Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 22: Problem 85

(a) What is the oxidation state of \(\mathrm{P}\) in \(\mathrm{PF}_{6}^{-}\) and of \(\mathrm{N}\) in \(\mathrm{NF}_{3} ?\) (b) Why doesn't \(\mathrm{N}\) form \(\mathrm{NF}_{6}^{-}\) ion analogous to \(\mathrm{P} ?\)

Short Answer

Expert verified
The oxidation state of P in PF6- is +5, and the oxidation state of N in NF3 is +3. Nitrogen does not form an NF6- ion because it belongs to the 2nd period and has only three unpaired electrons in its 2p orbital. Unlike phosphorus, nitrogen does not have access to empty d orbitals to accommodate extra electrons and extend its octet, allowing it to form only three bonds with three fluorine atoms in NF3.

Step by step solution

01

Determine the oxidation state of P in PF6-

To determine the oxidation state of P in PF6-, we have to note that the oxidation state of F is always -1. Since in the compound PF6-, there are 6 fluorine atoms, the total negative charge is -6. Since PF6- ion has a total negative charge of -1, the oxidation state of P has to be +5 in order to achieve the net charge of -1. Therefore, the oxidation state of P in PF6- is +5.
02

Determine the oxidation state of N in NF3

To determine the oxidation state of N in NF3, we have to consider that the oxidation state of F is -1. Since there are 3 fluorine atoms, the total negative charge is -3. Nitrogen, being the more electronegative element, has a lower oxidation state, which must be +3 in order to balance the net charge of the compound. Therefore, the oxidation state of N in NF3 is +3.
03

Analyze the electronic configuration of P and N

In order to understand why N doesn't form an NF6- ion analogous to P, let's analyze their electronic configurations. The electronic configuration of N is 1s² 2s² 2p³, and that of P is 1s² 2s² 2p⁶ 3s² 3p³. Nitrogen belongs to the 2nd period and has three unpaired electrons in the 2p orbital. On the other hand, phosphorus belongs to the 3rd period and has three unpaired electrons in the 3p orbital. However, phosphorus also has empty 3d orbitals, which can be used for the expansion of its octet by forming additional bonds. This is not the case for nitrogen, as it lacks available empty d orbitals for extra bond formation.
04

Explain why N doesn't form NF6- ion

Based on the electronic configuration analysis, nitrogen doesn't form the NF6- ion because it can only make three bonds with its three unpaired electrons in the 2p orbital. Since nitrogen belongs to the 2nd period, it doesn't have access to empty d orbitals to accommodate extra electrons that could extend its octet, as phosphorus does. Nitrogen can form three bonds with three fluorine atoms, resulting in an NF3 molecule, but it cannot form additional bonds, like phosphorus, to create an NF6- ion. In conclusion, the oxidation state of P in PF6- is +5, the oxidation state of N in NF3 is +3, and N doesn't form NF6- ion due to its electronic configuration preventing the extension of its octet.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) How does the structure of diborane $\left(\mathrm{B}_{2} \mathrm{H}_{6}\right)\( differ from that of ethane \)\left(\mathrm{C}_{2} \mathrm{H}_{6}\right) ?(\mathbf{b})$ Explain why diborane adopts the geometry that it does. (c) What is the significance of the statement that the hydrogen atoms in diborane are described as "hydridic"?

(a) Draw the Lewis structures for at least four species that have the general formula $$[: \mathrm{X} \equiv \mathrm{Y}:]^{n}$$ where \(X\) and Y may be the same or different, and \(n\) may have a value from +1 to \(-2 .\) (b) Which of the compounds is likely to be the strongest Bronsted base? Explain. [Sections 22.1, 22.7, and 22.9]

Write complete balanced half-reactions for (a) oxidation of nitrous acid to nitrate ion in acidic solution, (b) oxidation of \(\mathrm{N}_{2}\) to \(\mathrm{N}_{2} \mathrm{O}\) in acidic solution.

Consider the elements $\mathrm{Ba}, \mathrm{Na}, \mathrm{O}, \mathrm{B}, \mathrm{P},\( and \)\mathrm{Kr}$. From this list, select the element that (a) is most electronegative, (b) has the greatest metallic character, \((\mathbf{c})\) most readily forms a positive ion, \((\mathbf{d})\) exhibits a maximum oxidation sate of +5 , (e) exists as monoatomic gas at room temperature, (f) has multiple allotropes.

Complete and balance the following equations: (a) \(\mathrm{NaOCH}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (b) \(\mathrm{CuO}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow\) (c) \(\mathrm{WO}_{3}(s)+\mathrm{H}_{2}(g) \stackrel{\Delta}{\longrightarrow}\) (d) \(\mathrm{NH}_{2} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow\) (e) $\mathrm{Al}_{4} \mathrm{C}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow$
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free