Calculate the fourth ionization energy of beryllium in J atom^-1 & kJ mol^-1 units
There's an easier way to do this but I don't have it yet so here is the long way.
1/wavelength = RZ^2(1/n^2 - 1/n^2)
Solve for wavelength.
R = 1.0973E7
Z = 4
The first n^2 in the formula is 1
The second n&2 in the formula is 0
So 1.0973E7*4^2((1-0) and that gives you 1/wavelength in meters. Plug wavelength into and solve for E in J/atom
Multiply that by 6.022E23 to convert to J/mol and convert to kJ/mol.
Then E = hc/wavelength
To calculate the fourth ionization energy of beryllium (Be), you need to refer to its electron configuration and apply the concept of ionization energy.
The electron configuration of beryllium (Be) is 1s^2 2s^2.
In the ionization process, Be loses one electron at a time. The fourth ionization energy refers to the energy required to remove the fourth electron from a beryllium atom.
The ionization energies for the first, second, and third ionizations of beryllium are known values:
1st ionization energy (IE1) = 899.5 kJ/mol
2nd ionization energy (IE2) = 1,759.5 kJ/mol
3rd ionization energy (IE3) = 14,848.7 kJ/mol
To calculate the fourth ionization energy (IE4), we start with the electron configuration after the third ionization:
1s^2 (2nd shell completely empty)
Now, we remove the fourth (last) electron:
1s^1
To find the energy required to remove this electron, we need to consult the ionization energy for hydrogen (H). The fourth ionization energy is usually similar to the ionization energy for hydrogen.
The ionization energy for hydrogen is about 1312 kJ/mol. Therefore, we can roughly estimate the fourth ionization energy of beryllium to be close to this value.
So, the estimated value for the fourth ionization energy of beryllium is approximately 1312 kJ/mol or 1,312,000 J/mol.
To calculate the fourth ionization energy of beryllium, we need to consider its electron configuration and the number of valence electrons.
Beryllium (Be) has an atomic number of 4, meaning it has four electrons. The electron configuration of beryllium is 1s² 2s², indicating that it has two electrons in the 1s orbital and two electrons in the 2s orbital.
To calculate the fourth ionization energy, we need to remove one electron from the 2s orbital. Since the fourth ionization energy represents the energy required to remove the fourth electron, we need to look at the ion that has already lost three electrons.
If we remove the fourth electron from a Be³⁺ ion (which has already lost three electrons), the electron configuration would be 1s².
The ionization energy can be calculated using the equation:
Ionization energy = (final energy) - (initial energy)
The initial energy is the energy of a Be³⁺ ion with an electron configuration of 1s². Since the electron configuration is the same as that of helium (He), we can take the ionization energy of helium as a reference.
So, we have to look up the ionization energy of helium to find the initial energy. The ionization energy of helium is approximately 2.37 x 10⁶ J mol^-1 or 2.37 x 10³ kJ mol^-1.
Now, let's calculate the fourth ionization energy:
Fourth ionization energy = (final energy) - (initial energy)
= 0 J mol^-1 (since we have a Be⁴⁺ ion) -2.37 x 10⁶ J mol^-1
= -2.37 x 10⁶ J mol^-1 (or -2.37 x 10³ kJ mol^-1)
In this case, the fourth ionization energy of beryllium is -2.37 x 10⁶ J mol^-1 or -2.37 x 10³ kJ mol^-1. The negative sign indicates that energy is released when the fourth electron is removed.