Q.Why are low spin tetrahedral complexes not formed?

1. Understand the coordination geometries and isomerism rules

   • For octahedral complexes with two identical bidentate ligands (like oxalate, ox²⁻) and two monodentate ligands, the two monodentate ligands can be adjacent (cis) or opposite (trans).
   • Optical activity arises when the complex lacks a plane of symmetry — the cis isomer of such a complex is often chiral (non-superimposable mirror image), while the trans isomer is usually achiral.

2. Analyze option (A): $\left[\mathrm{CrCl}_2(\mathrm{ox})_2\right]^{3-}$

   • Chromium(III) is octahedral. Two oxalate (ox²⁻) ligands are bidentate, occupying four coordination sites. The two chloride ligands occupy the remaining two sites.
   • The two Cl⁻ can be cis (adjacent) or trans (opposite). So cis-trans isomerism exists.
   • The cis isomer has no plane of symmetry (the two oxalate rings create a chiral arrangement), so it is optically active. The trans isomer has a plane of symmetry and is optically inactive.
   • This matches the statement exactly.

3. Analyze option (B): $\left[\mathrm{PtCl}_2\left(\mathrm{NH}_3\right)_2\right]$

   • Platinum(II) is square planar. It exhibits cis-trans isomerism (cis and trans isomers are well-known).
   • However, square planar complexes with two identical monodentate ligands and two identical other monodentate ligands are never optically active — they have a plane of symmetry in both isomers.
   • So the claim “exhibits both cis-trans and optical isomerism” is false.

4. Analyze option (C): $\left[\mathrm{Fe}(\mathrm{CN})_4\left(\mathrm{NH}_3\right)_2\right]^{-}$

   • Iron in this complex is likely low-spin Fe(II) or Fe(III), octahedral. Four CN⁻ and two NH₃ ligands.
   • The two NH₃ can be cis or trans, so cis-trans isomerism is possible.
   • However, the cis isomer of an octahedral complex with four identical monodentate ligands and two identical others does have a plane of symmetry (the plane containing the two NH₃ and two opposite CN⁻). Thus, neither cis nor trans is chiral.
   • The statement “both exhibit optical isomerism” is false. …

Concept & Intuition

Stereoisomerism in coordination compounds depends on the coordination number, geometry, and the symmetry of the ligand arrangement.

• Square planar complexes (d⁸ metals like Pt²⁺) with two identical bidentate or two pairs of monodentate ligands can show cis‑trans isomerism.
• Tetrahedral complexes (like Ni²⁺ with four monodentate ligands) are almost always achiral and show no stereoisomerism because all vertices are equivalent.
• Octahedral complexes with three identical and three different monodentate ligands (MA₃B₃ type) exhibit fac‑mer isomerism.
• Octahedral complexes with a bidentate ligand (like en) and two identical monodentate ligands can show cis‑trans isomerism, and only the cis form is chiral (optical isomerism).

Step‑by‑Step Reasoning

1. Complex A: [Pt(NH₃)₂Cl₂]

   • Pt²⁺ is d⁸, so the geometry is square planar.
   • Two NH₃ and two Cl⁻ ligands can be arranged adjacent (cis) or opposite (trans).
   • This is classic cis‑trans isomerism.
   • Match: A → Q.

2. Complex B: Ni[(NH₃)₂Cl₂]

   • Ni²⁺ in this formulation (no charge shown, but neutral) is typically tetrahedral (common for Ni²⁺ with four monodentate ligands).
   • In a tetrahedron, all four positions are equivalent; swapping two identical ligands gives the same compound.
   • No cis/trans, no fac/mer, no optical activity.
   • Match: B → S (does not exhibit isomerism).

3. Complex C: [Co(NH₃)₃(NO₂)₃]

   • Co³⁺ is d⁶, almost always octahedral.
   • Three NH₃ and three NO₂⁻ ligands: this is an MA₃B₃ type.
   • The three identical ligands can occupy a face of the octahedron (fac) or a meridian (mer).
   • Match: C → P (fac‑mer).

