Metallic radii of some transition elements are given below. Which of these elements will have highest density? Element: Fe, Co, Ni, Cu Metallic radii/pm: Fe $= 126$, Co $= 125$, Ni $= 125$, Cu $= 128$ (i) Fe (ii) Ni (iii) Co (iv) Cu

Metallic radii of some transition elements are given below. Which of these elements will have highest density?

NCERT Exemplar · Q3

= 126

(i) Fe

(ii) Ni

(iii) Co

(iv) Cu

Yanam CbseMCQ· 1mImportance★★★★★

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Concept understanding — Ionization Energy Trends

Ionization Energy: The First Meeting

Imagine you're holding onto something precious — say, a favourite pen. How hard would someone have to pull to take it from your hand? That's the core idea behind ionization energy. In an atom, the "something precious" is an electron, and the "pulling force" is energy.

Ionization energy (IE) is the minimum energy required to remove the most loosely bound electron from a gaseous atom in its ground state.

Why "gaseous" and "ground state"? Because we want a fair comparison — no extra energy from neighbours or from the atom being already excited. We measure how tightly the atom holds its outermost electron when it's alone and calm.

The unit you'll see most often in exams: kJ/mol (kilojoules per mole of atoms).

The Intuition: What Controls the Grip?

Two factors decide how hard an atom holds its outermost electron:

Nuclear charge — more protons in the nucleus means a stronger pull on the electron. Simple: bigger positive charge, tighter grip.
Distance from the nucleus — the farther the electron is, the weaker the pull. Think of a magnet: it holds a paperclip strongly up close, but barely at arm's length.

But there's a subtle twist: shielding (or screening). Inner electrons partially block the nuclear charge from reaching the outer electron. The outermost electron doesn't "feel" the full nuclear charge — it feels only the effective nuclear charge ( $Z_{\text{eff}}$ ).

Note

$Z_{\text{eff}} = Z - S$ , where $Z$ is the atomic number and $S$ is the shielding constant (roughly the number of inner electrons). This is the net positive charge pulling on the outer electron.

So the real question becomes: How large is $Z_{\text{eff}}$ for the outermost electron, and how far away is it?

The Precise Trend: Across a Period

As you move left to right across a period (say, from Li to Ne in period 2):

Nuclear charge increases steadily (more protons).
Electrons are added to the same shell — no new inner layers.
Shielding stays roughly constant (same number of inner electrons).
Result: $Z_{\text{eff}}$ increases → the outer electron is pulled in tighter → ionization energy increases.

Across a period: IE increases (generally).

Example:

Li (IE = 520 kJ/mol) → Be (900) → B (801) → C (1086) → N (1402) → O (1314) → F (1681) → Ne (2081)

Wait — why does B have lower IE than Be? And O lower than N? That's the exception, not the rule. We'll come back to it.

The Precise Trend: Down a Group

As you move down a group (say, from Li to Cs in group 1):

Nuclear charge increases (more protons).
But electrons are added to new, higher shells — the outermost electron is much farther from the nucleus.
Shielding also increases significantly (more inner electrons).
Result: distance dominates → the outer electron is held more loosely → ionization energy decreases.

Down a group: IE decreases.

Example:

Li (520) → Na (496) → K (419) → Rb (403) → Cs (376) — all in kJ/mol.

The Two Exceptions (and Why They Matter)

Exception 1: Group 13 vs Group 2 (e.g., B vs Be)

Be has a full $2s^2$ subshell. B has $2s^2 2p^1$ . The $2p$ electron is slightly higher in energy and slightly better shielded by the $2s$ electrons than the $2s$ electrons shield each other. So removing the $2p$ electron from B takes less energy than removing a $2s$ electron from Be.

Watch out

Don't memorise "IE increases across a period" blindly. Group 13 always has lower IE than Group 2 in the same period.

Exception 2: Group 16 vs Group 15 (e.g., O vs N)

N has a half-filled $2p^3$ subshell — each $2p$ orbital has one electron. This is an especially stable arrangement (exchange energy stabilisation). O has $2p^4$ — one orbital gets a second electron. That extra electron experiences electron-electron repulsion, making it easier to remove. So O has lower IE than N.

Tip

|--------|--------------|--------------|------------------|

| 2 | N (1402) | O (1314) | N > O |

| 3 | P (1012) | S (1000) | P > S |

| 4 | As (947) | Se (941) | As > Se |

The pattern holds for all periods.

The Big Picture: What You Must Remember

Important

Ionization energy increases across a period (with two dips) and decreases down a group.

