The second ionization energy of He is almost exactly four times the ionization energy of H, and the third ionization energy of Li is almost exactly nine times the ionization energy of H:


IE (MJ mol-1)
H(g) --> H +(g) + e- (1.3120)
He+(g) --> He2+(g) + e- (5.2504)
Li2+(g) --> Li3+(g) + e- (11.8149)

Explain this trend on the basis of the Bohr equation for energy levels of single-electron systems.

Well, isn't it fascinating how these ionization energies seem to follow a pattern? It's almost like they're playing a game of "Simon Says," but with electrons instead. Let's break it down!

First, we have hydrogen (H), the little one-electron superstar. When it loses an electron, it needs to overcome an ionization energy of 1.3120 MJ mol-1. Effortful, but doable.

Now, let's move on to helium (He), the cool-headed gas. Since it has two electrons, the first electron peels off quite easily with an ionization energy of 1.3120 MJ mol-1. But then it becomes a one-electron species (He+), oh how lonely! It wants to hold on to that precious electron, so when it loses it, it needs to exert four times more energy - a whopping 5.2504 MJ mol-1. Talk about not letting go!

Last on this electrifying journey, we have lithium (Li). It's the Frankenstein of the group, with three electrons. Similar to He, the first and second electrons are relatively simple to remove. But once it reaches one electron left (Li2+), it clings to it like a lifebuoy during a storm. Therefore, to pry that last electron loose and transform Li2+ into Li3+, it takes a jaw-dropping 11.8149 MJ mol-1. Nine times more than H's ionization energy!

So, what's the connection here? Well, let's refer to our good friend, the Bohr equation for energy levels of single-electron systems. As you go from one electron species to the next, the number of electrons decreases, causing a more significant attraction between the remaining electrons and the positively charged nucleus. This increased attraction makes it harder and harder to remove each successive electron.

In essence, it's like trying to peel off layers of clothing from someone who's really attached to them. The more layers you remove, the more that person clings onto those remaining layers. And it takes a ton of effort to pry off that last piece!

So, thanks to the Bohr equation and the stubborn nature of electrons, we see this delightful trend where the second ionization energy of He is four times the ionization energy of H, and the third ionization energy of Li is nine times the ionization energy of H. It's a game of electron attractions that has us all captivated!

The Bohr equation describes the energy levels of single-electron systems, such as hydrogen atoms. It states that the energy of an electron in a particular energy level can be calculated using the formula:

E = -Rh / n^2

where E is the energy, Rh is the Rydberg constant (which depends on the nucleus being considered), and n is the principal quantum number representing the energy level.

In the case of hydrogen (H), the first ionization energy is the energy required to remove one electron from a hydrogen atom to form a hydrogen ion (H+). The equation for this process can be written as:

H(g) --> H+(g) + e-

The Ionization energy of hydrogen is given as 1.3120 MJ mol-1.

Now, let's consider helium (He), which has two electrons. The second ionization energy of helium is the energy required to remove one more electron from the helium ion (He+) to form the helium cation (He2+). The equation for this process can be written as:

He+(g) --> He2+(g) + e-

The second ionization energy of helium is given as 5.2504 MJ mol-1, which is almost exactly four times the ionization energy of hydrogen.

Next, let's consider lithium (Li), which has three electrons. The third ionization energy of lithium is the energy required to remove one more electron from the lithium ion (Li2+) to form the lithium cation (Li3+). The equation for this process can be written as:

Li2+(g) --> Li3+(g) + e-

The third ionization energy of lithium is given as 11.8149 MJ mol-1, which is almost exactly nine times the ionization energy of hydrogen.

This trend can be explained based on the Bohr equation. The energy levels of single-electron systems are determined by the principal quantum number (n). As the number of electrons in an atom increases, the effective nuclear charge experienced by each electron also increases. This increased nuclear charge attracts electrons more strongly, causing the energy levels to become closer together.

As a result, it requires more energy to remove an electron from an atom with multiple electrons compared to an atom with only one electron. This is why the second ionization energy of helium is four times greater than the ionization energy of hydrogen, and the third ionization energy of lithium is nine times greater than the ionization energy of hydrogen. The increasing number of electrons and their increased attraction to the nucleus make it energetically unfavorable to remove additional electrons from these atoms.

The Bohr equation for energy levels of single-electron systems can help explain the trend in the second and third ionization energies of He and Li, respectively.

According to the Bohr model, the energy of an electron in a hydrogen-like atom can be calculated using the equation:

E = -(2.18 x 10^-18 J)/(n^2)
where E is the energy, n is the principal quantum number, and the negative sign indicates the energy is bound.

In hydrogen (H), the electron is in the first energy level (n=1). Therefore, the energy of the electron in H is:

E(H) = -(2.18 x 10^-18 J)/(1^2) = -2.18 x 10^-18 J

When H loses an electron to form H+, the energy required can be calculated as the difference between the energy of the electron in the initial and final states. Thus, the first ionization energy of H is:

IE(H) = 0 - (-2.18 x 10^-18 J) = 2.18 x 10^-18 J

Now let's consider the second ionization energy of helium (He). In He+, the remaining electron is in the second energy level (n=2). Therefore, the energy of the electron in He+ is:

E(He+) = -(2.18 x 10^-18 J)/(2^2) = -0.545 x 10^-18 J

When He+ loses an electron to form He2+, the second ionization energy can be calculated as the difference between the energy of the electron in the initial and final states. Thus, the second ionization energy of He is:

IE(He) = E(He2+) - E(He+) = 0 - (-0.545 x 10^-18 J)= 0.545 x 10^-18 J

Comparing the first ionization energy of H and the second ionization energy of He, we find:

IE(He) / IE(H) = (0.545 x 10^-18 J) / (2.18 x 10^-18 J) ≈ 0.25

This implies that the second ionization energy of He is approximately four times the first ionization energy of H. This observation is consistent with the Bohr equation and supports the trend mentioned.

Similarly, we can apply the same logic to explain the trend in the third ionization energy of Li. In Li2+, the remaining electron is in the third energy level (n=3). Using the Bohr equation, we can calculate the energy of the electron in Li2+ and the difference in energy between Li3+ and Li2+, giving the third ionization energy of Li.

By comparing the relative ionization energies based on the Bohr equation, we see that the third ionization energy of Li is approximately nine times the first ionization energy of H:

IE(Li) / IE(H) = (11.8149 MJ mol^-1) / (1.3120 MJ mol^-1) ≈ 9

In summary, the trend in the second and third ionization energies of He and Li, respectively, can be explained based on the Bohr equation for energy levels of single-electron systems. The ionization energies increase because electrons are being removed from higher energy levels with increasing principal quantum numbers, which require more energy to remove them.

The Bohr model is a function of RZ^2(1/n^2 - 1/n^2) where the first n is n1 and the second is n2

Z is the atomic number.
He is 2 and 2^2 = 4
Li is 3 and 3^2 = 9