/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 71 Determine the electron configura... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 71

Determine the electron configurations for \(\mathrm{CN}^{+}, \mathrm{CN}\), and \(\mathrm{CN}^{-}\). (a) Which species has the strongest \(\mathrm{C}-\mathrm{N}\) bond? (b) Which species, if any, has unpaired electrons?

Short Answer

Expert verified
(a) The strongest C-N bond is present in CN-. (b) CN is the only species with an unpaired electron.

Step by step solution

01

Determine the total number of electrons for each molecule

For CN+: C has 6 electrons, N has 7 electrons, and 1 electron is removed due to the positive charge, resulting in 12 electrons. For CN: C has 6 electrons, and N has 7 electrons, resulting in 13 electrons. For CN-: C has 6 electrons, N has 7 electrons, and 1 electron is added because of the negative charge, resulting in 14 electrons. Step 2: Assign Electrons to Molecular Orbitals
02

Fill the molecular orbitals with given electrons

The molecular orbital diagram for CN molecules has 蟽 and 蟺 orbitals. The order of energy levels for these orbitals is: 蟽1s < 蟽*1s < 蟽2s < 蟽*2s < 蟺2px = 蟺2py < 蟽2pz < 蟺*2px = 蟺*2py < 蟽*2pz. For CN+ (12 electrons): 蟽1s^2 蟽*1s^2 蟽2s^2 蟽*2s^2 蟺2px^2 蟺2py^2 (All orbitals are filled) For CN (13 electrons): 蟽1s^2 蟽*1s^2 蟽2s^2 蟽*2s^2 蟺2px^2 蟺2py^2 蟽2pz^1 (蟺*2px and 蟺*2py orbitals are unfilled) For CN- (14 electrons): 蟽1s^2 蟽*1s^2 蟽2s^2 蟽*2s^2 蟺2px^2 蟺2py^2 蟽2pz^2 (Just the 蟺*2px and 蟺*2py orbitals are unfilled) Step 3: Evaluate C-N Bond Strength
03

Compare the bond orders for different species

Bond order is calculated as (number of electrons in bonding orbitals - number of electrons in anti-bonding orbitals)/2. For CN+: Bond order = (10 - 4)/2 = 3 (strongest bond) For CN: Bond order = (10 - 4 + 1)/2 = 3.5 For CN-: Bond order = (12 - 4)/2 = 4 Therefore, based on bond order, CN- has the strongest C-N bond. Step 4: Identify Unpaired Electrons
04

Identify molecules with unpaired electrons

We can examine the molecular orbitals of each molecule to identify unpaired electrons. For CN+: All orbitals are filled, with no unpaired electrons. For CN: 蟽2pz orbital has 1 unpaired electron. For CN-: All orbitals are filled, with no unpaired electrons. Only CN has an unpaired electron. In summary: (a) CN- has the strongest C-N bond. (b) CN is the only species with an unpaired electron.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

The Lewis structure for allene is Make a sketch of the structure of this molecule that is analogous to Figure \(9.27 .\) In addition, answer the following three questions: (a) Is the molecule planar? (b) Does it have a nonzero dipole moment? (c) Would the bonding in allene be described as delocalized? Explain.

Sulfur tetrafluoride (SF \(_{4}\) ) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide \(\left(\mathrm{OSF}_{4}\right)\) according to the following unbalanced reaction: $$ \mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \operatorname{OSF}_{4}(g) $$ The \(\mathrm{O}\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central \(S\) atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.4) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electrondomain geometry of \(\mathrm{OSF}_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) Which of the molecular geometries in part (d) is more likely to be observed for the molecule? Explain.

(a) Explain why the following ions have different bond angles: \(\mathrm{ClO}_{2}^{-}\) and \(\mathrm{NO}_{2}^{-}\). Predict the bond angle in each case. (b) Explain why the \(\mathrm{XeF}_{2}\) molecule is linear and not bent.

(a) What is the difference between hybrid orbitals and molecular orbitals? (b) How many electrons can be placed into each MO of a molecule? (c) Can antibonding molecular orbitals have electrons in them?

The molecule shown here is difluoromethane \(\left(\mathrm{CH}_{2} \mathrm{~F}_{2}\right)\), which is used as a refrigerant called \(\mathrm{R}\) -32. (a) Based on the structure, how many electron domains surround the \(C\) atom in this molecule? (b) Would the molecule have a nonzero dipole moment? (c) If the molecule is polar, in what direction will the overall dipole moment vector point in the molecule? [Sections \(9.2\) and 9.3]
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

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.