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.

COMEDK 2026Set 2026-A1 markMCQ
$\mathrm{Z}=23$
›Reveal solutionSolution
The key is to compare the electronic configurations of the third ionisation (removing the 3rd electron) for each element. The element with the highest third ionisation enthalpy is the one whose third removal disrupts a particularly stable configuration — here, that is Manganese (Mn), giving the answer (B).

Concept & Intuition

Ionisation enthalpy jumps sharply when you try to remove an electron from a stable, half-filled or fully-filled subshell. For transition metals, the third ionisation often involves removing an electron from the 3d subshell. If the +2 ion already has a half-filled 3d⁵ configuration, removing a third electron destroys that stability, requiring a lot of energy. So we look for the element whose +2 ion has exactly 3d⁵.

Step-by-step reasoning

Write the ground-state electronic configurations (using the Aufbau principle, remembering that 4s fills before 3d but empties first in ionisation):

V (Z=23): [Ar] 3d³ 4s²

Cr (Z=24): [Ar] 3d⁵ 4s¹ (special stability of half-filled 3d)

Mn (Z=25): [Ar] 3d⁵ 4s²

Fe (Z=26): [Ar] 3d⁶ 4s²

First two ionisations remove the 4s electrons (and possibly one 3d for Cr, because its 4s¹ is easily lost):

V → V²⁺: [Ar] 3d³

Cr → Cr²⁺: [Ar] 3d⁴ (loses the 4s¹ and one 3d)

Mn → Mn²⁺: [Ar] 3d⁵ (half-filled, very stable)

Fe → Fe²⁺: [Ar] 3d⁶

Third ionisation removes one more electron from the 3d subshell:

V³⁺: 3d² (removing from 3d³ — no special stability lost)

Cr³⁺: 3d³ (removing from 3d⁴ — no special stability lost)

Mn³⁺: 3d⁴ (removing from 3d⁵ — destroys the half-filled stability)

Fe³⁺: 3d⁵ (removing from 3d⁶ — actually gains half-filled stability, so this ionisation is lower than expected)

Compare the energy cost: …
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KCET 2025Set D-41 markMCQ
$Q. A member of the Lanthanoid series which is well known to exhibit +4 oxidation state is (A) Samarium (B) Europium (C) Erbium (D) Cerium$
›Reveal solutionSolution
Among the lanthanoids the +4 state is stable only when the resulting ion reaches an especially stable f-configuration — Cerium's $\ce{Ce^{4+}}$ is $4f^0$ , the bare xenon core.

Step 1 — The concept: why lanthanoids are normally +3

The characteristic oxidation state of the entire lanthanoid series is +3, formed by losing the two $6s$ electrons and one $4f$ (or $5d$ ) electron. Any state other than +3 appears only when there is a special stability to be gained, namely an

empty f-subshell, $4f^{0}$ ,

half-filled f-subshell, $4f^{7}$ , or

completely filled f-subshell, $4f^{14}$ .

This single rule explains every "anomalous" lanthanoid oxidation state you are asked to remember.

Step 2 — Apply the rule to Cerium (Z = 58)

Ground-state configuration:
$\ce{Ce}: [\text{Xe}]\,4f^{1}\,5d^{1}\,6s^{2}$
Removing all four of those outer electrons ( $1 + 1 + 2 = 4$ ) gives:
$\ce{Ce^{4+}}: [\text{Xe}]\,4f^{0}$
The ion is left with the complete, noble-gas xenon core — an $f^0$ configuration. That is a strong thermodynamic incentive, so Ce readily shows +4. In practice $\ce{Ce^{4+}}$ salts (ceric ammonium nitrate/sulphate) are stable, commercially used oxidising agents:
$\ce{Ce^{4+} + e^- -> Ce^{3+}} \qquad E^\circ \approx +1.74~\text{V}$
(The large positive $E^\circ$ tells us $\ce{Ce^{4+}}$ does still want to revert to the +3 state — it is a good oxidant — but it is kinetically stable enough to bottle and use, which is what "well known to exhibit +4" means.)

