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

NCERT Exemplar · Q62

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

(i) An element which can show +8 oxidation state

(ii) 3d block element that can show upto +7 oxidation state

(iii) 3d block element with highest melting point
Column II (Metal):

(a) Mn

(b) Cr

(d) Fe

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Concept understanding — Lanthanide Contraction

Lanthanide Contraction: The Intuition

Imagine you are walking through a dense forest. With every step forward, you push through thick undergrowth. The deeper you go, the more tired you become — each step feels a little harder, and you find yourself hunching forward, your shoulders pulling inward. That inward pull is exactly what happens inside the lanthanide atoms.

The lanthanides are the 14 elements from cerium ( $Ce$ , atomic number 58) to lutetium ( $Lu$ , atomic number 71). As you move from one element to the next, you add one proton to the nucleus and one electron to the atom. The new electron goes into a 4f orbital — a set of orbitals that are shaped like clover leaves and sit deep inside the atom, close to the nucleus.

Here is the key: 4f orbitals are poorly shielded. They do not spread out far from the nucleus, and they do not block the nuclear charge from pulling on the outer electrons. So when you add a proton, the nucleus gets stronger, and the 4f electrons do almost nothing to stop that extra pull. The result? The entire electron cloud — especially the outermost electrons — gets pulled inward. The atom shrinks.

Note

Shielding is the ability of inner electrons to "block" the outer electrons from feeling the full positive charge of the nucleus. Electrons in s and p orbitals shield well; 4f electrons shield very poorly.

The Precise Statement

Lanthanide contraction is the steady and significant decrease in the atomic and ionic radii of the lanthanide elements as atomic number increases from 58 (Ce) to 71 (Lu).

\text{Atomic radius} \propto \frac{1}{Z_{\text{eff}}}

The total contraction across the entire series is about 15–20 picometers — roughly 10–15% of the initial radius. That is a substantial shrinkage for a single row of the periodic table.

Why It Matters

This contraction has two enormous consequences in chemistry:

Similarity of post-lanthanide elements: After lutetium, the next elements are hafnium ( $Hf$ , 72), tantalum ( $Ta$ , 73), and tungsten ( $W$ , 74). Because the lanthanide contraction has made the atoms so small, these elements have almost identical atomic and ionic radii to their counterparts directly above them in the periodic table — zirconium ( $Zr$ ), niobium ( $Nb$ ), and molybdenum ( $Mo$ ). This is why zirconium and hafnium are chemically almost inseparable — they are the same size.
Difficulty in separating lanthanides: All lanthanide ions ( $Ln^{3+}$ ) have nearly identical chemical properties because their radii change so gradually. Separating them requires hundreds of repeated steps (ion-exchange chromatography, solvent extraction) — a painstaking process that was a major challenge in early nuclear chemistry.

Watch out

A common mistake is to think lanthanide contraction means the atoms get smaller because the 4f orbitals are "full" or because of some repulsion effect. It is purely due to poor shielding of the 4f electrons, which lets the nuclear charge pull everything inward.

The Numbers (for reference)

Element	Atomic Number	Ionic Radius ( $Ln^{3+}$ , pm)
Ce	58	103.4
Pr	59	101.3
Nd	60	99.5

Why this formula?

Lanthanide Contraction: Why It Happens

The Lanthanide Contraction is the steady decrease in atomic and ionic radii of the lanthanide elements (Ce to Lu) as atomic number increases. The key observation: the radii shrink by about 1–2 pm per element, despite adding electrons to the 4f subshell.

The Core Question

Why does adding electrons not increase the size, but instead decrease it?

The Formula That Governs It

The effective nuclear charge ( $Z_{\text{eff}}$ ) experienced by an electron is:

Z_{\text{eff}} = Z - S

Where:

$Z$ = atomic number (protons in nucleus)
$S$ = shielding constant (screening by inner electrons)

The key formula for the trend in ionic radii ( $r$ ) across the lanthanides is:

r \propto \frac{n^2}{Z_{\text{eff}}}

Where $n$ is the principal quantum number of the outermost electron (here, $n=6$ for the 6s orbital).

The Derivation: Step by Step

1. What happens when you add a proton and an electron?

Each lanthanide adds:

+1 proton to the nucleus (increases $Z$ by 1)
+1 electron to the 4f subshell

2. The 4f orbital is "penetrating" but poorly shielding

The 4f orbital has a radial distribution that peaks close to the nucleus (inside the 5s and 5p shells).
However, 4f electrons are very poor at shielding the outer 6s electrons from the nuclear charge.

Why?

The 4f orbital is diffuse and deeply buried — it does not effectively screen the outer electrons because:

Its shape (complex, multi-lobed) means it doesn't occupy the space between the nucleus and the 6s electrons efficiently.
The 4f electrons are inside the 5s/5p shells, so they don't block the nuclear pull on the 6s electrons.

3. The net effect on $Z_{\text{eff}}$

When you add one proton ( $\Delta Z = +1$ ) and one 4f electron ( $\Delta S \approx 0.85$ to 0.95), the change in effective nuclear charge is:

\Delta Z_{\text{eff}} \approx +1 - 0.85 = +0.15 \text{ to } +0.05

Result: $Z_{\text{eff}}$ increases slightly with each element.

