Chemistry: The Central Science, Books a la Carte Plus Mastering Chemistry with Pearson eText -- Access Card Package (14th Edition)
Chemistry: The Central Science, Books a la Carte Plus Mastering Chemistry with Pearson eText -- Access Card Package (14th Edition)
14th Edition
ISBN: 9780134557328
Author: Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, Catherine Murphy, Patrick Woodward, Matthew E. Stoltzfus
Publisher: PEARSON
Question
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Chapter 22, Problem 1DE

(a)

Interpretation Introduction

To determine: An experiment that identifies the given substances.

(a)

Expert Solution
Check Mark

Answer to Problem 1DE

Solution:

  • NF3 exist as gas.
  • PF3 exist as gas.
  • PF5 exist as gas.
  • PCl3 exist as liquid.
  • PCl5 exist as solid.

Explanation of Solution

The physical state of the substance depends upon the size of the substance.

The size of the atoms decreases from left to left along the period and increases down the group.

Nitrogen and fluorine belongs to second period. Phosphorous and chlorine belongs to third period. Therefore, the compounds to nitrogen and fluorine and phosphorous and fluorine are smaller in size and hence are in gaseous state. The compounds of phosphorous and chlorine are relatively larger in size. Phosphorous pentachloride is heavier since it contains a greater number of atoms and therefore, it exists as solid and phosphorous trichloride exist as liquid.

Thus,

  • NF3 exist as gas.
  • PF3 exist as gas.
  • PF5 exist as gas.
  • PCl3 exist as liquid.
  • PCl5 exist as solid.
Conclusion
  • NF3 exist as gas.
  • PF3 exist as gas.
  • PF5 exist as gas.
  • PCl3 exist as liquid.
  • PCl5 exist as solid.

(b)

Interpretation Introduction

To determine: The approach to determine the physical state of the given compounds using internet.

(b)

Expert Solution
Check Mark

Answer to Problem 1DE

Solution: The physical state of the substance is estimated by knowing the melting and boiling point of the substance.

Explanation of Solution

The physical state is the state of the substance at room temperature. Therefore, by determining the melting point and boiling point of the substance, the physical state is estimated.

If the substance boils at the temperature less than the room temperature, than the physical state of the substance is gas.

If the substance boils at the temperature more than the room temperature, than the physical state of the substance is liquid.

If the substance melts at the temperature more than the room temperature, than the physical state of the substance is solid.

Conclusion

The physical state of the substance is estimated by knowing the melting and boiling point of the substance.

(c)

Interpretation Introduction

To determine: The substance that undergoes the reaction to add more atoms around the central atom.

(c)

Expert Solution
Check Mark

Answer to Problem 1DE

Solution: The substance that undergoes the reaction to add more atoms around the central atom are PF3 and PCl3 .

Explanation of Solution

The atom that exhibit dorbital should have the principle quantum number of three or more.

Phosphorous possess the principle quantum number of three and nitrogen possess the principle quantum number of two.

Therefore, only phosphorous show the presence of vacant dorbital .

Nitrogen shares only three valence electrons out of five due to absence of vacant dorbital but phosphorous has the ability to share all the five valence electrons by undergoing hybridization of the five orbitals including vacant dorbital .

Thus, PF3 reacts with fluorine molecule to form PF5 and PCl3 reacts with chlorine molecule to form PCl5 .

Conclusion

The substance that undergoes the reaction to add more atoms around the central atom are PF3 and PCl3 due to the presence of vacant dorbital .

(d)

Interpretation Introduction

To determine: The substances that is likely to be solid.

(d)

Expert Solution
Check Mark

Answer to Problem 1DE

Solution: The substances that is likely to be solid are PF5 and PCl5 .

Explanation of Solution

The physical state depends upon the intermolecular force of attraction between the molecules.

The more the polarity of the substance, the stronger is the intermolecular force of attraction.

PF5 and PCl5 have five polar centers. Therefore, these substances are likely to be solid.

Conclusion

PF5 and PCl5 are likely to be solid.