Step 3 — Why the other three fail

| Element | Z | Configuration | Anomalous state it does show | Reason |

|---|---|---|---|---| …
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COMEDK 2025Set 2025-A1 markMCQ
$\mathrm{kJ} \mathrm{mol}^{-1}$
›Reveal solutionSolution
The first ionisation enthalpy of Al is estimated by interpolating between Mg and Si, accounting for the drop due to the p-orbital electron. The value is 578 kJ mol⁻¹, option (B).

The key concept here is periodic trends in ionisation enthalpy and the anomalous drop at Group 13. Normally, ionisation enthalpy increases across a period (left to right) as nuclear charge increases and atomic radius decreases. However, there is a well-known dip at aluminium (Group 13) compared to magnesium (Group 2) because aluminium’s outermost electron is in a 3p orbital, which is slightly higher in energy and more shielded than the 3s orbital of magnesium. This makes it easier to remove.

We are given:

Na (Group 1): 497 kJ mol⁻¹

Mg (Group 2): 738 kJ mol⁻¹

Si (Group 14): 787 kJ mol⁻¹

We need Al (Group 13). Let’s reason step by step.

Observe the trend from Na to Mg to Si

From Na to Mg, the increase is $738 - 497 = 241$ kJ mol⁻¹.

From Mg to Si, the increase is $787 - 738 = 49$ kJ mol⁻¹.

The jump from Na to Mg is large because Mg has a full 3s subshell and higher nuclear charge. The jump from Mg to Si is smaller because Si’s electron is in the same shell but with higher nuclear charge.

Estimate where Al should lie

If the trend were perfectly linear across the period, Al (between Mg and Si) would have an ionisation enthalpy roughly halfway between 738 and 787:

$\frac{738 + 787}{2} = 762.5 \ \text{kJ mol}^{-1}$
But we know this is not the case — Al has a lower value than this linear estimate because of the s-to-p orbital change.

Account for the drop due to the p-orbital electron

The drop from Mg to Al is a classic exception. Typically, the first ionisation enthalpy of Al is about 10–15% lower than that of Mg. A common empirical observation is that Al’s value is close to that of Na plus a small increment, but here we can use the given data more directly.

Notice that the increase from Na to Mg is 241, while from Mg to Si is only 49. The drop at Al means its value should be less than Mg’s value of 738. How much less? A reasonable estimate: the difference between Mg and Al is often around 150–160 kJ mol⁻¹.
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KCET 2022Set B-31 markMCQ

^{x+}

›Reveal solution

Invert the spin-only formula to get $n = 4$ unpaired electrons, then find which $\mathrm{Fe}^{x+}$ ion has exactly four.

Step 1 — The spin-only formula.

For a transition-metal ion whose magnetism comes from electron spin alone (orbital contribution quenched, as is usual for first-row ions):

\mu = \sqrt{n(n+2)}\ \text{BM}

where $n$ is the number of unpaired electrons. So a measured $\mu$ can be inverted to count the unpaired electrons — that is the whole strategy here.

Step 2 — Solve for $n$ .

\sqrt{n(n+2)} = \sqrt{24} \quad\Longrightarrow\quad n(n+2) = 24

n^2 + 2n - 24 = 0

(n + 6)(n - 4) = 0 \quad\Longrightarrow\quad n = 4 \ \text{or}\ n = -6

A count of electrons cannot be negative, so

\boxed{n = 4\ \text{unpaired electrons}}

(Quick check: $\mu = \sqrt{4 \times 6} = \sqrt{24} = 4.90$ BM ✓.)

Step 3 — Write the configurations of the candidate iron ions.

