Periodic Trend — Atomic Radius

What Does 'Atomic Radius' Even Mean?

At first glance, asking "what is the radius of an atom?" feels like asking for the radius of a cloud — electrons don't stop at a sharp boundary. Quantum mechanics tells us the electron wavefunction $\psi$ tails off exponentially, and there is never a hard edge. So how do chemists assign a definite number like "the atomic radius of carbon is 77 pm"?

The answer: we measure the distance between two atomic nuclei in a real compound and divide by 2. Since a nucleus is a well-defined point, internuclear distance is measurable by X-ray or electron diffraction to high precision. The radius we report is an operational quantity — it tells you how two atoms of the same kind sit next to each other when bonded.

Three operational definitions

But atoms can be "next to each other" in different ways, and each gives a different number. Class 11 chemistry recognises three standard types of radii:

1. Covalent radius ($r_{\text{cov}}$): half the distance between two identical atoms joined by a single covalent bond, as in $\text{H}_{2}, \text{Cl}_{2}$, or $\text{O}_{2}$.
2. Metallic radius ($r_{\text{met}}$): half the distance between two adjacent atoms in a metallic crystal — i.e. half the shortest inter-atomic distance in a pure metal like Na, Cu or Fe.
3. Van der Waals radius ($r_{\text{vdW}}$): half the distance of closest approach between two non-bonded atoms of the same element in a molecular solid — for example, two $\text{I}_{2}$ molecules side by side in an iodine crystal.

Why the three values differ

For the same atom, the ordering is always:

$$r_{\text{cov}} < r_{\text{met}} < r_{\text{vdW}}$$

Covalent radii are smallest because bonded atoms share electron density — their nuclei are pulled closest together. Van der Waals radii are largest because non-bonded atoms only contact each other at distances where their electron clouds just begin to touch, with no attractive pulling-in. Metallic radii sit in between: the metallic bond is weaker than a covalent bond but stronger than a van der Waals interaction.

Typical values to carry in your head

Atom	$r_{\text{cov}}$ / pm	$r_{\text{met}}$ / pm	$r_{\text{vdW}}$ / pm
H	37	–	120
C	77	–	170
O	66	–	152
Cl	99	–	180
Na	–	186	227
Cu	–	128	140
Xe	–	–	216

Noble gases are measured only by $r_{\text{vdW}}$ — they don't form covalent bonds (except in rare Xe compounds), and they are never metallic.

[JEE Tip] Always check what kind of radius a numerical problem is using. Mixing $r_{\text{cov}}$ values for metals with $r_{\text{met}}$ values of other metals is a common trap — the resulting "trend" you plot can be misleading.

Trend Across a Period — Radius Decreases Left to Right

Take any period of the periodic table, walk from left to right, and you will see atoms getting smaller. This is one of the cleanest trends in all of inorganic chemistry.

The numbers

Here are covalent / metallic radii (in pm) across Periods 2 and 3:

Group	1	2	13	14	15	16	17
Period 2	Li 152	Be 111	B 88	C 77	N 74	O 66	F 64
Period 3	Na 186	Mg 160	Al 143	Si 117	P 110	S 104	Cl 99

Across Period 2, radius drops by a factor of $152/64 \approx 2.4$. Across Period 3, the drop is similar in magnitude — $186/99 \approx 1.9$. The trend holds in every period of the table.

Why does radius shrink?

The central concept here is effective nuclear charge ($Z_{\text{eff}}$). As we move across a period:

• Nuclear charge $Z$ increases by 1 at each step (one more proton goes into the nucleus).
• A new electron is added — but it enters the same valence shell as all the others in that period.

Electrons in the same shell shield each other poorly because they are at roughly the same average distance from the nucleus. So the shielding parameter $\sigma$ increases by only ≈ 0.35 per added valence electron (Slater's rule), while $Z$ increases by 1. The net effect:

$$Z_{\text{eff}} = Z - \sigma \;\;\text{increases across the period}$$

A larger $Z_{\text{eff}}$ pulls the valence electrons closer to the nucleus, so the atom gets smaller. Same shell, more positive charge pulling on it — smaller orbit.

