Why atomic size is hard to define
Measuring the size of an atom is not like measuring the radius of a ball. An atom is exceedingly small — roughly $1.2\ \text{Å}$ ($1.2 \times 10^{-10}\ \text{m}$) in radius — and, more fundamentally, the electron cloud surrounding the nucleus has no sharp boundary at which we can say the atom ends. NCERT is explicit on this point: there is no practical way to measure the size of an isolated atom directly. What chemists do instead is estimate the size from the distance between atoms in the combined state, then assign each atom a share of that distance.
Because the estimate depends on how the atoms are bound — by a covalent bond, packed in a metal, or merely touching as neighbouring molecules — three operational radii arise: the covalent radius, the metallic radius and the van der Waals radius. They are not interchangeable, and a NEET stem that quotes a value usually implies one specific kind. The first task, then, is to keep the three definitions distinct.
Covalent, metallic and van der Waals radius
The covalent radius is obtained from a non-metallic element by measuring the bond length between two like atoms joined by a single covalent bond and taking half of it. In the chlorine molecule $\ce{Cl2}$ the internuclear distance is $198\ \text{pm}$, so the covalent radius of chlorine is $99\ \text{pm}$. The metallic radius is defined the same way for a metal: it is half the internuclear distance between adjacent atoms in the metallic crystal. In solid copper the nearest neighbours are $256\ \text{pm}$ apart, giving a metallic radius of $128\ \text{pm}$.
NCERT, for simplicity, uses the single term atomic radius to mean the covalent radius for non-metals and the metallic radius for metals. The van der Waals radius is different in kind: it is half the distance between the nuclei of two non-bonded atoms of adjacent molecules in close contact. Because non-bonded atoms cannot approach as closely as bonded ones, the van der Waals radius is the largest of the three for a given element. Atomic radii of all kinds are measured by X-ray or other spectroscopic methods.
Covalent radius from a bond length — chlorine as the worked case.
Half the bond distance in $\ce{Cl2}$ is taken as the atomic (covalent) radius of chlorine. The same halving applied to the $256\ \text{pm}$ separation in copper metal gives a metallic radius of $128\ \text{pm}$.
| Type of radius | How it is measured | Applies to | NCERT example |
|---|---|---|---|
| Covalent radius | Half the single-bond internuclear distance between two like atoms | Non-metals | Cl₂: 198 → 99 pm |
| Metallic radius | Half the internuclear distance between adjacent atoms in a metal crystal | Metals | Cu: 256 → 128 pm |
| van der Waals radius | Half the distance between non-bonded atoms of neighbouring molecules in contact | Noble gases; non-bonded contacts | Largest of the three |
Do not compare a noble gas using a covalent radius.
Noble gases are monoatomic and form no normal covalent bonds, so no covalent radius can be defined for them. NCERT notes that their non-bonded radii are very large, and they must be compared with the van der Waals radii of other elements — never with covalent radii. A stem that lists a "very large" radius for argon or neon is using the van der Waals scale.
For noble gases, size is a van der Waals radius — always larger than the covalent radius of a neighbouring element.
Trends across a period and down a group
Two trends emerge clearly from the tabulated radii. Across a period, atomic size generally decreases. Moving from lithium to fluorine in the second period, the outer electrons all occupy the same valence shell, but the nuclear charge rises with each successive element. The growing positive charge draws the valence electrons inward, so the atom contracts steadily from left to right. NIOS states the same rule plainly: the first member of each period — the group 1 atom — is the largest in its row, and size falls as we move rightward.
Down a group, atomic size increases. Descending group 1 from lithium to caesium, or group 17 from fluorine to iodine, the principal quantum number $n$ increases at each step and a new electron shell is added. The valence electrons lie progressively farther from the nucleus, and the filled inner shells shield them from the nuclear pull, so the radius grows regularly with atomic number. This is why a 2025 NEET option testing "the atomic radius of Cs is greater than that of Li and Rb" is correct on the down-group rule.
The two directions of the trend on a periodic-table grid.
The largest atoms cluster at the bottom-left (low nuclear charge, high $n$); the smallest at the top-right (high nuclear charge, same shell). The largest atom of any period is its group 1 member.
| Direction | What changes | Why | Effect on radius |
|---|---|---|---|
| Left → right (period) | Nuclear charge rises; same valence shell | Effective nuclear charge increases, pulling electrons in | Decreases |
| Top → bottom (group) | New shell added; $n$ increases | Valence electrons farther out; inner shells shield the nucleus | Increases |
Effective nuclear charge and shielding
Both trends are governed by one concept: the effective nuclear charge ($Z_{\text{eff}}$) — the net positive charge a valence electron actually experiences after the inner electrons partially screen the full nuclear charge. Across a period, each added proton is only weakly offset by an electron entering the same shell, so $Z_{\text{eff}}$ rises and the atom contracts. Down a group, the addition of a complete inner shell sharply increases the shielding, so $Z_{\text{eff}}$ on the outermost electrons grows only slowly while $n$ jumps, and the radius expands.
This single lens also explains the few irregularities NEET likes to probe. In group 13, the order is $\ce{B} < \ce{Ga} < \ce{Al} < \ce{In} < \ce{Tl}$ — gallium is slightly smaller than aluminium because the newly filled $3d$ electrons in Ga shield poorly, letting $Z_{\text{eff}}$ rise enough to contract the atom below Al. The same poor $d$-shielding logic recurs throughout the p-block.
A smaller radius means valence electrons are held more tightly — the direct cause of higher ionisation enthalpy across a period. Read size and energy together.
