Atomic radii, ionic radii, and isoelectronic ions are key concepts in understanding periodic trends in the periodic table. These concepts explain how the size of atoms and ions changes across periods and down groups, and how gain or loss of electrons affects ionic size. This section includes conceptual questions and answers that help clarify differences between atomic and ionic radii, as well as size variations in isoelectronic species based on nuclear charge, making it essential for building strong fundamentals in chemistry.
Conceptual Questions Based On Atomic Radii, Ionic Radii and Isoelectronic Ions Trends of Periodic Table
Q: Find one species that is isoelectronic with each :
(i) F⁻ (ii) Ar (iii) Mg²⁺ (iv) Rb⁺ [NCERT]
Answer :
(i) F⁻ :
Fluorine has atomic number 9, so a neutral atom has 9 electrons. The F⁻ ion gains one electron, making the total number of electrons 10. Any species with 10 electrons will be isoelectronic with F⁻.
Hence, Ne (atomic number 10) is isoelectronic with F⁻.
(ii) Ar :
Argon has atomic number 18, so it contains 18 electrons. Any species having 18 electrons will be isoelectronic with Ar.
For example, K⁺ is formed when potassium (atomic number 19) loses one electron, resulting in 18 electrons.
Hence, K⁺ is isoelectronic with Ar.
(iii) Mg²⁺ :
Magnesium has atomic number 12, so it has 12 electrons in the neutral state. Mg²⁺ loses two electrons, leaving 10 electrons.
Any species with 10 electrons will be isoelectronic.
For example, Na⁺ (11 – 1 = 10 electrons).
Hence, Na⁺ is isoelectronic with Mg²⁺.
(iv) Rb⁺ :
Rubidium has atomic number 37, so it has 37 electrons. Rb⁺ loses one electron, leaving 36 electrons.
Any species with 36 electrons will be isoelectronic.
For example, Kr (atomic number 36).
Hence, Kr is isoelectronic with Rb⁺.
Q: Which of the following species are isoelectronic?
(i) O²⁻ (ii) Mg²⁺ (iii) Na (iv) F (v) Cl⁻ (vi) Al³⁺ (vii) Ne (viii) S²⁻ (ix) Ca²⁺ (x) K [NCERT]
Answer :
Isoelectronic species are atoms or ions that have the same number of electrons.
Neutral atoms : Number of electrons = Atomic number (Z).
Anions (negative ions) : Electrons = Z + (number of extra electrons gained).
Cations (positive ions) : Electrons = Z − (number of electrons lost).
| Species | Atomic Number (Z) | Charge | Electrons Calculation | Total Electrons |
|---|---|---|---|---|
| O²⁻ | 8 | –2 | 8 + 2 | 10 |
| Mg²⁺ | 12 | +2 | 12 – 2 | 10 |
| Na | 11 | 0 | 11 | 11 |
| F | 9 | 0 | 9 | 9 |
| Cl⁻ | 17 | –1 | 17 + 1 | 18 |
| Al³⁺ | 13 | +3 | 13 – 3 | 10 |
| Ne | 10 | 0 | 10 | 10 |
| S²⁻ | 16 | –2 | 16 + 2 | 18 |
| Ca²⁺ | 20 | +2 | 20 – 2 | 18 |
| K | 19 | 0 | 19 | 19 |
Oxide ion O²⁻, magnesium ion Mg²⁺, aluminium ion Al³⁺, and neon atom each have 10 electrons, and hence they form one isoelectronic group.
Similarly, chloride ion Cl⁻, sulfide ion S²⁻, and calcium ion Ca²⁺ each contain 18 electrons and form another isoelectronic group.
The remaining species such as sodium (Na), fluorine (F), and potassium (K) do not share their electron count with any other species in the list and are therefore not isoelectronic with the others.
Although isoelectronic species have the same number of electrons, their sizes are not the same. The size depends on the nuclear charge, that is, the number of protons present in the nucleus. In an isoelectronic series, as the nuclear charge increases, the attraction between the nucleus and the electrons becomes stronger, pulling the electron cloud closer and thereby reducing the size.
Thus, among the species having 10 electrons, the order of increasing size is Al³⁺ < Mg²⁺ < Ne < O²⁻.
Similarly, for the species with 18 electrons, the size increases in the order Ca²⁺ < Cl⁻ < S²⁻.
This explains how both electron count and nuclear charge together determine the structure and size of atoms and ions.
