Q.Amongst the following, the most stable complex is

Concept and Intuition

We need to check three properties for each complex:

• Paramagnetic: has unpaired electrons (check electron configuration).
• High-spin: occurs when ligands are weak-field (small Δ), so electrons fill orbitals singly before pairing.
• Octahedral geometry: coordination number 6, with ligands at 90° angles.

We’ll determine the oxidation state, d-electron count, ligand field strength, and geometry for each option.

1. Option (A): $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$

   • The complex ion is $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+}$.
   • Oxidation state: NH₃ is neutral, Cl⁻ is –1, so Co + 0×5 + (–1) = +2 → Co²⁺.
   • Co²⁺ has electron configuration [Ar] 3d⁷.
   • Ligands: NH₃ (moderate field) and Cl⁻ (weak field). However, NH₃ is strong enough to cause pairing in Co³⁺ but for Co²⁺ it often gives low-spin? Actually, NH₃ is borderline; for Co²⁺, it typically gives low-spin d⁷ (one unpaired electron) because Δ is large enough. But here we have one Cl⁻, which is weak, but overall the field is still moderate. In practice, $\left[\mathrm{Co(NH_3)_5Cl}\right]^{2+}$ is low-spin (one unpaired electron) — paramagnetic, but not high-spin. So fails the high-spin condition.

2. Option (B): $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_2\left(\mathrm{C}_2 \mathrm{O}_4\right)_2\right]^{2-}$

   • The complex ion is $\left[\mathrm{Ni(H_2O)_2(C_2O_4)_2}\right]^{2-}$.
   • Oxidation state: H₂O neutral, oxalate (C₂O₄²⁻) each –2, so Ni + 0×2 + (–2)×2 = –2 → Ni – (–4) = +2? Wait: total charge –2, so Ni + 0 + (–4) = –2 → Ni = +2. Yes, Ni²⁺.
   • Ni²⁺ has [Ar] 3d⁸.
   • Ligands: H₂O (weak field) and oxalate (weak to moderate, but generally weak). Both are weak-field ligands, so Δ is small → high-spin configuration.
   • For d⁸ in octahedral field: high-spin means t₂g⁶ e_g² (two unpaired electrons in e_g) → paramagnetic.
   • Geometry: coordination number 6 (two water + two bidentate oxalates = 2 + 4 = 6) → octahedral.
   • So (B) satisfies all three.

3. Option (C): $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)(\mathrm{Cl})(\mathrm{en})_2\right]^{2+}$

   • Note: "CO" is likely a typo for "Co". The complex is $\left[\mathrm{Co(NH_3)(Cl)(en)_2}\right]^{2+}$.
   • Oxidation state: NH₃ neutral, Cl⁻ –1, en (ethylenediamine) neutral, so Co + 0 + (–1) + 0 = +2 → Co²⁺ again. …

Concept & Intuition

Crystal Field Stabilisation Energy (CFSE) measures the net energy gained (or lost) when d-electrons occupy the split $t_{2g}$ and $e_g$ orbitals in an octahedral field. The key is knowing:

• The metal ion’s d-electron count.
• Whether the ligand is weak-field (high-spin) or strong-field (low-spin).
• For weak-field ligands, electrons fill all five d-orbitals singly before pairing (Hund’s rule), minimising pairing energy. For strong-field ligands, electrons pair in the lower $t_{2g}$ set first.

Here, both $\ce{F-}$ and $\ce{H2O}$ are weak-field ligands (F⁻ is a weak-field halide; H₂O is intermediate but for Fe³⁺ it gives high-spin). So both complexes are high-spin.

Step-by-step reasoning

1. Determine the metal ion and d-electron count

   • (i) $\ce{[CoF6]^{3-}}$: Co in +3 oxidation state. Co atomic number = 27, Co³⁺ = 27 – 3 = 24 electrons. Argon core = 18, so d-electrons = 24 – 18 = 6 (d⁶).
   • (ii) $\ce{[Fe(H2O)6]^{3+}}$: Fe in +3 state. Fe atomic number = 26, Fe³⁺ = 26 – 3 = 23 electrons. Argon core = 18, so d-electrons = 23 – 18 = 5 (d⁵).

2. Identify spin state from ligand field strength

   • F⁻ is a weak-field ligand → small $\Delta_o$ → high-spin.
   • H₂O is a weak-to-intermediate field ligand; for Fe³⁺ (d⁵), it gives high-spin (all five electrons unpaired). So both are high-spin.

