Write Equations That Show the Process for the First Two Ionization Energies of Lead
Write An Equation That Shows The Process For The First Two Ionization Energies For Nickel, Ni.
If we wish to remove an electron from a gaseous atom the energy that is required is known as Ionization Energy.
It is defined as follows:
The amount of energy that is required to remove the most loosely bound electrons form an isolated gaseous atom to form a gaseous ion is called the ionization enthalpy or ionization energy.
It is usually represented in the units of kJ mol-1
When one electron is removed there is often a chance that other electrons can also be removed from the orbitals. That is when the successive ionization energy comes into force.
As the name suggests that the second Ionization energy is the energy that is required to remove an electron from a unipositive cation. An ion that only has lost a single electron due to the first ionization energy, has a single positive charge that is why it is known as a unipositive cation.
The dipostive cation will further be subjected to third ionization energy that will remove the third electron form the orbital.
And so the process continues.
When we look at the element at hand, Nickel. It has the atomic number 28. With the electronic configuration of [Ar] 3d8 4s1
General Formula for Successive Ionization Energy
M (g) + IE1 → M+ (g) + e-
M+ (g) + IE2 → M2+ (g) + e-
M2+ (g) + IE3 → M3+ (g) + e-
The second, third and fourth etc. ionization energies are known as successive ionization energies.
IE 3 > IE 2 > IE 1
So we will be using the above mentioned general formulae to solve the question.
Ni (g) + IE1 → Ni+ (g) + e- Ni+ (g) + IE2 → Ni2+ (g) + e-
If we talk in terms of numbers:
The first ionization energy of Nickel is about 737.1 kJ mol-1.
The second ionization energy of Nickel is 1753 kJ mol-1.
There is another very popular question that is generally paired with the question solved above.
It is the question on Zirconium.
The table of the values of ionization energies of Nickel, ionization energies of Lead and ionization energies of Zirconium is given below.
Write An Equation That Shows The Process For The Fourth Ionization Energies For Zirconium, Zr?
Zirconium as we all know is a transition metal with the atomic number 40. It has an electronic configuration:
[Kr] 4d² 5s²
So by using the general formula we get the following equation as the answer to the question.
Zr+ (g) + IE4 → Zr4+ (g) + e-
Numerically, it is equal to : 3313 kJ mol -1
Write An Equation That Shows The Process For The First Two Ionization Energies For Lead, Pb?
Lead is a heavy metal with the atomic number 82. It has an electronic configuration:
[Xe] 4f 14 5d 10 6s 2 6p 2
The general formula gives us an idea on how to solve this question.
Pb (g) + IE1 → Pb+ (g) + e-
Pb+ (g) + IE2 → Pb2+ (g) + e-
Numerically the first and second ionization energies of Lead, Pb is equal to:
First = 715.6 kJ mol -1
Second = 1450.5 kJ mol -1
Write a Chemical Reaction Representing The Second Ionization Energy For Lithium?
Lithium is a member of the group Alkali Metals and is has the highest second ionization energy.
It has the atomic number 3
And has the electronic configuration [He] 2s 1
Li+ (g) + IE2 → Li2+ (g) + e-
The equation given above is the equation which represents the second Ionization Energy for Lithium. Numerically, IE2 of Lithium is equal to 7298.1 kJ mol-1.
Conclusion
Here is the table given in HTML format.
IE | Nickel kJ mol-1 | Lead kJ mol-1 | Zirconium kJmol-1 |
---|---|---|---|
1 | 737.1 | 715.6 | 640.1 |
2 | 1753 | 1450.5 | 1270 |
3 | 3395 | 3081.5 | 2218 |
4 | 5300 | 4083 | 3313 |
5 | 7339 | 6640 | 7752 |
6 | 10400 | 9500 | |
7 | 12800 | ||
8 | 15600 | ||
9 | 18600 | ||
10 | 21670 |
Write Equations That Show the Process for the First Two Ionization Energies of Lead
Source: https://www.learningtopnotch.com/2020/11/write-equations-that-show-process-for.html
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