Q: Which of the following species will have the largest and the smallest size ? : Mg, Mg²⁺, Al, Al³⁺ [NCERT]
Answer :
In comparing the sizes of Mg, Mg²⁺, Al, and Al³⁺, we must consider both periodic trends and the effect of ion formation. Across a period, atomic size decreases from left to right due to increasing nuclear charge, which pulls electrons closer to the nucleus. Since magnesium lies to the left of aluminium in the same period, magnesium has a larger atomic size than aluminium.
When atoms form cations, they lose electrons while the number of protons remains unchanged. This results in an increased effective nuclear charge acting on fewer electrons, causing the electron cloud to contract. Therefore, Mg²⁺ is smaller than Mg, and Al³⁺ is smaller than Al.
A more precise comparison can be made using the concept of isoelectronic species. Both Mg²⁺ and Al³⁺ have the same number of electrons (10), similar to neon. In such a case, the species with the higher nuclear charge will have a smaller size because the electrons are pulled more strongly toward the nucleus. Since aluminium has a higher atomic number than magnesium, Al³⁺ is smaller than Mg²⁺.
Taking all these factors together, the correct order of size is :
$$\text{Al}^{3+} < \text{Mg}^{2+} < \text{Al} < \text{Mg}$$
Thus, Mg has the largest size, while Al³⁺ has the smallest size among the given species.
Q : Which of the following atoms and ions are isoelectronic?
(i) Al³⁺ (ii) F (iii) Cl⁻ (iv) O²⁻ (v) Na (vi) Mg²⁺.
Arrange the isoelectronic ions in the decreasing order of their size.
Answer :
| Species | Atomic Number (Z) | Change due to Ion Formation | No. of Electrons |
|---|---|---|---|
| Al³⁺ | 13 | Lost 3 e⁻ | 10 |
| F | 9 | Neutral atom | 9 |
| Cl⁻ | 17 | Gained 1 e⁻ | 18 |
| O²⁻ | 8 | Gained 2 e⁻ | 10 |
| Na | 11 | Neutral atom | 11 |
| Mg²⁺ | 12 | Lost 2 e⁻ | 10 |
Aluminium ion Al³⁺ is formed by the loss of three electrons from atomic number 13, giving 10 electrons. Oxide ion O²⁻ is formed by gaining two electrons over atomic number 8, also giving 10 electrons. Magnesium ion Mg²⁺ loses two electrons from atomic number 12, again resulting in 10 electrons. Thus, Al³⁺, O²⁻, and Mg²⁺ are isoelectronic species since they all contain 10 electrons.
The other species, F (9 electrons), Na (11 electrons), and Cl⁻ (18 electrons) are not isoelectronic with them.
Although these three ions have the same number of electrons, their sizes are different because of varying nuclear charge. In an isoelectronic series, the ion with a higher nuclear charge exerts a stronger attractive force on the same number of electrons, pulling them closer and reducing the size.
Oxygen has the lowest atomic number (Z = 8), magnesium has Z = 12, and aluminium has Z = 13. Therefore, the effective nuclear attraction increases in the order O²⁻ < Mg²⁺ < Al³⁺, which means size decreases in that direction.
Hence, the correct decreasing order of size is :
$$O^{2-} > Mg^{2+} > Al^{3+}$$
Q: Arrange the following ions in the order of increasing size :
Be²⁺, Cl⁻, S²⁻, Na⁺, Mg²⁺, Br⁻ and Why ?
Answer :
To arrange the given ions in order of increasing size, we must consider the effect of charge, number of electron shells, and nuclear charge.
Cations are generally smaller than their parent atoms because they lose electrons, which reduces electron–electron repulsion and increases the effective nuclear pull on the remaining electrons.
Anions, on the other hand, are larger because they gain electrons, increasing repulsion and expanding the electron cloud.
Among the given ions, Be²⁺, Mg²⁺, and Na⁺ are cations and therefore smaller than the anions Cl⁻, S²⁻, and Br⁻.
Within the cations, Be²⁺ has the smallest size because it has the least number of electron shells. Magnesium ion Mg²⁺ has one more shell than Be²⁺, making it larger. Sodium ion Na⁺, although isoelectronic with Mg²⁺ (both have 10 electrons), has a lower nuclear charge than Mg²⁺, so its electrons are held less tightly, making Na⁺ larger than Mg²⁺.
Among the anions, Cl⁻ and S²⁻ are isoelectronic with 18 electrons. Between them, Cl⁻ has a higher nuclear charge and hence a smaller size, while S²⁻, with lower nuclear charge, is larger. Bromide ion Br⁻ is the largest because it belongs to a lower period and has an additional electron shell, which increases its size significantly.
