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Ch.8 - Basic Concepts of Chemical Bonding
Chapter 8, Problem 98c

The Ti2+ ion is isoelectronic with the Ca atom. (c) What charge would Ti have to be isoelectronic with Ca2+ ?

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Identify the number of electrons in a neutral calcium (Ca) atom. Since calcium is element number 20 on the periodic table, a neutral Ca atom has 20 electrons.
Understand that Ca<sup>2+</sup> indicates a calcium ion with a charge of +2, meaning it has lost two electrons compared to the neutral atom. Therefore, Ca<sup>2+</sup> has 18 electrons.
Recognize that the Ti<sup>2+</sup> ion, having a charge of +2, implies it has lost two electrons from its neutral state. Titanium (Ti) is element number 22 on the periodic table, so a neutral Ti atom has 22 electrons. Thus, Ti<sup>2+</sup> has 20 electrons.
To find the charge Ti needs to have the same number of electrons as Ca<sup>2+</sup> (18 electrons), calculate how many electrons need to be lost from a neutral Ti atom (22 electrons) to reach 18 electrons.
Determine the charge on the titanium ion when it has lost the calculated number of electrons. This charge will make it isoelectronic with Ca<sup>2+</sup>.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Isoelectronic Species

Isoelectronic species are atoms or ions that have the same number of electrons, resulting in identical electron configurations. For example, the Ti<sup>2+</sup> ion and Ca atom both have 20 electrons, making them isoelectronic. Understanding this concept is crucial for comparing the electronic structures of different elements and ions.
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Charge of Ions

The charge of an ion is determined by the loss or gain of electrons relative to its neutral atom. A positive charge indicates the loss of electrons, while a negative charge indicates the gain. In the case of titanium (Ti), to become isoelectronic with Ca<sup>2+</sup>, which has lost two electrons, Ti must also lose electrons to match the electron count.
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Electron Configuration

Electron configuration describes the distribution of electrons in an atom's orbitals. For Ti, the neutral atom has an electron configuration of [Ar] 4s<sup>2</sup> 3d<sup>2</sup>. When forming ions, the configuration changes based on the number of electrons lost or gained, which is essential for determining how many electrons Ti must lose to be isoelectronic with Ca<sup>2+</sup>.
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Related Practice
Textbook Question

Consider the hypothetical molecule B-A=B. Are the following statements true or false? (a) This molecule cannot exist. (b) If resonance was important, the molecule would have identical A–B bond lengths.

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Textbook Question

An important reaction for the conversion of natural gas to other useful hydrocarbons is the conversion of methane to ethane. 2 CH4(g) → C2H6(g) + H2(g) In practice, this reaction is carried out in the presence of oxygen, which converts the hydrogen produced into water. 2 CH4(g) + 12 O2(g) → C2H6(g) + H2O(g) Use Table 8.3 to estimate H for these two reactions. Why is the conversion of methane to ethane more favorable when oxygen is used? Why is the conversion of methane to ethane more favorable when oxygen is used?

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Textbook Question

Two compounds are isomers if they have the same chemical formula but different arrangements of atoms. Use Table 8.3 to estimate H for each of the following gas-phase isomerization reactions and indicate which isomer has the lower enthalpy. (d) Methyl isocyanide → Acetonitrile

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Textbook Question

Consider the molecule C4H5N, which has the connectivity shown below. (a) After the Lewis structure for the molecule is completed, how many s and how many p bonds are there in this molecule?

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Textbook Question

The electron affinity of oxygen is -141 kJ/mol, corresponding to the reaction O(g) + e- → O-(g). The lattice energy of K2O(s) is 2238 kJ/mol. Use these data along with data in Appendix C and Figure 7.10 to calculate the 'second electron affinity' of oxygen, corresponding to the reaction O-(g) + e- → O2-(g)

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Open Question
You and a partner are asked to complete a lab entitled “Carbonates of Group 2 metal” 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 and determining the identity of the Group 2 metal (M). In the second lab, you are to determine the melting point of this compound. Upon going to the lab, you find two unlabeled vials containing white powder. You also find the following notes in your partner’s notebook—Compound 1: 40.04% M, 12.00% C, and 47.96% O (by mass); Compound 2: 69.59% M, 6.09% C, and 24.32% O (by mass). (a) What is the empirical formula for Compound 1 and the identity of M? (b) What is the empirical formula for Compound 2 and the identity of M? Upon determining the melting points of these two compounds, you find that both compounds do not melt up to the maximum temperature of your apparatus; instead, the compounds decompose and liberate a colorless gas. (c) What is the identity of the colorless gas?