Match the properties given in Column I with the metals given in Column II. Column I (Property): (i) Element wi

NCERT Exemplar · Q66

Q. Match the properties given in Column I with the metals given in Column II. Column I (Property):

(i) Element with highest second ionisation enthalpy

(ii) Element with highest third ionisation enthalpy

(iii) M in

M(CO)_6

(iv) Element with highest heat of atomisation
Column II (Metal):

(a) Co

(b) Cr

(d) Zn

(e) Ni

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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

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. …

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Correct matches: (i) → (c) Cu, (ii) → (d) Zn, (iii) → (b) Cr, (iv) → (e) Ni.

Highest second ionisation enthalpy → Cu. $IE_2$ removes an electron from $M^{+}$ . For copper, $Cu^{+} = [Ar]3d^{10}$ — a stable, fully-filled $d$ subshell — so removing the next electron is exceptionally difficult. Cu has the highest $IE_2$ of the 3d series.
Highest third ionisation enthalpy → Zn. $IE_3$ removes an electron from $M^{2+}$ . For zinc, $Zn^{2+} = [Ar]3d^{10}$ (stable filled $d$ subshell), so its $IE_3$ is the highest of the series. (iii) M in $M(CO)_6$ → Cr. By the 18-electron rule, six CO ligands donate $6\times2 = 12$ electrons, so the metal must supply $18-12 = 6$ . Chromium ( $[Ar]3d^5\,4s^1$ , 6 valence electrons) meets this, giving the stable $Cr(CO)_6$ . …

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Method: Electronic Configuration & Periodic Trend Analysis

This method uses electronic configurations and periodic trends (ionization enthalpy, stability of half-filled/d orbitals, and metallic bonding strength) to match properties with metals.

Step 1: Write electronic configurations of all metals

Metal	Atomic No.	Configuration
Co	27	$[Ar]\,3d^7\,4s^2$
Cr	24	$[Ar]\,3d^5\,4s^1$
Cu	29	$[Ar]\,3d^{10}\,4s^1$
Zn	30	$[Ar]\,3d^{10}\,4s^2$
Ni	28	$[Ar]\,3d^8\,4s^2$

Step 2: Match (i) — Highest second ionisation enthalpy

Second IE = energy to remove one electron from $M^+$ ion.
After losing one electron, Cu becomes $[Ar]\,3d^{10}$ — fully filled, very stable.
Removing a second electron from this filled shell requires very high energy, giving Cu the highest second IE among the given metals.
Result: (i) → (c) Cu

Step 3: Match (ii) — Highest third ionisation enthalpy

After losing two electrons, Zn becomes $[Ar]\,3d^{10}$ — fully filled, very stable.
Removing a third electron from this stable $d^{10}$ core needs extremely high energy.
Result: (ii) → (d) Zn

Step 4: Match (iii) — M in $M(CO)_6$

Metal carbonyls follow the 18-electron rule.
For $M(CO)_6$ , each CO donates 2 electrons → 12 from CO.
M must contribute 6 electrons to reach 18.
Cr has configuration $3d^5\,4s^1$ — total 6 valence electrons.
Result: (iii) → (b) Cr

Step 5: Match (iv) — Highest heat of atomisation

Heat of atomisation depends on metallic bond strength; the actual experimental trend does not simply track the half-filled/fully-filled stability rule used for ionisation enthalpy. …

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Here are the common mistakes students make when solving this specific question on ionization enthalpy trends, along with how to avoid each.

Mistake 1: Confusing "Highest Second IE" with "Highest First IE"

The Error:

Students often pick Zn for (i) because Zn has a high first ionization enthalpy due to its stable $3d^{10}4s^2$ configuration. However, the question asks for second ionization enthalpy.

Why It’s Wrong:

Zn’s second IE is low because after losing one electron, it becomes $3d^{10}4s^1$ — losing the second electron gives a stable $3d^{10}$ configuration, which is easy.
The element with the highest second IE is Cu.
- Cu: $[Ar] 3d^{10}4s^1$ → after losing one electron → $3d^{10}$ (stable). Removing a second electron from a filled d-subshell requires a huge amount of energy.

How to Avoid:

Always write the electronic configuration of the atom and the ion after the first removal.
Look for the stability of the resulting configuration — a filled or half-filled d-subshell makes the next removal very hard.

Correct match: (i) → (c) Cu

Mistake 2: Forgetting that Third IE depends on Core Stability

The Error:

Students sometimes pick Cu again for (iii) or guess Ni without checking the configuration after two removals.

Why It’s Wrong:

After losing two electrons, Cu becomes $3d^9$ — not particularly stable.
The element with the highest third IE is Zn.
- Zn: $[Ar] 3d^{10}4s^2$ → after losing two electrons → $3d^{10}$ (stable). Removing a third electron from a filled d-subshell is extremely difficult.

How to Avoid:

Track the ion after each removal.
For third IE, check which element reaches a noble gas core or a filled d-subshell after two removals.

Correct match: (ii) → (d) Zn

Mistake 3: Misidentifying the Metal in $M(CO)_6$

The Error:

Students often pick Co or Ni because they are common in carbonyl complexes, but they forget the 18-electron rule.

Why It’s Wrong:

$M(CO)_6$ means the metal is bonded to 6 CO ligands. Each CO donates 2 electrons → total 12 electrons from ligands.
For the complex to be stable, the metal must contribute 6 electrons to reach 18.
Cr has atomic number 24: $[Ar] 3d^5 4s^1$ → it contributes 6 electrons (5 from 3d + 1 from 4s).
Co and Ni would contribute 9 and 10 electrons respectively, leading to electron counts >18, which is unstable for this geometry.

How to Avoid:

Memorize the 18-electron rule for carbonyls.
For $M(CO)_6$ , the metal must be in zero oxidation state and have 6 valence electrons.

Correct match: (iii) → (b) Cr

Mistake 4: Assuming "Highest Heat of Atomisation" means "Highest Melting Point"

The Error:

Students pick Cr because it has a very high melting point, but they don't check the actual trend in atomisation enthalpy.

Why It’s Wrong:

Heat of atomisation depends on metallic bond strength, which is influenced by the number of unpaired electrons in the d-subshell. …

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