Give examples and suggest reasons for the following features of the transition metal chemistry: (i) The lowest

Exercises · 4.25

Q. Give examples and suggest reasons for the following features of the transition metal chemistry:

(i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.

(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.

(iii) The highest oxidation state is exhibited in oxoanions of a metal.

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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 that the variable oxidation states of transition metals arise from the involvement of both ns and (n–1)d electrons in bonding. The highest oxidation state is stabilised by small, highly electronegative atoms like oxygen and fluorine, which can accept electron density and form strong π-bonds. The amphoteric/acidic nature of higher oxides follows from the increased polarising power of the metal cation at high oxidation states.

Let’s unpack each of these three features one by one, starting with the concept that ties them all together.

The Core Concept: Why Transition Metals Have Variable Oxidation States

Unlike s-block metals which only lose their outermost s-electrons, transition metals can lose electrons from both the ns and the (n–1)d subshells. The energy difference between these subshells is small, so removing a few d-electrons is not prohibitively costly. This gives them a range of oxidation states, from +2 (losing only the two 4s electrons) up to +8 (losing almost all valence electrons).

The stability of a particular high oxidation state depends critically on the ligand or anion it is bonded to. This is where the three features come from.

(i) The lowest oxide is basic, the highest is amphoteric/acidic.

1. The trend in ionic character.

Consider a metal M forming two oxides: $M_2O$ (low oxidation state) and $M_2O_5$ or $MO_3$ (high oxidation state). In the low oxide, the metal ion has a small charge and a relatively large radius. According to Fajan’s rules, this means low polarising power — the metal cation does not distort the electron cloud of the oxide ( $O^{2-}$ ) ion much. The bond remains largely ionic, and the oxide behaves like a basic oxide: it reacts with acids to form salts and water.

2. What happens at high oxidation states?

When the metal is in a very high oxidation state (e.g., +6 or +7), the cation is tiny and carries a huge positive charge. Its polarising power (charge/radius ratio) is enormous. It pulls electron density so strongly from the oxide ions that the M–O bond becomes significantly covalent. More importantly, the oxygen atoms themselves become electron-deficient. The oxide now behaves like an acidic oxide — it reacts with bases to form oxoanions (like chromate or permanganate).

3. The amphoteric middle ground.

Oxides in intermediate oxidation states (e.g., $Cr_2O_3$ , $V_2O_5$ ) are amphoteric: they can react with both acids and bases. This is because the polarising power is just enough to make the oxide weakly acidic, but not enough to destroy its basic character entirely.

Watch out

A common mistake is to think that all oxides of a given metal follow this trend strictly. For example, $MnO$ is basic, $Mn_2O_3$ is weakly basic, $MnO_2$ is amphoteric, $Mn_2O_7$ is strongly acidic. But $FeO$ is basic, $Fe_2O_3$ is weakly basic/amphoteric — iron never reaches a high enough oxidation state to form a purely acidic oxide. The trend is real, but the range depends on the metal.

(ii) A transition metal exhibits its highest oxidation state in oxides and fluorides.

1. Why oxygen and fluorine are special.

Both oxygen and fluorine are the most electronegative elements. When a transition metal forms a compound with them, the metal can surrender a large number of its electrons to the ligand. This is because the ligand is so electronegative that it can accommodate a high negative charge density.

2. The role of π-bonding (the real key).

For oxides, there is an additional stabilisation: oxygen can form π-bonds with the metal using its lone pairs. When a metal is in a very high oxidation state, it is electron-deficient. Oxygen’s filled 2p orbitals can donate electron density into the metal’s empty d-orbitals, forming a dπ–pπ back-bond. This delocalises the positive charge on the metal and stabilises the high oxidation state enormously.

Fluorine, being the most electronegative, is excellent at withdrawing electron density, but it cannot form π-bonds (it has no available d-orbitals and its p-orbitals are too contracted). So fluorine stabilises high oxidation states mainly through its high electronegativity and small size. For the very highest oxidation states (like +7 in $Mn_2O_7$ or +8 in $OsO_4$ ), oxygen is essential because of π-bonding.

