Ionic radius is:
A. inversely proportional to effective nuclear charge.
B. inversely proportional to square of effective nuclear charge.
C. directly proportional to effective nuclear charge.
D. directly proportional to square of effective nuclear charge.
Answer
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Hint: Ionic radius and effective nuclear charge has a very strong relation between them. In a period, effective nuclear charge increases due to the increase in the number of electrons in the outermost shell. Thus it is more attracted to the nucleus.
Complete step by step solution:
Ionic radius is the radius of an atom’s ion. It is actually the distance between the nucleus’ centre of ion and the outermost shell of atom. Generally ionic radius is 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. More the shells, greater the shielding effect experienced by the valence electrons since it is very far from the nucleus. Effective nuclear charge is the net positive charge experienced by an electron. It is given by the formula:
${{\text{Z}}_{{\text{eff}}}} = {\text{Z}} - {\text{S}}$, where \[{\text{Z}} \to \]number of protons in nucleus or atomic number
\[{\text{S}} \to \]number of electrons between nuclei.
When electrons are given out, the effective nuclear charge gets increased. The electrons are more 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.
Thus ionic radii is inversely proportional to effective nuclear charge.
Hence option A is correct.
Additional information:
When the proton to electron ratio is large, the radius will be small. Atomic radius increases on moving down the group. This is due to the presence of an extra shell.
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.
Complete step by step solution:
Ionic radius is the radius of an atom’s ion. It is actually the distance between the nucleus’ centre of ion and the outermost shell of atom. Generally ionic radius is 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. More the shells, greater the shielding effect experienced by the valence electrons since it is very far from the nucleus. Effective nuclear charge is the net positive charge experienced by an electron. It is given by the formula:
${{\text{Z}}_{{\text{eff}}}} = {\text{Z}} - {\text{S}}$, where \[{\text{Z}} \to \]number of protons in nucleus or atomic number
\[{\text{S}} \to \]number of electrons between nuclei.
When electrons are given out, the effective nuclear charge gets increased. The electrons are more 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.
Thus ionic radii is inversely proportional to effective nuclear charge.
Hence option A is correct.
Additional information:
When the proton to electron ratio is large, the radius will be small. Atomic radius increases on moving down the group. This is due to the presence of an extra shell.
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.
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