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Using orbital box diagrams and noble gas notation, depict the electron configurations of (a) Ti, (b) \(\mathrm{Ti}^{2+},\) and \((\mathrm{c}) \mathrm{Ti}^{4+} .\) Is the element or any of the ions paramagnetic?

Short Answer

Expert verified
Ti and \(\text{Ti}^{2+}\) are paramagnetic; \(\text{Ti}^{4+}\) is diamagnetic.

Step by step solution

01

Determine Atomic Number and Electron Configuration

Titanium (Ti) has an atomic number of 22, which means it has 22 electrons when it is neutral. Using noble gas notation, the electron configuration of Ti is \([\text{Ar}] \, 3d^2 \, 4s^2\).
02

Draw Orbital Box Diagrams for Neutral Ti

In the orbital box diagram, the 3d sublevel has two electrons, and the 4s sublevel also has two electrons. This can be visualized as: \([\text{Ar}] \) \[ \begin{array}{c c} 3d: & \Box{\uparrow} \enspace\Box{\uparrow} \enspace\Box \enspace\Box \enspace\Box \ 4s: & \Box{\uparrow\downarrow} \end{array} \]
03

Determine Electron Configuration of \( \text{Ti}^{2+} \) Ion

When Ti loses two electrons to form \( \text{Ti}^{2+} \), they are removed first from the 4s sublevel. Thus, the electron configuration becomes \([\text{Ar}] \, 3d^2\).
04

Draw Orbital Box Diagram for \( \text{Ti}^{2+} \) Ion

For \( \text{Ti}^{2+} \), the orbital configuration is: \([\text{Ar}] \) \[ \begin{array}{c} 3d: \Box{\uparrow} \enspace\Box{\uparrow} \enspace\Box \enspace\Box \enspace\Box \end{array} \] All 4s electrons are now absent.
05

Determine Electron Configuration of \( \text{Ti}^{4+} \) Ion

To form \( \text{Ti}^{4+} \), two more electrons are removed from the \(3d\) sublevel. The electron configuration becomes \([\text{Ar}]\).
06

Assess Paramagnetism

An element or ion is paramagnetic if it has unpaired electrons. Ti has unpaired electrons in the 3d sublevel, and \(\text{Ti}^{2+}\) also has unpaired electrons in 3d. Thus, both are paramagnetic. On the other hand, \( \text{Ti}^{4+} \) has no unpaired electrons, making it diamagnetic.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Paramagnetism
Paramagnetism is an important property of elements and ions, related to their unpaired electrons. If a substance has one or more unpaired electrons in its electron configuration, it is attracted slightly to a magnetic field. This is due to the spin of the unpaired electrons, which creates a tiny magnetic dipole around the atom. However, if all electrons are paired, the material is diamagnetic and slightly repelled by a magnetic field.
In the case of Titanium (Ti) and its ions, Ti and Ti^{2+} both have unpaired electrons in the 3d orbital, making them paramagnetic. Ti^{4+}, on the other hand, lacks unpaired electrons, making it diamagnetic. It is essential to look at the electron configuration diagram to identify the presence of unpaired electrons in the d-orbitals.
Orbital Box Diagrams
Orbital box diagrams are a visual representation of an atom's electron configuration, showing each sublevel as a series of boxes. Each box represents an orbital, and arrows within represent electrons, indicating their spin direction (up or down). This visualization helps determine the pairing of electrons and assess whether the atom is paramagnetic.
For titanium, the orbital box diagram shows the 3d and 4s orbitals with the appropriate distribution of electrons. In its ground state, Ti has the electron configuration \[ \text{[Ar]} \, 3d^2 \, 4s^2\], which can be depicted as two arrows in the 3d box and a pair in the 4s box. Understanding these diagrams is key to determining how the electrons distribute themselves across orbitals.
Noble Gas Notation
Noble gas notation is a shorthand for writing electron configurations. It shortens the long string of energy levels by representing them through the previous noble gas, enclosed in brackets. For instance, for titanium (Ti) with atomic number 22, the electron configuration of the noble gas Argon (Ar) is used as a basis. Thus, the configuration is written as \[ \text{[Ar]} \, 3d^2 \, 4s^2\].
This makes it easier to understand and quickly write electron configurations for atoms and ions, omitting the need to repeat all filled sublevel information. When ions are formed, we modify the shorthand accordingly, removing electrons from the outermost orbitals as required, leading computations such as for Ti^{2+} and Ti^{4+} ions.
Transition Metals
Transition metals are elements found in the d-block of the periodic table. They are characterized by their ability to form various oxidation states due to the filling of d-orbitals. Titanium (Ti) is a transition metal and exhibits typical transition metal behavior. It can lose different numbers of electrons from its 3d and 4s orbitals.
One key property of transition metals is their colorful compound formation, often due to d-d electron transitions. Titanium can exist in multiple oxidation states; Ti^{2+} arises when two electrons are removed from the 4s orbital, while Ti^{4+} results from further removal from the 3d orbital. These properties make transition metals versatile and essential in various applications, including chemical catalysis and alloy formation.
Ion Formation
Ion formation involves the loss or gain of electrons by an atom to achieve a more stable electron configuration. In the context of transition metals like titanium (Ti), the process often involves removing electrons from s and d orbitals to form cations.
When Ti forms the Ti^{2+} ion, it loses the two electrons from its 4s orbital, resulting in \[ \text{[Ar]} \, 3d^2\]. As Ti^{4+}, it loses two additional electrons from the 3d orbital, yielding \[ \text{[Ar]}\]. These processes highlight the stability achieved by reaching an electron configuration similar to that of a noble gas. Understanding ion formation is key to predicting the chemical behavior and reactivity of transition metals in ionic compounds.

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Most popular questions from this chapter

The discovery of two new elements (atomic numbers 113 and 115 ) was announced in February 2004. (a) Use spáf and noble gas notations to give the electron configurations of these two elements. (b) For each of these elements, name another element in the same periodic group. (c) Element 113 was made by firing the nucleus of a light atom at a heavy americium atom. The two nuclei combine to give a nucleus with 113 protons. What light atom was used as a projectile?

Select the atom or ion in each pair that has the larger radius. (a) Cs or Rb (b) \(\mathrm{O}^{2-}\) or \(\mathrm{O}\) (c) Br or As

Manganese is found as \(\mathrm{MnO}_{2}\) in deep ocean deposits. (a) Depict the electron configuration of this element using the noble gas notation and an orbital box diagram. (b) Using an orbital box diagram, show the electrons beyond those of the preceding noble gas for \(\mathrm{Mn}^{4+}\) (c) Is \(\mathrm{Mn}^{4+}\) paramagnetic? (d) How many unpaired electrons does the \(\mathrm{Mn}^{4+}\) ion have?

The lanthanides, once called the rare earth elements, are really only "medium rare." Using noble gas and spdf notations, depict reasonable electron configurations for the following elements. (a) Samarium, Sm. This lanthanide is used in magnetic materials. (b) Ytterbium, Yb. This element was named for the village of Ytterby in Sweden, where a mineral source of the element was found.

Why is the radius of \(L i^{+}\) so much smaller than the radius of Li? Why is the radius of \(F^{-}\) so much larger than the radius of \(\mathrm{F} ?\)

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