Combining all these factors, the correct order of increasing size is :
$$\text{Be}^{2+} < \text{Mg}^{2+} < \text{Na}^{+} < \text{Cl}^{-} < \text{S}^{2-} < \text{Br}^{-}$$
Q: In each of the following pairs, which species has a larger size and why ?
(i) Br or Br⁻ (ii) O²⁻ or F⁻ (iii) K or K⁺ (iv) Li⁺ or Na⁺ (v) P or As (vi) Na⁺ or Mg²⁺
Answer
To compare the sizes of the given species, we use fundamental concepts such as gain or loss of electrons, number of electron shells, and effective nuclear charge. These factors together determine the size of atoms and ions.
(i) Br or Br⁻
Larger : Br⁻
When bromine gains one electron to form Br⁻, the number of electrons increases while the nuclear charge remains the same. This increases electron–electron repulsion and reduces the effective attraction per electron, causing the electron cloud to expand. Hence, Br⁻ is larger than Br.
(ii) O²⁻ or F⁻
Larger : O²⁻
Both O²⁻ and F⁻ are anions, but O²⁻ has gained two electrons whereas F⁻ has gained only one. This results in greater electron–electron repulsion in O²⁻, making it larger. Also, both are isoelectronic (10 electrons), but oxygen has a lower nuclear charge than fluorine, so O²⁻ holds electrons less tightly and is therefore larger.
(iii) K or K⁺
Larger : K
Potassium loses one electron to form K⁺, and in the process, it loses its outermost shell. This significantly reduces its size. Therefore, the neutral atom K is much larger than K⁺.
(iv) Li⁺ or Na⁺
Larger : Na⁺
Both are cations, but Na⁺ has one more electron shell than Li⁺. Since size increases with the number of shells, Na⁺ is larger than Li⁺.
(v) P or As
Larger : As
Arsenic lies below phosphorus in the periodic table. As we move down a group, new electron shells are added, increasing the size. Hence, As is larger than P.
(vi) Na⁺ or Mg²⁺
Larger : Na⁺
Na⁺ and Mg²⁺ are isoelectronic species (both have 10 electrons). In such cases, size depends on nuclear charge. Magnesium has a higher nuclear charge than sodium, so Mg²⁺ pulls electrons more strongly and is smaller. Therefore, Na⁺ is larger than Mg²⁺.
Q: Arrange the following in order of increasing radii with reason :
(i) I, I⁺, I⁻
(ii) N, O, P
(iii) F, Cl, Br
Answer :
(i) I, I⁺, I⁻
When iodine loses an electron to form I⁺, the number of electrons decreases while the nuclear charge remains the same. This increases the effective nuclear attraction on the remaining electrons, pulling them closer and reducing the size. In contrast, when iodine gains an electron to form I⁻, electron–electron repulsion increases, causing expansion of the electron cloud. The neutral atom lies between these two cases.
Hence, the order of increasing size is :
$$\text{I}^+ < \text{I} < \text{I}^-$$
(ii) N, O, P
Nitrogen and oxygen belong to the same period, and atomic size decreases from left to right due to increasing nuclear charge. Therefore, oxygen is smaller than nitrogen. Phosphorus lies below nitrogen in the same group, so it has an additional electron shell, making it larger than both.
Thus, the order of increasing size is :
$$\text{O} < \text{N} < \text{P}$$
(iii) F, Cl, Br
Fluorine, chlorine, and bromine belong to the same group (halogens). As we move down the group, new electron shells are added, increasing the distance of the outer electrons from the nucleus. Although nuclear charge also increases, the effect of added shells dominates, leading to an increase in size.
Therefore, the order of increasing size is :
$$\text{F} < \text{Cl} < \text{Br}$$
Q: For each of the following pairs, state which one is larger in size and why :
(a) Li, F (b) O, Se (c) Fe²⁺, Fe³⁺ (d) Br, Br⁻ (e) Na⁺, F⁻ (f) K, K⁺
Answer :
(a) Li and F
Larger : Li
Lithium and fluorine lie in the same period. As we move from left to right across a period, the nuclear charge increases, pulling electrons closer to the nucleus and decreasing size. Since Li is on the left and F is on the right, Li is larger.
(b) O and Se
Larger : Se
Oxygen and selenium belong to the same group. As we move down a group, new electron shells are added, increasing the distance between the nucleus and the outer electrons. Therefore, Se is larger than O.
(c) Fe²⁺ and Fe³⁺
Larger : Fe²⁺
Both are cations of iron, but Fe³⁺ has lost one more electron than Fe²⁺. This results in a higher effective nuclear charge acting on fewer electrons, pulling them closer and reducing size. Hence, Fe²⁺ is larger than Fe³⁺.