Ionic radius: cations and anions
When an atom loses or gains electrons it becomes an ion, and its radius changes in a direction that follows directly from the electron count. Ionic radii are estimated from the distances between cations and anions in ionic crystals, and in general they follow the same period and group trends as atomic radii.
A cation is always smaller than its parent atom. Removing electrons leaves the nuclear charge unchanged but reduces the number of electrons, so each remaining electron feels a stronger net pull and the cloud contracts; frequently an entire outer shell is removed. NCERT's figures make the magnitude clear: the sodium atom has a radius of $186\ \text{pm}$, but the sodium ion $\ce{Na+}$ is only $95\ \text{pm}$. An anion is always larger than its parent atom. Adding electrons increases electron–electron repulsion and lowers the effective nuclear charge per electron, so the cloud swells: the fluorine atom is $64\ \text{pm}$, while the fluoride ion $\ce{F-}$ is $136\ \text{pm}$.
Cation shrinks, anion expands — the same nucleus, a different electron count.
Losing an electron contracts sodium to roughly half its radius; gaining one expands fluorine to more than double. The nucleus is the same in each pair — only the electron count and the resulting repulsion change.
Isoelectronic species and their ordering
Atoms and ions that contain the same number of electrons are called isoelectronic species. NCERT's standard example is the ten-electron set $\ce{O^2-, F-, Na+, Mg^2+}$, to which $\ce{Ne}$, $\ce{N^3-}$ and $\ce{Al^3+}$ also belong. Since every member carries the identical electron cloud, the only thing distinguishing their sizes is the nuclear charge holding that cloud.
The rule is therefore sharp: among isoelectronic species, the greater the nuclear charge (number of protons), the smaller the radius. A cation with a larger positive charge pulls the fixed cloud in tightest; an anion with a larger negative charge has the fewest protons relative to its electrons, so net repulsion outweighs nuclear attraction and the species expands. Ordering by increasing atomic number gives decreasing size in lockstep.
| Species (10 electrons) | Protons (Z) | Net charge | Relative size |
|---|---|---|---|
| $\ce{N^3-}$ | 7 | 3− | Largest |
| $\ce{O^2-}$ | 8 | 2− | ↑ |
| $\ce{F-}$ | 9 | 1− | ↑ |
| $\ce{Ne}$ | 10 | 0 | middle |
| $\ce{Na+}$ | 11 | 1+ | ↓ |
| $\ce{Mg^2+}$ | 12 | 2+ | ↓ |
| $\ce{Al^3+}$ | 13 | 3+ | Smallest |
Two different rules — keep them apart.
Confusing the cation/anion rule with the isoelectronic rule is the single most common error here. For a parent atom versus its own ions, the comparison is by electron count: cation < atom < anion. For an isoelectronic set, the electron count is fixed and the comparison is by nuclear charge: more protons means smaller. Always check first whether you are comparing one element's species or many species with equal electrons.
Same element, different electrons → use charge sign. Same electrons, different elements → use proton count: more Z, smaller size.
Worked ordering example
The reasoning becomes mechanical once the right rule is selected. The two NCERT-style problems below combine both ideas, exactly as NEET stems do.
Arrange the following in increasing order of size: $\ce{Mg, Mg^2+, Al, Al^3+}$.
Step 1. Mg and Al are neutral atoms in the same period; atomic radius decreases left to right, so $\ce{Al} < \ce{Mg}$.
Step 2. Each cation is smaller than its parent atom: $\ce{Mg^2+} < \ce{Mg}$ and $\ce{Al^3+} < \ce{Al}$.
Step 3. $\ce{Mg^2+}$ and $\ce{Al^3+}$ are isoelectronic (10 electrons each); the larger nuclear charge of $\ce{Al^3+}$ (Z = 13) makes it smaller than $\ce{Mg^2+}$ (Z = 12).
Result: $\ce{Al^3+} < \ce{Mg^2+} < \ce{Al} < \ce{Mg}$. The largest species is $\ce{Mg}$; the smallest is $\ce{Al^3+}$.
Order the ions $\ce{Al^3+, Mg^2+, Na+, F-}$ by increasing ionic size.
Step 1. Check electrons: each ion has 10 electrons, so this is an isoelectronic set — order by nuclear charge alone.
Step 2. Proton counts are Al = 13, Mg = 12, Na = 11, F = 9. More protons pull the fixed cloud tighter, so size rises as Z falls.
Result: $\ce{Al^3+} < \ce{Mg^2+} < \ce{Na+} < \ce{F-}$ — precisely the NEET 2016 answer key for increasing ionic size.
Atomic & ionic radius in one screen
- Atomic radius is estimated from bonded distances: covalent radius (half a single-bond length, e.g. $\ce{Cl2}$ → 99 pm), metallic radius (half the crystal separation, e.g. Cu → 128 pm), and van der Waals radius (non-bonded, largest, used for noble gases).
- Across a period, radius decreases as effective nuclear charge rises; down a group it increases as new shells are added and shielding grows.
- A cation is smaller than its parent atom ($\ce{Na+}$ 95 pm vs Na 186 pm); an anion is larger ($\ce{F-}$ 136 pm vs F 64 pm).
- Isoelectronic species share the same electron count, so size is set by nuclear charge: $\ce{N^3- > O^2- > F- > Ne > Na+ > Mg^2+ > Al^3+}$.
- Watch group 13: $\ce{B < Ga < Al < In < Tl}$ — Ga dips below Al because poor $3d$ shielding raises $Z_{\text{eff}}$.