Iron has $Z = 26$ : $\mathrm{Fe} = [\mathrm{Ar}]\,3d^6 4s^2$ . Remove the $4s$ electrons first, then $3d$ . Fill the five $d$ orbitals by Hund's rule (singly first, then pair up):

Ion	Configuration	$d$ -orbital filling	Unpaired $n$	$\mu = \sqrt{n(n+2)}$
$\mathrm{Fe}$ ( $x=0$ )	$3d^6 4s^2$	—	4 (but $4s$ paired; neutral atom, not an ion)	—
$\mathrm{Fe^{+}}$	$3d^{6}4s^{1}$	—	5	5.92
$\mathrm{Fe^{2+}}$	$[\mathrm{Ar}]\,3d^{6}$	↿⇂ ↿ ↿ ↿ ↿	4 ✓	4.90 = $\sqrt{24}$ ✓

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KCET 2022Set B-31 markMCQ
$Q. The property of halogens which is not correctly matched is (A) I > Br > Cl > F (density) (B) F > Cl > Br > I (electron gain enthalpy) (C) F > Cl > Br > I (ionization enthalpy) (D) F > Cl > Br > I (electronegativity)$
›Reveal solutionSolution

Every option follows the normal group trend except electron gain enthalpy, whose correct (magnitude) order is $\mathrm{Cl>F>Br>I}$ — fluorine's small size makes it the famous exception.

Step 1 — Check (A): density, $\mathrm{I>Br>Cl>F}$ .

Down group 17 the atomic mass rises much faster than the atomic volume, so density increases down the group ( $\mathrm{F_2}$ and $\mathrm{Cl_2}$ are gases, $\mathrm{Br_2}$ a liquid, $\mathrm{I_2}$ a solid). The order $\mathrm{I>Br>Cl>F}$ is correct.

Step 2 — Check (C): ionisation enthalpy, $\mathrm{F>Cl>Br>I}$ .

Ionisation enthalpy decreases down a group: the outermost electron sits in a shell of larger $n$ , further from the nucleus and better shielded, so it is easier to pull out. $\mathrm{F>Cl>Br>I}$ is correct (halogens have very high IE anyway — they want to gain, not lose, an electron).

Step 3 — Check (D): electronegativity, $\mathrm{F>Cl>Br>I}$ .

Fluorine is the most electronegative element in the periodic table (4.0 on the Pauling scale), and electronegativity falls down the group as size grows: F (4.0) > Cl (3.2) > Br (3.0) > I (2.7). Correct.

Step 4 — Check (B): electron gain enthalpy — the exception.

Halogens have the most negative $\Delta_{eg}H$ of all groups (one electron completes the octet). But the value is not most negative for fluorine. The observed order of magnitude is

$\mathrm{Cl\;(-349)\;>\;F\;(-328)\;>\;Br\;(-325)\;>\;I\;(-295)\ \ kJ\,mol^{-1}}$ …
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KCET 2021Set B-21 markMCQ

^{3+}

›Reveal solution

Write the d-electron configuration of every ion; an aquated ion is coloured only when the d sub-shell is partially filled ( $d^1$ – $d^9$ ), so the pair in which both ions satisfy that is the answer.

1. Why d-electron count decides the colour

In an octahedral aqua complex $[\mathrm{M(H_2O)_6}]^{n+}$ the five degenerate d orbitals split into a lower $t_{2g}$ set and an upper $e_g$ set, separated by the crystal-field splitting energy $\Delta_o$ . Absorption of a visible photon promotes an electron $t_{2g}\rightarrow e_g$ (a d–d transition), and the ion appears in the complementary colour of the light absorbed.