(d) Br and Br⁻
Larger : Br⁻
Bromine gains one electron to form Br⁻, increasing electron–electron repulsion and expanding the electron cloud. Therefore, Br⁻ is larger than neutral Br.
(e) Na⁺ and F⁻
Larger : F⁻
Na⁺ loses an electron and becomes much smaller, while F⁻ gains an electron, increasing repulsion and size. Thus, F⁻ is larger than Na⁺.
(f) K and K⁺
Larger : K
Potassium loses one electron to form K⁺ and, in doing so, loses its outermost shell. This causes a significant decrease in size. Therefore, the neutral atom K is larger than K⁺.
Q : Account for the difference in size of Na⁺ (0.095 nm) and Mg²⁺ (0.065 nm), both of which have the same noble gas configuration.
Answer :
Na⁺ and Mg²⁺ are isoelectronic species, both having the same electronic configuration as neon $(1s^2,2s^2,2p^6)$ and thus the same number of electrons (10). However, their sizes are different because the nuclear charge acting on these electrons is not the same.
In Na⁺, the nucleus contains 11 protons, whereas in Mg²⁺, it contains 12 protons. Since both ions have the same number of electrons but Mg²⁺ has a higher nuclear charge, the attractive force between the nucleus and the electron cloud is stronger in Mg²⁺. This increased electrostatic attraction pulls the electrons closer to the nucleus, resulting in a smaller ionic radius.
Therefore, Mg²⁺ has a smaller size (0.065 nm) compared to Na⁺ (0.095 nm), even though both have the same noble gas configuration.
Q: A boy has reported the radii of Cu, Cu⁺, and Cu²⁺ as 0.096 nm, 0.122 nm, and 0.072 nm respectively. However, he interchanged the values by mistake. Assign the correct values to the different species.
Answer :
For a given element, the size decreases as it forms positive ions because electrons are removed while the nuclear charge remains the same. This increases the effective nuclear attraction on the remaining electrons, pulling them closer to the nucleus. Hence, the general order is :
$$
\text{Cu} > \text{Cu}^+ > \text{Cu}^{2+}
$$
Among the given radii, the largest value (0.122 nm) must correspond to the neutral atom Cu. The next smaller value (0.096 nm) corresponds to Cu⁺, and the smallest value (0.072 nm) corresponds to Cu²⁺.
Thus, the correct assignment is :
$$
\text{Cu} = 0.122 ,\text{nm}, \quad \text{Cu}^+ = 0.096 ,\text{nm}, \quad \text{Cu}^{2+} = 0.072 ,\text{nm}
$$
Q : Arrange the following ions in the order of increasing size :
Be²⁺, Cl⁻, S²⁻, Na⁺, Mg²⁺, Br⁻ and why ? [NCERT]
Answer :
Among the given ions, Be²⁺, Mg²⁺, and Na⁺ are cations, while Cl⁻, S²⁻, and Br⁻ are anions. Hence, all cations will be smaller than all anions.
Within the cations, Be²⁺ is the smallest because it has the fewest electron shells and a high effective nuclear charge acting on a very small electron cloud. Mg²⁺ has one more shell than Be²⁺, so it is larger. Na⁺ and Mg²⁺ are isoelectronic (both have 10 electrons), but Mg²⁺ has a higher nuclear charge and is therefore smaller than Na⁺. Hence, Na⁺ is larger than Mg²⁺.
Among the anions, Cl⁻ and S²⁻ are isoelectronic (18 electrons). Since Cl⁻ has a higher nuclear charge than S²⁻, it is smaller, while S²⁻ is larger. Br⁻ is the largest because it lies in a lower period and has an additional electron shell.
Combining all these comparisons, the correct order of increasing size is :
$$\text{Be}^{2+} < \text{Mg}^{2+} < \text{Na}^{+} < \text{Cl}^{-} < \text{S}^{2-} < \text{Br}^{-}$$
Q : Select from each group the species which has the smallest radius :
(a) O, O⁻, O²⁻
(b) K⁺, Sr²⁺, Ar
(c) Si, P, Cl [NCERT]
Answer :
(a) O, O⁻, O²⁻
When oxygen gains electrons to form O⁻ and O²⁻, electron–electron repulsion increases, causing expansion of the electron cloud. Thus, size increases from O to O⁻ to O²⁻. The neutral atom has the least repulsion and the highest effective nuclear attraction on its electrons.
Therefore, the smallest species is O.
