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Chapter 8: Problem 12

(a) Write the electron configuration for the element titanium, Ti. How many valence electrons does this atom possess? (b) Hafnium, Hf, is also found in group \(4 \mathrm{~B}\). Write the electron configuration for Hf. (c) Ti and Hf behave as though they possess the same number of valence electrons. Which of the subshells in the electron configuration of Hf behave as valence orbitals? Which behave as core orbitals?

Short Answer

Expert verified
The electron configuration of titanium (Ti) is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2\), and it has 2 valence electrons. For hafnium (Hf), the electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4f^14 5d^2 \). The valence orbitals for Hf are \(6s^2, 5d^2\) and the rest are core orbitals.

Step by step solution

01

(a) Electron configuration for titanium (Ti)

To find the electron configuration for titanium (Ti), which has an atomic number of 22, we will follow the Aufbau principle and fill the orbitals in the increasing order of energy levels. Here's the electron configuration for Ti: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2\) Now, let's find out the number of valence electrons.
02

Valence electrons for titanium (Ti)

In the electron configuration above, the highest principal quantum number (n) is 4. Thus, the electrons in the 4s orbital are the valence electrons. Titanium has 2 valence electrons.
03

(b) Electron configuration for hafnium (Hf)

Hafnium (Hf) has an atomic number of 72. To determine its electron configuration, we'll follow the Aufbau principle again and fill the orbitals in increasing order of energy levels. The electron configuration for hafnium is: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4f^14 5d^2 \)
04

(c) Valence and core orbitals for hafnium (Hf)

Valence orbitals are the orbitals containing the outermost electrons of an atom. Thus, for Hf, the valence orbitals are the ones with the highest principal quantum number, which is 6 (6s^2) and also the partially filled 5d subshell (5d^2). The rest of the orbitals are core orbitals, as they contain the inner core electrons: Valence orbitals: \(6s^2, 5d^2\) Core orbitals: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 4f^14 \)

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Most popular questions from this chapter

Write the electron configuration for each of the following ions, and determine which ones possess noble-gas configura- tions: (a) \(\mathrm{Sr}^{2+},(\mathbf{b}) \mathrm{Ti}^{2+},(\mathrm{c}) \mathrm{Se}^{2-},(\mathrm{d}) \mathrm{Ni}^{2+},(\mathrm{e}) \mathrm{Br}^{-},\) (f) \(\mathrm{Mn}^{3+}\)

For elements in the third row of the periodic table and beyond, the octet rule is often not obeyed. What factors are usually cited to explain this fact?

Consider the following statement: "For some molecules and ions, a Lewis structure that satisfies the octet rule does not lead to the lowest formal charges, and a Lewis structure that leads to the lowest formal charges does not satisfy the octet rule." Illustrate this statement using the hydrogen sulfite ion, \(\mathrm{HSO}_{3}^{-}\), as an example (the \(\mathrm{H}\) atom is bonded to one of the \(\mathrm{O}\) atoms).

(a) What is the trend in electronegativity going from left to right in a row of the periodic table? (b) How do electronegativity values generally vary going down a column in the periodic table? (c) How do periodic trends in electronegativity relate to those for ionization energy and electron affinity?

The following three Lewis structures can be drawn for \(\mathrm{N}_{2} \mathrm{O}:\) \(\mathrm{N} \equiv \mathrm{N}-\ddot{\mathrm{O}}: \longleftrightarrow: \mathrm{N}^{*}-\mathrm{N} \equiv \mathrm{O}: \longleftrightarrow: \ddot{\mathrm{N}}=\mathrm{N}=\ddot{\mathrm{O}}:\) (a) Using formal charges, which of these three resonance forms is likely to be the most important? (b) The \(\mathrm{N}-\mathrm{N}\) bond length in \(\mathrm{N}_{2} \mathrm{O}\) is \(1.12 \mathrm{~A}\), slightly longer than a typical \(\mathrm{N} \equiv \mathrm{N}\) bond; and the \(\mathrm{N}-\mathrm{O}\) bond length is \(1.19 \AA\), slightly shorter than a typical \(\mathrm{N}=\mathrm{O}\) bond. (See Table \(8.5 .\) ) Rationalize these observations in terms of the resonance structures shown previously and your conclusion for part (a).
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