Such a transition needs (i) at least one electron in the d sub-shell and (ii) at least one vacancy in it. Hence:

\text{coloured} \iff d^1 \text{ to } d^9 \qquad\qquad \text{colourless} \iff d^0 \text{ or } d^{10}

2. Configurations of the ions offered

Use $Z$ and remove the 4s electrons first (4s is lost before 3d on ionisation):

Ion	$Z$	Neutral atom	Configuration of the ion	d-count	Colour
$\mathrm{Sc^{3+}}$	21	$[\mathrm{Ar}]3d^1 4s^2$	$[\mathrm{Ar}]3d^0$	$d^0$	colourless
$\mathrm{Ti^{4+}}$	22	$[\mathrm{Ar}]3d^2 4s^2$	$[\mathrm{Ar}]3d^0$	$d^0$	colourless
$\mathrm{Ti^{3+}}$	22	$[\mathrm{Ar}]3d^2 4s^2$	$[\mathrm{Ar}]3d^1$	$d^1$	coloured (purple)
$\mathrm{Mn^{2+}}$	25	$[\mathrm{Ar}]3d^5 4s^2$	$[\mathrm{Ar}]3d^5$	$d^5$	coloured (pale pink)
$\mathrm{Ni^{2+}}$	28	$[\mathrm{Ar}]3d^8 4s^2$	$[\mathrm{Ar}]3d^8$	$d^8$	coloured (green)
$\mathrm{Cu^{+}}$	29	$[\mathrm{Ar}]3d^{10}4s^1$	$[\mathrm{Ar}]3d^{10}$	$d^{10}$	colourless

3. Test each pair …

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COMEDK 2021Set 2021-B1 markMCQ
$Q. The graph given represents the variation in first Ionization enthalpy with change in Z value of Group 1 elements of the Periodic table represented as P, Q, R, S and T (not the actual symbols of the elements). Identify the actual Group 1 element which has minimum IE1 value. [FIGURE: plot of Ionisation enthalpy (kJ mol^-1, ~350-550) vs Z, with points P, Q, R, S, T; the curve falls steeply from P (highest) through Q, R and levels off to S and T (lowest)] (A) Rb (B) Na (C) K (D) Cs$
›Reveal solutionSolution
[!TLDR]

Down Group 1 (Li → Cs) the outer electron is farther out and more shielded, so IE1 falls steadily; the element with minimum IE1 among the choices is Cs.

Concept

For the alkali metals (a standard CBSE/NCERT Class 11 periodic-trends idea), moving down a group increases the principal quantum number of the valence electron. The atomic radius grows and inner-shell shielding rises faster than the nuclear charge is felt, so the single ns¹ electron is removed more easily — first ionization enthalpy decreases: Li > Na > K > Rb > Cs.

Solution …
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KCET 2019Set A-11 markMCQ
$\text{Cr}_2\text{O}_7^{2-}$
›Reveal solutionSolution
The reduction of $\text{Cr}_2\text{O}_7^{2-}$ to $\text{Cr}^{3+}$ involves a change in oxidation state from +6 to +3 per chromium atom, requiring 3 electrons per Cr. With 2 Cr atoms per dichromate ion, 6 electrons are needed per mole of $\text{Cr}_2\text{O}_7^{2-}$ . For 0.2 mole, that’s $0.2 \times 6 = 1.2$ moles of electrons.

The key is to track the change in oxidation number of chromium. In $\text{Cr}_2\text{O}_7^{2-}$ , each Cr is in the +6 oxidation state (since oxygen is -2, and the overall charge is -2: $2x + 7(-2) = -2$ gives $x = +6$ ). In $\text{Cr}^{3+}$ , the oxidation state is +3. So each chromium atom gains 3 electrons during reduction.

Since one $\text{Cr}_2\text{O}_7^{2-}$ ion contains two chromium atoms, the total electrons required per ion is $2 \times 3 = 6$ .

Now we scale to the given amount:

Find electrons per mole of dichromate

For 1 mole of $\text{Cr}_2\text{O}_7^{2-}$ , the reduction to $\text{Cr}^{3+}$ consumes 6 moles of electrons.