A quick $Z_{\text{eff}}$ sanity check (Period 2)

Using Slater's approximation for the last-entering $2p$ electron:

Atom	$Z$	$\sigma$ (Slater)	$Z_{\text{eff}}$
Li	3	2 × 0.85 = 1.70	1.30
C	6	3 × 0.35 + 2 × 0.85 = 2.75	3.25
F	9	6 × 0.35 + 2 × 0.85 = 3.80	5.20
Ne	10	7 × 0.35 + 2 × 0.85 = 4.15	5.85

So $Z_{\text{eff}}$ goes from 1.30 at Li to 5.85 at Ne — the outer electron in neon feels roughly four times the pull that the outer electron in lithium does. No wonder the atom shrinks.

[Board Level] CBSE expects you to be able to state: "Radius decreases across a period because $Z_{\text{eff}}$ increases while electrons are added to the same shell." Write exactly this one-liner and you get full marks for the "explain the trend" part.

One tiny exception

Within a given d-block row (Sc → Zn in Period 4, for example), radii do not decrease as smoothly as in the s- and p-blocks. After an initial small decrease, they become nearly constant, and by the time we reach Cu and Zn the radii even increase slightly. This happens because added $d$-electrons shield the outer electrons poorly, so the contraction is modest, and later electron-electron repulsion within the increasingly filled d-subshell causes a slight expansion.

Trend Down a Group — Radius Increases Top to Bottom

The second great trend is equally clean but runs in the opposite direction: atomic radius grows as we descend a group. Here are the numbers for Groups 1 and 17:

Group 1 (alkali metals)	Group 17 (halogens)
Li 152 pm	F 64 pm
Na 186 pm	Cl 99 pm
K 227 pm	Br 114 pm
Rb 248 pm	I 133 pm
Cs 265 pm	At 150 pm

Caesium has a radius 1.7 times that of lithium. Iodine has a radius 2 times that of fluorine. The growth is steady, if not uniform.

Why does radius grow?

Going down a group, two things happen at once:

1. A brand-new electron shell is added at each step (Li's valence is in $n=2$, Na's in $n=3$, K's in $n=4$, and so on). Each new shell is, on average, farther from the nucleus than the previous one — for hydrogenic orbitals the mean radius grows as $\langle r \rangle \propto n^{2} / Z$.
2. Nuclear charge $Z$ also grows — by a lot. From Li ($Z=3$) to Cs ($Z=55$), $Z$ jumps by 52. Naively, more nuclear charge means smaller radius.

So why does the radius still grow? Because the newly added shells also contain a large number of core electrons that shield the outer electron almost completely. By Slater's rules, electrons two or more shells below contribute $\sigma = 1$ each — they cancel out the extra nuclear charge perfectly.

The key inequality: the distance effect from adding a new shell (which scales as $n^{2}$) beats the charge effect from the extra protons (which is mostly cancelled by shielding). So the net motion is outward.

A quantitative taste

For an alkali metal, valence is $ns^{1}$. Using Slater:

Metal	$Z$	$\sigma$	$Z_{\text{eff}}$	Expected $r \propto n^{2} / Z_{\text{eff}}$
Li	3	2 × 0.85 = 1.70	1.30	$4 / 1.30 \approx 3.08$
Na	11	2 + 8 × 0.85 = 8.80	2.20	$9 / 2.20 \approx 4.09$
K	19	10 + 8 × 0.85 = 16.80	2.20	$16 / 2.20 \approx 7.27$

The predicted ratio Li:Na:K ≈ 1 : 1.3 : 2.4 matches quite well with the observed Li:Na:K ≈ 1 : 1.2 : 1.5 (the match isn't perfect because Slater is a simple approximation).

