NCERT Solutions · Class 12 Chemistry Coordination Compounds

Werner's central idea is that a metal ion in a complex satisfies two different kinds of valency at the same time — one that ions can neutralize, and one that only fixed positions in space can satisfy. Once you separate those two valencies, every observation about coordination compounds — why some chloride is precipitated by \(\displaystyle \text{AgNO}_3 \) and some is not, why the compounds have definite geometric shapes — falls into place.Postulate $\displaystyle 1$ — Primary and secondary valency are distinct. Every metal ion has: Postulate $\displaystyle 2$ — Every metal has a fixed secondary valency. The secondary valency (coordination number) is a fixed characteristic of the metal ion — for example \(\displaystyle \text{Co}^{3+} \) almost always shows a secondary valency of $\displaystyle 6$, \(\displaystyle \text{Pt}^{2+} \) shows $\displaystyle 4$ — and the metal tries to satisfy both its primary and secondary valencies.Postulate $\displaystyle 3$ — Secondary valencies point in fixed directions in space. Because the secondary valencies are directional, the groups attached by secondary valency (ligands) occupy fixed positions around the metal, giving the complex a definite geometry: a coordination number of $\displaystyle 6$ gives an octahedral shape, $\displaystyle 4$ gives tetrahedral or square planar. Primary valencies, by contrast, are non-directional.Applying this to a real series of compounds. Werner explained the puzzling behaviour of the compounds of \(\displaystyle \text{CoCl}_3 \) with \(\displaystyle \text{NH}_3 \) using exactly this idea. Experimentally, when treated with excess \(\displaystyle \text{AgNO}_3 \), the number of \(\displaystyle \text{Cl}^- \) ions precipitated as \(\displaystyle \text{AgCl} \) per formula unit was found to be:\[\text{CoCl}_3\cdot 6\text{NH}_3 \;\to\; 3\ \text{Cl}^-\text{ precipitated} \] \[\text{CoCl}_3\cdot 5\text{NH}_3 \;\to\; 2\ \text{Cl}^-\text{ precipitated} \] \[\text{CoCl}_3\cdot 4\text{NH}_3 \;\to\; 1\ \text{Cl}^-\text{ precipitated} \]This is exactly the observation that a simple ionic formula cannot explain — all three compounds contain the same \(\displaystyle \text{Co}^{3+} \) and the same total \(\displaystyle \text{Cl}^- \), yet different fractions of the chloride behave as free ions. Werner resolved this by saying \(\displaystyle \text{Co}^{3+} \) has a primary valency of $\displaystyle 3$ and a secondary valency of $\displaystyle 6$, and that \(\displaystyle \text{NH}_3 \) molecules preferentially occupy the six secondary-valency positions, with \(\displaystyle \text{Cl}^- \) ions filling any secondary positions left over. A \(\displaystyle \text{Cl}^- \) held by secondary valency sits directly on the metal and is not free to ionize, while a \(\displaystyle \text{Cl}^- \) held only by primary valency is outside this fixed arrangement and ionizes in solution.This gives the constitutions \[[\text{Co(NH}_3)_6]\text{Cl}_3,\qquad [\text{Co(NH}_3)_5\text{Cl}]\text{Cl}_2,\qquad [\text{Co(NH}_3)_4\text{Cl}_2]\text{Cl} \]The species inside the square brackets is the coordination entity, held together by the six secondary valencies of cobalt (a fixed octahedral arrangement, per Postulate $\displaystyle 3$) — these ligands do not dissociate. The \(\displaystyle \text{Cl}^- \) ions written outside the bracket are held only by primary valency, remain as free ions in solution, and are the ones that precipitate with \(\displaystyle \text{AgNO}_3 \). This is why the count of precipitable chloride drops from $\displaystyle 3$ to $\displaystyle 2$ to $\displaystyle 1$ as more \(\displaystyle \text{NH}_3 \) molecules (up to the fixed secondary valency of $\displaystyle 6$) take up positions directly on the cobalt and push chloride ions out of the primary-valency role into the secondary-valency role inside the bracket.Answer: Werner's postulates explain bonding in coordination compounds by proposing that a metal has a fixed primary (ionizable) valency equal to its oxidation state and a fixed, spatially directed secondary (non-ionizable) valency equal to its coordination number; ligands occupying the secondary-valency positions form a fixed geometric coordination entity (e.g., octahedral for \(\displaystyle [\text{Co(NH}_3)_6]^{3+} \)) and do not ionize, while only groups satisfying primary valency outside this entity dissociate as free ions — as confirmed by the graded \(\displaystyle \text{AgCl} \) precipitation from \(\displaystyle \text{CoCl}_3\cdot 6\text{NH}_3 \), \(\displaystyle \text{CoCl}_3\cdot 5\text{NH}_3 \), and \(\displaystyle \text{CoCl}_3\cdot 4\text{NH}_3 \).

