Q.Which of the following complexes formed by $Cu^{2+}$ ions is most stable?

Classify each species by its unpaired electrons:

• $\text{O}_2$ — molecular-orbital theory places 2 electrons singly in the degenerate $\pi^*$ orbitals $\Rightarrow$ paramagnetic.
• $\text{C}_6\text{H}_6$ (benzene) — all bonding and $\pi$ electrons paired $\Rightarrow$ diamagnetic.
• $\text{Cu}^{2+}$ — $[\text{Ar}]3d^9$, one unpaired electron $\Rightarrow$ paramagnetic.
• $\text{Cu}^{+}$ — $[\text{Ar}]3d^{10}$, fully filled $\Rightarrow$ diamagnetic.
• $\text{NaCl}$ — $\text{Na}^+$ ($[\text{Ne}]$) and $\text{Cl}^-$ ($[\text{Ar}]$), both closed shells $\Rightarrow$ diamagnetic.
• $\text{O}_2^+$ — removing one $\pi^*$ electron leaves one unpaired electron (bond order 2.5) $\Rightarrow$ paramagnetic.
• $\text{MnO}$ ($\text{Mn}^{2+}$, $3d^5$), $\text{Fe}_3\text{O}_4$, $\text{MgFe}_2\text{O}_4$ — these oxides show cooperative magnetic ordering ($\text{MnO}$ is antiferromagnetic; $\text{Fe}_3\text{O}_4$ and $\text{MgFe}_2\text{O}_4$ are ferrimagnetic), so they are not counted among the simple paramagnetic species.

Count:

Concept and intuition

Group‑14 elements (C, Si, Ge, Sn, Pb) can expand their coordination number beyond four by using empty d‑orbitals. However, forming a stable octahedral $\mathrm{[MX_6]^{2-}}$ complex requires the central atom to be large enough to sterically accommodate six bulky ligands. Fluorine is very small, so even silicon (atomic radius ~111 pm) can host six fluorides. Chlorine is much larger (ionic radius ~181 pm), so only larger central atoms like Ge or Sn can fit six chlorides around them. Silicon is too small for six chlorides — the ligand‑ligand repulsion destabilises the complex.

Step‑by‑step reasoning

1. Identify the central atoms and their sizes

   The central atoms are Si, Ge, and Sn. Atomic radii increase down the group:

   Si ≈ 111 pm, Ge ≈ 125 pm, Sn ≈ 145 pm. Larger central atoms can accommodate more or larger ligands.

2. Consider the ligand sizes

   Fluoride ion (F⁻) is very small (ionic radius ~133 pm). Chloride ion (Cl⁻) is much larger (~181 pm). Hydroxide (OH⁻) is intermediate but flexible.

3. Check each complex

   • (B) $\left[\mathrm{SiF}_6\right]^{2-}$: Si is small, but F⁻ is tiny. Six fluorides fit comfortably — this complex is well‑known (e.g., hexafluorosilicate).
   • (A) $\left[\mathrm{GeCl}_6\right]^{2-}$: Ge is larger than Si, so it can accommodate six chlorides. This complex exists (e.g., as the salt $\mathrm{(NH_4)_2[GeCl_6]}$).
   • (D) $\left[\mathrm{Sn(OH)}_6\right]^{2-}$: Sn is even larger, and OH⁻ is not too bulky. This complex exists (e.g., in stannate salts).
   • (C) $\left[\mathrm{SiCl}_6\right]^{2-}$: Si is too small to host six large Cl⁻ ions. The chloride‑chloride repulsion is too strong, and the complex is unstable. It has never been isolated under normal conditions.

4. Conclusion

   The only complex that does not exist is $\left[\mathrm{SiCl}_6\right]^{2-}$.

Step 1 — How many unpaired electrons give $2.86$ BM?

So the target species must have $n=2$.