Summary of the two trends

The modern periodic table has a diagonal-arrow symbol system for trends:

• → (left-to-right, across a period): radius decreases
• ↓ (top-to-bottom, down a group): radius increases

Memorise this "arrow" picture — it's the template for every trend question on every exam.

[JEE Tip] Many JEE problems give you an unfamiliar pair of elements and ask you to compare their radii. Always use the arrow rule first: same period → rightmost is smallest; same group → lowest is largest; diagonal → use the trend that the pair lies along more strongly. This simple rule answers about 70% of JEE radius questions in 5 seconds.

Effective Nuclear Charge and Slater's Rules

We have mentioned $Z_{\text{eff}}$ several times now — let's make it rigorous. Effective nuclear charge is the net positive charge felt by an electron, after other electrons partially cancel (shield) the full nuclear charge:

$$Z_{\text{eff}} = Z - \sigma$$

where $Z$ is the atomic number and $\sigma$ (sigma) is the shielding constant — the portion of $Z$ that is effectively cancelled out by all the other electrons.

Where does shielding come from?

Picture an atom from the point of view of one specific ("test") valence electron. It feels:

• The full attraction of the nucleus (a pull of $+Z$).
• The repulsion from every other electron in the atom (a push away).

The shielding constant $\sigma$ packages those repulsions into a single number we can subtract from $Z$. Slater's rules give a clever hand-calculation recipe for $\sigma$, which is accurate enough for Class 11 / JEE-level predictions.

Slater's rules (the full list)

To find $\sigma$ for a chosen "test" electron, first group the orbitals into these seven sets:

$$(1s)\,(2s,\,2p)\,(3s,\,3p)\,(3d)\,(4s,\,4p)\,(4d)\,(4f)\,(5s,\,5p)\,\ldots$$

Then, for a test electron that sits in an $ns$ or $np$ orbital:

• Rule A. Electrons in groups to the right of the test electron's group contribute $\sigma = 0$ (they don't shield at all).
• Rule B. Electrons in the same group as the test electron each contribute $\sigma = 0.35$. (Exception: for $1s$, the other $1s$ electron contributes $\sigma = 0.30$.)
• Rule C. Electrons in the $(n-1)$ shell each contribute $\sigma = 0.85$.
• Rule D. Electrons in $(n-2)$ or deeper shells each contribute $\sigma = 1.00$.

For a test electron in an $nd$ or $nf$ orbital, Rules A and B stay the same, but:

• Rule E. All electrons in groups to the left contribute $\sigma = 1.00$ (d and f orbitals shield very poorly, so inner electrons are fully effective).

Worked computation — $Z_{\text{eff}}$ for a $3p$ electron in sulfur

Sulfur: $Z = 16$, configuration $1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{4}$.

Group the electrons: $(1s^{2})(2s^{2}\,2p^{6})(3s^{2}\,3p^{4})$.

The test electron is in the $(3s, 3p)$ group. Apply Slater:

• Same group, excluding the test itself: $2 + 4 - 1 = 5$ electrons. Contribution: $5 \times 0.35 = 1.75$.
• $(n-1)$ group $(2s, 2p)$: 8 electrons × 0.85 = 6.80.
• $(n-2)$ group $(1s)$: 2 electrons × 1.00 = 2.00.

Total $\sigma = 1.75 + 6.80 + 2.00 = 10.55$.

$$Z_{\text{eff}}(\text{S}, 3p) = 16 - 10.55 = \mathbf{5.45}$$

For comparison, the same calculation for phosphorus ($Z = 15$, one fewer $3p$ electron) gives $\sigma = (4 \times 0.35) + 6.80 + 2.00 = 10.20$ and $Z_{\text{eff}} = 4.80$. So moving from P to S, $Z_{\text{eff}}$ jumps from 4.80 to 5.45 — the outer electrons are held more tightly, and the atom is smaller (P covalent radius 110 pm vs S 104 pm).