A double salt breaks apart completely into its simple ions the moment it dissolves; a coordination compound keeps its metal ion locked inside a complex ion. That difference is the whole answer here.What happens with \(\displaystyle FeSO_4 + (NH_4)_2SO_4 \) ($\displaystyle 1$:$\displaystyle 1$)Mixing these in a $\displaystyle 1$:$\displaystyle 1$ molar ratio and crystallising gives Mohr's salt, \[FeSO_4\cdot(NH_4)_2SO_4\cdot 6H_2O \] This is a double salt — a solid formed by two simple salts crystallising together, with no separate chemical identity of its own. As soon as it is dissolved in water it dissociates completely into its simple, free ions: \[FeSO_4\cdot(NH_4)_2SO_4\cdot 6H_2O \xrightarrow{\text{water}} Fe^{2+} + 2NH_4^{+} + 2SO_4^{2-} + 6H_2O \]Aside — this is the step people gloss over: a double salt is only a double salt as a solid. In solution it is nothing more than a mixture of the ions of the two parent salts, exactly as if you had dissolved \(\displaystyle FeSO_4 \) and \(\displaystyle (NH_4)_2SO_4 \) separately.Because the solution genuinely contains free \(\displaystyle Fe^{2+} \) ions, it gives every characteristic qualitative test of \(\displaystyle Fe^{2+} \) — for instance the brown-ring test, or a precipitate with potassium ferricyanide. The ammonium and sulphate ions are just spectators; they don't interfere.What happens with \(\displaystyle CuSO_4 + \) aqueous \(\displaystyle NH_3 \) ($\displaystyle 1$:$\displaystyle 4$)Here the $\displaystyle 1$:$\displaystyle 4$ ratio is deliberate — it is exactly enough \(\displaystyle NH_3 \) to saturate the coordination number of \(\displaystyle Cu^{2+} \), forming a coordination compound, tetraamminecopper(II) sulphate: \[CuSO_4 + 4NH_3 \longrightarrow [Cu(NH_3)_4]SO_4 \]Aside — the key structural difference: the four \(\displaystyle NH_3 \) molecules are not free spectator molecules the way \(\displaystyle NH_4^+ \) was above. Each \(\displaystyle NH_3 \) donates its lone pair to \(\displaystyle Cu^{2+} \) through a coordinate (dative) bond, and that bond does not break on dissolving.When \(\displaystyle [Cu(NH_3)_4]SO_4 \) dissolves, it dissociates only as far as separating the ionisable sulphate from the complex ion: \[[Cu(NH_3)_4]SO_4 \longrightarrow [Cu(NH_3)_4]^{2+} + SO_4^{2-} \]The complex ion itself is held together strongly (it has a large formation constant, i.e. a very small dissociation/instability constant), so it does not break down further into free \(\displaystyle Cu^{2+} \) and \(\displaystyle NH_3 \) to any measurable extent: \[[Cu(NH_3)_4]^{2+} \rightleftharpoons Cu^{2+} + 4NH_3 \qquad (K_{\text{inst}} \text{ extremely small}) \]Only a vanishingly small equilibrium concentration of free \(\displaystyle Cu^{2+} \) exists in this solution — far too little to trigger the usual tests for \(\displaystyle Cu^{2+} \) (such as the black precipitate of \(\displaystyle CuS \) with \(\displaystyle H_2S\), or the pale-blue precipitate with \(\displaystyle NaOH \)). Almost all of the copper is "hidden" as the deep-blue \(\displaystyle [Cu(NH_3)_4]^{2+} \) complex ion, which is itself what gives the solution its characteristic intense blue colour, not free \(\displaystyle Cu^{2+} \).Aside — this is the exact test students misapply: seeing "\(\displaystyle Cu^{2+} \) in solution" and expecting the standard \(\displaystyle Cu^{2+} \) reactions to work, without checking whether the copper is present as a free ion or locked in a complex.The underlying rule: a double salt loses its separate identity completely on dissolving — its constituent ions behave independently and give their individual tests. A coordination compound retains its identity in solution — the