Q.Why is $Cr^{2+}$ reducing and $Mn^{3+}$ oxidising when both have $d^4$ configuration?

Step 1 — Oxidation state and d-configuration of nickel

Let the oxidation state of Ni be $x$. Chloride carries $-1$ each, and the overall charge is $-2$:

So the metal is $\ce{Ni^{2+}}$.

Nickel is $Z = 28$: $\;\ce{Ni} = [\text{Ar}]\,3d^{8}\,4s^{2}$. Removing the two $4s$ electrons:

Step 2 — Coordination number

Chloride is a unidentate ligand (one donor atom, Cl). With four of them:

Step 3 — This already settles statement (b)

(b) "Co-ordination number of Ni is 2 and oxidation state is +4"

Both halves are wrong — the CN is 4 (Step 2) and the oxidation state is +2 (Step 1). (b) is FALSE.

This is decisive: every option containing (b) — namely (B), (C) and (D) — is eliminated at once. Only (A) a, c, d and e remains. Let us verify that all four of those statements are indeed true.

Step 4 — Ligand field strength decides everything else

In the spectrochemical series, chloride sits near the weak-field end:

$\ce{Cl-}$ is a weak-field ligand, so the splitting energy $\Delta$ it produces is small — smaller than the electron pairing energy $P$:

When $\Delta < P$, it costs less energy for an electron to occupy a higher orbital than to pair up in a lower one. Therefore no pairing occurs — the $3d^8$ configuration is left untouched, with its two unpaired electrons intact.

Since the $3d$ orbitals are not vacated, no inner $3d$ orbital is available for hybridisation.

Step 5 — Hybridisation and geometry ⇒ statements (a) and (c)

With the $3d$ set unavailable, Ni²⁺ must use its outer orbitals: one $4s$ and three $4p$.

Four $sp^3$ orbitals point to the corners of a tetrahedron ($109.5^\circ$).

• (a) "The complex has tetrahedral geometry" — TRUE ✓
• (c) "The complex is $sp^3$ hybridised" — TRUE ✓ …

The question asks why actinoids exhibit a larger number of oxidation states than lanthanoids. This is a classic comparison in f-block chemistry, rooted in the electronic structure of the two series.

Concept and Intuition

Oxidation states arise when an atom loses electrons. For f-block elements, the electrons lost can come from both the f and d orbitals. The ease of removing f-electrons depends on how tightly they are held, which is related to the energy gap between the f and d orbitals. A smaller gap means f-electrons can be promoted to d orbitals more readily, allowing a wider range of oxidation states. Actinoids (5f series) have a smaller 5f–6d energy difference than lanthanoids (4f–5d), so they can access more oxidation states.

Let’s break it down step by step.

1. Understand the orbital energy trends

   In lanthanoids, the 4f orbitals are deeply buried inside the atom, shielded by outer electrons. The 5d orbitals are at a significantly higher energy. This large 4f–5d energy gap makes it difficult to remove or promote 4f electrons, so lanthanoids typically show only +3 (and occasionally +2 or +4) oxidation states.

   In actinoids, the 5f orbitals are less shielded and more extended (diffuse). The 5f and 6d orbitals are much closer in energy. This small energy difference allows 5f electrons to be easily promoted to 6d orbitals or directly lost, enabling a variety of oxidation states (e.g., +3, +4, +5, +6, and even +7 in some cases like neptunium and plutonium).

2. Evaluate the options

   • (A) Lesser energy difference between 5f and 6d than between 4f and 5d orbitals — This matches the explanation above.
   • (B) More energy difference — This would make it harder to involve f-electrons, reducing oxidation states, so incorrect.
   • (C) 4f orbitals are more diffused than 5f orbitals — Actually, 5f orbitals are more diffused (less tightly held) due to poorer shielding, so this is false. …

Let’s examine each statement carefully, using chemical principles to decide which one is accurate.