(b) K⁺, Sr²⁺, Ar
K⁺ and Sr²⁺ are cations, while Ar is a neutral atom. Cations are smaller due to loss of electrons and increased effective nuclear charge. Between K⁺ and Sr²⁺, Sr²⁺ has a higher positive charge and hence a stronger nuclear pull on its electrons, making it even smaller.
Therefore, the smallest species is Sr²⁺.
(c) Si, P, Cl
All three are neutral atoms in the same period (Period 3). As we move from left to right across a period, nuclear charge increases and atomic size decreases. Thus, silicon is largest, followed by phosphorus, and chlorine is the smallest.
Therefore, the smallest species is Cl.
FAQs – Frequently Asked Very Short Atomic and Ionic Radii Conceptual Questions
Between Al³⁺ and Al, which has a smaller radius and why?
Al³⁺ has a smaller radius than Al because it loses three valence electrons, resulting in a stable configuration similar to neon. With fewer electrons and the same nuclear charge, the effective nuclear attraction increases, pulling the remaining electrons closer and reducing the size.
What is the correct order of radii for F⁻, F, and F⁺?
The correct order is : F+ < F < F–
F⁺ is smallest due to loss of an electron, increasing nuclear attraction. F is neutral, while F⁻ is largest due to gain of an electron, which increases electron–electron repulsion.
Which ion is larger : O²⁻ or S²⁻? Explain.
S²⁻ is larger than O²⁻ because sulfur lies below oxygen in the periodic table and has more electron shells. The increase in shells leads to a larger atomic size despite having the same charge.
Arrange Ca²⁺, K⁺, and Cl⁻ in decreasing order of size.
The correct order is :
$$Cl^- > K^+ > Ca^{2+}$$
Cl⁻ is largest due to gain of an electron, while Ca²⁺ is smallest because of higher positive charge and stronger nuclear attraction.
Which has the smallest radius among Li⁺, Be²⁺, and B³⁺?
B³⁺ has the smallest radius because it has the highest positive charge. Greater positive charge results in stronger attraction between nucleus and electrons, leading to a smaller size.
Arrange Na, Na⁺, and Mg²⁺ in increasing order of size.
The correct order is :
$$Mg^{2+} < Na^+ < Na$$
Mg²⁺ is smallest due to highest positive charge, Na⁺ is intermediate, and neutral Na is largest.
Between Cl and Ar, which atom is smaller and why?
Ar is smaller than Cl because both lie in the same period, but argon has a higher nuclear charge, which pulls electrons closer to the nucleus, reducing size.
Arrange N³⁻, O²⁻, and F⁻ in increasing order of size.
The correct order is :
$$F^- < O^{2-} < N^{3-}$$
All are isoelectronic, but nitrogen has the lowest nuclear charge, so N³⁻ is largest, while F⁻ is smallest.
Which ion is larger: Zn²⁺ or Cu²⁺?
Zn²⁺ is slightly larger than Cu²⁺ because Cu²⁺ has a higher effective nuclear charge, which pulls its electrons closer, resulting in a smaller size.
Arrange Rb, K, and Na in decreasing order of size.
The correct order is :
$$Rb > K > Na$$
All are alkali metals, and atomic size increases down the group due to the addition of electron shells.
Important Chapter Interlinks
This section provides a complete and interconnected study of Classification of Elements and Periodicity in Properties, starting with detailed theory and notes for Class 11 Chemistry to build a strong conceptual foundation. You can explore atomic radius and its types including covalent, van der Waals, metallic, and ionic radii to understand periodic trends in atomic size. It also includes Screening Effect (Shielding Effect) : Calculation of Effective or Reduced Nuclear Charge (Slater’s Rules), which explains how inner electrons reduce the nuclear attraction on outer electrons and influence periodic trends. In addition, topics like Radius of Cation is Less and Anion is More Than Its Parent Atom, Size Variation in Isoelectronic Series help explain how ionic size changes due to gain or loss of electrons and how nuclear charge affects size in species with the same number of electrons. The causes of periodicity explain why elements show repeating properties based on electronic configuration, which is further supported by the modern periodic law and structure of the modern periodic table including groups, periods, and blocks for elements even beyond atomic number 100. The historical development is covered through Mendeleev’s periodic law and table, leading to the modern classification of elements into s, p, d, and f blocks with prediction of period, group, and block. To strengthen exam preparation, you can practice JEE Main PYQs, IMU CET PYQs and Merchant Navy sponsorship exam MCQs, and other previous year questions with solutions, along with solved examples, conceptual questions, and practice problems on the modern periodic table. Additionally, complete study material, mock tests, and guidance are provided under Anand Classes Chemistry notes, along with expert support from Er Neeraj Anand, making this section a comprehensive resource for competitive exam preparation.