central metal ion stays bonded to its ligands through coordinate bonds, so the "combined" metal ion no longer behaves like the free ion and fails its normal tests.Answer: \(\displaystyle FeSO_4\cdot(NH_4)_2SO_4\cdot 6H_2O \) is a double salt that dissociates fully in water into free \(\displaystyle Fe^{2+}, NH_4^{+}, SO_4^{2-} \) ions, so free \(\displaystyle Fe^{2+} \) gives its test; \(\displaystyle [Cu(NH_3)_4]SO_4 \) is a coordination compound whose complex ion \(\displaystyle [Cu(NH_3)_4]^{2+} \) does not dissociate appreciably (\(\displaystyle Cu^{2+} \) is held by coordinate bonds to four \(\displaystyle NH_3 \) ligands), so essentially no free \(\displaystyle Cu^{2+} \) is present and the \(\displaystyle Cu^{2+} \) test fails.

A coordination entity is the whole bracket — the metal plus everything directly attached to it. In a formula like \(\displaystyle \left[\text{Co(NH}_3)_6\right]\text{Cl}_3\), the coordination entity is only what's inside the square brackets, \(\displaystyle \left[\text{Co(NH}_3)_6\right]^{3+}\) — the \(\displaystyle \text{Co}^{3+}\) ion together with the six \(\displaystyle \text{NH}_3\) molecules bonded to it. The three \(\displaystyle \text{Cl}^-\) outside the bracket are counter ions, not part of the entity.Two examples: A ligand is any ion or molecule directly bonded to the central atom, and it must have a lone pair (or a π system) to donate. The ligand is the electron-pair donor; the metal is the acceptor. Ligands can be neutral, anionic, monodentate (one donor atom) or polydentate (several donor atoms from one ligand binding at once).Two examples: Coordination number counts donor atoms attached to the metal, not the number of ligands. This is the step people get wrong with polydentate ligands: three molecules of en give a coordination number of $\displaystyle 6$ (each en contributes $\displaystyle 2$ donor N atoms), not 3.Two examples: Coordination polyhedron is the geometric shape traced out by joining the donor atoms around the central atom. It is a $\displaystyle 3$-D shape, fixed by the coordination number: $\displaystyle 4$ gives tetrahedral or square planar, $\displaystyle 6$ gives octahedral, and so on. Naming the coordination number does not by itself fix which of these shapes occurs — that depends on the metal, its oxidation state, and the ligands.Two examples: (A third useful contrast: \(\displaystyle \left[\text{PtCl}_4\right]^{2-}\) also has coordination number $\displaystyle 4$, but its polyhedron is square planar, not tetrahedral — the same coordination number does not force the same polyhedron.)Homoleptic means every ligand attached to the metal is of the same kind; heteroleptic means at least two different kinds of ligand are present. This is purely about how many distinct donor species surround the metal — it says nothing about the coordination number or the charge.Homoleptic examples: Heteroleptic examples: Answer: Coordination entity — \(\displaystyle \left[\text{Co(NH}_3)_6\right]^{3+}\), \(\displaystyle \left[\text{Ni(CO)}_4\right]\); Ligand — \(\displaystyle \text{Cl}^-\), ethylenediamine; Coordination number — $\displaystyle 6$ in \(\displaystyle \left[\text{PtCl}_6\right]^{2-}\), $\displaystyle 4$ in \(\displaystyle \left[\text{Ni(NH}_3)_4\right]^{2+}\); Coordination polyhedron — octahedral for \(\displaystyle \left[\text{Co(NH}_3)_6\right]^{3+}\), tetrahedral for \(\displaystyle \left[\text{Ni(CO)}_4\right]\); Homoleptic — \(\displaystyle \left[\text{Co(NH}_3)_6\right]^{3+}\), \(\displaystyle \left[\text{Fe(CN)}_6\right]^{4-}\); Heteroleptic — \(\displaystyle \left[\text{Co(NH}_3)_4\text{Cl}_2\right]^{+}\), \(\displaystyle \left[\text{Pt(NH}_3)_2\text{Cl}_2\right]\).

