Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidatio

Exercises · 4.6

Q. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

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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 for first-row transition metals, the highest stable oxidation state often equals the group number (number of valence electrons). The oxometal anions where this occurs are vanadate ( $\text{VO}_4^{3-}$ ), chromate ( $\text{CrO}_4^{2-}$ ), and permanganate ( $\text{MnO}_4^-$ ).

Why This Question Matters

The question asks you to connect two fundamental ideas: the periodic table's group numbers and the oxidation states metals can achieve in their oxyanions. For the first transition series (Sc through Zn), the group number tells you the total number of 4s and 3d electrons available. When a metal uses all of these electrons in bonding to oxygen, it reaches its maximum oxidation state — equal to its group number. But not every metal can do this stably in an aqueous oxyanion.

Step-by-Step Reasoning

1. Identify the first transition series and their group numbers

The first transition series runs from scandium (Sc, atomic number 21) to zinc (Zn, atomic number 30). Their group numbers in the modern periodic table are:

Element	Group Number	Valence electrons (4s + 3d)
Sc	3	3
Ti	4	4
V	5	5
Cr	6	6
Mn	7	7
Fe	8	8
Co	9	9
Ni	10	10
Cu	11	11
Zn	12	12

2. Understand what "oxidation state equal to group number" means

For a metal M in an oxoanion $\text{MO}_x^{n-}$ , the oxidation state of M is + (group number). This means the metal has lost all its valence electrons to oxygen. For example, vanadium (group 5) in $\text{VO}_4^{3-}$ has oxidation state +5.

3. Check which metals can actually achieve this state in a stable oxoanion

Not all metals can. The stability of high oxidation states decreases across a period due to increasing effective nuclear charge. Let's examine each:

Scandium (group 3): Sc forms $\text{Sc}_2\text{O}_3$ but not a discrete oxoanion like $\text{ScO}_3^{3-}$ . Sc(III) is the only stable state, but it doesn't form a well-defined oxoanion in water.
Titanium (group 4): Ti(IV) exists in $\text{TiO}_2$ and in titanates like $\text{TiO}_3^{2-}$ (metatitanate) or $\text{Ti}_2\text{O}_5^{2-}$ , but the simple orthotitanate $\text{TiO}_4^{4-}$ is not stable in aqueous solution. So no common oxoanion with Ti in +4 state.
Vanadium (group 5): V(V) forms the very stable orthovanadate ion $\text{VO}_4^{3-}$ . This is a classic example. Yes.
Chromium (group 6): Cr(VI) forms chromate $\text{CrO}_4^{2-}$ and dichromate $\text{Cr}_2\text{O}_7^{2-}$ . Both are stable in water. Yes.
Manganese (group 7): Mn(VII) forms permanganate $\text{MnO}_4^-$ , a deep purple, stable oxoanion. Yes.
Iron (group 8): Fe(VIII) is unknown in aqueous solution. The highest stable oxidation state for Fe is +6 (ferrate, $\text{FeO}_4^{2-}$ ), but that's not equal to its group number (8). So no.
Cobalt (group 9): Co(IX) is impossible in water. Highest is Co(IV) or Co(V) in special complexes. No.
Nickel (group 10): Ni(X) is not known. Highest is Ni(IV). No.
Copper (group 11): Cu(XI) is absurd. Highest is Cu(III) in rare cases. No.
Zinc (group 12): Zn(II) is the only state; Zn(XII) is impossible. No. …

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Method: Periodic Trend Analysis + Electronic Configuration Mapping

This method uses the relationship between group number, oxidation state, and oxoanion formation across the first transition series.