A visual rule of thumb

Shielding is approximately:

• Very effective ($\sigma \approx 1$): electrons two or more shells deeper than the test.
• Moderately effective ($\sigma \approx 0.85$): electrons one shell deeper than the test.
• Poorly effective ($\sigma \approx 0.35$): electrons in the same shell as the test.
• Not at all ($\sigma = 0$): electrons in higher shells than the test (which is physically reasonable — an electron further out can't shield one closer in).

[JEE Tip] Slater calculations are a favourite of JEE Advanced. The recipe is entirely algorithmic — if you memorise the four numerical coefficients (0.35, 0.85, 1.00, 0.00), you can compute $Z_{\text{eff}}$ for any ground-state electron in about 45 seconds.

Exceptions and Wrinkles You Should Know

Like every periodic trend, atomic radius has interesting exceptions. Here are the ones you need to know for Boards and JEE/NEET.

Exception 1 — Noble gas radii appear "too large"

Look up atomic radii and you may see that noble-gas radii (Ne 160 pm, Ar 188 pm) look much larger than the atom just before them (F 64 pm, Cl 99 pm) in the same period — apparently breaking the "shrinks to the right" rule.

Resolution: noble gases are not measured by covalent or metallic radii. They have no bonded neighbours to hand — so the only radius available is the van der Waals radius, which is intrinsically larger than $r_{\text{cov}}$. When you compare like with like (all covalent, or all van der Waals), the across-the-period trend is restored.

[Board Level] On CBSE Boards, if a question shows noble-gas radii looking "anomalously large", the expected explanation is exactly: "because those values are van der Waals radii, not covalent radii."

Exception 2 — Transition-series contractions

Within a d-block period, radii fall sharply for the first 4–5 elements and then become almost constant:

Period-4 d-block	Sc	Ti	V	Cr	Mn	Fe	Co	Ni	Cu	Zn
Metallic radius / pm	162	147	134	128	127	126	125	124	128	134

Why? Adding $d$-electrons causes only a modest increase in $Z_{\text{eff}}$ felt by the outer electrons, because $d$-electrons shield poorly. After the middle of the series, increased electron-electron repulsion within the crowded d-subshell causes the radius to level off and even rise slightly toward Cu and Zn.

Exception 3 — The lanthanoid contraction

The 14-step filling of the $4f$ sub-shell (Ce → Lu) shrinks the radius by only about 10 pm, but that contraction accumulates so that elements just after the lanthanoids (Hf, Ta, W …) end up almost identical in size to their Period-5 counterparts (Zr, Nb, Mo):

Pair	Period-5	Period-6	Difference
Zr / Hf (Group 4)	160 pm	159 pm	≈ 1 pm
Nb / Ta (Group 5)	146 pm	146 pm	0 pm
Mo / W (Group 6)	139 pm	139 pm	0 pm

This is the lanthanoid contraction — worth 1 dedicated MCQ on every JEE paper. It is the electronic reason why second- and third-row transition metals of the same group (e.g. Zr/Hf) behave almost identically chemically and are hard to separate.

Exception 4 — d-block vs p-block crossover

Gallium (Z = 31) actually has a covalent radius smaller than aluminium (Al 143 pm, Ga 141 pm) despite being one period lower! The reason is that Ga sits just after the first d-block, and the intervening $3d^{10}$ electrons shield the $4p$ electrons poorly — so $Z_{\text{eff}}$ for Ga's outer electron is unusually high, and the atom is pulled in.

The general pattern: after the first d-block period, the expected Group-13 expansion (Al → Ga) is absent or reversed because of d-block contraction effects. This is why Ga is denser than Al, and why Ga's melting point is so low (its metallic bonds are mildly compressed).

