Log out Go to App
Go to App

Learning Materials

Features

Discover

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 \(\mathrm{Hf}\) behave as valence orbitals? Which behave as core orbitals?

Short Answer

Expert verified
The electron configuration of Titanium (Ti) is 1s²2s²2p⁶3s²3p⁶4s²3d², and it has 4 valence electrons (located in 4s²3d² orbitals). Hafnium (Hf) has the electron configuration 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d², and also has 4 valence electrons (located in 5d²6s² orbitals). Both Ti and Hf have the same number of valence electrons, which are found in their respective outermost orbitals. The valence orbitals for Hf are 5d and 6s, while all other orbitals are considered core orbitals.

Step by step solution

01

Determine the electron configuration of Titanium (Ti)

To find the electron configuration of Titanium (Ti), we first look at its atomic number which is 22. This means that Ti has 22 electrons. We will fill the orbitals following the Aufbau principle, Hund's rule, and the Pauli exclusion principle. The electron configuration of Ti is given by the arrangement: 1s²2s²2p⁶3s²3p⁶4s²3d²
02

Determine the number of valence electrons for Titanium (Ti)

Valence electrons are the electrons found in the outermost energy level of an atom. In the case of Ti, the outermost energy level is the 4th energy level. Therefore, Ti's valence electrons are present in the 4s and 3d orbitals. There are 2 electrons in the 4s orbital and 2 electrons in the 3d orbital, so Ti has a total of 4 valence electrons.
03

Determine the electron configuration of Hafnium (Hf)

To find the electron configuration of Hafnium (Hf), we need to look at its atomic number, which is 72. This means that Hf has 72 electrons. We will fill the orbitals following the same rules as for Ti. The electron configuration of Hf is as follows: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d²
04

Compare the valence electrons of Titanium (Ti) and Hafnium (Hf)

Observing the electron configuration of Hf, we can see that the valence electron levels are the 5d and 6s orbitals. There are 2 electrons in the 5d orbital and 2 electrons in the 6s orbital, so Hf also has a total of 4 valence electrons, just like Ti.
05

Identify the valence and core orbitals of Hafnium (Hf)

In Hf's electron configuration, the valence orbitals are those that are the outermost and contain valence electrons, which are the 5d and 6s orbitals. All other orbitals in Hf's electron configuration are considered core orbitals as they contain the core electrons. These are the orbitals: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶4f¹⁴.

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

Draw the Lewis structure for \(\mathrm{NO}^{+} .\) Is the nitrogen-oxygen bond in \(\mathrm{NO}^{+}\) longer, shorter, or the same length as the nitrogen- oxygen bond in NO? Explain.

Consider the ionic compounds KF, NaCl, NaBr, and LiCl. (a) Use ionic radil (Figure 7.8) to estimate the cation-anion distance for each compound. (b) Based on your answer to part (a), arrange these four compounds in order of decreasing lattice energy. (c) Check your predictions in part (b) with the experimental values of lattice energy from Table \(8.1 .\) Are the predictions from ionic radii correct?

By referring only to the periodic table, select (a) the most electronegative element in group \(6 \mathrm{A} ;(\mathbf{b})\) the least electronegative element in the group Al, Si, P; (c) the most electronegative element in the group \(\mathrm{Ga}, \mathrm{P}, \mathrm{Cl}, \mathrm{Na} ;(\mathbf{d})\) the element in the group \(\mathrm{K}\) \(\mathrm{C}, \mathrm{Zn}, \mathrm{F}\) that is most likely to form an ionic compound with Ba.

List the individual steps used in constructing a Born-Haber cycle for the formation of \(\mathrm{Bal}_{2}\) from the elements. Which of the steps would you expect to be exothermic?

Ammonia reacts with boron trifluoride to form a stable compound, as we saw in Section 8.7 . (a) Draw the Lewis structure of the ammonia-boron trifluoride reaction product. (b) The B-N bond is obviously more polar than the \(\mathrm{C}-\mathrm{C}\) bond. Draw the charge distribution you expect on the \(\mathrm{B}-\mathrm{N}\) bond within the molecule (using the delta plus and delta minus symbols mentioned in Section 8.4\()\) . ( ) Boron trichloride also reacts with ammonia in a similar way to the trifluoride. Predict whether the \(B-N\) bond in the trichloride reaction product would be more or less polar than the \(B-N\) bond in the trifluoride product, and justify your reasoning.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical 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