Steps

Recall the first transition series

Elements: Sc (21), Ti (22), V (23), Cr (24), Mn (25), Fe (26), Co (27), Ni (28), Cu (29), Zn (30).
Identify group numbers
- Group 3: Sc
- Group 4: Ti
- Group 5: V
- Group 6: Cr
- Group 7: Mn
- Group 8: Fe
- Group 9: Co
- Group 10: Ni
- Group 11: Cu
- Group 12: Zn
Apply the condition: oxidation state = group number, in a stable, discrete oxoanion

Only metals that can both (a) reach an oxidation state equal to their group number, AND (b) do so in a well-defined, stable, discrete oxoanion, satisfy the question.
Check each element
- Sc (Group 3): $Sc^{3+}$ exists, but scandium does not form a discrete oxoanion — its oxide, $\ce{Sc2O3}$ , is basic, not anionic.
- Ti (Group 4): $Ti^{4+}$ is titanium's common oxidation state, but titanium does not form a simple, discrete aqueous oxoanion analogous to vanadate or chromate — its oxo-compounds (e.g. $\ce{BaTiO3}$ ) are extended solid-state structures (linked $\ce{TiO6}$ octahedra), not isolable $\ce{TiO4^{4-}}$ -type ions. So titanium does not qualify.
- V (Group 5): $V^{5+}$ exists in the well-known, stable vanadate ion, $\ce{VO4^{3-}}$ . Qualifies.
- Cr (Group 6): $Cr^{6+}$ exists in the well-known, stable chromate ( $\ce{CrO4^{2-}}$ ) and dichromate ( $\ce{Cr2O7^{2-}}$ ) ions. Qualifies.
- Mn (Group 7): $Mn^{7+}$ exists in the well-known, stable permanganate ion, $\ce{MnO4^-}$ . Qualifies.
- Fe, Co, Ni, Cu, Zn (Groups 8-12): oxidation states equal to the group number (+8 to +12) are not observed in any stable oxoanion — the highest oxo-species known for iron, for example, is ferrate, $\ce{FeO4^{2-}}$ , which is only +6, not +8.
List the valid oxometal anions

| Metal | Group | Oxidation state | Oxometal anion | Name | …

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Common Mistakes: Oxometal Anions with Metal Oxidation State = Group Number

Mistake 1: Confusing "Group Number" with "Period Number"

The error: Students often pick elements from the wrong period or confuse group number (vertical column) with period number (horizontal row).

How to avoid: Remember that group number for transition metals (Groups 3–12) tells you the maximum possible oxidation state. For the first series (Sc to Zn, Period 4), you must only consider these elements.

Group 5 → V (oxidation state +5)
Group 6 → Cr (oxidation state +6)
Group 7 → Mn (oxidation state +7)

Mistake 2: Including Elements That Don't Form Stable Oxoanions

The error: Students list Sc, Ti, Fe, Co, Ni, Cu, or Zn — but these either:

Don't form oxometal anions at all, or
Don't achieve an oxidation state equal to their group number in an oxoanion.

How to avoid: Know the stable oxoanions for the first transition series:

Group	Element	Oxidation State	Oxoanion
5	V	+5	$\text{VO}_4^{3-}$ (vanadate)
6	Cr	+6	$\text{CrO}_4^{2-}$ (chromate) or $\text{Cr}_2\text{O}_7^{2-}$ (dichromate)
7	Mn	+7	$\text{MnO}_4^{-}$ (permanganate)

Only these three satisfy the condition.

Mistake 3: Forgetting the Charge on the Anion

The error: Writing the formula without the correct negative charge, e.g., writing $\text{VO}_4$ instead of $\text{VO}_4^{3-}$ .

How to avoid: Calculate the charge systematically:

Vanadium in $\text{VO}_4^{3-}$ : V is +5, each O is –2 → total = $+5 + 4(-2) = -3$
Chromium in $\text{CrO}_4^{2-}$ : Cr is +6 → $+6 + 4(-2) = -2$
Manganese in $\text{MnO}_4^{-}$ : Mn is +7 → $+7 + 4(-2) = -1$

Mistake 4: Confusing Oxidation State with Charge on the Ion

The error: Thinking that if the metal has oxidation state +6, the anion must have charge –6. …

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Q.Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Why This Question Matters

Step-by-Step Reasoning

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