Denticity is the count of donor atoms from one single ligand that actually bond to the central metal ion at the same time — not the number of atoms in the ligand, and not the ligand's overall charge.Unidentate (monodentate) ligands attach to the metal through only one donor atom, so the ligand occupies a single position in the coordination sphere.Examples: (Other equally valid examples: \(\displaystyle \text{H}_2\text{O} \), \(\displaystyle \text{CN}^- \), \(\displaystyle \text{OH}^- \).)Didentate (bidentate) ligands carry two donor atoms in the same molecule/ion, and both of them coordinate to the same metal simultaneously, closing a ring (a chelate) with the metal at one vertex. The aside worth flagging: two donor atoms does not mean two separate ligand molecules attaching — it is one ligand wrapping back to bond twice.Examples: Ambidentate ligands are ligands that have two different donor atoms, but only one of them bonds to the metal in any given complex — so the ligand is still monodentate in each individual complex, but which atom it uses can switch depending on the metal and conditions. The point people miss: this is not the same idea as didentate — an ambidentate ligand never uses both donor atoms on the same metal at once; it picks one or the other.Examples: Answer: Unidentate ligands bind through one donor atom (e.g. \(\displaystyle \text{Cl}^- \), \(\displaystyle \text{NH}_3 \)); didentate ligands bind through two donor atoms of the same ligand simultaneously (e.g. ethane-$\displaystyle 1,2$-diamine, \(\displaystyle \text{C}_2\text{O}_4^{2-} \)); ambidentate ligands have two different possible donor atoms but bind through only one at a time, which one depending on conditions (e.g. \(\displaystyle \text{NO}_2^- \), \(\displaystyle \text{SCN}^- \)).