1. Statement (A): “The green manganate ion shows diamagnetic nature but the permanganate ion exhibits paramagnetic nature.”

   • The manganate ion is $\text{MnO}_4^{2-}$, where manganese is in the +6 oxidation state. Electronic configuration of Mn in +6: $[\text{Ar}]\,3d^1$. That’s one unpaired electron → paramagnetic, not diamagnetic.
   • The permanganate ion is $\text{MnO}_4^{-}$, with Mn in +7: $[\text{Ar}]\,3d^0$. No unpaired electrons → diamagnetic.
   • So the statement gets both magnetic natures backwards. False.

2. Statement (B): “Interstitial compounds of transition metals have lower melting points than that of pure transition metals and their compounds are chemically reactive.”

   • Interstitial compounds (e.g., carbides, nitrides, hydrides) form when small atoms like C, N, or H occupy holes in the metal lattice. This usually increases hardness and melting point (often very high), and they are chemically inert (not reactive).
   • Both claims here are opposite to reality. False.

3. Statement (C): “Cerium is a lanthanoid metal which exists in a stable oxidation state of +4, besides exhibiting an oxidation state of +3.”

   • Cerium (Ce, atomic number 58) has the electron configuration $[Xe]\,4f^1\,5d^1\,6s^2$. The common +3 state arises from losing the 5d and 6s electrons.
   • The +4 state is also stable because losing one more electron gives a $4f^0$ configuration (empty f-subshell), which is especially stable. Ce(IV) is well-known in compounds like $\text{CeO}_2$ and ceric ammonium nitrate.
   • This is a textbook fact. True. …

Let’s go through each option carefully, building the reasoning step by step.

1. Option (A): “The ability of Fluorine to stabilise the higher oxidation states of transition metals exceeds that of Oxygen”

   • In transition-metal oxyanions (like $\mathrm{CrO_4^{2-}}$, $\mathrm{MnO_4^-}$), oxygen stabilises high oxidation states via strong $\pi$-bonding (O donates electron density to the metal, reducing its effective charge).
   • Fluorine is more electronegative than oxygen, but it is a poor $\pi$-donor (it has no available d-orbitals for back-bonding) and forms weaker multiple bonds.
   • For example, $\mathrm{Mn_2O_7}$ (Mn in +7) is stable, but $\mathrm{MnF_7}$ does not exist; the highest fluoride of Mn is $\mathrm{MnF_4}$ (+4).
   • Conclusion: Oxygen stabilises high oxidation states better than fluorine. So statement (A) is false.

2. Option (B): “Cu(I) compounds in aqueous medium undergo disproportionation reaction”

   • Disproportionation: $2\mathrm{Cu^+} \rightarrow \mathrm{Cu} + \mathrm{Cu^{2+}}$.
   • In water, the standard reduction potentials: $\mathrm{Cu^+ + e^- \rightarrow Cu}$ ($E^\circ = +0.52\,\mathrm{V}$) and $\mathrm{Cu^{2+} + e^- \rightarrow Cu^+}$ ($E^\circ = +0.16\,\mathrm{V}$).
   • The net cell potential for disproportionation is $E^\circ_{\text{cell}} = 0.52 - 0.16 = +0.36\,\mathrm{V} > 0$, so it is spontaneous.
   • Conclusion: Statement (B) is true.

3. Option (C): “$\mathrm{Cr^{2+}}$ is a stronger reducing agent than $\mathrm{Fe^{2+}}$”

   • Standard reduction potentials: $\mathrm{Cr^{3+} + e^- \rightarrow Cr^{2+}}$ ($E^\circ = -0.41\,\mathrm{V}$) $\mathrm{Fe^{3+} + e^- \rightarrow Fe^{2+}}$ ($E^\circ = +0.77\,\mathrm{V}$)
   • A more negative reduction potential means the reduced form (here $\mathrm{Cr^{2+}}$) is more easily oxidised — i.e., a stronger reducing agent. …

The pattern $(n-1)d^{10}ns^2$ means a filled $(n-1)d$ subshell together with a filled $ns$ subshell. This is the hallmark of the group-12 elements — Zn, Cd, Hg and Cn — in their ground state. To find the pair that breaks the pattern, we check each element's configuration.

