Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 101

The \(\mathrm{Ti}^{2+}\) ion is isoelectronic with the Ca atom. (a) Are there any differences in the electron configurations of \(\mathrm{Ti}^{2+}\) and Ca? (b) With reference to Figure 6.24 , comment on the changes in the ordering of the \(4 s\) and \(3 d\) subshells in Ca and \(\mathrm{Ti}^{2+}\), (c) Will Ca and \(\mathrm{Ti}^{2+}\) have the same number of unpaired electrons? Explain.

Short Answer

Expert verified
The electron configurations of Ti²⁺ and Ca are different: Ti²⁺ has a configuration of \( [\mathrm{Ar}] 3d^2 \) while Ca has a configuration of \( [\mathrm{Ar}] 4s^2 \). The energy ordering of the 4s and 3d subshells changes in Ti²⁺, making the 3d subshell lower in energy than the 4s subshell. Ca has 0 unpaired electrons and Ti²⁺ has 2 unpaired electrons, so they do not have the same number of unpaired electrons.

Step by step solution

01

(a) Electron configurations of Ti²⁺ and Ca

To determine the electron configurations of Ti²⁺ and Ca, let's first look at their atomic numbers. Ti has an atomic number of 22 while Ca has an atomic number of 20. The electron configuration of a neutral Ti atom is \( [\mathrm{Ar}] 3d^2 4s^2 \). Since Ti²⁺ has lost 2 electrons, its electron configuration becomes \( [\mathrm{Ar}] 3d^2 \). The electron configuration of Ca is \( [\mathrm{Ar}] 4s^2 \). Notice that both Ti²⁺ and Ca have the same number of electrons, 18.
02

(b) Changes in the ordering of 4s and 3d subshells

Referring to Figure 6.24, we can observe that the energy order of the subshells differs in Ca and Ti²⁺. In Ca, the 4s subshell is lower in energy than the 3d subshell, thus filling in 4s first. The electron configuration of Ca is \( [\mathrm{Ar}] 4s^2 \). For Ti²⁺, however, the electron configuration is \( [\mathrm{Ar}] 3d^2 \). This means that after losing 2 electrons from the 4s subshell, it becomes energetically more favorable for the remaining electrons to occupy the 3d subshell instead. So the ordering of subshells changes in Ti²⁺ as the 3d subshell now becomes lower in energy than the 4s subshell.
03

(c) Number of unpaired electrons

In order to find out if Ca and Ti²⁺ have the same number of unpaired electrons, we need to look at their electron configurations. In Ca, the electron configuration is \( [\mathrm{Ar}] 4s^2 \). We can see that all the electrons are paired in the 4s subshell. In Ti²⁺, the electron configuration is \( [\mathrm{Ar}] 3d^2 \). The 3d subshell can accommodate up to 10 electrons, but Ti²⁺ only has 2 electrons in the 3d subshell, making them unpaired. Therefore, Ca has 0 unpaired electrons, while Ti²⁺ has 2 unpaired electrons. They do not have the same number of unpaired electrons.

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) Write one or more appropriate Lewis structures for the nitrite ion, \(\mathrm{NO}_{2}^{-}\). (b) With what allotrope of oxygen is it isoelectronic? (c) What would you predict for the lengths of the bonds in \(\mathrm{NO}_{2}^{-}\) relative to \(\mathrm{N}-\mathrm{O}\) single bonds and double bonds?

Construct a Born-Haber cycle for the formation of the hypothetical compound \(\mathrm{NaCl}_{2}\), where the sodium ion has a \(2+\) charge (the second ionization energy for sodium is given in Table 7.2). (a) How large would the lattice energy need to be for the formation of \(\mathrm{NaCl}_{2}\) to be exothermic? (b) If we were to estimate the lattice energy of \(\mathrm{NaCl}_{2}\) to be roughly equal to that of \(\mathrm{MgCl}_{2}\) ( \(2326 \mathrm{~kJ} / \mathrm{mol}\) from Table 8.2 ), what value would you obtain for the standard enthalpy of formation, \(\Delta H_{j}^{9}\), of \(\mathrm{NaCl}_{2}\) ?

You and a partner are asked to complete a lab entitled "Oxides of Ruthenium" that is scheduled to extend over two lab periods. The first lab, which is to be completed by your partner, is devoted to carrying out compositional analysis. In the second lab, you are to determine melting points. Upon going to lab you find two unlabeled vials, one containing a soft yellow substance and the other a black powder. You also find the following notes in your partner's notebook-Compound 1: \(76.0 \%\) \(\mathrm{Ru}\) and \(24.0 \% \mathrm{O}\) (by mass), Compound 2: \(61.2 \% \mathrm{Ru}\) and \(38.8 \%\) O (by mass). (a) What is the empirical formula for Compound \(1 ?\) (b) What is the empirical formula for Compound \(2 ?\) (c) Upon determining the melting points of these two compounds, you find that the yellow compound melts at \(25^{\circ} \mathrm{C},\) while the black powder does not melt up to the maximum temperature of your apparatus, \(1200^{\circ} \mathrm{C}\). What is the identity of the yellow compound? What is the identity of the black compound? Be sure to use the appropriate naming convention depending on whether the compound is better described as a molecular or ionic compound.

Use Table 8.4 to estimate the enthalpy change for each of the following reactions: (a) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{O}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{O}-\mathrm{Cl}(g)\) (b) \(\mathrm{H}_{2} \mathrm{O}_{2}(g)+2 \mathrm{CO}(g) \longrightarrow \mathrm{H}_{2}(g)+2 \mathrm{CO}_{2}(g)\) (c) \(3 \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12}(g)\) (the six carbon atoms form a six-membered ring with two \(\mathrm{H}\) atoms on each \(\mathrm{C}\) atom \()\)

The substance chlorine monoxide, \(\mathrm{ClO}(g)\), is important in atmospheric processes that lead to depletion of the ozone layer. The ClO molecule has a dipole moment of \(1.24 \mathrm{D}\) and the (a) Determine the magnitude of \(\mathrm{Cl}-\mathrm{O}\) bond length is \(1.60 \mathrm{~A}\). the charges on the \(\mathrm{Cl}\) and \(\mathrm{O}\) atoms in units of the electronic charge, e. (b) Based on the electronegativities of the elements, which atom would you expect to have a negative charge in the ClO molecule? (c) By using formal charges as a guide, propose the dominant Lewis structure for the molecule. Are the formal charges consistent with your answers to parts (a) and (b)? Can you reconcile any differences you find?
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Inorganic 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.

Sign-up for free