4. Complex D: [PtCl₂(en)₂]²⁺ …

Concept & Intuition

Optical activity arises when a molecule lacks an improper rotation axis (i.e., it is chiral). For coordination compounds, chirality often comes from chelate rings that create a helical or propeller‑like arrangement with no plane or center of symmetry. The key is to check whether the complex has a mirror plane or a center of inversion — if it does, it cannot be optically active. Here, all complexes contain the bidentate ligand ethylenediamine (en), which forms five‑membered chelate rings. The cis/trans geometry and the number of chelate rings determine symmetry.

Step‑by‑step reasoning

1. Analyze (A): cis‑$\left[\mathrm{PtCl}_2(\mathrm{en})_2\right]^{2+}$

   • Pt(IV) is octahedral. Two en ligands and two Cl⁻ ions. In the cis isomer, the two Cl⁻ are adjacent.
   • The two en chelate rings create a chiral arrangement: the complex has no plane of symmetry (the chelate rings are not coplanar) and no center of inversion.
   • It exists as a pair of non‑superimposable mirror images (Δ and Λ isomers).
   • Conclusion: This complex is optically active.

2. Analyze (B): $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$

   • Co(III) is octahedral with three bidentate en ligands. This is a classic example of a chiral complex.
   • The three chelate rings form a propeller shape (like a three‑bladed fan). There is no plane or center of symmetry.
   • It is well‑known to exist as Δ and Λ enantiomers.
   • Conclusion: This complex is optically active.

3. Analyze (C): trans‑$\left[\mathrm{CoCl}_2(\mathrm{en})_2\right]^{+}$

   • Co(III) octahedral with two en ligands and two Cl⁻ ions. In the trans isomer, the two Cl⁻ are opposite each other (180° apart).
   • The two en ligands lie in a plane; the Cl⁻ atoms are on the axis perpendicular to that plane.
   • The molecule has a plane of symmetry that bisects the en ligands and contains the Co–Cl bonds. …

Concept & Intuition

Geometrical isomerism arises when ligands can occupy different positions around the metal (e.g., cis vs. trans in an octahedral complex). Optical isomerism requires the complex to lack a plane of symmetry, so its mirror image is non-superimposable — like left and right hands. For a complex to show both, it must have at least two different arrangements of ligands (geometrical isomers), and at least one of those arrangements must be chiral. We check each option for these conditions.

Step-by-step reasoning

1. Option (A): $\left[\operatorname{Cr}(\mathrm{ox})_3\right]^{3-}$

   • ox = oxalate ($\mathrm{C}_2\mathrm{O}_4^{2-}$), a bidentate ligand.
   • The complex is octahedral with three identical bidentate ligands.
   • All three oxalate ligands are equivalent; there is no possibility of different ligand arrangements (no cis/trans).
   • However, the complex is chiral (like the $[\mathrm{M}(\mathrm{AA})_3]$ type) and shows optical isomerism.
   • Conclusion: Geometrical isomerism is absent, so this fails.

2. Option (B): $\left[\mathrm{Co}(\mathrm{en})_2 \mathrm{Cl}_2\right]^{+}$

   • en = ethylenediamine (bidentate), Cl = monodentate.
   • The two Cl ligands can be adjacent (cis) or opposite (trans) — this gives geometrical isomerism.
   • The trans isomer has a plane of symmetry (the Co–Cl–Co axis and the en ligands lie in a plane), so it is achiral.
   • The cis isomer has no plane of symmetry: the two en rings create a helical twist, and the mirror image cannot be rotated to match the original. Hence the cis form is chiral and shows optical isomerism.
   • Conclusion: Both geometrical and optical isomerism exist (cis isomer is optically active). This works.

3. Option (C): $[\mathrm{Co}(\mathrm{CN})_6]^{3-}$

   • All six ligands are identical monodentate CN⁻.
   • Only one possible arrangement (no geometrical isomers).
   • The complex has many planes of symmetry (e.g., through opposite CN groups), so it is achiral.
   • Conclusion: Neither type of isomerism occurs.