Exception 5 — Size of $\text{H}^{+}$ is essentially zero

Hydrogen loses its only electron to form $\text{H}^{+}$ — a bare proton. A proton's radius is on the order of $10^{-15}$ m = $10^{-6}$ pm — a million times smaller than any other ion. In solution, $\text{H}^{+}$ never exists alone; it immediately associates with a water molecule to give the hydronium ion $\text{H}_{3}\text{O}^{+}$ (radius ≈ 102 pm, similar to an atom).

Summary of exceptions

Where	What happens	Why
End of period (noble gas)	Radius jumps up	vdW not covalent
Within a d-row	Radius plateaus at middle	poor d-electron shielding + e−e repulsion
Across lanthanoids	Radius contracts steadily	$f$-electrons shield poorly
Just after first d-row (Al → Ga)	Radius doesn't grow	d-block contraction
Hydrogen cation $\text{H}^{+}$	Radius ≈ 0 (bare proton)	No electrons left

[NEET Important] Questions of the form "which element is larger — Hf or Zr?" lean on the lanthanoid contraction. Correct answer: they are nearly identical, with Zr very slightly larger (160 pm vs 159 pm). Don't just mechanically apply the "down-a-group-means-bigger" rule.

Putting It All Together — The Master Trend Map

Here is the one-paragraph summary you should be able to recite in your sleep.

The four pieces

• Atomic radius decreases across a period (left to right). Reason: $Z_{\text{eff}}$ increases while electrons are added to the same shell.
• Atomic radius increases down a group (top to bottom). Reason: a whole new shell is added at each step; the distance effect overwhelms the nuclear-charge effect.
• $Z_{\text{eff}}$ is the quantitative handle. Compute it with Slater's rules; it explains almost every trend you will meet for the rest of this chapter.
• Exceptions cluster at the end of each period (noble gases, vdW), within d-block rows (d-contraction), and around the lanthanoids (lanthanoid contraction leading to Zr/Hf, Nb/Ta, Mo/W similarity).

The universal consequence

Because all the other periodic trends (ionisation enthalpy, electron affinity, electronegativity, metallic character, acid-base character of oxides) are ultimately driven by atomic size and $Z_{\text{eff}}$, understanding the radius trend is the single most valuable investment for the rest of this chapter.

• Smaller atom + higher $Z_{\text{eff}}$ → valence electrons held tighter → higher IE, more negative EA, higher electronegativity, more non-metallic behaviour.
• Larger atom + lower $Z_{\text{eff}}$ → valence electrons held loosely → lower IE, smaller EA, lower electronegativity, more metallic behaviour.

Memorise this chain of implications. It turns 5 apparently independent "trend" questions into one single question — is the atom smaller or larger? — to which we already know the answer from the arrow map.

Exam tactics — a quick checklist

When you face a radius question in an exam:

• Step 1. Locate both atoms on the periodic table mentally. Same period? Same group? Or diagonally placed?
• Step 2. Apply the arrow rule: across → smaller to the right; down → larger at the bottom.
• Step 3. Watch for the exception zones: if either atom is a noble gas, d-block metal past group 8, or post-lanthanoid element (Hf, Ta, W), use the exception rule instead of the naive trend.
• Step 4. If the question asks why, write the $Z_{\text{eff}}$ explanation: "across a period, $Z_{\text{eff}}$ grows; down a group, shell number grows faster than $Z_{\text{eff}}$."

Follow that checklist and 95% of radius questions fall in 30 seconds flat.

[JEE Tip] The trap question: "Arrange in increasing order of atomic radius: F, Cl, Br, I, At." (Answer: F < Cl < Br < I < At — pure "down a group" trend.) The real trap: "Arrange O, F, Ne in increasing order of atomic radius." — the expected answer is F < O < Ne, but only if you realise Ne's quoted radius is vdW while O and F are covalent. This level of trap is worth full marks in JEE Advanced.

Solved Examples

Example 1: Arrange in order of increasing atomic radius

Arrange the following in order of increasing atomic radius: Li, Na, K, Rb

Solution:

All four elements are Group 1 (alkali metals). Going down a group, atomic radius increases because each new element has its valence electron in a new, larger shell.