Step 2 — Test each complex

• (A) $[\text{CoF}_6]^{3-}$: $\text{Co}^{3+}$ is $d^6$; $\text{F}^-$ is a weak-field ligand → high-spin octahedral $t_{2g}^{4}e_g^{2}$ → 4 unpaired ($\mu\approx4.9$ BM).
• (B) $[\text{Co(NH}_3)_6]^{3+}$: $\text{Co}^{3+}$ ($d^6$); $\text{NH}_3$ is strong-field → low-spin $t_{2g}^{6}$ → 0 unpaired ($\mu=0$).
• (C) $[\text{NiCl}_4]^{2-}$: $\text{Ni}^{2+}$ is $d^8$; $\text{Cl}^-$ weak-field, tetrahedral → $e^{4}t_2^{4}$ with the $t_2$ set as $\uparrow\downarrow\ \uparrow\ \uparrow$ → 2 unpaired ($\mu\approx2.83$ BM). ✓
• (D) $[\text{Ni(CN)}_4]^{2-}$: $\text{Ni}^{2+}$ ($d^8$); $\text{CN}^-$ strong-field → square planar, all paired → 0 unpaired ($\mu=0$).

Only $[\text{NiCl}_4]^{2-}$ matches $2$ unpaired electrons.

Concept and intuition:

Unpaired electrons in a complex depend on the metal ion's d-electron configuration and the ligand field strength. Weak-field ligands (like $\mathrm{F}^-$ and $\mathrm{Cl}^-$) cause a small splitting, so electrons fill all five d-orbitals singly before pairing (high-spin, Hund's rule maximized across all five orbitals). Strong-field ligands cause a large splitting, forcing pairing in the lower-energy $t_{2g}$ orbitals first (low-spin). We find the oxidation state of the metal, then its d-electron count, then apply the appropriate spin state.

1. $[\mathrm{FeF}_6]^{3-}$

   • Iron is +3 (since $x + 6(-1) = -3 \Rightarrow x = +3$).
   • $\mathrm{Fe}^{3+}$: $[\mathrm{Ar}]\,3d^5$.
   • $\mathrm{F}^-$ is weak-field → high-spin: all five d-electrons occupy the five orbitals singly.
   • Unpaired electrons = 5.

2. $[\mathrm{MnCl}_6]^{3-}$

   • Manganese is +3 ($x + 6(-1) = -3 \Rightarrow x = +3$).
   • $\mathrm{Mn}^{3+}$: $[\mathrm{Ar}]\,3d^4$.
   • $\mathrm{Cl}^-$ is weak-field → high-spin: the four electrons occupy four different orbitals ($t_{2g}^3 e_g^1$), all unpaired.
   • Unpaired electrons = 4.

3. $[\mathrm{CoF}_6]^{3-}$

   • Cobalt is +3 ($x + 6(-1) = -3 \Rightarrow x = +3$).
   • $\mathrm{Co}^{3+}$: $[\mathrm{Ar}]\,3d^6$.
   • $\mathrm{F}^-$ is weak-field → high-spin: $t_{2g}^4 e_g^2$ — $t_{2g}$ has one paired orbital and two singly occupied (2 unpaired), $e_g$ has two singly occupied orbitals (2 unpaired).
   • Unpaired electrons = 4.

4. $[\mathrm{Co(C_2O_4)_3}]^{3-}$

   • Same $\mathrm{Co}^{3+}$, $3d^6$, but oxalate is a stronger-field ligand that favours pairing for $\mathrm{Co}^{3+}$ → low-spin: $t_{2g}^6 e_g^0$.
   • Unpaired electrons = 0.

5. Compare the pairs

   • (A) $[\mathrm{FeF}_6]^{3-}$ (5) and $[\mathrm{MnCl}_6]^{3-}$ (4) — not equal.
   • (B) $[\mathrm{CoF}_6]^{3-}$ (4) and $[\mathrm{Co(C_2O_4)_3}]^{3-}$ (0) — not equal.
   • (C) $[\mathrm{MnCl}_6]^{3-}$ (4) and $[\mathrm{CoF}_6]^{3-}$ (4) — equal.
   • (D) $[\mathrm{FeF}_6]^{3-}$ (5) and $[\mathrm{CoF}_6]^{3-}$ (4) — not equal.

The concept here is inner orbital vs. outer orbital complexes, a classification based on whether the metal ion uses its (n−1)d orbitals (inner) or nd orbitals (outer) for hybridisation with ligands. Inner orbital complexes are typically low-spin (strong-field ligands) and can be paramagnetic if unpaired electrons remain. Outer orbital complexes are high-spin (weak-field ligands). We must check each complex’s electronic configuration, ligand field strength, and resulting hybridisation.

