Chapter 8, Problem 8.1P

Interpretation Introduction

To determine:The shape of ${\text{H}}_{\text{3}}{\text{O}}^{+}$.

Concept Introduction:

• To predict the shapes, we have to considerer the special distribution of the atoms and the pairs of free electrons.
• The pairs of free electrons are those do not form part of bonds between atoms.
• The electron dot structure shows the valence electrons distribution around the atoms.

Some characteristic shapes are:

Number of bonds	Number of pairs of free electrons	Shapes
2	0	Linear
2	3	Linear
2	1	Bent
4	0	Tetrahedral
4	2	Square planar
3	0	Trigonal planar
5	1	Square pyramidal
3	1	Trigonal pyramidal
6	0	Octahedral

Answer to Problem 8.1P

Solution:

Trigonal pyramid.

Explanation of Solution

The electron dot structure of ${\text{H}}_{\text{3}}{\text{O}}^{+}$is,

It has three bonds and one pair of lone electrons. So, its shape is trigonal pyramid.

(c)

Interpretation Introduction

To determine:The shape of ${\text{XeF}}_{\text{2}}$.

Concept Introduction:

• To predict the shapes, we have to considerer the special distribution of the atoms and the pairs of free electrons.
• The pairs of free electrons are those do not form part of bonds between atoms.
• The electron dot structure shows the valence electrons distribution around the atoms.

Some characteristic shapes are:

Number of bonds	Number of pairs of free electrons	Shapes
2	0	Linear
2	3	Linear
2	1	Bent
4	0	Tetrahedral
4	2	Square planar
3	0	Trigonal planar
5	1	Square pyramidal
3	1	Trigonal pyramidal
6	0	Octahedral

Answer to Problem 8.1P

Solution:

Linear.

Explanation of Solution

The electron dot structure of ${\text{XeF}}_{\text{2}}$is,

It has two bonds and three pairs of lone electrons. So, its shape is linear.

(d)

Interpretation Introduction

To determine:The shape of ${\text{PF}}_6^{-}$

. Concept Introduction:

• To predict the shapes, we have to considerer the special distribution of the atoms and the pairs of free electrons.
• The pairs of free electrons are those do not form part of bonds between atoms.
• The electron dot structure shows the valence electrons distribution around the atoms.

Some characteristic shapes are:

Number of bonds	Number of pairs of free electrons	Shapes
2	0	Linear
2	3	Linear
2	1	Bent
4	0	Tetrahedral
4	2	Square planar
3	0	Trigonal planar
5	1	Square pyramidal
3	1	Trigonal pyramidal
6	0	Octahedral

Answer to Problem 8.1P

Solution:

Octahedral.

Explanation of Solution

The electron dot structure of ${\text{PF}}_6^{-}$s,

It has six bonds and no lone pair of electrons. So, its shape is octahedral.

(e)

Interpretation Introduction

To determine:The shape of ${\text{XeOF}}_4$

. Concept Introduction:

• To predict the shapes, we have to considerer the special distribution of the atoms and the pairs of free electrons.
• The pairs of free electrons are those do not form part of bonds between atoms.
• The electron dot structure shows the valence electrons distribution around the atoms.

Some characteristic shapes are:

Number of bonds	Number of pairs of free electrons	Shapes
2	0	Linear
2	3	Linear
2	1	Bent
4	0	Tetrahedral
4	2	Square planar
3	0	Trigonal planar
5	1	Square pyramidal
3	1	Trigonal pyramidal
6	0	Octahedral

Answer to Problem 8.1P

Solution:

Square pyramidal.

Explanation of Solution

The electron dot structure of ${\text{XeOF}}_4$s,

It has five bonds and one pair of lone electrons. So, its shape is square pyramidal.

(f)

Interpretation Introduction

To determine:The shape of ${\text{AlH}}_4^{-}$. .

Concept Introduction:

• To predict the shapes, we have to considerer the special distribution of the atoms and the pairs of free electrons.
• The pairs of free electrons are those do not form part of bonds between atoms.
• The electron dot structure shows the valence electrons distribution around the atoms.

