Question

A space probe identifies a new element in a sample collected from an asteroid. Successive ionization energies (in attojoules per atom) for the new element are shown below.
I1 I2 I3 I4 I5 I6 I7
0.507 1.017 4.108 5.074 6.147 7.903 8.294

To what family of the periodic table does this new element probably belong?
1(1A)
2(2A)
13(3A)
14(4A)
15(5A)
16(6A)
17(7A)
18(8A)

Answer 1

Ionization energy (IE) is the amount of energy required to remove an electron.
 
If you observe the IEs sequentially, there is a large gap between the 2nd and 3rd. This suggests it is difficult to remove more than 2 two electrons. Elements that lose two electrons to become more stable are found in the Group 2A (2 representing the number of electrons in the outermost valence shell).

Answer 2

The new element identified by a space probe belongs to $\boxed{{\text{2A}}}$ family of the periodic table.

Further explanation:

Ionization energy:

The amount of energy that is required to remove the most loosely bound valence electrons from the isolated neutral gaseous atom is known as ionization energy. It is represented by IE. Its value is related to the ease of removing the outermost valence electrons. If these electrons are removed so easily, small ionization energy is required and vice-versa. It is inversely proportional to the size of the atom.

Successive ionization energies are evaluated when the electrons are to be removed from successive shells. When the first electron is removed from the isolated neutral gaseous atom, ionization energy is termed as the first ionization energy $\left( {{\text{I}}{{\text{E}}_{\text{1}}}} \right)$. Similarly when the second electron is removed from the monoatomic cation, ionization energy is called the second ionization energy $\left( {{\text{I}}{{\text{E}}_{\text{2}}}} \right)$ and so on…

Ionization energy trends in the periodic table:

While moving from left to right in a period, IE increases due to the decrease in the atomic size of the succeeding members. This results in the strong attraction of electrons and hence are difficult to remove.

While moving from top to bottom in a group, IE decreases due to the increase in the atomic size of the succeeding members. This results in the lesser attraction of electrons and hence are easy to remove.

The first ionization energy of the new element is 0.507 and its second ionization energy is 1.017. There is only a small difference in the first and second ionization energies. But when we go from second ionization energy to the third one, there occurs a large difference in the ionization energy of the element. This indicates the removal of two electrons occurs with great ease but third electron is removed with high difficulty. So the new element can easily lose two electrons and therefore it belongs to 2A family of the periodic table.

Learn more:

1. Rank the elements according to first ionization energy: brainly.com/question/1550767

2. Write the chemical equation for the first ionization energy of lithium: brainly.com/question/5880605

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Periodic classification of elements

Keywords: ionization energy, isolated, neutral gaseous atom, valence electrons, successive ionization energy, IE, IE1, IE2, first ionization energy, second ionization energy, 2A, group, period, periodic table.