The oxidation number of the metal is whatever makes the sum of all charges equal the charge written outside the bracket. Ligands carry fixed charges — halides and \(\displaystyle CN^-\) are \(\displaystyle -1\), water and ammine (\(\displaystyle NH_3\)) are neutral, and ethylenediamine (en) is a neutral bidentate ligand contributing \(\displaystyle 0\) — so set up \(\displaystyle x + (\text{sum of ligand charges}) = (\text{overall charge of the entity})\) and solve for \(\displaystyle x\), the oxidation number of the metal.(i) \(\displaystyle [Co(H_2O)(CN)(en)_2]^{2+}\)Ligand charges: \(\displaystyle H_2O = 0\), \(\displaystyle CN^- = -1\), each en \(\displaystyle = 0\) (two of them \(\displaystyle = 0\)).\[x + 0 + (-1) + 2(0) = +2 \] \[x - 1 = +2 \implies x = +3 \]Oxidation number of Co \(\displaystyle = +3\).(ii) \(\displaystyle [CoBr_2(en)_2]^{+}\)Ligand charges: two \(\displaystyle Br^-\) give \(\displaystyle 2 \times (-1) = -2\); two en give \(\displaystyle 0\).\[x + (-2) + 0 = +1 \] \[x = +1 + 2 = +3 \]Oxidation number of Co \(\displaystyle = +3\).(iii) \(\displaystyle [PtCl_4]^{2-}\)Four \(\displaystyle Cl^-\) give \(\displaystyle 4 \times (-1) = -4\).\[x + (-4) = -2 \] \[x = -2 + 4 = +2 \]Oxidation number of Pt \(\displaystyle = +2\).(iv) \(\displaystyle K_3[Fe(CN)_6]\)This is the step people trip on: the charge you balance against is the charge on the complex ion, not zero — the three \(\displaystyle K^+\) ions outside the bracket must balance the complex ion's charge, so \(\displaystyle [Fe(CN)_6]\) carries a charge of \(\displaystyle -3\).Six \(\displaystyle CN^-\) give \(\displaystyle 6 \times (-1) = -6\).\[x + (-6) = -3 \] \[x = -3 + 6 = +3 \]Oxidation number of Fe \(\displaystyle = +3\).(v) \(\displaystyle [Cr(NH_3)_3Cl_3]\)No ions are written outside the bracket, so the whole entity is neutral: overall charge \(\displaystyle = 0\).Three \(\displaystyle NH_3\) give \(\displaystyle 0\); three \(\displaystyle Cl^-\) give \(\displaystyle 3 \times (-1) = -3\).\[x + 0 + (-3) = 0 \] \[x = +3 \]Oxidation number of Cr \(\displaystyle = +3\).Answer: (i) Co = +$\displaystyle 3$, (ii) Co = +$\displaystyle 3$, (iii) Pt = +$\displaystyle 2$, (iv) Fe = +$\displaystyle 3$, (v) Cr = +$\displaystyle 3$

Coordination compounds show two fundamentally different kinds of isomerism: structural isomerism, where the atoms are connected differently (a different bond is broken/made), and stereoisomerism, where the connectivity is identical and only the spatial arrangement around the metal differs.A. Structural isomerism1. Linkage isomerism — occurs when an ambidentate ligand (one with two different donor atoms) attaches to the metal through either atom. The nitrite ion \(\displaystyle NO_2^- \) can bond through N (nitro) or through O (nitrito): \[[Co(NH_3)_5(NO_2)]^{2+} \quad \text{(nitro, Co–N bond)} \] \[[Co(NH_3)_5(ONO)]^{2+} \quad \text{(nitrito, Co–O bond)} \] The molecular formula is identical in both — only which atom of \(\displaystyle NO_2^- \) touches the metal changes. \(\displaystyle SCN^- \) (thiocyanato/isothiocyanato) is another common example.2. Coordination isomerism — occurs in salts built from two complex ions (one cationic, one anionic), where the ligands swap between the two metal centres: \[[Co(NH_3)_6][Cr(CN)_6] \quad \rightleftharpoons \quad [Cr(NH_3)_6][Co(CN)_6] \] Both compounds have the same overall