40.3k views
3 votes
The successive ionization energies for an unknown element are listed below. To which family in the periodic table does the unknown element likely belong. Explain your reasoning please.

I = 896 kj/mol
I2 = 1752 kj/mol
I3 = 14807 kj/mol
I4 = 17948 kj/mol

User Forage
by
7.6k points

2 Answers

1 vote
The jump from 1966 to 16347 is the largest one or simply we can say it is hard to lose the 3rd electron.Whereas, it is relatively easy to lose the first two electrons.

So there will be only 2 electrons in the outer most shell.

According to the information mentioned above we can conclude the unknown element likely belongs to the second group.
I2 = 1752 kj/mol
User Shivid
by
6.5k points
5 votes

Answer:

Belongs to the group 2A

Step-by-step explanation:

As you can see, the first two ionization energies are close and low, meaning that this element ionizates easily.

Not only loses easily the first electron, but the second too

To remove the third electron you requiered a huge amount of energy

Now, elements easily ionizable are the ones from group IA, group 2A and transition metals.

The last ones have mixed characteristics in matter of how many electrons you can remove from them, so they are not a family.

Now the question: group I or group II ?

The elements of group I have low ionization energies for the first electron but high energies for the second ones.

Being all that said, the unknown element belongs to the Group 2A

User Afshin
by
7.0k points