Ionic Radii - Ionic Radii of Cation and Anion, Trends Across the Periodic Table, Comparison of Radii of Isoelectronic Species, Practice Problems and FAQ

When a tree sheds its leaves in the autumn, the amount of shade it casts on its surroundings decreases. When a tree grows new leaves in the spring, the radius of its shadow increases, and so does the size of its shadow!

Consider the tree as an atom and the spread of its leaves as atomic size. When it loses some leaves (electrons), it forms ions! The size of this new form of the tree (i.e., ion) is analogous to the ionic radii.

It is important to study the periodic trends of ionic radii for analysing and predicting molecular properties and interactions.

Ions are formed when an atom loses or gains electrons. When an atom loses an electron it forms a cation and when it gains an electron it forms an anion. The ionic radius can be found by measuring the distance between the cation and the anion in an ionic crystal.

Ionic radius is the distance from the nucleus of an ion up to which it has an influence on its electron cloud.

rc and ra are the radius of cation and anion, respectively.

TABLE OF CONTENTS

Ionic Radii of Cation
Ionic Radii of Anion
Trends in Ionic Radii Down the Group
Trends in Ionic Radii Across a Period
Comparison of Radii of Isoelectronic Species
Important Facts on Ionic Radii
Practice Problems
Frequently Asked Questions - FAQ

Ionic Radii of Cation

A cation is smaller than its parent atom because it has fewer electrons while its nuclear charge remains the same. So, an increase in effective nuclear charge decreases the size.

For example, the atomic radius of Sodium (Na) is 186 pm compared to the ionic radius of 95 pm for Na⁺.

Ionic Radii of Anion

The size of an anion is larger than its parent atom because the addition of one or more electrons results in increased repulsion among the electrons and a decrease in the effective nuclear charge.

For example, the ionic radius of fluoride ion F^- is 136 pm, whereas the atomic radius of fluorine is only 64 pm.

The relation between internuclear distance: d_A-B= r_c+ r_a

Trends in Ionic Radii Down the Group

On moving down a group in the modern periodic table, atoms add an extra shell due to which the ionic radius of elements increases down a group.

For example, Li⁺< Na⁺ < K⁺ < Rb⁺ and Be²⁺< Mg²⁺ < Ca²⁺ < Sr²⁺

Ions	Configuration	Ionic radii (nm)	Ions	Configuration	Ionic radii (nm)
Li⁺	2	0.076	F^-	2,8	0.133
Na⁺	2,8	0.102	Cl^-	2,8,8	0.181
K⁺	2,8,8	0.138	Br^-	2,8,18,8	0.196

Trends in Ionic Radii Across a Period

We shall try to understand the trends in the ionic radius of elements across a period with an example.

In period 3, we find that the atomic radius first decreases and then suddenly increases and then again it slowly decreases.

This is because the starting elements in a period tend to form cations, and the elements towards the end of a period tend to form anions.

Trends in Ionic Radii across Period 3

Period 3	Na⁺	Mg²⁺	Al³⁺	P^3-	S^2-	Cl^-
Number of protons	11	12	13	15	16	17
Electronic configuration	2,8	2,8	2,8	2,8,8	2,8,8	2,8,8
Ionic radius	0.102	0.072	0.054	0.212	0.184	0.181

Comparison of Radii of Isoelectronic Species

Atoms/Ions having the same number of electrons are called isoelectronic species. Their radii would be different because of different nuclear charges.

In a set of species having the same number of electrons (isoelectronic), the size decreases with an increase in the magnitude of nuclear charge.

Cations with a greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus.

An anion with a greater negative charge will have a larger radius as the net repulsion of the electrons will outweigh the nuclear charge and the ion will expand in size.

The ionic radii of isoelectronic species increase with a decrease in the magnitude of nuclear charge.

Important facts on Ionic radii

The radius of a cation is always smaller than that of the parent atom.

The radius of an anion is always larger than that of the parent atom.

In a certain group, the ionic radii increases down the group.

In a set of given species that are isoelectronic, the size decreases with an increase in the magnitude of effective nuclear charge.

The size of the cations of the same element decreases with the increase in positive charge.

For the same atomic number - Size of anion > Size of atom > Size of cation.

