoms and have two lone pairs aside. The lone pairs exert greater repulsion, pushing the Hydrogen atoms closer and forming a bent structure. Hence, the shape of this molecule using VSEPR theory is \(Bent\).
03

BeF2

This compound is made up of one Beryllium (Be) and two Fluorine (F) atoms. Be has 2 valence electrons and F has 7, totalling to 16. Insert these around F atoms surrounding Be, with no lone pairs on Be. As a result, the shape of the molecule is \(Linear\).
04

SF6

This compound consists of one Sulfur (S) and six Fluorine (F) atoms. S has 6 valence electrons and F has 7, totalling to 48. Surround S, the central atom, with F atoms and there are no lone pairs on S. The shape of the molecule is \(Octahedral\).
05

IF5

This compound is made up of one Iodine (I) and five Fluorine (F) atoms. I has 7 valence electrons and F has 7, totalling to 42. Surround I, the central atom, with F atoms and one lone pair on I. The lone pair of electrons will alter the shape, resulting in a \(Square Pyramidal\) structure.

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Most popular questions from this chapter

Convert the following structural formulae to bond-line formulae: (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}\), (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Cl}\) (c) \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{2} \mathrm{C}=\mathrm{O}\) (d) \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{3}\) (e)

Convert the following abbreviated formulae to structural formulae: (a) \(\mathrm{BuOH}\) (b) \(\mathrm{PhCH}_{2} \mathrm{Br}\) (c) \(\mathrm{Pr}^{\mathrm{i}} \mathrm{NH}_{2}\) (d) \(\mathrm{AcOEt}\) (e) \(\mathrm{TsCl}\)
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