The dips occur at Group 13 (lower than Group 2) and Group 16 (lower than Group 15).

The underlying reason is always the same: effective nuclear charge and distance. When $Z_{\text{eff}}$ is high and the electron is close, IE is high. When the electron is far or repulsion helps it leave, IE is low.

A Final Check: First vs Second Ionization Energy

Removing one electron from an atom leaves a positive ion. Removing a second electron from that ion is always harder — the ion has a higher positive charge pulling on the remaining electrons.

Second IE > First IE — always. For example, Na: first IE = 496 kJ/mol, second IE = 4562 kJ/mol. That's nearly 10 times larger. This huge jump tells you that the second electron comes from a different shell (closer to the nucleus).

In exams, this jump is used to identify the group of an element — a sudden large increase in successive ionization energies indicates you've stripped off all valence electrons and are now pulling from a core shell.

"Ionization energy trends periodic table" and "periodicity class 11 chemistry important questions" are extremely common searches, both anchored in the Classification of Elements and Periodicity chapter of the NCERT/CBSE Class 11 Chemistry curriculum. The Group 13 and Group 16 exceptions in particular are a favourite trap question in board exams and JEE Main.

Why this formula?

Ionization Energy Trends: The Why Behind the Trends

Ionization energy (IE) is the minimum energy required to remove the most loosely bound electron from a gaseous atom in its ground state.

It is measured in kJ/mol or eV/atom.

The key trend is:

IE increases across a period (left → right) and decreases down a group (top → bottom).

But why? Let’s break down the reasoning step-by-step.

1. The Core Formula: Coulomb’s Law

The energy needed to remove an electron is fundamentally governed by the electrostatic attraction between the electron and the nucleus.

F = \frac{k \cdot Z_{\text{eff}} \cdot e^2}{r^2}

Where:

$Z_{\text{eff}}$ = effective nuclear charge (net positive charge felt by the electron)
$r$ = distance of the electron from the nucleus
$e$ = charge of electron
$k$ = Coulomb constant

Key insight: The stronger the attraction, the higher the ionization energy.

2. Why IE Increases Across a Period

Reasoning:

As you move left → right, protons increase in the nucleus.
Electrons are added to the same principal energy level (same shell).
Shielding by inner electrons remains roughly constant (same number of inner shells).
Therefore, $Z_{\text{eff}}$ increases — the outer electrons feel a stronger pull.

Result:

\text{IE} \propto Z_{\text{eff}}

So IE increases across a period.

Example:

Na (Z=11): IE = 496 kJ/mol
Mg (Z=12): IE = 738 kJ/mol
Al (Z=13): IE = 578 kJ/mol (slight dip due to p-orbital shielding — see exception below)

3. Why IE Decreases Down a Group

Reasoning:

As you move down a group, principal quantum number $n$ increases.
The outermost electron is farther from the nucleus ( $r$ increases).
Shielding increases because more inner electron shells are present.
$Z_{\text{eff}}$ increases only slightly (not enough to compensate for distance).

Result:

\text{IE} \propto \frac{1}{r^2}

So IE decreases down a group.

Example:

Li (n=2): IE = 520 kJ/mol
Na (n=3): IE = 496 kJ/mol
K (n=4): IE = 419 kJ/mol

4. The Mathematical Expression (Approximation)

For a hydrogen-like atom (single electron), the ionization energy is given by:

\text{IE} = \frac{13.6 \, \text{eV} \cdot Z^2}{n^2}

For multi-electron atoms, we replace $Z$ with $Z_{\text{eff}}$ :

\text{IE} \approx \frac{13.6 \, \text{eV} \cdot Z_{\text{eff}}^2}{n^2}

Why this holds:

$Z_{\text{eff}}$ accounts for shielding by inner electrons.
$n$ is the principal quantum number of the electron being removed.
The $1/n^2$ dependence comes from the Bohr model — energy levels scale as $E_n \propto -Z^2/n^2$ .

5. Exceptions (Why the Trend Isn’t Perfect)

a) Group 13 vs Group 2 (e.g., Al vs Mg)

Al has a p-orbital electron (higher energy, easier to remove) than Mg’s s-orbital.
Also, p-orbitals are more shielded by s- and p-electrons.

b) Group 16 vs Group 15 (e.g., O vs N)

N has a half-filled p-subshell (extra stability).
O has one paired electron — electron-electron repulsion makes removal easier.

