Answer
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Hint: The ionic radius can be defined as the distance from the nucleus of the ion to the outermost electron in an ion, whereas the atomic radius can be defined as the distance from the nucleus of the atom to the normal valence electron in the outermost orbit.
Complete step by step answer:
Coming to given options, option A, Ionic radius is proportional to atomic number.
The ionic radius increases as the atomic number increases. Both trends exist because as we move from top to bottom in the groups of periodic tables, the size of the atom increases because addition of electrons takes place in new shells. As the number of electrons increases the shielding effect also increases. So, the ion is not held together with the nucleus as its size increases. Therefore the outermost electrons are not attracted strongly by the nucleus.
But when we are moving from left to right in the periodic table ionic size decreases then ionic radius is inversely proportional to atomic number in periods.
So, option A and B are wrong, because those options are not going to validate throughout the periodic table.
Coming to option C, Ionic radius is inversely proportional to effective nuclear charge.
The radius of the ion is subject to the effective nuclear charge, more the effective nuclear charge, more strongly the electrons are attracted towards the nucleus and the size of the ion decreases. Therefore the ionic radius is inversely proportional to effective nuclear charge.
\[Ionic\text{ }radius\propto \dfrac{1}{{{Z}_{eff}}\text{(effective nuclear charge)}}\]
So, the correct answer is “Option C”.
Note: When we are moving from top to bottom in the periodic table ionic radius is inversely proportional to effective nuclear charge and as we are moving from right to left in the periodic table ionic radius is inversely proportional to effective nuclear charge.
Complete step by step answer:
Coming to given options, option A, Ionic radius is proportional to atomic number.
The ionic radius increases as the atomic number increases. Both trends exist because as we move from top to bottom in the groups of periodic tables, the size of the atom increases because addition of electrons takes place in new shells. As the number of electrons increases the shielding effect also increases. So, the ion is not held together with the nucleus as its size increases. Therefore the outermost electrons are not attracted strongly by the nucleus.
But when we are moving from left to right in the periodic table ionic size decreases then ionic radius is inversely proportional to atomic number in periods.
So, option A and B are wrong, because those options are not going to validate throughout the periodic table.
Coming to option C, Ionic radius is inversely proportional to effective nuclear charge.
The radius of the ion is subject to the effective nuclear charge, more the effective nuclear charge, more strongly the electrons are attracted towards the nucleus and the size of the ion decreases. Therefore the ionic radius is inversely proportional to effective nuclear charge.
\[Ionic\text{ }radius\propto \dfrac{1}{{{Z}_{eff}}\text{(effective nuclear charge)}}\]
So, the correct answer is “Option C”.
Note: When we are moving from top to bottom in the periodic table ionic radius is inversely proportional to effective nuclear charge and as we are moving from right to left in the periodic table ionic radius is inversely proportional to effective nuclear charge.
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