3. Fill the d-orbitals for high-spin d⁶ and d⁵

   • For high-spin d⁶: $t_{2g}$: ↑ ↑ ↑ (three electrons) $e_g$: ↑ ↑ (two electrons) Then the sixth electron goes into $t_{2g}$ pairing: $t_{2g}$: ↑↓ ↑ ↑ → total $t_{2g}^4 e_g^2$.
   • For high-spin d⁵: $t_{2g}$: ↑ ↑ ↑ (three) $e_g$: ↑ ↑ (two) → $t_{2g}^3 e_g^2$ (all unpaired).

4. Calculate CFSE

   CFSE formula: $(\text{number of } t_{2g} \text{ electrons}) \times (-0.4\Delta_o) + (\text{number of } e_g \text{ electrons}) \times (+0.6\Delta_o)$. …

Step 1 — Ligand field strength

In the spectrochemical series, $CN^-$ is one of the strongest field ligands, far stronger than $CO$ acting on Fe(0), $C_2O_4^{2-}$ (oxalate, a moderately weak-field chelating ligand), or $H_2O$ (a weak-field ligand).

Step 2 — Effect on Fe(II)

With $Fe^{2+}$ ($d^6$), the strong field of six $CN^-$ ligands forces a low-spin $t_{2g}^6 e_g^0$ configuration, giving the maximum possible CFSE for a $d^6$ ion and a very high formation constant for $[Fe(CN)_6]^{4-}$.

Step 3 — Comparing the others …

Let’s break down each statement carefully.

1. Statement A: Co²⁺ is easily oxidised to Co³⁺ in the presence of a strong ligand like CN⁻.

   • Co²⁺ has a $d^7$ configuration. In the presence of a strong field ligand like CN⁻, the crystal field splitting is large, causing electrons to pair up in the lower-energy $t_{2g}$ orbitals. For Co²⁺ in an octahedral field, this gives $t_{2g}^6 e_g^1$.
   • Co³⁺ has a $d^6$ configuration. With strong ligands, it becomes low-spin $t_{2g}^6$, which is very stable (like the inert [Co(CN)₆]³⁻).
   • The oxidation of Co²⁺ to Co³⁺ is thus favoured because the resulting low-spin $d^6$ complex gains extra stabilisation (large CFSE). So this statement is correct.

2. Statement B: [Fe(CN)₆]⁴⁻ is an octahedral complex ion which is paramagnetic.

   • Fe²⁺ in [Fe(CN)₆]⁴⁻ has a $d^6$ configuration. CN⁻ is a strong field ligand, causing large splitting. The six electrons all pair up in the $t_{2g}$ set: $t_{2g}^6$.
   • With all electrons paired, the complex is diamagnetic, not paramagnetic. So this statement is false.

3. Statement C: Removal of H₂O molecules from [Ti(H₂O)₆]Cl₃ on strong heating converts it to a colourless compound.

   • [Ti(H₂O)₆]Cl₃ contains Ti³⁺, which has a $d^1$ configuration. The complex is coloured (typically violet) due to d-d transitions.
   • On strong heating, water molecules are lost, and the compound decomposes to TiCl₃ (or further to TiCl₄ and other products). In the absence of water ligands, the d-d transitions are no longer possible in the same way, and the resulting anhydrous compound (e.g., TiCl₃) is often colourless or pale.
   • More precisely, removal of water destroys the octahedral aqua complex, and the resulting solid does not show the characteristic d-d absorption. So the statement is correct. …

Concept & Intuition

Hydration enthalpy ($\Delta H_\text{hyd}$) is the energy released when gaseous ions are surrounded by water molecules. For a given charge, the key factors are:

• Ionic radius: smaller ion → stronger ion-dipole attraction → more exothermic (more negative) $\Delta H_\text{hyd}$.
• Crystal field stabilization energy (CFSE): In octahedral hydration complexes, $d$-electrons in lower-energy $t_{2g}$ orbitals provide extra stabilization, making hydration more exothermic. Ions with higher CFSE (more electrons in $t_{2g}$) have more negative $\Delta H_\text{hyd}$.

For $\mathrm{Cr}^{2+}$, $\mathrm{Mn}^{2+}$, $\mathrm{Co}^{2+}$, $\mathrm{Ni}^{2+}$, the ionic radii decrease from left to right across the period (due to increasing nuclear charge), so we expect $\Delta H_\text{hyd}$ to become more negative. However, $\mathrm{Mn}^{2+}$ ($d^5$, high-spin) has zero CFSE in a weak-field octahedral environment, making its hydration less exothermic than expected from size alone. This creates a "dip" in the trend.