4. How this shrinks the radius

From the formula $r \propto \frac{n^2}{Z_{\text{eff}}}$ :

$n$ (the principal quantum number of the 6s orbital) stays constant at 6.
$Z_{\text{eff}}$ increases.
Therefore, $r$ decreases. …

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The key idea is to match each property to the correct metal by reasoning from electronic configuration, oxidation state trends, and periodic properties. The final matches are: (i) → (c) Os, (ii) → (a) Mn, (iii) → (b) Cr.

Let’s unpack each property one by one. The question tests your understanding of transition metal chemistry — specifically, how the d-block elements vary in their ability to show high oxidation states and how physical properties like melting point relate to electronic structure.

1. Property (i): An element which can show +8 oxidation state

This is a rare and extreme oxidation state. In the entire periodic table, only a handful of elements can reach +8, and they all belong to the second and third transition series — specifically, the iron, cobalt, and nickel triads (Groups 8, 9, 10). The +8 state is possible only when the element has enough d-electrons to lose all of them, plus some from the s-orbital, and still remain stable.

Among the given options, Os (osmium) is famous for forming OsO₄, where osmium is in the +8 oxidation state. Ruthenium (Ru) also shows +8, but it’s not in the list. Manganese (Mn) can go up to +7 (in permanganate, MnO₄⁻), but never +8. Chromium (Cr) maxes out at +6 (in chromate/dichromate). Iron (Fe) reaches +6 only in rare compounds like ferrate(VI).

Watch out

A common mistake is to think Mn can show +8 because it’s in Group 7. But +8 requires losing all 8 valence electrons (3d⁵4s² → 8 electrons), which is impossible for a 3d element — the 3d orbitals are too compact and the nuclear charge too low to stabilize such a high positive charge. Only 4d and 5d elements (like Ru and Os) have diffuse enough d-orbitals to handle it.

So, (i) matches with (c) Os.

2. Property (ii): 3d block element that can show up to +7 oxidation state

Now we’re restricted to the 3d series (Sc through Zn). Among these, the maximum oxidation state increases from left to right, peaks around Group 7, then drops. The 3d element that reaches +7 is manganese (Mn). Its electronic configuration is [Ar] 3d⁵ 4s². In the +7 state, it loses all 7 valence electrons, giving a 3d⁰ configuration — which is stable in oxoanions like MnO₄⁻ (permanganate).

No other 3d element reaches +7. Chromium (Cr) maxes at +6, iron (Fe) at +6 (rare), and cobalt (Co) at +5. So this is a clean match. …

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Method: Oxidation State & Periodic Trends Analysis

This method uses two key ideas:

Maximum oxidation state = sum of ns and (n-1)d electrons (for transition metals)
Melting point trends in 3d series: peaks at half-filled d⁵ configuration (Mn) and maximum at Cr due to maximum number of unpaired electrons in 3d⁵4s¹

Step 1: Identify element with `+8` oxidation state

+8 is the highest known oxidation state among transition metals.
Only Os (Osmium, 5d⁶6s²) and Ru (Ruthenium, 4d⁷5s¹) can reach +8 (e.g., OsO₄, RuO₄).
From Column II, Os is present.

Match: (i) → (c) Os

Step 2: Identify 3d block element with `+7` oxidation state

3d series: Sc to Zn.
Maximum +7 is achieved by Mn (Manganese, 3d⁵4s²).
- All 7 valence electrons can be lost: Mn²⁺ to Mn⁷⁺ (e.g., MnO₄⁻).
No other 3d element reaches +7 (Fe max +6, Cr max +6).

Match: (ii) → (a) Mn

Step 3: Identify 3d block element with highest melting point

Melting point in 3d series: …

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Here are the common mistakes students make on this matching question, along with how to avoid each.

Mistake 1: Confusing `+8` oxidation state with a 3d element

The error:

Students see +8 and immediately think of Mn or Fe, because they are familiar 3d metals. They forget that +8 is not possible for any 3d element (maximum in 3d is +7 for Mn).

How to avoid:

Memorise the maximum oxidation states across the d-block:
- 3d series: max +7 (Mn)
- 4d and 5d series: can reach +8 (e.g., Os in OsO₄, Ru in RuO₄)
Key fact: Only Os and Ru show stable +8 oxidation state.

✓ Correct match: (i) → (c) Os

Mistake 2: Thinking `+7` is shown by Cr or Fe

The error:

Students often recall that Cr shows +6 (in CrO₃, K₂Cr₂O₇) and Fe shows +6 (in K₂FeO₄), and assume +7 is also possible for them. This is wrong.

How to avoid:

Remember the trend:
- Sc to Zn: max O.S. increases up to Mn, then decreases.
- Mn is the only 3d element that shows +7 (in KMnO₄).
Cr max = +6, Fe max = +6, Co max = +5, Ni max = +4.

✓ Correct match: (ii) → (a) Mn

Mistake 3: Assuming Cr has the highest melting point among 3d metals

The error:

Cr has a very high melting point (1907 °C), but it is not the highest. Students often pick Cr because it is "hard" and has many unpaired electrons.

How to avoid:

Know the melting point trend in the 3d series: it rises from Sc to a maximum around the middle of the series, then falls (and $W$ in 5d is the highest overall).
Data check:
- V: 1910 °C
- Cr: 1907 °C
- Fe: 1538 °C
- Mn: 1246 °C
Strictly, V (Vanadium) has the highest melting point in the 3d series (1910 °C), just above Cr. But V is not among the options here, so among the given options (Mn, Cr, Os, Fe), Cr has the highest melting point. …

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

1. Property (i): An element which can show +8 oxidation state

2. Property (ii): 3d block element that can show up to +7 oxidation state

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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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