formula, but which metal holds the ammines and which holds the cyanides is reversed.3. Ionisation isomerism — occurs when the counter ion sitting outside the coordination sphere can itself act as a ligand, so it trades places with a ligand already inside the sphere: \[[Co(NH_3)_5Br]SO_4 \quad \text{(sulphate is the counter ion, Br}^- \text{ is coordinated)} \] \[[Co(NH_3)_5SO_4]Br \quad \text{(bromide is the counter ion, } SO_4^{2-} \text{ is coordinated)} \] The aside worth remembering: these two give different ions in solution — the first gives a precipitate with \(\displaystyle BaCl_2 \) (free \(\displaystyle SO_4^{2-} \)) but not with \(\displaystyle AgNO_3 \); the second does the opposite. That test is exactly how you'd tell them apart experimentally.4. Solvate (hydrate) isomerism — occurs when a solvent molecule (usually water) is either coordinated directly to the metal or sitting free in the crystal lattice: \[[Cr(H_2O)_6]Cl_3 \quad \text{(violet)} \] \[[Cr(H_2O)_5Cl]Cl_2\cdot H_2O \quad \text{(grey-green)} \] \[[Cr(H_2O)_4Cl_2]Cl\cdot 2H_2O \quad \text{(dark green)} \] All three have the formula \(\displaystyle CrCl_3\cdot 6H_2O \), but the number of \(\displaystyle H_2O \) molecules actually bonded to \(\displaystyle Cr^{3+} \) — versus merely present as water of crystallisation — is different in each, which is also why they are different colours.B. Stereoisomerism5. Geometrical (cis–trans) isomerism — arises in square planar (coordination number $\displaystyle 4$) and octahedral (coordination number $\displaystyle 6$) complexes, where two identical ligands can sit adjacent to each other (cis) or opposite each other (trans). Square planar example: \[[Pt(NH_3)_2Cl_2] \;\;\to\;\; \text{cis (Cl's adjacent) and trans (Cl's opposite)} \] Octahedral example: \(\displaystyle [Co(NH_3)_4Cl_2]^+ \) also exists as cis and trans forms. This is a connectivity-preserving isomerism — every bond is the same Co–N or Co–Cl bond in both isomers, only the geometry around the metal changes, which is the detail people conflate with structural isomerism.6. Optical isomerism — arises when a complex is chiral, i.e., its mirror image cannot be superimposed on the original. This is common in octahedral complexes with three symmetric bidentate ligands: \[[Co(en)_3]^{3+} \quad \text{(en = ethylenediamine)} \] This ion exists as a pair of non-superimposable mirror images, the \(\displaystyle d \)- and \(\displaystyle l \)-isomers (enantiomers), which rotate plane-polarised light in opposite directions but are otherwise chemically identical.Answer: Coordination compounds show six recognised types of isomerism — structural: ($\displaystyle 1$) linkage (e.g., \(\displaystyle [Co(NH_3)_5(NO_2)]^{2+} \) vs \(\displaystyle [Co(NH_3)_5(ONO)]^{2+} \)), ($\displaystyle 2$) coordination (e.g., \(\displaystyle [Co(NH_3)_6][Cr(CN)_6] \) vs \(\displaystyle [Cr(NH_3)_6][Co(CN)_6] \)), ($\displaystyle 3$) ionisation (e.g., \(\displaystyle [Co(NH_3)_5Br]SO_4 \) vs \(\displaystyle [Co(NH_3)_5SO_4]Br \)), ($\displaystyle 4$) solvate/hydrate (e.g., the three hydrates of \(\displaystyle CrCl_3\cdot 6H_2O \)) — and stereoisomerism: ($\displaystyle 5$) geometrical/cis-trans (e.g., cis- and trans-\(\displaystyle [Pt(NH_3)_2Cl_2] \)) and ($\displaystyle 6$) optical (e.g., \(\displaystyle d \)- and \(\displaystyle l \)-\(\displaystyle [Co(en)_3]^{3+} \)).

