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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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set 56/1/11 markMCQ
$Q. Which one of the following first row transition elements is expected to have the highest third ionization enthalpy? (A) Iron (Z = 26) (B) Manganese (Z = 25) (C) Chromium (Z = 24) (D) Vanadium (Z = 23)$
›Reveal solutionSolution
The third ionization enthalpy is highest for the element whose +2 ion has the most stable electronic configuration (half-filled or fully filled d-subshell). Among Fe, Mn, Cr, and V, the +2 ion of manganese (Mn²⁺) has a half-filled 3d⁵ configuration, making it exceptionally stable and hardest to remove an electron from. Thus, Mn has the highest third ionization enthalpy.

The key to this question lies not in memorizing numbers but in understanding what the third ionization enthalpy actually measures. It is the energy required to remove the third electron from a gaseous atom — that is, to go from the +2 ion to the +3 ion. So we are really comparing the stability of the M²⁺ ions of these elements. The more stable the M²⁺ ion, the harder it is to pull off another electron, and the higher the third ionization enthalpy.

Now, stability in transition metal ions is heavily influenced by the d-electron configuration. A half-filled d⁵ or fully filled d¹⁰ subshell confers exceptional stability. Let us examine each element’s +2 ion.

Vanadium (Z = 23)

Electronic configuration of V: $[\text{Ar}]\,3d^3\,4s^2$

V²⁺: remove two 4s electrons → $[\text{Ar}]\,3d^3$

This is neither half-filled nor fully filled. It is a relatively ordinary configuration.

Chromium (Z = 24)

Cr has a special ground state: $[\text{Ar}]\,3d^5\,4s^1$ (half-filled d gives extra stability).

Cr²⁺: remove the 4s electron and one 3d electron → $[\text{Ar}]\,3d^4$

This is not half-filled. The half-filled stability of Cr atom is lost in Cr²⁺.

Manganese (Z = 25)

Mn: $[\text{Ar}]\,3d^5\,4s^2$

Mn²⁺: remove two 4s electrons → $[\text{Ar}]\,3d^5$

This is exactly half-filled! The d⁵ configuration is exceptionally stable. Removing a third electron would break this stable half-filled shell, requiring a large amount of energy.

Iron (Z = 26)

Fe: $[\text{Ar}]\,3d^6\,4s^2$

Fe²⁺: remove two 4s electrons → $[\text{Ar}]\,3d^6$ …
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CBSE 2024Set 56/1/11 markMCQ
$Q. Assertion (A): Separation of Zr and Hf is difficult. Reason (R): Zr and Hf have similar radii due to lanthanoid contraction. [Codes (A)-(D) as in the Assertion-Reason instruction.]$
›Reveal solutionSolution
Lanthanoid contraction causes Zr and Hf to have nearly identical radii despite being in different periods, making their chemical properties so similar that separation becomes extremely difficult. Both statements are true and R correctly explains A.

The question tests your understanding of how the lanthanoid contraction affects the chemistry of post-lanthanoid elements, particularly the 4d and 5d transition metals.

Why Zr and Hf are Chemical Twins

Zirconium (Zr, atomic number 40) sits in the second transition series (4d block), while hafnium (Hf, atomic number 72) belongs to the third transition series (5d block). Normally, when you move down a group in the periodic table, atomic and ionic radii increase because you're adding entire electron shells. This is why sodium is larger than lithium, and potassium larger than sodium.

But something unusual happens between the 4d and 5d series. Between Zr and Hf, the periodic table inserts the entire lanthanoid series (elements 57–71) — fourteen f-block elements. As electrons fill the poorly shielding 4f orbitals across the lanthanoids, the effective nuclear charge experienced by outer electrons increases steadily. This pulls all the electron shells inward, causing a cumulative contraction in atomic size. By the time we reach Hf, this lanthanoid contraction has almost exactly compensated for the addition of an extra shell.

The result? The atomic radius of Zr is approximately 160 pm, while Hf is about 159 pm — virtually identical. Their ionic radii ( $\text{Zr}^{4+}$ ≈ 72 pm, $\text{Hf}^{4+}$ ≈ 71 pm) are equally similar.

Why Similar Radii Make Separation Difficult

Chemical behavior depends heavily on ionic size and charge. When two elements have:

The same oxidation states (both commonly +4)

Nearly identical ionic radii

The same coordination preferences

...they form compounds with almost indistinguishable properties. Their oxides, halides, and complexes have similar solubilities, crystal structures, and stabilities. Traditional separation methods like fractional crystallization or precipitation rely on differences in these properties, so when the properties are nearly identical, separation becomes extraordinarily challenging.