4. Option (D): $\left[\mathrm{Co}\left(\mathrm{NO}_3\right)_3\left(\mathrm{NH}_3\right)_3\right]$

   • Here $\mathrm{NO}_3^-$ is monodentate (bonded through oxygen). …

Concept & Intuition

Geometrical isomerism arises when ligands can occupy different spatial positions around a metal center (e.g., cis/trans in octahedral complexes with two identical ligands). Structural isomerism here refers to ionization isomerism — where the counter ion and a ligand exchange places, giving different ions in solution. We must check each complex for both possibilities.

Step-by-step reasoning

1. Complex A: $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{NO}_2$

   • The coordination sphere is $\left[\mathrm{Co}(\mathrm{NH}_3)_4\mathrm{Cl}_2\right]^+$ with $\mathrm{NO}_2^-$ as counter ion.
   • Geometrical isomerism: With four $\mathrm{NH}_3$ and two $\mathrm{Cl}$ ligands in an octahedral geometry, the two chlorines can be cis (adjacent) or trans (opposite). So yes, geometrical isomers exist.
   • Structural (ionization) isomerism: The $\mathrm{NO}_2^-$ could swap with a $\mathrm{Cl}^-$ to give $\left[\mathrm{Co}(\mathrm{NH}_3)_4(\mathrm{NO}_2)\mathrm{Cl}\right]\mathrm{Cl}$ — a different compound with different ions. So ionization isomerism is possible.
   • Conclusion: A exhibits both.

2. Complex B: $\left[\mathrm{Co}\left(\mathrm{NH}_3\right) \mathrm{Br}\right] \mathrm{SO}_4$

   • The formula suggests a coordination sphere with one $\mathrm{NH}_3$ and one $\mathrm{Br}^-$ (likely $\left[\mathrm{Co}(\mathrm{NH}_3)\mathrm{Br}\right]^+$), but cobalt(III) typically has coordination number 6. This formula is unrealistic — it likely means $\left[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Br}\right]\mathrm{SO}_4$ (a common typo in such problems).
   • Assuming the intended formula is $\left[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Br}\right]\mathrm{SO}_4$:
     • Geometrical isomerism: With five identical $\mathrm{NH}_3$ and one $\mathrm{Br}$, no cis/trans isomerism is possible (only one arrangement).
     • Structural isomerism: $\mathrm{SO}_4^{2-}$ could replace $\mathrm{Br}^-$ to give $\left[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{SO}_4\right]\mathrm{Br}$ — ionization isomerism is possible.
   • So B shows only structural isomerism, not geometrical.

3. Complex C: $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3\left(\mathrm{NO}_2\right)_3\right]$

   • This is a neutral complex (no counter ion).
   • Geometrical isomerism: With three $\mathrm{NH}_3$ and three $\mathrm{NO}_2$ ligands, two geometrical isomers exist: fac (all three identical ligands on one face) and mer (three in a meridian). So yes. …

The question tests your understanding of isomerism in coordination compounds — specifically, when geometrical (cis/trans, fac/mer) and optical isomers are possible. Let’s examine each statement carefully.

1. Statement (A): A tetrahedral complex of the type $[Ma_2b_2]$ does not show geometrical isomerism.

   Tetrahedral complexes have all four ligands at the vertices of a tetrahedron. In $[Ma_2b_2]$, any two identical ligands are always adjacent (there is no “opposite” position as in a square). Rotating the molecule gives the same arrangement — all possible placements are equivalent. So no geometrical isomers exist. This statement is correct.

2. Statement (B): Coordination entities of the type $[Ma_3b_3]$ do not exhibit geometrical isomerism.

   For an octahedral complex with three identical ligands (a) and three others (b), two distinct arrangements are possible:

   • fac (facial): the three a ligands occupy one face of the octahedron (all mutually adjacent).
   • mer (meridional): the three a ligands lie in a plane that goes through the metal, with two opposite and one in between. These are geometrical isomers. So this statement is false — it claims they do not exhibit geometrical isomerism, but they do.