Geometrical (cis–trans) isomerism needs two different kinds of ligand positions to swap — a symmetric tris-chelate has no such choice, while an \(\displaystyle MA_3B_3\) octahedral complex does.(i) \(\displaystyle [Cr(C_2O_4)_3]^{3-}\)The oxalate ion \(\displaystyle C_2O_4^{2-}\) is a symmetric bidentate ligand — both donor oxygen atoms on each oxalate are equivalent, so swapping the two ends of one oxalate group gives back the same structure. In this complex the chromium is octahedrally surrounded by three such oxalate rings, each occupying two adjacent (cis) coordination positions.Because a chelating ring can only span cis positions (its two ends are too close to reach across the octahedron to a trans position), there is no alternative "trans" arrangement to compare against — every way of placing three identical symmetric bidentate ligands around the octahedron gives the same connectivity. So there is no geometrical isomerism possible here.This complex is not isomer-free, though — it is chiral. The whole \(\displaystyle [M(AA)_3]\) framework (like \(\displaystyle [Co(en)_3]^{3+}\)) exists as a pair of non-superimposable mirror images, so \(\displaystyle [Cr(C_2O_4)_3]^{3-}\) shows optical isomerism (\(\displaystyle d\) and \(\displaystyle l\) forms), just not geometrical isomerism. Do not confuse the two: optical isomers differ only in chirality, geometrical isomers differ in which positions are cis or trans to each other.Number of geometrical isomers of \(\displaystyle [Cr(C_2O_4)_3]^{3-}\): $\displaystyle 0$(ii) \(\displaystyle [Co(NH_3)_3Cl_3]\)This is an octahedral complex of the type \(\displaystyle MA_3B_3\), with three \(\displaystyle NH_3\) and three \(\displaystyle Cl^-\) as monodentate ligands (unlike part (i), these are six independent, unlinked ligands, so all six octahedral vertices are individually swappable). Label the six octahedral positions as three pairs of trans sites. The three identical ligands of one kind (say, the three \(\displaystyle Cl^-\)) can be arranged in exactly two distinct ways: These two arrangements are genuinely different geometrical (positional) isomers — they cannot be interconverted without breaking and re-forming bonds — and no further distinct arrangement exists for three identical ligands of each kind on an octahedron.Number of geometrical isomers of \(\displaystyle [Co(NH_3)_3Cl_3]\): $\displaystyle 2$ (fac and mer)Answer: (i) \(\displaystyle [Cr(C_2O_4)_3]^{3-}\) shows no geometrical isomers ($\displaystyle 0$) — only optical isomerism (\(\displaystyle d\), \(\displaystyle l\)); (ii) \(\displaystyle [Co(NH_3)_3Cl_3]\) shows $\displaystyle 2$ geometrical isomers — facial (fac) and meridional (mer).

A structure and its mirror image are optical isomers (enantiomers) only when they cannot be superimposed on each other — which happens exactly when the complex has no plane of symmetry, no centre of symmetry, and no improper rotation axis. Geometrical (cis/trans) isomerism has to be sorted out first in parts (ii) and (iii), because in those two complexes it is only the cis form that turns out to be chiral.(i) \(\displaystyle \left[Cr(C_2O_4)_3\right]^{3-}\)Oxalate, \(\displaystyle C_2O_4^{2-}\), is a bidentate ligand, so this ion is an octahedral tris-chelate complex — the oxidation state check confirms Cr is +$\displaystyle 3$: \[x + 3(-2) = -3 \;\Rightarrow\; x = +3 \]With three identical bidentate ligands each spanning two mutually cis coordination sites, the complex is isostructural with \(\displaystyle \left[Co(en)_3\right]^{3+}\): looking straight down the pseudo-three-fold axis, the six oxygen donors resolve into two triangles of three, staggered $\displaystyle 60$° from each other (a trigonal-antiprism projection of the octahedron), and each O–Cr–O chelate ring runs from a vertex of the near triangle to an adjacent vertex of the far triangle. Traced this way, the three rings wind around the axis like the blades of a propeller.There are exactly two ways to make that winding go — clockwise or anticlockwise as you trace it from the near triangle to the far one — and these two windings are mirror images of each other that cannot be rotated into coincidence, because the molecule has no \(\displaystyle \sigma\), no \(\displaystyle i\), and no \(\displaystyle S_n\). They are labelled \(\displaystyle \Delta\) (right-handed twist) and \(\displaystyle \Lambda\) (left-handed twist). This is the same Δ/Λ pair a student would draw for \(\displaystyle \left[Co(en)_3\right]^{3+}\), with oxalate's O–C–C–O backbone standing in for en's N–C–C–N backbone.