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

Short Answer

Expert verified
Ti and \(\mathrm{Ti}^{2+}\) are paramagnetic; \(\mathrm{Ti}^{4+}\) is not.

Step by step solution

01

Determine the Electron Configuration of Neutral Ti

Titanium (Ti) has an atomic number of 22. This means it has 22 electrons. The electron configuration in electron box notation for Ti is as follows:1s虏 2s虏 2p鈦 3s虏 3p鈦 4s虏 3d虏. Using noble gas notation, this can be abbreviated as: \[ \text{[Ar]} 4s^2 3d^2 \] The orbital box diagram would have:- Two arrows in the 4s box (鈫戔啌)- Two separate arrows (one in each) in the 3d boxes (鈫慱 鈫慱 ).
02

Determine the Electron Configuration of \(\mathrm{Ti}^{2+}\)

The \(\mathrm{Ti}^{2+}\) ion is formed by removing two electrons from the neutral titanium atom. Electrons are removed first from the outermost shell (4s orbital) and then from the 3d if needed. Hence, we remove the two electrons from the 4s orbital:Using noble gas notation: \[ \text{[Ar]} 3d^2 \] Orbital box diagram:- Two arrows (one in each) in the 3d boxes (鈫慱 鈫慱 ).
03

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

The \(\mathrm{Ti}^{4+}\) ion is formed by removing a total of four electrons. The two electrons removed from the 4s were the first step, and two more electrons are removed from the 3d orbital:Using noble gas notation: \[ \text{[Ar]} \] Orbital box diagram:- No arrows in the 4s or 3d boxes, as all valence electrons are removed.
04

Determine if Configurations are Paramagnetic

Paramagnetism occurs when an atom or ion possesses unpaired electrons. - **Ti**: Has unpaired electrons in the 3d orbitals, thus is paramagnetic.- **\(\mathrm{Ti}^{2+}\)**: Still has unpaired electrons, hence also paramagnetic.- **\(\mathrm{Ti}^{4+}\)**: No unpaired electrons (because all are removed), so it is not paramagnetic.

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Key Concepts

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

Paramagnetism
In the world of chemistry, paramagnetism is a crucial concept that describes how substances respond to magnetic fields. This property arises when there are unpaired electrons in an atom or ion. Electrons spin and create a tiny magnetic field. When electrons are unpaired, their magnetic fields don't cancel out. This results in a net magnetic moment, making the atom or ion attracted to an external magnetic field.

For instance, let's look at titanium (Ti). For neutral titanium, there are unpaired electrons in the 3d orbitals, making it paramagnetic. The same applies to the \(\mathrm{Ti}^{2+}\) ion because it also retains unpaired electrons in its 3d orbitals. However, when you remove all 3d electrons, as in \(\mathrm{Ti}^{4+}\), all the spins are paired or absent, resulting in no net magnetic field, thus, making it not paramagnetic.

Understanding paramagnetism helps us predict and explain behaviors in magnetic fields, especially useful in fields like material science and quantum chemistry.
Orbital Box Diagrams
Orbital box diagrams serve as visual representations of electron configurations within atoms. They provide a simple way to understand how electrons fill different orbitals according to the Aufbau principle, Hund's rule, and the Pauli exclusion principle.

Here's how you can visualize titanium's electron configuration using orbital boxes:
  • The 1s, 2s, and 2p orbitals are filled first, followed by 3s and 3p.
  • Then the 4s orbital gets filled before the 3d orbitals, even though 3d are lower in energy.
  • In the 4s orbital, you place two opposite spins (shown as arrows 鈫戔啌) in a single box representing paired electrons.
  • In the 3d orbitals, you have one electron per box until they start pairing, illustrating unpaired electrons in neutral titanium as 鈫 _ 鈫 _.
This representation helps students visually grasp the concept of electron pairing and the resulting magnetic properties of an atom or ion. It can simplify identifying unpaired electrons to assess paramagnetism.
Noble Gas Notation
In writing electron configurations efficiently, noble gas notation provides a succinct method. It simplifies the notation process by allowing us to start from the nearest noble gas configuration, rather than writing out everything from the very beginning.

Titanium, with an atomic number of 22, can use argon \(\text{[Ar]}\) as its noble gas starting point because argon has 18 electrons. From there, you continue to account for the remaining electrons: for neutral titanium, it becomes \(\text{[Ar]} 4s^2 3d^2\).