Tip

You can remember this: Fluorine gets the metal to a high state, but oxygen keeps it there. For example, Mn can be oxidised to $MnF_4$ (Mn in +4) and $MnF_3$ , but to get Mn to +7, you need oxygen ( $MnO_4^-$ ). Similarly, Pt can form $PtF_6$ (Pt in +6), but $PtO_3$ is less stable.

3. Why not with chlorine or sulphur?

Chlorine and sulphur are less electronegative and larger. They cannot stabilise the same high charge density. Manganese makes the point sharply: fluorine takes it up to $MnF_4$ (+4), but with chlorine it stops at $MnCl_2$ (+2) — and $MnCl_7$ certainly does not exist (Table 4.5). The larger, less electronegative chlorine atoms simply form bonds too weak to hold a high positive charge on the metal.

(iii) The highest oxidation state is exhibited in oxoanions of a metal.

1. Oxoanions are the ultimate stabilisers.

An oxoanion like $MnO_4^-$ (permanganate) or $CrO_4^{2-}$ (chromate) has the metal at its centre, surrounded by four oxygen atoms. The metal is in its highest possible oxidation state (+7 for Mn, +6 for Cr). Why is this even more stable than a simple oxide like $Mn_2O_7$ ?

2. Charge delocalisation and resonance. …

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(i) Lowest oxide is basic, highest is amphoteric/acidic

Reason: This follows the variable oxidation state and electronegativity trend.

Low oxidation state: Metal has low charge density, behaves like a metal — forms basic oxides (e.g., $\text{CrO}$ — basic).
High oxidation state: Metal has high charge density, pulls electron density from oxygen, making the oxide covalent and amphoteric or acidic (e.g., $\text{Cr}_2\text{O}_3$ — amphoteric; $\text{CrO}_3$ — acidic).

Examples:

Metal	Low oxide	Nature	High oxide	Nature
Mn	MnO	Basic	Mn $_2$ O $_7$	Acidic
V	VO	Basic	V $_2$ O $_5$	Amphoteric

(ii) Highest oxidation state in oxides and fluorides

Reason: Oxygen and fluorine are the most electronegative elements. They can stabilize high oxidation states by:

Strong $\pi$ -bonding (O) or high electronegativity (F)
Removing electrons from the metal to a large extent

Examples:

$\text{Mn}_2\text{O}_7$ (Mn in +7)
$\text{OsO}_4$ (Os in +8)
$\text{CrF}_6$ (Cr in +6) — the highest oxidation numbers in halides are reached in TiX $_4$ , VF $_5$ and CrF $_6$ (Table 4.5)
$\text{VF}_5$ (V in +5) …

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Here are the common mistakes students make on this question, broken down by each sub-part, along with the conceptual fixes to avoid them.

General Mistake: Memorizing Without Understanding the "Why"

Students often memorize the statements (e.g., "highest oxide is acidic") but fail to link them to the core concept: variable oxidation states and the stabilization of high oxidation states.

How to avoid: For every property, ask "What is the electron configuration doing here?" and "Which element (ligand) is stabilizing the metal?"

(i) The lowest oxide is basic, the highest is amphoteric/acidic.

Common Mistake 1: Confusing "Basic" with "Metallic"

Students think "lowest oxide" means the oxide of the metal in its metallic (zero) state. They forget that "lowest oxide" means the oxide where the metal is in its lowest positive oxidation state (e.g., CrO where Cr is +2, not Cr metal).

How to avoid: Always write the oxidation state of the metal in the oxide.

Lowest oxide: Metal is in a low positive state (e.g., $+2$ ). The oxide is basic (reacts with acids).
Highest oxide: Metal is in a high positive state (e.g., $+6$ or $+7$ ). The oxide is acidic (reacts with bases) or amphoteric (reacts with both).

Common Mistake 2: Forgetting the "Charge Density" Rule

Students fail to connect basic/acidic character to the polarizing power of the metal ion.

