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 given in the table. 1 12 13 | 0.507 | 1.017 | 4.108 14 5.074 15 16 6.147 | 7.903 17 8.294 | To what family of the periodic table does this new element probably belong? 18 (8A) O 1 (1A) 16 (6A) 14 (4A) 15 (5A) 13 (3A) 17 (7A) O 2 (2A)

Answer 1

We know that if an element has n electrons, it has n ionization energies namely I_{​​​​​​1} , I_{​​​​​​2} , I_{​​​​​​3}, ....

We also know that

I​​​​​_{​​​​​​1}< I_{​​​​​​2} < I_​​3 < I_{​​​​​​4 .......}

Frist ionization energy is less than second ionization energy.

This is due to in neutral atom has equal number of protons and electrons. Where as in uni positively charged ion has more number of protons than electrons.

In first ionisation energy, electron has to be removed from the neutral atom where as for second ionization energy, electron has to be removed from the unipositively charged ion.

As more nuclear charge, more attraction, hence more energy is required to remove an electron.

If we observed the trend, there is very high increase for I_{​​​​​​2} to I_{​​​​​​3.}

It means that after losing 2 electrons, it is getting stable electronic configuration. Hence to remove third electron, more energy is required.

From this we can conclude that the atom may 2 electrons in its valency shell.

The element may be 2nd group element ie alkaline earth metal.

Last option is correct.

Answer 2

answer is 2(2A)