3. Statement (C): Square planar complex of the type $[Mabcd]$ has 3 possible geometrical isomers.

   In a square planar complex with four different ligands, the number of distinct arrangements (ignoring mirror images) is given by $(4-1)! / 2 = 3$ (since rotations and reflections are considered the same). These correspond to placing each ligand in turn at a fixed reference position and arranging the other three. So 3 isomers is correct.

4. Statement (D): In a coordination entity of the type $[Ma_2(bb)_2]^{2+}$ only the cis-isomer is optically active. …

Option (A) — incorrect. For square-planar $[\mathrm{Mabcd}]$ there are three geometrical isomers, obtained by placing each of b, c, d trans to a. Describing them as "one cis-form and two trans-forms" mis-states the isomerism, so this statement is wrong.

Option (B) — correct. $[\mathrm{Co(NH_3)_5NO_2}]\mathrm{Cl_2}$ exists as the yellow nitro ($-\mathrm{NO_2}$, N-bonded) and red nitrito ($-\mathrm{ONO}$, O-bonded) linkage isomers.

Option (C) — correct. $\mathrm{CN^-}$ is a strong-field ligand; with $\mathrm{Fe^{3+}}$ ($d^5$) it forces electron pairing, giving a low-spin complex. …

fac–mer isomerism occurs in octahedral complexes of type $\mathrm{Ma_3b_3}$: the three like ligands can occupy one triangular face (facial) or a meridian (meridional).

• (A) $[\mathrm{Co(H_2O)_3(NO_2)_3}]$ — $\mathrm{Ma_3b_3}$ type $\Rightarrow$ shows fac and mer isomers. ✓
• (B) $[\mathrm{Co(en)_2Cl_2}]^+$ — $\mathrm{M(AA)_2b_2}$ $\Rightarrow$ cis/trans (and optical), not fac/mer. …

The complex has Pt bonded to two $\mathrm{Cl^-}$ ligands and two bidentate ethylenediamine (en) ligands, giving a coordination number of 6 (octahedral).

• Geometrical isomerism: the two $\mathrm{Cl^-}$ can be adjacent (cis) or opposite (trans). …

A complex shows optical isomerism only if it lacks any improper symmetry element (plane / centre of symmetry) — typically $cis$-$[M(AA)_2X_2]$ or $cis$-$[M(AA)_2XY]$ types with bidentate chelates.

1. $cis$-$[\mathrm{CoCl(en)_2(NH_3)}]^{2+}$: two $\mathrm{en}$ chelates in a cis, unsymmetrical arrangement $\Rightarrow$ chiral. ✓
2. $cis$-$[\mathrm{CrCl_2(ox)_2}]^{3-}$: two oxalate chelates cis to each other $\Rightarrow$ chiral (classic $cis$-$[M(AA)_2X_2]$). ✓ …

Concept: Optical isomerism in octahedral complexes requires the absence of a plane (or centre) of symmetry. Bidentate chelate rings (en, ox) in the cis arrangement destroy the mirror plane; complexes with only monodentate ligands of type [M(A)4B2] retain a plane.

(A) cis-[Co(en)2Cl2]+ : cis-bis(chelate) - non-superimposable on its mirror image -> OPTICALLY ACTIVE.

(B) cis-[CrCl2(ox)2]^3- : cis-bis(chelate) -> OPTICALLY ACTIVE. …

Step 1 — The void-counting rule for close packing

In any close-packed arrangement (ccp/fcc or hcp) of $N$ spheres:

• number of octahedral voids $= N$
• number of tetrahedral voids $= 2N$

This is the single fact the whole problem rests on. (Each sphere contributes one octahedral void and two tetrahedral voids to the lattice.)

Step 2 — Set up with $N$ oxide ions

Let the ccp lattice contain $N$ ions of $O^{2-}$.

Step 3 — Fill the voids as stated

Aluminium occupies $\dfrac14$ of the tetrahedral voids:

Beryllium occupies $\dfrac14$ of the octahedral voids:

Step 4 — Form the ratio

Multiply throughout by $\dfrac{4}{N}$ to clear fractions:

Step 5 — Check electrical neutrality (an independent confirmation) …