1. Determine the oxidation state and d-electron count for each metal.

   • I: $[\mathrm{Mn(CN)_6}]^{3-}$ — Mn is +3 (since CN⁻ is −1, total charge −3: Mn + 6(−1) = −3 → Mn = +3). Mn³⁺ has $d^4$ configuration.
   • II: $[\mathrm{FeF_6}]^{3-}$ — Fe is +3 (F⁻ is −1, total −3: Fe + 6(−1) = −3 → Fe = +3). Fe³⁺ has $d^5$.
   • III: $[\mathrm{MnCl_6}]^{3-}$ — Mn is +3 again, $d^4$.
   • IV: $[\mathrm{CoF_6}]^{3-}$ — Co is +3 (F⁻ is −1, total −3: Co + 6(−1) = −3 → Co = +3). Co³⁺ has $d^6$.
   • V: $[\mathrm{Fe(CN)_6}]^{3-}$ — Fe is +3, $d^5$.
   • VI: $[\mathrm{Co(C_2O_4)_3}]^{3-}$ — Oxalate (C₂O₄²⁻) is −2 each, total −6; Co must be +3 to give overall −3. Co³⁺, $d^6$.
   • VII: $[\mathrm{Fe(CN)_6}]^{4-}$ — Fe is +2 (CN⁻ six gives −6, total −4 → Fe = +2). Fe²⁺ has $d^6$.

2. Classify ligands as strong-field or weak-field.

   • CN⁻ is a strong-field ligand (causes large splitting, favours low-spin).
   • F⁻ and Cl⁻ are weak-field ligands (small splitting, favour high-spin).
   • C₂O₄²⁻ (oxalate) is intermediate but generally considered moderate to strong; for Co³⁺ it is strong enough to cause low-spin.

3. Determine spin state and hybridisation for each.

   • I: $d^4$, strong-field CN⁻ → low-spin. Configuration: $t_{2g}^4 e_g^0$. All four electrons paired in $t_{2g}$? No — $t_{2g}$ holds 6 electrons max, so with 4 electrons, two are paired and two unpaired. Actually, $t_{2g}^4$ means one orbital doubly occupied, two singly occupied → 2 unpaired electrons. Hybridisation: inner $d^2sp^3$ (uses two 3d orbitals). Inner orbital, paramagnetic.
   • II: $d^5$, weak-field F⁻ → high-spin. Configuration: $t_{2g}^3 e_g^2$ → 5 unpaired electrons. Hybridisation: $sp^3d^2$ (uses outer 4d orbitals). Outer orbital, paramagnetic.
   • III: $d^4$, weak-field Cl⁻ → high-spin. Configuration: $t_{2g}^3 e_g^1$ → 4 unpaired electrons. Hybridisation: $sp^3d^2$. Outer orbital, paramagnetic.
   • IV: $d^6$, weak-field F⁻ → high-spin. Configuration: $t_{2g}^4 e_g^2$ → 4 unpaired electrons. Hybridisation: $sp^3d^2$. Outer orbital, paramagnetic.
   • V: $d^5$, strong-field CN⁻ → low-spin. Configuration: $t_{2g}^5 e_g^0$ → one unpaired electron. Hybridisation: $d^2sp^3$ (inner). Inner orbital, paramagnetic.
   • VI: $d^6$, strong-field oxalate → low-spin. Configuration: $t_{2g}^6 e_g^0$ → zero unpaired electrons. Hybridisation: $d^2sp^3$ (inner). Inner orbital, diamagnetic (not paramagnetic).
   • VII: $d^6$, strong-field CN⁻ → low-spin. Configuration: $t_{2g}^6 e_g^0$ → zero unpaired electrons. Hybridisation: $d^2sp^3$ (inner). Inner orbital, diamagnetic (not paramagnetic).

4. Select only those that are both inner orbital AND paramagnetic. From above: I (paramagnetic, inner), V (paramagnetic, inner). VI and VII are inner but diamagnetic, so they do not qualify. II, III, IV are outer orbital. Thus the set is I and V only.

The question tests your understanding of how ligand field strength influences geometry and magnetic properties in nickel and cobalt complexes. The key is to determine the oxidation state and d-electron configuration of the central metal ion, then see how the ligand (strong or weak field) affects electron pairing and geometry.