Step-by-step reasoning

1. Identify electronic configurations and CFSE

   All these ions are high-spin in aqueous solution (water is a weak-field ligand).

   • $\mathrm{Cr}^{2+}$: $d^4$ → $t_{2g}^3 e_g^1$ → CFSE = $3 \times (-0.4\Delta_o) + 1 \times (0.6\Delta_o) = -0.6\Delta_o$
   • $\mathrm{Mn}^{2+}$: $d^5$ → $t_{2g}^3 e_g^2$ → CFSE = $3 \times (-0.4\Delta_o) + 2 \times (0.6\Delta_o) = 0$
   • $\mathrm{Co}^{2+}$: $d^7$ → $t_{2g}^5 e_g^2$ → CFSE = $5 \times (-0.4\Delta_o) + 2 \times (0.6\Delta_o) = -0.8\Delta_o$
   • $\mathrm{Ni}^{2+}$: $d^8$ → $t_{2g}^6 e_g^2$ → CFSE = $6 \times (-0.4\Delta_o) + 2 \times (0.6\Delta_o) = -1.2\Delta_o$
   Tip

   CFSE is negative (stabilizing) except for $d^0$, $d^5$ (high-spin), and $d^{10}$, where it is zero. The more negative the CFSE, the more exothermic the hydration.

2. Consider ionic radii

   Across the series $\mathrm{Cr}^{2+} \rightarrow \mathrm{Mn}^{2+} \rightarrow \mathrm{Co}^{2+} \rightarrow \mathrm{Ni}^{2+}$, ionic radii decrease (due to increasing effective nuclear charge). Smaller radius → stronger ion-water attraction → more negative $\Delta H_\text{hyd}$. So, based on size alone, the order would be:

   $\mathrm{Cr}^{2+} > \mathrm{Mn}^{2+} > \mathrm{Co}^{2+} > \mathrm{Ni}^{2+}$ in radius, meaning $\Delta H_\text{hyd}$ becomes more negative from $\mathrm{Cr}^{2+}$ to $\mathrm{Ni}^{2+}$. …

Step 1 — State the criteria for aromaticity.

A species is aromatic only if all four hold:

1. It is cyclic.
2. It is planar.
3. It is completely conjugated — every ring atom must be $sp^2$ (or carry a lone pair in a p orbital), so the p orbitals form an unbroken loop. A single $sp^3$ centre breaks the ring of overlap.
4. It obeys Hückel's rule: the delocalised π system contains $(4n+2)$ π-electrons, $n = 0, 1, 2, \ldots$ (i.e. 2, 6, 10, 14 …).

Step 2 — I: Benzene.

Six-membered ring, three alternating C=C ⇒ all six carbons $sp^2$, planar, $6\ \pi$ electrons $= 4(1)+2$. Aromatic ✓ (the archetype).

Step 3 — II: Naphthalene.

Two fused six-membered rings, fully conjugated and planar, with $5$ double bonds $\Rightarrow 10\ \pi$ electrons $= 4(2)+2$ with $n = 2$. Aromatic ✓ (a polynuclear/benzenoid aromatic hydrocarbon).

Step 4 — III: Cyclopentadiene.

The five-membered ring has two C=C and one $\mathrm{CH_2}$ carbon, which is $sp^3$. That saturated carbon has no p orbital, so the conjugation loop is broken — criterion 3 fails. (Its π count is only 4 anyway.) Not aromatic ✗

Step 5 — IV: Cyclopentadienyl anion.

Removing the $\mathrm{H^+}$ from that $\mathrm{CH_2}$ leaves a carbanion whose lone pair sits in a p orbital, so that carbon becomes $sp^2$ and the ring is now fully conjugated and planar. π-electron count:

$2\ \text{C=C} \times 2\ \pi\ \text{each} \;+\; 2\ \text{(the lone pair)} \;=\; 6\ \pi \;=\; 4(1)+2 .$ …

Analyse each:

• (A) $[\mathrm{Ni(CN)_4}]^{2-}$: $\mathrm{Ni^{2+}} = d^8$; $\mathrm{CN^-}$ is a strong-field ligand $\Rightarrow$ electrons pair up, square-planar ($dsp^2$), 0 unpaired $\Rightarrow$ diamagnetic.
• (B) $[\mathrm{CoF_6}]^{3-}$: $\mathrm{Co^{3+}} = d^6$; $\mathrm{F^-}$ weak-field $\Rightarrow$ high-spin, 4 unpaired $\Rightarrow$ paramagnetic. …

The question is about adsorption — the adhesion of atoms, ions, or molecules from a gas, liquid, or dissolved solid to a surface. Hydrogen gas is a small, diatomic molecule that can be adsorbed onto the surface of certain solids. The amount of gas adsorbed depends critically on the surface area available and the nature of the adsorbent.