(ii) \(\displaystyle \left[PtCl_2(en)_2\right]^{2+}\)Two Cl⁻ plus two neutral bidentate en ligands give an octahedral Pt(IV) centre: \[x + 2(-1) + 2(0) = +2 \;\Rightarrow\; x = +4 \]With the two en ligands together occupying four of the six sites, the two Cl atoms can sit either opposite each other (trans) or next to each other (cis) — a geometrical-isomer choice that has to be made before optical isomerism can even be asked about.In trans-\(\displaystyle \left[PtCl_2(en)_2\right]^{2+}\), the two Cl occupy the axial positions and the two en rings sit in the equatorial girdle, related to each other by the axial \(\displaystyle C_2\). Reflecting through the equatorial plane swaps top-Cl with bottom-Cl and maps each en ring onto itself — the molecule is superimposable on its own mirror image (it has a \(\displaystyle \sigma_h\)). Trans-\(\displaystyle \left[PtCl_2(en)_2\right]^{2+}\) is achiral: it has no optical isomer.In cis-\(\displaystyle \left[PtCl_2(en)_2\right]^{2+}\), the two Cl are adjacent, and no plane can be drawn through the ion that reflects it onto itself — one en ring is always left sticking out on the "wrong" side of any candidate mirror plane. With only a bare \(\displaystyle C_2\) axis and no \(\displaystyle \sigma\), cis-\(\displaystyle \left[PtCl_2(en)_2\right]^{2+}\) is chiral, and it is this cis form — not the trans form — that exists as a non-superimposable mirror-image pair, again conventionally labelled \(\displaystyle \Delta\) and \(\displaystyle \Lambda\).(iii) \(\displaystyle \left[Cr(NH_3)_2Cl_2(en)\right]^{+}\)Here en occupies two cis sites (its short "bite" physically cannot reach across a $\displaystyle 180$° pair of trans sites), leaving two NH₃ and two Cl to fill the remaining four positions. Oxidation state: \[x + 2(0) + 2(-1) + 1(0) = +1 \;\Rightarrow\; x = +3 \]Again the two Cl can end up trans or cis to each other, and again that choice decides whether an optical isomer exists at all.Trans-\(\displaystyle \left[Cr(NH_3)_2Cl_2(en)\right]^{+}\) (the two Cl opposite each other) has a mirror plane running through both Cl atoms, the Cr, and the midpoint of the en ring — reflection in that plane swaps the two NH₃ groups with each other and leaves everything else fixed, so the molecule maps onto itself. It is achiral: no optical isomer.Cis-\(\displaystyle \left[Cr(NH_3)_2Cl_2(en)\right]^{+}\) (the two Cl adjacent, which also forces the two NH₃ to be adjacent to each other) has four different "directions" around the Cr with no internal symmetry left to exploit — there is no plane, centre, or improper axis that maps the ion onto its mirror image. This lower-symmetry cis form is chiral and is the one that exists as a pair of non-superimposable enantiomers; the trans form does not.Answer: (i) \(\displaystyle \left[Cr(C_2O_4)_3\right]^{3-}\) is chiral and exists as a non-superimposable mirror-image pair, the right-handed (\(\displaystyle \Delta\)) and left-handed (\(\displaystyle \Lambda\)) propeller forms, exactly as in \(\displaystyle \left[Co(en)_3\right]^{3+}\). (ii) Of the two geometrical isomers of \(\displaystyle \left[PtCl_2(en)_2\right]^{2+}\), only cis-\(\displaystyle \left[PtCl_2(en)_2\right]^{2+}\) is chiral and gives a \(\displaystyle \Delta/\Lambda\) enantiomer pair; trans-\(\displaystyle \left[PtCl_2(en)_2\right]^{2+}\) has a mirror plane and is optically inactive. (iii) Of the two geometrical isomers of \(\displaystyle \left[Cr(NH_3)_2Cl_2(en)\right]^{+}\), only cis-\(\displaystyle \left[Cr(NH_3)_2Cl_2(en)\right]^{+}\) is chiral and gives a pair of enantiomers; trans-\(\displaystyle \left[Cr(NH_3)_2Cl_2(en)\right]^{+}\) has a mirror plane and is optically inactive.