Using noble gas notation for ions like \(\mathrm{Ti}^{2+}\), we have \(\text{[Ar]} 3d^2\), as we remove the 4s electrons first. For \(\mathrm{Ti}^{4+}\), it is simply \(\text{[Ar]}\) since all valence electrons are removed.

This abbreviated form not only makes it easier to write configurations but also reflects the stability of electron arrangements based on completed orbitals, aligning with the chemical inertness exhibited by noble gases. It is particularly useful for large atoms and ions, where a full electron configuration would be tedious and overly complicated.

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Most popular questions from this chapter

Using orbital box diagrams, depict an electron configuration for each of the following ions: (a) \(\mathrm{Na}^{+}\) (b) \(\mathrm{Al}^{3+},\) (c) \(\mathrm{Ge}^{2+},\) and \((\mathrm{d}) \mathrm{F}^{-}\).

What is the maximum number of electrons that can be identified with each of the following sets of quantum numbers? In some cases, the answer may be "none." In such cases, explain why "none" is the correct answer. (a) \(n=3\) (b) \(n=3\) and \(\ell=2\) (c) \(n=4, \ell=1, m_{\ell}=-1,\) and \(m_{s}=+1 / 2\) (d) \(n=5, \ell=0, m_{\ell}=-1, m_{\mathrm{s}}=+1 / 2\)

Slater's rules are a simple way to estimate the effective nuclear charge experienced by an electron. In this approach, the "shielding constant," \(\mathrm{S}\), is calculated. The effective nuclear charge is then the difference between S and the atomic number, \(Z\). (Note that the results in Table 7.2 and Figure 7.2 were calculated in a slightly different way.) $$ Z^{*}=Z-\mathrm{S} $$ The shielding constant, \(\mathrm{S}\), is calculated using the following rules: 1\. The electrons of an atom are grouped as follows: (1s) \((2 s, 2 p)(3 s, 3 p)\) (3d) \((4 s, 4 p)\) (4d), and so on. 2\. Electrons in higher groups (to the right) do not shield those in the lower groups. 3\. For \(n s\) and \(n p\) valence electrons (a) Electrons in the same \(n s, n p\) group contribute 0.35 (for \(1 s 0.30\) works better). (b) Electrons in the \(n-1\) group contribute 0.85 (c) Electrons in the \(n-2\) group (and lower) contribute 1.00 4\. For \(n d\) and \(n f\) electrons, electrons in the same \(n d\) or \(n f\) group contribute \(0.35,\) and those in groups to the left contribute 1.00 As an example, let us calculate \(Z^{*}\) for the outermost electron of oxygen: $$ \begin{array}{c} \mathrm{S}=(2 \times 0.85)+(5 \times 0.35)=3.45 \\ Z^{*}=8-3.45=4.55 \end{array} $$ Here is a calculation for a \(d\) electron in \(\mathrm{Ni}\) : $$ Z^{*}=28-[18 \times 1.00]-[7 \times 0.35]=7.55 $$ and for an s electron in \(\mathrm{Ni}\) : $$ \begin{aligned} Z^{*}=28-[10 \times 1.00]-[16 \times 0.85] & \\ -[1 \times 0.35] &=4.05 \end{aligned} $$ (Here \(3 s, 3 p,\) and \(3 d\) electrons are in the \((n-1)\) groups.) (a) Calculate \(Z^{*}\) for \(F\) and \(N e\). Relate the \(Z^{*}\) values for O, F, and Ne to their relative atomic radii and ionization energies. (b) Calculate \(Z^{* *}\) for one of the \(3 d\) electrons of \(\mathrm{Mn}\), and compare this with \(Z^{*}\) for one of the \(4 s\) electrons of the element. Do the \(Z^{*}\) values give us some insight into the ionization of Mn to give the cation?

The actinide americium, Am, is a radioactive element that has found use in home smoke detectors. Depict its electron configuration using noble gas and spdf notations.

Use noble gas and spdf notations to depict electron configurations for the following metals of the third transition series. (a) Tantalum, Ta. The metal and its alloys resist corrosion and are often used in surgical and dental tools. (b) Platinum, Pt. This metal was used by pre-Columbian Indians in jewelry. Now it is still used in jewelry but it is also the basis for anticancer drugs and catalysts (such as those in automobile exhaust systems).

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