How to avoid: Use the Fajan's rule logic:

A small, highly charged cation (high oxidation state) pulls electron density from oxygen, making the O–H bond in water more polar. This releases $H^+$ → acidic.
A large, low-charge cation (low oxidation state) does not polarize oxygen strongly. The oxide accepts $H^+$ → basic.

Example:

MnO (Mn in +2): Large Mn $^{2+}$ ion. Basic.
Mn $_2$ O $_7$ (Mn in +7): Tiny Mn $^{7+}$ ion. Acidic (explosive, gives permanganic acid).

(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.

Common Mistake 1: Thinking "Highest" Means "Most Stable"

Students assume that because the oxidation state is high, the compound is stable. In reality, high oxidation states are strongly oxidizing and often unstable in water.

How to avoid: Focus on stabilization by ligands. Oxygen and fluorine are the only common ligands that can stabilize the highest oxidation states because:

They are highly electronegative.
They are small (especially F), allowing close approach and strong $\pi$ -bonding (for oxygen) or strong $\sigma$ -bonding (for fluorine).

Common Mistake 2: Forgetting Why Chlorine or Bromine Don't Work

Students wonder: "Why not highest in chlorides?" They forget that Cl is larger and less electronegative, so it cannot pull electrons away from the metal as effectively.

How to avoid: Compare the electronegativity and size:

F (electronegativity 4.0) and O (3.5) are the best at oxidizing the metal to its highest state.
Cl (3.0) is weaker. Example: Mn in +7 exists as Mn $_2$ O $_7$ or KMnO $_4$ , but not as MnCl $_7$ (which doesn't exist).

Example:

Cr in +6: both CrO $_3$ (oxide) and CrF $_6$ (fluoride) exist — chromium reaches its highest oxidation state with oxygen and with fluorine alike (Table 4.5). Manganese shows the contrast better: its +7 state appears only with oxygen ( $Mn_2O_7$ , $MnO_4^-$ ); the highest simple manganese fluoride is MnF $_4$ (only the oxofluoride MnO $_3$ F carries Mn to +7 with F).
Cr in +6: CrCl $_6$ does not exist at all (Cl cannot stabilize +6).

(iii) The highest oxidation state is exhibited in oxoanions of a metal.

Common Mistake 1: Confusing "Oxoanion" with "Oxide"

Students think this is the same as point (ii). It is not. An oxoanion is a negatively charged polyatomic ion containing oxygen (e.g., $MnO_4^-$ , $CrO_4^{2-}$ , $VO_4^{3-}$ ). An oxide is a neutral compound (e.g., $MnO_2$ , $CrO_3$ ). …

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Q.Give examples and suggest reasons for the following features of the transition metal chemistry:

Concept understanding — Lanthanide Contraction

Lanthanide Contraction: The Intuition

The Precise Statement

Why It Matters

The Numbers (for reference)

Why this formula?

Lanthanide Contraction: Why It Happens

The Core Question

The Formula That Governs It

The Derivation: Step by Step

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

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

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

4. How this shrinks the radius

Key Idea: Variable Oxidation States and Their Stabilisation

The Core Concept: Why Transition Metals Have Variable Oxidation States

(i) The lowest oxide is basic, the highest is amphoteric/acidic.

(ii) A transition metal exhibits its highest oxidation state in oxides and fluorides.

(iii) The highest oxidation state is exhibited in oxoanions of a metal.

(i) Lowest oxide is basic, highest is amphoteric/acidic

(ii) Highest oxidation state in oxides and fluorides

General Mistake: Memorizing Without Understanding the "Why"

(i) The lowest oxide is basic, the highest is amphoteric/acidic.

(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.

(iii) The highest oxidation state is exhibited in oxoanions of a metal.

Q.Give examples and suggest reasons for the following features of the transition metal chemistry:

The Core Concept: Why Transition Metals Have Variable Oxidation States

(i) The lowest oxide is basic, the highest is amphoteric/acidic.

(ii) A transition metal exhibits its highest oxidation state in oxides and fluorides.

(iii) The highest oxidation state is exhibited in oxoanions of a metal.

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