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Chapter 22 Solutions

Chemistry: The Central Science, Books a la Carte Plus Mastering Chemistry with Pearson eText -- Access Card Package (14th Edition)

Ch. 22.10 - Prob. 21.10.1PECh. 22.10 - Prob. 21.10.2PECh. 22 - Prob. 1DECh. 22 - Prob. 1ECh. 22 - Prob. 2ECh. 22 - Prob. 3ECh. 22 - Prob. 4ECh. 22 - Prob. 5ECh. 22 - Prob. 6ECh. 22 - Prob. 7ECh. 22 - Prob. 8ECh. 22 - Prob. 9ECh. 22 - Prob. 10ECh. 22 - Prob. 11ECh. 22 - Prob. 12ECh. 22 - Prob. 13ECh. 22 - Prob. 14ECh. 22 - Prob. 15ECh. 22 - Prob. 16ECh. 22 - Prob. 17ECh. 22 - Prob. 18ECh. 22 - Prob. 19ECh. 22 - Prob. 20ECh. 22 - Prob. 21ECh. 22 - Prob. 22ECh. 22 - Prob. 23ECh. 22 - Prob. 24ECh. 22 - Prob. 25ECh. 22 - Prob. 26ECh. 22 - Prob. 27ECh. 22 - Prob. 28ECh. 22 - Why does xenon form stable compounds with...Ch. 22 - Prob. 30ECh. 22 - Prob. 31ECh. 22 - Prob. 32ECh. 22 - Prob. 33ECh. 22 - Prob. 34ECh. 22 - Prob. 35ECh. 22 - Prob. 36ECh. 22 - Prob. 37ECh. 22 - Prob. 38ECh. 22 - Prob. 39ECh. 22 - Prob. 40ECh. 22 - Prob. 41ECh. 22 - Prob. 42ECh. 22 - Prob. 43ECh. 22 - Prob. 44ECh. 22 - Prob. 45ECh. 22 - Prob. 46ECh. 22 - Prob. 47ECh. 22 - Prob. 48ECh. 22 - Prob. 49ECh. 22 - Prob. 50ECh. 22 - Prob. 51ECh. 22 - Prob. 52ECh. 22 - Prob. 53ECh. 22 - Prob. 54ECh. 22 - Prob. 55ECh. 22 - Prob. 56ECh. 22 - Prob. 57ECh. 22 - Write a chemical formula for each compound or ion,...Ch. 22 - Prob. 59ECh. 22 - Prob. 60ECh. 22 - Prob. 61ECh. 22 - Prob. 62ECh. 22 - Prob. 63ECh. 22 - Prob. 64ECh. 22 - Prob. 65ECh. 22 - Prob. 66ECh. 22 - Prob. 67ECh. 22 - Prob. 68ECh. 22 - Prob. 69ECh. 22 - Write the formulas for the following compounds,...Ch. 22 - Prob. 71ECh. 22 - Prob. 72ECh. 22 - Prob. 73ECh. 22 - Prob. 74ECh. 22 - Prob. 75ECh. 22 - Prob. 76ECh. 22 - Prob. 77ECh. 22 - Prob. 78ECh. 22 - Prob. 79AECh. 22 - Prob. 80AECh. 22 - Prob. 81AECh. 22 - Prob. 82AECh. 22 - Prob. 83AECh. 22 - Prob. 84AECh. 22 - Prob. 85AECh. 22 - Prob. 86AECh. 22 - Prob. 87AECh. 22 - Prob. 88AECh. 22 - Prob. 89AECh. 22 - Prob. 90AECh. 22 - Prob. 91IECh. 22 - Prob. 92IECh. 22 - Prob. 93IECh. 22 - Prob. 94IECh. 22 - Prob. 95IECh. 22 - Prob. 96IECh. 22 - Prob. 97IECh. 22 - Prob. 98IECh. 22 - Prob. 99IECh. 22 - Prob. 100IECh. 22 - Prob. 101IECh. 22 - Prob. 102IECh. 22 - Prob. 103IECh. 22 - Prob. 104IE
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