Colloidal particles are extremely small (1–100 nm in diameter), so a given mass of a colloidal substance has an enormous total surface area compared to the same mass in bulk form. This is why colloidal solutions are often excellent adsorbents. However, not all colloidal particles are equally good at adsorbing hydrogen — the chemical affinity between the adsorbent and hydrogen matters just as much.

1. Finely divided platinum (Option A) and finely divided nickel (Option C) are both good catalysts for hydrogenation reactions because they adsorb hydrogen. But "finely divided" means a powder, not a colloid. While the surface area is large, it is still far smaller than that of a true colloidal solution. For example, 1 gram of finely divided platinum might have a surface area of a few square meters, whereas 1 gram of a colloidal metal can have hundreds of square meters.

2. Colloidal Fe(OH)$_3$ (Option B) is a common colloid, but iron(III) hydroxide does not have a strong affinity for hydrogen gas. Its surface is polar and hydrated, and hydrogen is nonpolar — so adsorption is weak. This option is a distractor. …

The coordination number of a metal in a complex ion is not simply the number of ligand molecules or ions attached — it is the number of donor atoms from the ligands that are directly bonded to the central metal. This is a crucial distinction.

A ligand may be monodentate (one donor atom, like $\text{SCN}^-$) or polydentate (multiple donor atoms, like $\text{C}_2\text{O}_4^{2-}$). Counting the number of ligands alone can mislead you; you must count the number of bonds each ligand forms.

Let’s examine each complex separately.

1. Complex $[\text{Fe}(\text{C}_2\text{O}_4)_3]^{3-}$

   The ligand here is the oxalate ion, $\text{C}_2\text{O}_4^{2-}$. Each oxalate ion has two negatively charged oxygen atoms that can each donate a lone pair to the metal. So each oxalate ligand is bidentate — it forms two coordinate bonds with the central Fe$^{3+}$ ion.

   With three such ligands, the total number of donor atoms (and hence coordinate bonds) is $3 \times 2 = 6$.

   Therefore, the coordination number of Fe in this complex is 6.

2. Complex $[\text{Co}(\text{SCN})_4]^{2-}$

   The ligand here is the thiocyanate ion, $\text{SCN}^-$. Thiocyanate is monodentate — it binds through either the sulfur atom or the nitrogen atom, but only one atom per ligand coordinates to the metal.

   There are four such ligands, each contributing one donor atom. So the total number of coordinate bonds is $4 \times 1 = 4$. …

Step 1 — The two things that control C–Cl bond length.

1. Resonance / partial double-bond character. If the chlorine is attached to an sp$^2$ carbon that is part of a $\pi$ system (a benzene ring or a C=C), the Cl lone pair is delocalised:

The C–Cl bond thereby gains partial double-bond character $\Rightarrow$ it becomes shorter and stronger.

2. Hybridisation / s-character. An sp$^2$ carbon (33% s) holds its bonding electrons closer to the nucleus than an sp$^3$ carbon (25% s), so an sp$^2$C–Cl bond is shorter than an sp$^3$C–Cl bond.

Both effects push the same way: sp$^3$, non-conjugated C–Cl = longest.

Step 2 — Examine each option.

• (A) Chlorobenzene: Cl on an sp$^2$ ring carbon, lone pair delocalised into the ring $\Rightarrow$ strong resonance, short C–Cl ($\approx 1.69$ Å).
• (B) 1-Chloro-4-nitrobenzene: same as (A), and the para $-NO_2$ withdraws electron density, increasing the C–Cl double-bond character further $\Rightarrow$ even shorter C–Cl. …

Why this reads differently from the standard NCERT question. The more common textbook question specifies "a corner atom" (shared by 8 unit cells) or "a face-centred atom" (shared by 2). This question's own printed wording is looser — "the unit cell is shared equally by how many unit cel …