Chapter 22, Problem 1DE

(a)

Interpretation Introduction

To determine:

The experiments that would allow the identification of the given substances.

Answer to Problem 1DE

Solution:

The given substances can be identified by reaction with lewis acid and lewis base.

Explanation of Solution

Samples of five substances are given. The substances are ${\text{NF}}_{\text{3}}$ , ${\text{PF}}_{\text{3}}$ , ${\text{PCl}}_{\text{3}}$ , ${\text{PF}}_{\text{5}}$ and ${\text{PCl}}_{\text{5}}$ . ${\text{NF}}_{\text{3}}$ is a lewis base whereas ${\text{PF}}_{\text{3}}$ and ${\text{PCl}}_{\text{3}}$ are lewis acids. The order of lewis acidity is ${\text{PF}}_{\text{3}}>{\text{PCl}}_{\text{3}}$ . ${\text{PF}}_{\text{5}}$ will not react with lewis acid and bases. Among the halides, one is solid. ${\text{PCl}}_{\text{5}}$ has the highest molecular weight.

Conclusion

The given substances can be identified by reaction with lewis acid and lewis base.

(b)

Interpretation Introduction

To determine:

The procedure for identification if data from internet is available.

Answer to Problem 1DE

Solution:

The given substances can be identified by their melting points.

Explanation of Solution

Samples of five substances are given. The substances are ${\text{NF}}_{\text{3}}$ , ${\text{PF}}_{\text{3}}$ , ${\text{PCl}}_{\text{3}}$ , ${\text{PF}}_{\text{5}}$ and ${\text{PCl}}_{\text{5}}$ .
These substances have different melting points. Therefore, they can be identified with the help of their melting points. ${\text{PCl}}_{\text{5}}$ has highest molecular weight. Therefore, its melting point will be highest.

Conclusion

The given substances can be identified by their melting points.

(c)

Interpretation Introduction

To determine:

The substance that could undergo reaction to add more atoms around the central atom.

Answer to Problem 1DE

Solution:

The substances that could undergo reaction to add more atoms around the central atom are ${\text{PCl}}_{\text{3}}$ and ${\text{PF}}_{\text{5}}$ .

Explanation of Solution

Among the given compounds, ${\text{PCl}}_{\text{3}}$ and ${\text{PF}}_{\text{5}}$ can undergo reaction to add more atoms around the central atom. In these two compounds, phosphorus has vacant d orbital. Reaction with the halogens can be done in these two cases. Phosphorus can accommodate more atoms around it. In ${\text{PCl}}_{\text{5}}$ , there will be steric crowding.

Conclusion

The substances that could undergo reaction to add more atoms around the central atom are ${\text{PCl}}_{\text{3}}$ and ${\text{PF}}_{\text{5}}$ .

(d)

Interpretation Introduction

To determine:

The substance which is likely to be solid.

Answer to Problem 1DE

Solution:

The substance which is likely to be solid is ${\text{PCl}}_{\text{5}}$ .

Explanation of Solution

Among the given substances, ${\text{PCl}}_{\text{5}}$ has the largest size and molecular weight. The Van der Waal force of attraction in ${\text{PCl}}_{\text{5}}$ will strongest. Since the intermolecular force of attraction is greatest in ${\text{PCl}}_{\text{5}}$ , therefore it will be solid.

Conclusion

The substance which is likely to be solid is ${\text{PCl}}_{\text{5}}$ .