Question

Write equations that show the process for (a) the first two ionization energies of tin and (b) the fourth ionization energy of titanium.

Short answer

a) The first two ionization energies of tin (Sn) are represented by the following equations: 1) $Sn(g)\to S{n}^{+}(g)+e^{-}$ 2) $S{n}^{+}(g)\to S{n}^{2+}(g)+e^{-}$ b) The fourth ionization energy of titanium (Ti) is represented by the equation: $T{i}^{3+}(g)\to T{i}^{4+}(g)+e^{-}$

Step-by-step solution

Write the general ionization energy equation

To calculate the first ionization energy (I1) of tin (Sn), we will use the following general ionization energy equation: $$A(g)\to A^{+}(g)+e^{-}$$ Where A is the atom whose ionization energy we want to find (in this case, Sn), and the superscripts denote the charge on the ion and electron, respectively.

Apply the equation to tin

For the first ionization energy of tin, we have: $$Sn(g)\to S{n}^{+}(g)+e^{-}$$ This equation represents the process of removing one electron from a gaseous tin atom to form a Sn(+1) ion and an electron. #b) Second Ionization Energy of Tin#

Write the general ionization energy equation

The general ionization energy equation holds true not only for the first ionization energy but also for any successive ionization. For the second ionization energy (I2) of tin (Sn), we will still use the equation: $$A^{+}(g)\to A^{2+}(g)+e^{-}$$

Apply the equation to tin

For the second ionization energy of tin, we need to consider the Sn(+1) ion created in the first ionization step: $$S{n}^{+}(g)\to S{n}^{2+}(g)+e^{-}$$ This equation represents the process of removing another electron from the Sn(+1) ion to form a Sn(+2) ion. #c) Fourth Ionization Energy of Titanium#

Write the general ionization energy equation

For the fourth ionization energy (I4) of titanium (Ti), we will use the following general ionization energy equation: $$A^{3+}(g)\to A^{4+}(g)+e^{-}$$

Apply the equation to titanium

For the fourth ionization energy of titanium, we have: $$T{i}^{3+}(g)\to T{i}^{4+}(g)+e^{-}$$ This equation represents the process of removing a fourth electron from a gaseous Ti(+3) ion to form a Ti(+4) ion and an electron.

Key concepts

Tin Ionization

Tin (Sn) is a chemical element that undergoes ionization to form cations by losing electrons. Ionization energy refers to the energy required to remove an electron from a gaseous atom or ion. For tin, the first ionization energy is the energy needed to remove the first electron from a neutral tin atom.
This process can be depicted by the equation: $$Sn(g)\to S{n}^{+}(g)+e^{-}$$ After removing the first electron, tin becomes a positively charged ion $S{n}^{+}$. To further ionize the $S{n}^{+}$ ion, removing a second electron, creates a $S{n}^{2+}$ ion. This can be shown by: $$S{n}^{+}(g)\to S{n}^{2+}(g)+e^{-}$$ Understanding these steps is crucial, as each ionization step requires progressively higher energy due to increased attraction between the remaining electrons and the positively charged nucleus.

Titanium Ionization

Titanium (Ti) also ionizes by losing electrons, like many other transition metals. In the context of its fourth ionization energy, we consider the removal of an electron from an already triply charged titanium ion. The equation depicting this transformation is: $$T{i}^{3+}(g)\to T{i}^{4+}(g)+e^{-}$$ The complexity and energy demand increase with each successive removal of an electron. For titanium, the fourth ionization energy indicates that three electrons have already been removed, and now another is taken away, making the atom more positive and stable in terms of electronic configuration.

Electron Removal

Removing electrons from atoms turns them into positively charged ions. This process involves overcoming the electrostatic force between the negative electrons and the positive nucleus. Each electron removed requires energy, known as ionization energy.
Factors influencing the ionization energy include:

• Atomic size: Larger atoms have electrons further from the nucleus, thus requiring less energy to remove.
• Nuclear charge: Atoms with more protons have a stronger nuclear attraction, needing more energy for ionization.
• Electron shielding: Inner shell electrons can shield outer electrons from the full effect of the nucleus, affecting ionization energy.

Each step of electron removal reveals a different ionization energy, emphasizing the unique energy required at each stage.

Successive Ionization

Successive ionization refers to the step-by-step removal of electrons, where each removal increases the ion's positive charge. As more electrons are removed, they are closer to the nucleus and more strongly attracted to it, resulting in higher ionization energies. For example, after removing the first electron from tin, the energy needed to remove a second electron increases because it is being removed from a positively charged ion rather than a neutral atom.
In general:

• The first ionization energy is the lowest.
• Each subsequent ionization energy is higher than the previous one.
• Successive ionizations reveal the shell structure and electronic configuration of an element.

Understanding these steps is vital in chemistry to predict the behavior of elements during reactions.

General Ionization Equation

The general ionization equation provides a framework for understanding how atoms convert into ions by losing electrons. This equation applies universally to any element and its electrons.
The formula for the first ionization is: $$A(g)\to A^{+}(g)+e^{-}$$ Where $A$ stands for any element undergoing ionization. For subsequent ionizations, we modify the charge of $A$ in the equation, such as: $$A^{+}(g)\to A^{2+}(g)+e^{-}$$ Key Points:

• The general form remains constant, adapting to any subsequent charge changes.
• Simplifies the understanding of ionization across different elements.
• Can be adapted for multiple ionizations easily, demonstrating the pattern of increasing energy requirements with each electron removal.

Being acclimated to these equations aids in understanding the energetic requirements for forming ions from atoms.