1. Option (A): Zn, Cd.

   Zn ($Z=30$) is $[\text{Ar}]3d^{10}4s^2$; Cd ($Z=48$) is $[\text{Kr}]4d^{10}5s^2$. Both group 12 — both fit the pattern.

2. Option (B): Cd, Hg.

   Cd fits; Hg ($Z=80$) is $[\text{Xe}]4f^{14}5d^{10}6s^2$. Both group 12 — both fit.

3. Option (C): Hg, Cn.

   Hg fits; Cn ($Z=112$) is the group-12 element $[\text{Rn}]5f^{14}6d^{10}7s^2$. Both fit.

4. Option (D): Cu, Zn.

   Zn is $3d^{10}4s^2$ (fits), but Cu ($Z=29$) is the classic exception: its ground state is $[\text{Ar}]3d^{10}4s^1$, not $3d^9 4s^2$, because a completely filled $d^{10}$ subshell is especially stable and pulls one electron out of $4s$. So Cu is $(n-1)d^{10}ns^1$ and does not show the $ns^2$ configuration. …

Option (A) — incorrect. Fluorine, being small and highly electronegative, forms fluorides with high lattice enthalpy (and high M–F bond enthalpy). It is this high lattice/bond enthalpy that lets fluorine stabilise the highest oxidation states of transition metals. The statement wrongly attributes it to "low lattice enthalpy."

Option (B) — correct. The irregular ionisation enthalpies in this series arise from the stability of the half-filled $d^5$ configurations produced ($\mathrm{Mn^{2+}}$, $\mathrm{Fe^{3+}}$), as noted in NCERT.

Option (C) — correct. Transition metals form alloys readily because their metallic radii are similar (within ~15%). …

Concept & Intuition

Cerium is the first element in the lanthanide series (Z=58). Its ground-state electronic configuration is $[Xe]\,4f^1\,5d^1\,6s^2$. The key to its chemistry lies in the stability of the empty $4f$ subshell ($4f^0$) and the half-filled $4f$ subshell ($4f^7$). For cerium, losing four electrons gives the $Ce^{4+}$ ion with a $[Xe]$ configuration — a noble gas core, which is exceptionally stable. This stability makes $Ce^{4+}$ a strong oxidising agent (it readily accepts electrons to go back to the more common $Ce^{3+}$ state). In contrast, $Ce^{3+}$ has a $[Xe]\,4f^1$ configuration and is the most stable oxidation state for cerium in aqueous solution.

Now let’s examine each statement.

1. Statement (A): $Ce^{4+}$ is a reducing agent.

   A reducing agent is a substance that donates electrons (gets oxidised itself). $Ce^{4+}$ has a strong tendency to gain one electron and become $Ce^{3+}$ (the $4f^1$ configuration is more stable than $4f^0$ in most chemical environments). This means $Ce^{4+}$ is an oxidising agent, not a reducing agent. In fact, $Ce^{4+}$ is a well-known oxidising agent in analytical chemistry (e.g., in cerimetric titrations). So this statement is false.

2. Statement (B): Atomic size of Ce is more than that of Lu.

   This is true. Across the lanthanide series (from Ce, Z=58, to Lu, Z=71), there is a steady decrease in atomic and ionic radii — the lanthanide contraction. The 4f electrons are poorly shielding, so as nuclear charge increases, the electron cloud is pulled inward. Ce is near the beginning of the series, Lu at the end, so Ce has a larger atomic radius than Lu.

3. Statement (C): Ce in +3 oxidation state is more stable than in +4. …