Let's examine each statement one by one.

1. Statement A: Geometry of $[Ni(CN)_4]^{2-}$ and $[NiCl_4]^{2-}$

   Nickel in both complexes is in the +2 oxidation state. Ni has atomic number 28, so Ni$^{2+}$ has the electron configuration $[Ar]\,3d^8$.

   For $[Ni(CN)_4]^{2-}$, CN$^-$ is a strong field ligand. With 8 d-electrons, a strong field causes pairing in the lower energy orbitals, leaving one d-orbital empty. This favours square planar geometry (dsp$^2$ hybridisation), not tetrahedral.

   For $[NiCl_4]^{2-}$, Cl$^-$ is a weak field ligand. The 8 d-electrons remain unpaired as much as possible, leading to tetrahedral geometry (sp$^3$ hybridisation).

   Since one is square planar and the other tetrahedral, statement A is false.

2. Statement B: Magnetic behaviour of $[Ni(CO)_4]$

   CO is a strong field ligand. Nickel in $[Ni(CO)_4]$ is in the 0 oxidation state. Ni(0) has configuration $[Ar]\,3d^8 4s^2$. In the presence of strong field CO, the 4s electrons pair with the 3d electrons, resulting in complete pairing of all 10 d-electrons (d$^{10}$ configuration). The complex is diamagnetic (no unpaired electrons), not paramagnetic.

   So statement B is false.

3. Statement C: Magnetic behaviour of $[Co(NH_3)_6]^{3+}$

   Cobalt here is in the +3 oxidation state. Co has atomic number 27, so Co$^{3+}$ has $[Ar]\,3d^6$ configuration. NH$_3$ is a moderately strong field ligand (though weaker than CN$^-$ or CO, it is strong enough to cause pairing in Co$^{3+}$).

   With a strong field, the 6 d-electrons pair completely in the three lower t$_{2g}$ orbitals, giving a low-spin d$^6$ configuration with zero unpaired electrons. Hence the complex is diamagnetic.

   Statement C is true.

Concept & Intuition

Bond order tells us the net number of chemical bonds between two atoms. For molecular oxygen species, we use the molecular orbital (MO) diagram for homonuclear diatomic molecules of the second period (with $2s$ and $2p$ orbitals). The key ordering for $O_2$ and its ions is:

For oxygen (Z=8), each atom contributes 8 electrons, so $O_2$ has 16 valence electrons. Ions add or remove electrons from the highest occupied molecular orbitals (HOMOs), which are the $\pi^*$ orbitals. The bond order formula is:

We are told that $O_2^+$ has bond order $x$. We compute the bond orders of $O_2^-$ and $O_2^{2+}$ numerically, then express them as multiples of $x$.

Step-by-step solution

1. Find $x$ (bond order of $O_2^+$)

   • $O_2^+$ has 15 valence electrons (lost one from neutral $O_2$).
   • Electron configuration: $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p})^4 (\pi_{2p}^*)^1$
   • Bonding electrons: $2+2+4 = 8$ (the $\sigma_{2s}$ and $\sigma_{2s}^*$ cancel each other, but we count all bonding and antibonding explicitly). Actually, careful: Bonding: $\sigma_{2s}$ (2), $\sigma_{2p_z}$ (2), $\pi_{2p}$ (4) → total 8 bonding. Antibonding: $\sigma_{2s}^*$ (2), $\pi_{2p}^*$ (1) → total 3 antibonding.
   • Bond order $= \frac{8 - 3}{2} = \frac{5}{2} = 2.5$.
   • So $x = 2.5$.

2. Bond order of $O_2^-$

   • $O_2^-$ has 17 valence electrons (one extra).
   • Configuration: $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p})^4 (\pi_{2p}^*)^3$
   • Bonding electrons: still 8.
   • Antibonding electrons: $\sigma_{2s}^*$ (2) + $\pi_{2p}^*$ (3) = 5.
   • Bond order $= \frac{8 - 5}{2} = \frac{3}{2} = 1.5$.

3. Bond order of $O_2^{2+}$

   • $O_2^{2+}$ has 14 valence electrons (lost two).
   • Configuration: $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p})^4 (\pi_{2p}^*)^0$
   • Bonding electrons: 8.
   • Antibonding electrons: $\sigma_{2s}^*$ (2) only = 2.
   • Bond order $= \frac{8 - 2}{2} = \frac{6}{2} = 3.0$.