Practice Problems

Q1. Mention the correct increasing order in the size for the ions: Li+, Mg2+, K+, Al3+

A) Mg²⁺< Al³⁺ < Li⁺< K⁺
B) Al³⁺< Mg²⁺< Li⁺< K⁺
C) Li⁺ < Mg²⁺< Al³⁺< K⁺
D) Al³⁺< K⁺< Li⁺< Mg²⁺

Answer: B)

Al³⁺and Mg²⁺are isoelectronic species, but due to the greater effective nuclear charge of Al³⁺, the size of Al³⁺ is less than that of Mg²⁺. The size of K⁺ is larger than Li⁺as size increases down the group. Li⁺ is the smallest in radii as it has only 1 shell.

Q2. Which of the following pairs is an example of isoelectronic species?

A) Na⁺and Si⁴⁺
B) Na⁺and Cl^-
C) Mg²⁺ and Cl^-
D) S^2- and O^2-

Answer: A)

The atomic number of silicon is 14 and Si⁴⁺ is a 10 electron species.
The atomic number of sodium is 11 and Na⁺ is a 10 electron species.
The atomic number of chlorine is 17 and Cl^- is a 18 electron species.
The atomic number of magnesium is 12 and Mg²⁺ is a 10 electron species.
The atomic number of sulphur is 16 and S^2- is a 18 electron species.
The atomic number of oxygen is 8, and O^2-is a 10 electron species.

From the above information, the pair of ions given in only option A) is an example of isoelectronic species.

Q3. Which has the highest ionic radii?

A) Sodium
B) Magnesium
C) Chlorine
D) Potassium

Answer: D)

Ionic radii decreases with the increase in effective nuclear charge as we move from left to right across a period.

Magnesium has dipositive charge, and lies next to sodium in the third period. Due to the greater effective nucelar charge in magnesium, the ionic radius of magnesium will be comparatively lesser than that of sodium. Chlorine lies to the extreme right in the third period. So, the ionic radius of chloride ion will be the least among all.

On moving down a group, atoms add an extra shell due to which the ionic radius of elements increases down a group. Potassium lies in the fourth period, just below sodium, and hence the ionic size of potassium is the highest among the given options.

Q4. Why is the ionic radii of Ga³⁺ lower than Al³⁺?

Answer: Gallium contains inner d - electrons which do not shield the nuclear charge effectively. Consequently, the outer electrons in gallium experience a greater force of attraction by the nucleus than in aluminium. Hence, the ionic radius of Ga³⁺ is slightly less than that of Al³⁺ l

Frequently Asked Questions - FAQ

Question 1. Which has the smallest radii, F^-or H^-?
Answer: Though fluorine has an extra shell as compared to hydrogen, the smallest anion is F^- and not H^-. This is because of the higher effective nuclear charge in F^- as compared to H^-.

Question 2. What is the impact of shielding effect on ionic radii?
Answer: The shielding effect causes a decrease in the force of attraction exerted by the nucleus on the valence electrons due to the presence of electrons in the inner shells and as such ionic radii increase with an increase in the screening effect.

Question 3. What is the penetration effect and what is its relation with ionic radii?
Answer: The penetration effect corresponds to the penetration of effective nuclear charge through the orbitals which attract the outermost shell electrons. It is the opposite of the shielding effect. Hence, the greater the penetration effect, the lower the ionic radii.

Question 4. How does d-orbital contraction impact ionic radii?
Answer: In d-block elements, on moving from left to right, the effective nuclear charge increases. Hence, the size decreases.

This trend can be observed till Mn, after which the atomic size remains unchanged and then increases for Zn.
Unpaired d-electrons increase the nuclear force of attraction and nuclear charge by one unit. Hence, the size decreases up to Mn, after this pairing of electrons starts.
This is due to the poor shielding effect of d-orbitals, i.e., the d-orbital screens the outer electrons (4s) from the inward pull of the nucleus, but not effectively. This is known as the d-orbital contraction.
Towards the end of the series, there is an increase in the interelectronic repulsion between the electrons added to the 3d orbitals.
This increase in repulsion becomes greater than that of the attraction between the nucleus and the 4s electrons. Because of the greater magnitude of electron-electron repulsion, the electron cloud of Zn expands. Hence, its size also becomes greater than that of Cu.

Related Topics

Atomic Radii- Comparison of Various atomic radii	Modern Periodic Table
Electron affinity and Electron gain Enthalpy	Factors affecting electron gain enthalpy
Ionization Enthalpy	Successive electron gain enthalpy