• Li (Period 2): 152 pm
• Na (Period 3): 186 pm
• K (Period 4): 227 pm
• Rb (Period 5): 248 pm

So the order of increasing radius is:

$$\mathbf{Li < Na < K < Rb}$$

This is the textbook "down a group" trend.

Example 2: Across a period

Which element has the largest atomic radius: Mg, Al, Si, P?

Solution:

All four are in Period 3. Moving from left to right across a period, atomic radius decreases because $Z_{\text{eff}}$ rises while electrons are added to the same shell. So the leftmost element has the largest radius.

Element	Group	Covalent radius / pm
Mg	2	160
Al	13	143
Si	14	117
P	15	110

Largest = Mg (160 pm).

Example 3: Diagonal comparison

Compare the atomic radii of Li (Period 2, Group 1) and Mg (Period 3, Group 2). Explain.

Solution:

Li → Mg is a diagonal move: one period down (radius grows) but also one group to the right (radius shrinks). Both effects compete.

• Li: 152 pm
• Mg: 160 pm

The values are very close — this is the diagonal near-equality that gives rise to diagonal relationships (Li-Mg, Be-Al, B-Si) that we study later in the chapter.

Radius: Mg > Li, but only slightly. The going-down effect just wins the tug-of-war.

Example 4: Slater's rules — $Z_{\text{eff}}$ for a $2p$ electron in oxygen

Using Slater's rules, calculate $Z_{\text{eff}}$ for a $2p$ electron in an oxygen atom (Z = 8).

Solution:

Oxygen: $Z = 8$, configuration $1s^{2}\,2s^{2}\,2p^{4}$.

Group: $(1s^{2})(2s^{2}\,2p^{4})$. Test electron in $(2s, 2p)$ group.

Apply Slater:

• Same group (excluding test): $2 + 4 - 1 = 5$ electrons. Contribution: $5 \times 0.35 = 1.75$.
• $(n-1)$ shell $(1s)$: 2 electrons × 0.85 = 1.70.
• No $(n-2)$ electrons.

Total $\sigma = 1.75 + 1.70 = 3.45$.

$$Z_{\text{eff}}(\text{O}, 2p) = 8 - 3.45 = \mathbf{4.55}$$

For comparison, $Z_{\text{eff}}$ of a $2p$ electron in nitrogen ($Z=7$): $\sigma = 4 \times 0.35 + 2 \times 0.85 = 3.10$, so $Z_{\text{eff}} = 3.90$.

O has higher $Z_{\text{eff}}$ than N → O's $2p$ electrons are held more tightly → O atom is smaller than N. Consistent with observation: $r(\text{O}) = 66$ pm < $r(\text{N}) = 74$ pm.

Example 5: Slater's rules — $Z_{\text{eff}}$ for a $3d$ electron

Calculate $Z_{\text{eff}}$ for a $3d$ electron in manganese (Z = 25).

Solution:

Manganese: $Z = 25$, configuration $1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,3d^{5}\,4s^{2}$.

Group ordering for Slater: $(1s)(2s,2p)(3s,3p)(3d)(4s,4p)$. The test electron is in the $3d$ group.

For a $3d$ test electron, the rules are:

• Same group $(3d)$ excluding test: $5 - 1 = 4$ electrons × 0.35 = 1.40.
• All electrons in groups to the left (so $1s$ through $3p$) contribute 1.00 each: $2 + 8 + 8 = 18$ electrons × 1.00 = 18.00.
• Electrons to the right (i.e. $4s^{2}$) contribute 0.

Total $\sigma = 1.40 + 18.00 = 19.40$.

$$Z_{\text{eff}}(\text{Mn}, 3d) = 25 - 19.40 = \mathbf{5.60}$$

Takeaway: $d$-electrons feel a much larger effective nuclear charge than one might naively guess, because inner electrons shield $d$-orbitals much better than same-shell electrons shield one another.