6. Summary Table

Factor	Across Period (→)	Down Group (↓)
$Z_{\text{eff}}$	Increases	Increases slightly
$r$ (distance)	Decreases slightly	Increases
Shielding	Constant	Increases
IE	Increases	Decreases

Final Takeaway

Ionization energy is not just a number — it’s a direct consequence of Coulomb’s law, modified by shielding and orbital shape.

The trend is driven by $Z_{\text{eff}}$ (across) and distance + shielding (down).

Always ask: “How strongly is this electron held?” — and the answer lies in the balance of nuclear charge, distance, and shielding.

Density depends on atomic mass and atomic volume (radius³). Among Fe, Co, Ni, and Cu, copper has the highest atomic mass and a relatively large radius, giving it the highest density.

Why density depends on radius and mass

Density is mass per unit volume. For a metallic crystal, the density of the element is proportional to:

\text{Density} \propto \frac{\text{Atomic mass}}{(\text{Metallic radius})^3}

The exact formula involves packing fraction and Avogadro’s number, but for comparing elements with the same crystal structure (all four are face-centered cubic at room temperature), the packing fraction cancels out. So we only need to compare:

\frac{\text{Atomic mass}}{r^3}

A higher atomic mass and a smaller radius both push density up. Let’s see which element wins.

Step-by-step comparison

1. List the given data

Element	Metallic radius (pm)	Atomic mass (g/mol)
Fe	126	55.85
Co	125	58.93
Ni	125	58.69
Cu	128	63.55

The radii are nearly equal — all within 3 pm of each other. So the atomic mass will be the deciding factor.

2. Compute $M / r^3$ for each

We can work with relative values since the constant factor (packing fraction, Avogadro’s number) is the same for all.

For Fe:

\frac{55.85}{126^3} = \frac{55.85}{2\,000\,376} \approx 2.79 \times 10^{-5}

For Co:

\frac{58.93}{125^3} = \frac{58.93}{1\,953\,125} \approx 3.02 \times 10^{-5}

For Ni:

\frac{58.69}{125^3} \approx 3.00 \times 10^{-5}

For Cu:

\frac{63.55}{128^3} = \frac{63.55}{2\,097\,152} \approx 3.03 \times 10^{-5}

3. Compare the values

Fe: $2.79 \times 10^{-5}$ — lowest, because Fe has the smallest atomic mass and a mid-sized radius.
Co: $3.02 \times 10^{-5}$ — higher than Fe.
Ni: $3.00 \times 10^{-5}$ — very close to Co, slightly lower.
Cu: $3.03 \times 10^{-5}$ — the highest value.

Copper’s atomic mass is about 7–8% higher than cobalt’s, and its radius is only 2.4% larger. The cube in the denominator means a 2.4% radius increase raises the volume by about 7.4%, but the mass increase of ~7.8% more than compensates. So Cu edges ahead.

Tip

You don’t need to compute the exact numbers. Just compare ratios:

For Co vs Cu: $\frac{58.93}{125^3}$ vs $\frac{63.55}{128^3}$ .

Notice $128/125 = 1.024$ , so $(128/125)^3 \approx 1.074$ .

The mass ratio $63.55/58.93 \approx 1.078$ . Since $1.078 > 1.074$ , Cu wins.

4. Final ranking

Cu > Co > Ni > Fe in density.

Watch out

A common mistake is to pick the element with the smallest radius (Co or Ni) thinking that smaller radius always means higher density. But density also depends on atomic mass — copper is heavier enough to overcome its slightly larger radius.

✓Final answer

The element with the highest density is (iv) Cu .

Method: Density Estimation from Metallic Radius and Atomic Mass

This problem uses the relationship between density, atomic mass, and atomic radius in a periodic table trend context.

Concept Behind the Method

Density ( $\rho$ ) is mass per unit volume. For metallic elements in the same period:

Mass depends on atomic mass (increases across a period)
Volume depends on atomic radius (generally decreases across a period)

Since density $\propto \frac{\text{mass}}{\text{volume}}$ , and volume $\propto r^3$ , we can compare density using:

\rho \propto \frac{\text{Atomic mass}}{r^3}

Steps to Solve

Step 1: List the given data

Element	Metallic radius (pm)	Atomic mass (g/mol)
Fe	126	55.85
Co	125	58.93
Ni	125	58.69
Cu	128	63.55

Step 2: Calculate $r^3$ for each element

Fe: $126^3 = 2,000,376$
Co: $125^3 = 1,953,125$
Ni: $125^3 = 1,953,125$
Cu: $128^3 = 2,097,152$

Step 3: Compute $\frac{\text{Atomic mass}}{r^3}$ for comparison

Fe: $\frac{55.85}{2,000,376} \approx 2.79 \times 10^{-5}$
Co: $\frac{58.93}{1,953,125} \approx 3.02 \times 10^{-5}$
Ni: $\frac{58.69}{1,953,125} \approx 3.00 \times 10^{-5}$
Cu: $\frac{63.55}{2,097,152} \approx 3.03 \times 10^{-5}$

Step 4: Compare the values

The highest value indicates the highest density.

