Question

The \(\mathrm{Ti}^{2+}\) ion is isoelectronic with the Ca atom. (a) Writ the electron configurations of \(\mathrm{Ti}^{2+}\) and Ca. (b) Calculate th number of unpaired electrons for \(\mathrm{Ca}\) and for \(\mathrm{Ti}^{2+}\). (c) Wha charge would Ti have to be isoelectronic with \(\mathrm{Ca}^{2+}\) ?

Short answer

The electron configurations for \(\mathrm{Ti}^{2+}\) and \(\mathrm{Ca}\) are: \(\mathrm{Ti}^{2+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^0 3d^2\) \(\mathrm{Ca}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2\) There are 0 unpaired electrons for \(\mathrm{Ca}\) and 2 unpaired electrons for \(\mathrm{Ti}^{2+}\). The charge on \(\mathrm{Ti}\) to be isoelectronic with \(\mathrm{Ca}^{2+}\) is +4 (so it becomes \(\mathrm{Ti}^{4+}\)).

Step-by-step solution

Determine the electron configurations for \(\mathrm{Ti}\) and \(\mathrm{Ca}\)

Using the periodic table, we can find that Titanium (Ti) has an atomic number of 22, which means it has 22 electrons in its neutral atom. Calcium (Ca), on the other hand, has an atomic number of 20, which means it has 20 electrons in its neutral atom. Thus, the electron configurations for the neutral atoms of these two elements are: Ti: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2\) Ca: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2\) Since \(\mathrm{Ti}^{2+}\) lost 2 electrons, the electron configuration becomes: \(\mathrm{Ti}^{2+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^0 3d^2\) Since \(\mathrm{Ca}\) is neutral, its electron configuration remains unchanged.

Calculate the number of unpaired electrons for \(\mathrm{Ti}^{2+}\) and \(\mathrm{Ca}\)

For \(\mathrm{Ca}\), the configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2\), and all electrons are paired. Therefore, the number of unpaired electrons for \(\mathrm{Ca}\) is 0. For \(\mathrm{Ti}^{2+}\), the configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^0 3d^2\), and there are 2 unpaired electrons in the 3d subshell. Therefore, the number of unpaired electrons for \(\mathrm{Ti}^{2+}\) is 2.

Determine the charge on \(\mathrm{Ti}\) that would make it isoelectronic with \(\mathrm{Ca}^{2+}\)

We know that \(\mathrm{Ca}\) has 20 electrons in its neutral state. Since \(\mathrm{Ca}^{2+}\) lost 2 electrons, it would have 18 electrons. In order for \(\mathrm{Ti}\) to be isoelectronic with \(\mathrm{Ca}^{2+}\), it must also have 18 electrons. Since the neutral \(\mathrm{Ti}\) atom has 22 electrons, it would need to lose 4 electrons to be isoelectronic with \(\mathrm{Ca}^{2+}\). Therefore, the required charge on \(\mathrm{Ti}\) is +4 (represented as \(\mathrm{Ti}^{4+}\)). In summary, the electron configurations for \(\mathrm{Ti}^{2+}\) and \(\mathrm{Ca}\) are: \(\mathrm{Ti}^{2+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^0 3d^2\) \(\mathrm{Ca}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2\) There are 0 unpaired electrons for \(\mathrm{Ca}\) and 2 unpaired electrons for \(\mathrm{Ti}^{2+}\). The charge on \(\mathrm{Ti}\) to be isoelectronic with \(\mathrm{Ca}^{2+}\) is +4 (so it becomes \(\mathrm{Ti}^{4+}\)).

Key concepts

Unpaired Electrons

Unpaired electrons play a crucial role in the magnetic properties of an atom or ion. When electrons fill the orbitals of an atom, they first occupy different orbitals before pairing up. If orbitals contain electrons that are not paired, these are called unpaired electrons.

In the exercise example, let's examine

• Titanium (Ti): The \[\mathrm{Ti}^{2+}\] ion, which means two electrons are lost, has an electron configuration of \[1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^2\]. This reveals that in the 3d subshell, there are two unpaired electrons.
• Calcium (Ca): The neutral atom of calcium has an electron configuration of \[1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\]. All electrons in this configuration are paired, resulting in zero unpaired electrons.

Understanding unpaired electrons is vital, as these can influence chemical reactivity and magnetic properties of substances. More unpaired electrons typically lead to greater magnetic attraction.

Isoelectronic Species

Isoelectronic species are atoms or ions that have the same number of electrons and, therefore, the same electron configuration. This similarity in electron number gives isoelectronic species similar chemical properties.

In the exercise above,

• The \[\mathrm{Ti}^{2+}\] ion has the same electron configuration as a neutral calcium atom, making them isoelectronic. Both species contain 20 electrons.
• For titanium to become isoelectronic with \[\mathrm{Ca}^{2+}\], it must lose a total of four electrons, forming a \[\mathrm{Ti}^{4+}\] ion. This means both \[\mathrm{Ti}^{4+}\] and \[\mathrm{Ca}^{2+}\] ions have 18 electrons, sharing the same electron arrangement.

The concept of isoelectronic species is important in predicting and understanding the behavior of ions in different chemical reactions. When ions are isoelectronic, they may have similar sizes and shared physical and chemical properties.

Atomic Number

The atomic number is a unique identifier for each chemical element in the periodic table. It represents the number of protons in the nucleus of an atom, which also equals the number of electrons in a neutral atom. The atomic number determines the element's place in the periodic table and its chemical properties.

In our exercise example:

• Titanium (Ti): Has an atomic number of 22, indicating it has 22 protons and, in its neutral state, 22 electrons. When it loses two electrons to form \[\mathrm{Ti}^{2+}\], it still maintains the 22 protons.
• Calcium (Ca): Has an atomic number of 20, indicating it has 20 protons and 20 electrons when neutral. It maintains the number of protons regardless of electron loss and gain.

The atomic number is central to understanding periodic trends and predicting the behavior of elements. It's essential for chemists to compare and contrast different atoms and ions based on their atomic structure.