Example 6: Explain the trend

Explain, in three sentences, why atomic radius decreases as we move across Period 2 from Li to F.

Solution:

Moving from Li ($Z=3$) to F ($Z=9$), the nuclear charge $Z$ increases by 6, while the added electrons all go into the same $n=2$ shell. Electrons in the same shell shield each other poorly (Slater: σ ≈ 0.35 per electron), so $Z_{\text{eff}} = Z - \sigma$ rises steeply across the period — from about 1.30 at Li to 5.20 at F. The higher $Z_{\text{eff}}$ pulls the valence electrons closer to the nucleus, so the atomic radius shrinks: Li 152 pm → F 64 pm.

Example 7: Why noble gas radii look anomalous

A student plots atomic radius across Period 3 and observes that Cl (99 pm) is followed by Ar (188 pm). Is this a violation of the across-the-period trend? Explain.

Solution:

No — this is only an apparent anomaly because Cl's tabulated value is a covalent radius, while Ar has no covalent bond and so its tabulated value is a van der Waals radius. Van der Waals radii are intrinsically larger than covalent radii, so the two numbers should not be directly compared.

Thus the apparent jump from Cl to Ar does not mean the periodic trend has failed — it means the definition of radius has changed. Across-the-period shrinkage is valid only when the same type of radius is being compared.

Takeaway: always check the radius type before declaring a trend broken.

Example 8: The lanthanoid contraction

Explain why zirconium (Zr, Z = 40) and hafnium (Hf, Z = 72) have almost identical atomic radii (≈ 160 pm and 159 pm respectively), despite Hf being one full period below Zr.

Solution:

Between Zr (Period 5) and Hf (Period 6), the 14 lanthanoid elements (Ce to Lu) have filled in the $4f$ sub-shell. The $4f$ electrons shield the outer electrons very poorly, because the $4f$ orbitals are tucked deep inside the atom with complicated angular shapes. So as we cross the lanthanoid series, $Z_{\text{eff}}$ for the outermost electrons increases far more than it would for a normal 14-step increase in $Z$.

The cumulative effect is a contraction of about 10 pm across the lanthanoid series — an amount almost exactly equal to the "natural" expansion Hf would have shown over Zr by moving one period down. The two effects cancel, giving:

$$r(\text{Zr}) \approx r(\text{Hf}) \approx 159\text{-}160\ \text{pm}$$

[JEE Tip] Zr/Hf, Nb/Ta, Mo/W are the three textbook pairs where the lanthanoid contraction shows up. Know all three, and know that they have nearly identical chemistries as a consequence — they are famously hard to separate industrially.

Example 9: Comparing two atoms in different periods and groups

Compare the atomic radii of potassium (K, Z = 19) and calcium (Ca, Z = 20). Predict which is larger and by roughly how much.

Solution:

K and Ca are both in Period 4. K is in Group 1 ($4s^{1}$) and Ca is in Group 2 ($4s^{2}$). Across a period, radius decreases, so Ca should be smaller than K.

Quantitatively:

• $Z_{\text{eff}}$ for K's $4s$ electron (Slater): $\sigma = 8 \times 0.85 + 10 \times 1.00 = 16.80$; $Z_{\text{eff}} = 19 - 16.80 = 2.20$.
• $Z_{\text{eff}}$ for Ca's $4s$ electron: $\sigma = 1 \times 0.35 + 8 \times 0.85 + 10 \times 1.00 = 17.15$; $Z_{\text{eff}} = 20 - 17.15 = 2.85$.

Ca has higher $Z_{\text{eff}}$ → smaller radius. Observed values: K 227 pm, Ca 197 pm. Ca is ≈ 30 pm smaller — consistent with the jump in $Z_{\text{eff}}$.

Example 10: Ordering in a mixed set

Arrange the following in order of increasing atomic radius: O, C, F, S, P, Si.