4. Express both in terms of $x = 2.5$

   • For $O_2^-$: $1.5 = \frac{3}{2} = \frac{3}{5} \times \frac{5}{2} = \frac{3}{5}x$.
   • For $O_2^{2+}$: $3.0 = 3 = \frac{6}{5} \times \frac{5}{2} = \frac{6}{5}x$.

Thus the pair is $\left( \frac{3}{5}x,\ \frac{6}{5}x \right)$, which matches option (B).

Concept & Intuition

Magnetic properties (paramagnetic vs. diamagnetic) arise from unpaired electrons. Paramagnetic compounds have at least one unpaired electron; diamagnetic compounds have all electrons paired.

To decide, we need:

• The oxidation state of the metal.
• The number of d‑electrons.
• Whether the ligands are strong‑field (causing low‑spin, more pairing) or weak‑field (causing high‑spin, fewer pairings). Carbonyl ($\mathrm{CO}$) is a very strong‑field ligand; cyanide ($\mathrm{CN}^-$) is also strong‑field; water ($\mathrm{H}_2\mathrm{O}$) is intermediate; chloride ($\mathrm{Cl}^-$) is weak‑field. We’ll check each complex step by step.

1. Option (A): $[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]\mathrm{Br}_2$

   • The complex cation is $[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ (since two $\mathrm{Br}^-$ counterions).
   • Chromium in +2 oxidation state: atomic number 24, $\mathrm{Cr}^{2+}$ has $3d^4$ configuration.
   • $\mathrm{H}_2\mathrm{O}$ is a weak‑field ligand (spectrochemical series: $\mathrm{H}_2\mathrm{O}$ < $\mathrm{NH}_3$ < $\mathrm{CN}^-$).
   • For $d^4$ in an octahedral weak field: high‑spin arrangement, $t_{2g}^3 e_g^1$ → 4 unpaired electrons.
   • Paramagnetic — matches the statement. ✓

2. Option (B): $\mathrm{Na}_4[\mathrm{Fe}(\mathrm{CN})_6]$

   • The complex anion is $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$ (four $\mathrm{Na}^+$ counterions).
   • Iron in +2 oxidation state: $\mathrm{Fe}^{2+}$ has $3d^6$ configuration.
   • $\mathrm{CN}^-$ is a very strong‑field ligand → low‑spin octahedral.
   • For $d^6$ low‑spin: $t_{2g}^6$ → all electrons paired (0 unpaired).
   • Diamagnetic — matches the statement. ✓

3. Option (C): $[\mathrm{Ni}(\mathrm{CO})_4]$

   • Nickel in zero oxidation state: $\mathrm{Ni}^0$ has $3d^8 4s^2$? Actually, neutral Ni is $[\mathrm{Ar}]\,3d^8 4s^2$, but in complexes the 4s electrons are lost first; for $\mathrm{Ni}(0)$, the effective configuration is $3d^{10}$ after reorganisation due to strong‑field ligands.
   • $\mathrm{CO}$ is an extremely strong‑field ligand, and the complex is tetrahedral (not octahedral).
   • In tetrahedral geometry, the splitting is smaller and inverted, but for $\mathrm{Ni}(\mathrm{CO})_4$, the strong field causes complete pairing: all 10 d‑electrons are paired → $d^{10}$ configuration.
   • Diamagnetic (no unpaired electrons).
   • The statement says “Paramagnetic” — this is incorrect. ✗

4. Option (D): $\mathrm{Na}_2[\mathrm{NiCl}_4]$

   • The complex anion is $[\mathrm{NiCl}_4]^{2-}$ (two $\mathrm{Na}^+$ counterions).
   • Nickel in +2 oxidation state: $\mathrm{Ni}^{2+}$ has $3d^8$ configuration.
   • $\mathrm{Cl}^-$ is a weak‑field ligand, and the geometry is tetrahedral (common for $\mathrm{Ni}^{2+}$ with $\mathrm{Cl}^-$).
   • For $d^8$ in tetrahedral field: the splitting is small, so high‑spin arrangement → two unpaired electrons.
   • Paramagnetic — matches the statement. ✓