Order: Cu > Co > Ni > Fe

Final Answer

Cu has the highest density.

\boxed{\text{(iv) Cu}}

Why This Works

Across the first transition series, atomic mass increases steadily while atomic radius remains nearly constant (due to poor shielding by d-electrons). Copper has the highest atomic mass among these four, with only a slightly larger radius — giving it the highest density.

Common Mistakes & How to Avoid Them

Mistake 1: Assuming the smallest radius always gives the highest density

Why students make it:

They see Co and Ni have the smallest radii (125 pm) and assume "smaller atom = more tightly packed = denser," picking (iii) Co or (ii) Ni without checking atomic mass.

Why it's wrong:

Density depends on mass per unit volume, not radius alone. A smaller radius does increase density for a fixed mass, but here the atomic masses are also different — and that difference decides the outcome.

How to avoid:

Always write the formula:

\text{Density} \propto \frac{\text{Atomic mass}}{\text{Atomic volume}}

For a spherical atom, volume $\propto r^3$ . So:

\text{Density} \propto \frac{\text{Atomic mass}}{r^3}

Check the numbers:

Element	Atomic mass (g/mol)	Radius (pm)	$r^3$ (× $10^6$ pm³)	Mass/ $r^3$ (relative)
Fe	55.85	126	2.00	27.9
Co	58.93	125	1.95	30.2
Ni	58.69	125	1.95	30.1
Cu	63.55	128	2.10	30.3

Result: Cu has the highest mass per unit volume → highest density, even though its radius is the largest of the four — its atomic mass is high enough to overcome the larger volume.

Correct answer: (iv) Cu

Mistake 2: Forgetting that density depends on atomic mass, not just radius

Why students make it:

They focus only on the given radii and ignore the atomic masses (which are not directly given but must be recalled from periodic table knowledge).

How to avoid:

Always recall or note the atomic masses of the elements in the series. For 3d transition series:

Fe ≈ 56
Co ≈ 59
Ni ≈ 58.7
Cu ≈ 63.5

Cu has both the largest radius and the largest mass among these four — so the ratio, not either number alone, decides the answer.

Mistake 3: Thinking density tracks radius alone across the series

Why students make it:

They remember that atomic radii change only slightly across Fe–Co–Ni–Cu, so they assume the element with the smallest radius must be densest.

Why it's wrong:

Across the 3d series, density is set by the combination of mass and volume. Cu's radius is about 2.4% larger than Co's/Ni's, which raises its volume by about 7.4% — but Cu's atomic mass is about 7.8% higher than Co's, which more than compensates.

How to avoid:

Compare ratios directly instead of eyeballing radius or mass alone:

Fe → Co: mass ↑ noticeably, radius ↓ slightly → density increases
Co → Ni: mass and radius both nearly unchanged → density nearly the same
Ni → Cu: mass ↑ more than radius ↑ → density increases again, reaching its highest value at Cu

So the peak is at Cu, not Co.

Mistake 4: Confusing metallic radius with density formula

Why students make it:

They try to use the metallic radius directly in density without converting to volume.

How to avoid:

Remember: density uses volume, which scales as $r^3$ . A 2% change in radius causes ~6-7% change in volume — significant enough to matter here.

Quick Summary — Do This Instead

Step	Action
1	Note atomic masses (from memory or periodic table)
2	Compute $r^3$ for each element
3	Compute mass/ $r^3$ (relative comparison is enough)
4	Pick the largest ratio

Final answer: (iv) Cu

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Q.Metallic radii of some transition elements are given below. Which of these elements will have highest density? Element: Fe, Co, Ni, Cu Metallic radii/pm: Fe =126= 126=126, Co =125= 125=125, Ni =125= 125=125, Cu =128= 128=128

Why density depends on radius and mass

Step-by-step comparison

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Q.Metallic radii of some transition elements are given below. Which of these elements will have highest density? Element: Fe, Co, Ni, Cu
Metallic radii/pm: Fe $= 126$ , Co $= 125$ , Ni $= 125$ , Cu $= 128$

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