Some characteristic shapes are:

Number of bonds	Number of pairs of free electrons	Shapes
2	0	Linear
2	3	Linear
2	1	Bent
4	0	Tetrahedral
4	2	Square planar
3	0	Trigonal planar
5	1	Square pyramidal
3	1	Trigonal pyramidal
6	0	Octahedral

Answer to Problem 8.1P

Solution:

Tetrahedral.

Explanation of Solution

The electron dot structure of ${\text{AlH}}_4^{-}$s,

It has four bonds and no lone pair of electrons. So, its shape is tetrahedral.

(g)

Interpretation Introduction

To determine:The shape of ${\text{BF}}_4^{-}$

. Concept Introduction:

• To predict the shapes, we have to considerer the special distribution of the atoms and the pairs of free electrons.
• The pairs of free electrons are those do not form part of bonds between atoms.
• The electron dot structure shows the valence electrons distribution around the atoms.

Some characteristic shapes are:

Number of bonds	Number of pairs of free electrons	Shapes
2	0	Linear
2	3	Linear
2	1	Bent
4	0	Tetrahedral
4	2	Square planar
3	0	Trigonal planar
5	1	Square pyramidal
3	1	Trigonal pyramidal
6	0	Octahedral

Answer to Problem 8.1P

Solution: Tetrahedral.

Explanation of Solution

The electron dot structure of ${\text{BF}}_4^{-}$s,

It has four bonds and no lone pair of electrons. So, its shape is tetrahedral.

(h)

Interpretation Introduction

To determine:The shape of ${\text{SiCl}}_{\text{4}}$.

Concept Introduction:

• To predict the shapes, we have to considerer the special distribution of the atoms and the pairs of free electrons.
• The pairs of free electrons are those do not form part of bonds between atoms.
• The electron dot structure shows the valence electrons distribution around the atoms.

Some characteristic shapes are:

Number of bonds	Number of pairs of free electrons	Shapes
2	0	Linear
2	3	Linear
2	1	Bent
4	0	Tetrahedral
4	2	Square planar
3	0	Trigonal planar
5	1	Square pyramidal
3	1	Trigonal pyramidal
6	0	Octahedral

Answer to Problem 8.1P

Solution: Tetrahedral.

Explanation of Solution

The electron dot structure of ${\text{SiCl}}_{\text{4}}$is,

It has four bonds and no lone pair of electrons. So, its shape is tetrahedral.

(i)

Interpretation Introduction

To determine:The shape of ${\text{ICl}}_4^{-}$

. .concept Introduction:

• To predict the shapes, we have to considerer the special distribution of the atoms and the pairs of free electrons.
• The pairs of free electrons are those do not form part of bonds between atoms.
• The electron dot structure shows the valence electrons distribution around the atoms.

Some characteristic shapes are:

Number of bonds	Number of pairs of free electrons	Shapes
2	0	Linear
2	3	Linear
2	1	Bent
4	0	Tetrahedral
4	2	Square planar
3	0	Trigonal planar
5	1	Square pyramidal
3	1	Trigonal pyramidal
6	0	Octahedral

Answer to Problem 8.1P

Solution: Square planar.

Explanation of Solution

The electron dot structure of ${\text{ICl}}_4^{-}$s,

It has four bonds and two pairs of lone electrons. So, its shape is square planar.

(j)

Interpretation Introduction

To determine:The shape of ${\text{AlCl}}_{\text{3}}$.

Concept Introduction:

• To predict the shapes, we have to considerer the special distribution of the atoms and the pairs of free electrons.
• The pairs of free electrons are those do not form part of bonds between atoms.
• The electron dot structure shows the valence electrons distribution around the atoms.

Some characteristic shapes are:

Number of bonds	Number of pairs of free electrons	Shapes
2	0	Linear
2	3	Linear
2	1	Bent
4	0	Tetrahedral
4	2	Square planar
3	0	Trigonal planar
5	1	Square pyramidal
3	1	Trigonal pyramidal
6	0	Octahedral

Answer to Problem 8.1P

Solution: Trigonal planar.

Explanation of Solution

The electron dot structure of ${\text{AlCl}}_{\text{3}}$is,

It has three bonds and no lone pair of electrons. So, its shape is trigonal planar.