Scale to 0.2 mole

Moles of electrons needed = $0.2 \times 6 = 1.2$ . …
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KCET 2018Set A-11 markMCQ
$\mathrm{SiO_2}$
›Reveal solutionSolution
Among the four oxides only $\mathrm{CrO_2}$ has a partially filled, delocalised $d$ -band, so only it conducts electricity like a metal.

Step 1 — The concept: metallic conduction needs mobile electrons.

A solid conducts like a metal when its electrons occupy a partially filled band (or when a filled band overlaps an empty one). Electrons can then move freely under an applied field, and — the metallic signature — conductivity decreases as temperature rises (lattice vibrations scatter the electrons).

Step 2 — Test each oxide.

(A) $\mathrm{SiO_2}$ — a giant covalent (network) solid. Every valence electron is locked in a localised $\mathrm{Si-O}$ $\sigma$ bond, and the band gap is very large. Quartz is a classic insulator.

(B) $\mathrm{MgO}$ — an ionic solid with the closed-shell ions $\mathrm{Mg^{2+}}$ ( $[\mathrm{Ne}]$ ) and $\mathrm{O^{2-}}$ ( $[\mathrm{Ne}]$ ). The ions are fixed in the lattice and no electrons are free, so solid MgO is an insulator (it conducts only when molten/aqueous, and then ionically, not metallically).

(C) $\mathrm{SO_2}$ (s) — a molecular solid of discrete $\mathrm{SO_2}$ molecules held by weak van der Waals forces. No free electrons or ions: an insulator. …
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KCET 2018Set A-11 markMCQ
$\mathrm{MnO}_4^-$
›Reveal solutionSolution

Find the change in oxidation number of Mn; the number of electrons per ion equals the number of faradays per mole.

Step 1 — Oxidation state of Mn in $\mathrm{MnO_4^-}$ .

Oxygen is $-2$ ; the overall charge is $-1$ :
$x + 4(-2) = -1 \;\Rightarrow\; x - 8 = -1 \;\Rightarrow\; x = +7$
Step 2 — Oxidation state of Mn in $\mathrm{MnO_2}$ .
$y + 2(-2) = 0 \;\Rightarrow\; y = +4$
Step 3 — Electrons transferred.

Going from $+7$ to $+4$ is a gain of $3$ electrons per Mn atom (reduction):
$\mathrm{MnO_4^-} + 4\mathrm{H^+} + 3e^- \rightarrow \mathrm{MnO_2} + 2\mathrm{H_2O}$
(The half-equation balances: charge on the left $= -1 + 4 - 3 = 0$ , and on the right $= 0$ . ✓)

Step 4 — Convert to charge (Faraday's first law).

One mole of electrons carries one faraday:

$1\ \mathrm{F} = N_A \times e = (6.022\times10^{23})(1.602\times10^{-19}\ \mathrm{C}) \approx 96500\ \mathrm{C\,mol^{-1}}$ …
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KCET 2018Set A-11 markMCQ
$_2$
›Reveal solutionSolution

Bauxite ore is contaminated mainly by ferric oxide (red mud), which is separated from alumina in the Bayer process because $\mathrm{Fe_2O_3}$ is not amphoteric.

Step 1 — What bauxite is.

Bauxite is the principal ore of aluminium, $\mathrm{Al_2O_3 \cdot xH_2O}$ . Freshly mined bauxite is reddish-brown, not white — the colour itself is the clue to the impurity.

Step 2 — Which impurities are actually present.

The standard impurities are $\mathrm{Fe_2O_3}$ (the major one, responsible for the red colour), $\mathrm{SiO_2}$ and $\mathrm{TiO_2}$ . Copper, zinc and chromium oxides are not associated with bauxite deposits.

Step 3 — Why the Bayer process works (the chemistry behind the answer).

Alumina is amphoteric, so hot concentrated NaOH dissolves it:

$\mathrm{Al_2O_3 + 2NaOH + 3H_2O \longrightarrow 2Na[Al(OH)_4]}$ …
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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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