Solution:

Three are in Period 2 (C, O, F) and three are in Period 3 (Si, P, S).

Step 1 — order within each period (radius decreases to the right):

• Period 2: C (77 pm) > O (66 pm) > F (64 pm)
• Period 3: Si (117) > P (110) > S (104)

Step 2 — compare across periods. Any Period-3 element is larger than the element directly above it (down a group → radius grows).

Putting the two pieces together:

$$\mathbf{F < O < C < S < P < Si}$$

Checking with actual values:

Element	r (pm)
F	64
O	66
C	77
S	104
P	110
Si	117

So the order of increasing radius is: $\mathbf{F < O < C < S < P < Si}$.

Example 11: Why is radius of Ga smaller than Al?

Aluminium (Al, Z = 13) has a covalent radius of 143 pm, while gallium (Ga, Z = 31) — directly below Al in the periodic table — has a radius of only 141 pm. Explain this apparent violation of the "down-a-group" rule.

Solution:

Between Al (Period 3) and Ga (Period 4), we cross the first d-block (Sc to Zn). Adding 10 electrons to the $3d$ sub-shell gives ten extra units of nuclear charge to the Ga nucleus — but the $3d$ electrons shield the outer $4p$ electron poorly. So the net increase in $Z_{\text{eff}}$ for Ga's valence electron is unexpectedly large.

This extra $Z_{\text{eff}}$ pulls Ga's outer electrons inward enough to cancel the expected radial expansion. Thus $r(\text{Ga}) \lesssim r(\text{Al})$ — a classic example of d-block contraction.

[JEE Tip] The same effect helps explain why Ga is denser than Al despite lying below it.

Example 12: CBSE Board 3-marker — state, compute, compare

(a) Define atomic radius. Mention two types of atomic radii with example. (b) Calculate $Z_{\text{eff}}$ for the outermost electron in sodium (Z = 11) using Slater's rules. (c) State, with reason, whether $Z_{\text{eff}}$ for sodium's outermost electron is greater or less than that for lithium's outermost electron.

Solution:

(a) Atomic radius is the effective size of an atom, measured operationally as half the distance between two bonded or non-bonded nuclei of the same element. Two types:

• Covalent radius — half the distance between two identical atoms covalently bonded, e.g. $r_{\text{cov}}(\text{Cl}) = 99$ pm from $\text{Cl}_{2}$.
• Metallic radius — half the shortest inter-atomic distance in a pure metal, e.g. $r_{\text{met}}(\text{Na}) = 186$ pm.

(Van der Waals radius also acceptable as a third type.)

(b) Sodium: $Z = 11$, configuration $1s^{2}\,2s^{2}\,2p^{6}\,3s^{1}$. Test electron is the lone $3s^{1}$.

• Same group $(3s)$, excluding test: 0 electrons. Contribution: 0.
• $(n-1)$ shell $(2s, 2p)$: 8 electrons × 0.85 = 6.80.
• $(n-2)$ shell $(1s)$: 2 electrons × 1.00 = 2.00.

$\sigma = 6.80 + 2.00 = 8.80$, so $Z_{\text{eff}}(\text{Na}, 3s) = 11 - 8.80 = \mathbf{2.20}$.

(c) For Li ($Z = 3$, config $1s^{2}\,2s^{1}$): $\sigma = 2 \times 0.85 = 1.70$; $Z_{\text{eff}} = 3 - 1.70 = 1.30$.

Na has $Z_{\text{eff}} = 2.20$, Li has $Z_{\text{eff}} = 1.30$.

$Z_{\text{eff}}$ is greater for Na. Reason: although Na is in a higher shell and therefore larger overall, its outer electron also experiences a larger net nuclear attraction than Li's outer electron because the increase in nuclear charge is only partly offset by shielding.

(Full 3-marker scheme: 1 mark for (a), 1 for (b) calculation, 1 for (c) comparison + reason.)