Question

A student reported the radii of Cu, ${Cu^{+}}$, and ${Cu^{2+}}$ as 96pm, 122pm, and 72pm respectively. Do you agree with the reported value? Justify your answer.

Answer 1

In order to obtain the answer to the question, we must get familiar with the relation between ionic radius and effective nuclear charge. Ionic radius is the distance between the nucleus and last electron in the shell and the attraction caused by the nucleus on the electrons is known as effective nuclear charge.

Complete answer:
- The ionic radius is generally smaller than atomic radius. The electrons can protect each other from the pull of the nucleus.
- This is called the shielding effect. It decreases the attraction between electron and nuclei.
- Ionic radius and effective nuclear charge have a very strong relation between them.
- In a period, the effective nuclear charge increases due to increase in the number of electrons in the outermost shell. Thus, it is more attracted to the nucleus.
- When electrons are given out, the effective nuclear charge gets increased. The electrons are attracted towards the nucleus. When anions are formed, electrons are gained by the atom. Thus, the nucleus has a lesser force towards the outermost shell.
- Hence, the ionic radii are inversely proportional to effective nuclear charge.
- As for the given question, the values to ionic radii are incorrect.
- In the question, the given values suggest that the order of radii should be
$Cu^{+} > Cu > Cu^{2+}$
- But, according to the statements mentioned above, the order of the ionic radii should be
$Cu^{+2} < Cu^{+} < Cu$

Note:
Ionic radius increases on moving down the group. Left to right ionic radii decreases initially then increases. This is because initially they form cations and finally, they form anions.