Question 1
Consider this set of ionization energies
- IE_1 = 578 kJ/mol
- IE_2 = 1820 kJ/mol
- IE_3 = 2750 kJ/mol
- IE_4 = 11, 600 kJ/mol
To which third-period element do these ionization values belong?
Question 2
Question 3
These ionization energies (KJ/mol) have been reported for element “ZZ”:
first: 500
second: 4560
third: 6910
fourth: 9540
fifth: 13400
Which of the following elements is most likely “ZZ”?
1. SiSi.
2. MgMg.
3. AlAl.
4. NaNa.
5. ClCl.
Answer to question 1
The given ionization energies of an element of group 3 are as follows:
IE1=578kJ mol−1IE1=578kJ mol−1
IE2=1820kJ mol−1IE2=1820kJ mol−1
IE3=2750kJ mol−1IE3=2750kJ mol−1
IE4=11,600kJ mol−1IE4=11,600kJ mol−1.
Ionization energies values are increasing as the number of electrons is removed. The value of fourth ionization energy is highest. It states that the element of the third period acquired the nearest noble gas configuration after the removal of the third electron. Its second ionization energy value is much higher than the first one which states that electron is removed from stable orbital.
Among the elements of period 3, aluminum is the only element which can easily lose its outer valence electron from 3p orbital as only one electron is present in this subshell. Next electron is removed from fully filled 3s orbital, so it requires more energy.
Therefore, the above given values of ionization energies belong to Al (aluminum ) of period 3 with atomic number 13.
Its electronic configuration is:
[Ne]3s23p1[Ne]3s23p1.
Answer to question 2
Answer to question 3
The first ionization energy is small and there is a large jump at the second ionization energy. The third and fourth ionization energies gradually increase. This suggests that the atom is in the noble gas conformation once the first electron has been removed and that a large amount of energy is required to remove the second electron since this conformation is very stable. This means the atom is likely an alkali metal. Therefore the identity of element Z is most likely 4.