Historically, chemists struggled for decades to separate Zr and Hf. Even today, industrial separation requires sophisticated techniques like solvent extraction or ion exchange with carefully chosen ligands that can exploit the tiny remaining differences. …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. The first ionisation enthalpy of Xenon is almost identical with that of:$
(a) Molecular oxygen
(b) Molecular Nitrogen
(c) Molecular Fluorine
(d) Molecular Hydrogen
›Reveal solutionSolution

Xenon's ionisation enthalpy is unusually low for a noble gas — close enough to that of O2 that Xe can be oxidised by the same species that oxidise O2.

Noble gases normally have very high ionisation enthalpies because of their stable, fully-filled valence shells. However, Xenon is a large atom with a valence shell far from the nucleus (weak nuclear hold on outer electrons), so its first ionisation enthalpy (~1170 kJ/mol) is unusually low — almost identical to that of molecular oxygen (~1175 kJ/mol). This numerical coincidence is historically important: Neil Bartlett …

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CBSE 2023Set 56/3/11 markMCQ
$3d^3 4s^2$
›Reveal solutionSolution
The highest oxidation state in transition metals is achieved when all electrons from both the 4s and 3d orbitals are removed. Among the given configurations, $3d^5 4s^2$ (option C) allows the removal of 7 electrons, giving a maximum oxidation state of +7.

The key to this question lies in understanding how transition metals exhibit variable oxidation states. Unlike main group elements where the outermost s and p electrons are the only ones involved, transition metals can use both the ns and (n−1)d electrons for bonding. The highest possible oxidation state for a given configuration is simply the total number of electrons in the outermost s and d orbitals — because in principle, all of them can be lost.

Let’s examine each option carefully.

Option A: $3d^3 4s^2$

Total electrons in the valence shell = $3 + 2 = 5$ . So the maximum oxidation state possible is +5. This is seen in elements like vanadium (V), which indeed shows +5 in compounds like $\text{V}_2\text{O}_5$ .

Option B: $3d^5 4s^1$

Total = $5 + 1 = 6$ . Maximum oxidation state = +6. Chromium (Cr) has this configuration and shows +6 in $\text{CrO}_3$ and dichromates. Notice that chromium’s actual ground state is $3d^5 4s^1$ , not $3d^4 4s^2$ , due to the extra stability of a half-filled d-subshell.

Option C: $3d^5 4s^2$

Total = $5 + 2 = 7$ . Maximum oxidation state = +7. Manganese (Mn) has this configuration and exhibits +7 in permanganate ion ( $\text{MnO}_4^-$ ). This is the highest among the given options.

Option D: $3d^6 4s^2$ …
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CBSE 2023Set ANNUAL1 mark
$Q. Which is the most electronegative element in periodic table?$
›Reveal solutionSolution

Electronegativity generally increases across a period and decreases down a group; fluorine, at the top-right of the periodic table (excluding noble gases), is the most electronegative element.

Electronegativity is the tendency of an atom to attract a shared pair of electrons towards itself in a bond. It increases left-to-right across a period (smaller atomic size, more nuclear charge) and decreases top-to-bottom down a group (larger atomic size, more shielding). Fluor …

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CBSE 2022Set HE2181 mark
$Q. Fill in the blank: The highest electron affinity has ______ element.$
›Reveal solutionSolution

Chlorine, not fluorine, has the largest electron affinity, because fluorine's very small atomic size causes strong electron-electron repulsion in its compact valence shell when an extra electron is added.

Electron affinity (electron gain enthalpy) generally becomes more negative (larger magnitude) across a period (increasing effective nuclear charge) and less negative down a group (increasing atomic size, the added electron feels a weaker pull). Naively this predicts fluorine (top of Group 17) should have the highest electron affinity. However, fluorine is anomalous: its atom is so small and its 2p subshell so compact that the incoming electron experiences significant inter-electron repulsion, which partly offsets the favourable nuclear attraction and reduces the energy released. …

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CBSE 2022Set ANNUAL1 markMCQ
$Q. Which of the following element does not show allotropy?$
(a) Nitrogen
(b) Bismuth
(c) Antimony
(d) Arsenic
›Reveal solutionSolution
Nitrogen exists only as the diatomic $N_2$ molecule and has no allotropes; the heavier Group-15 elements (As, Sb, Bi) do show allotropy. Option (A).

Allotropy is the existence of an element in two or more different physical forms in the same physical state.

Nitrogen — exists as the gas $N_2$ with a strong $N\equiv N$ triple bond; it does not form allotropes. …
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Q.Match the properties given in Column I with the metals given in Column II. Column I (Property):

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

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