Answer :
To determine the predicted order of first ionization energies from highest to lowest for lithium (Li), sodium (Na), potassium (K), and rubidium (Rb), we need to consider their positions in the periodic table and the general trend of ionization energy.
Ionization energy is the energy required to remove an electron from an atom in its gaseous state. As we move down a group in the periodic table, the ionization energy generally decreases. This happens because the outer electrons are farther from the nucleus and are therefore less tightly bound due to increased electron shielding.
Here’s a step-by-step analysis of the first ionization energies for the elements in question:
1. Lithium (Li):
- Lithium is in the second period and Group 1 of the periodic table.
- It has a relatively high ionization energy among Group 1 elements because it is closer to the nucleus and experiences less electron shielding.
2. Sodium (Na):
- Sodium is in the third period and Group 1 of the periodic table.
- Compared to lithium, sodium's outer electron is further from the nucleus and experiences more electron shielding, resulting in a lower ionization energy than lithium.
3. Potassium (K):
- Potassium is in the fourth period and Group 1 of the periodic table.
- Potassium's outer electron is even further from the nucleus than sodium's, leading to even more electron shielding and lower ionization energy.
4. Rubidium (Rb):
- Rubidium is in the fifth period and Group 1 of the periodic table.
- Rubidium's outer electron is the farthest from the nucleus among these four elements, resulting in the greatest electron shielding and the lowest first ionization energy.
Considering these points, we can predict the order of first ionization energies from highest to lowest as follows:
- Lithium (Li) has the highest ionization energy.
- Next comes Sodium (Na).
- Potassium (K) follows.
- Rubidium (Rb) has the lowest ionization energy.
So, the correct predicted order from highest to lowest ionization energy is:
Li > Na > K > Rb
Thus, the correct answer is:
[tex]\[ \boxed{Li > Na > K > Rb} \][/tex]
Ionization energy is the energy required to remove an electron from an atom in its gaseous state. As we move down a group in the periodic table, the ionization energy generally decreases. This happens because the outer electrons are farther from the nucleus and are therefore less tightly bound due to increased electron shielding.
Here’s a step-by-step analysis of the first ionization energies for the elements in question:
1. Lithium (Li):
- Lithium is in the second period and Group 1 of the periodic table.
- It has a relatively high ionization energy among Group 1 elements because it is closer to the nucleus and experiences less electron shielding.
2. Sodium (Na):
- Sodium is in the third period and Group 1 of the periodic table.
- Compared to lithium, sodium's outer electron is further from the nucleus and experiences more electron shielding, resulting in a lower ionization energy than lithium.
3. Potassium (K):
- Potassium is in the fourth period and Group 1 of the periodic table.
- Potassium's outer electron is even further from the nucleus than sodium's, leading to even more electron shielding and lower ionization energy.
4. Rubidium (Rb):
- Rubidium is in the fifth period and Group 1 of the periodic table.
- Rubidium's outer electron is the farthest from the nucleus among these four elements, resulting in the greatest electron shielding and the lowest first ionization energy.
Considering these points, we can predict the order of first ionization energies from highest to lowest as follows:
- Lithium (Li) has the highest ionization energy.
- Next comes Sodium (Na).
- Potassium (K) follows.
- Rubidium (Rb) has the lowest ionization energy.
So, the correct predicted order from highest to lowest ionization energy is:
Li > Na > K > Rb
Thus, the correct answer is:
[tex]\[ \boxed{Li > Na > K > Rb} \][/tex]
