NCERT Solutions · Class 12 Chemistry Coordination Compounds

Color in coordination compounds comes from d–d transitions, and the energy gap for that transition depends on the ligand's field strength — not just on which metal ion is present.Both complexes contain the same metal ion in the same oxidation state:\[\text{Fe in } [\text{Fe(CN)}_6]^{4-}: \quad +2 \text{ (since CN}^- \text{ is } -1 \text{ each, and } 6(-1) + x = -4 \Rightarrow x = +2) \] \[\text{Fe in } [\text{Fe(H}_2\text{O})_6]^{2+}: \quad +2 \text{ (H}_2\text{O is neutral, so the charge on the ion is the oxidation state)} \]So in both cases the central ion is \(\displaystyle \text{Fe}^{2+} \), a \(\displaystyle d^6 \) system. If the metal and its \(\displaystyle d \)-electron count were all that mattered, the two complexes would look the same — but they don't, because the ligand decides how those six \(\displaystyle d \)-electrons are arranged.Step $\displaystyle 1$: Where the electrons sit depends on the ligand's position in the spectrochemical series.The spectrochemical series ranks ligands by how strongly they split the metal's \(\displaystyle d \)-orbitals into the lower \(\displaystyle t_{2g} \) set and the higher \(\displaystyle e_g \) set (in an octahedral field). From this series:\[\text{CN}^- \; (\text{strong field}) \;>\; \text{H}_2\text{O} \; (\text{weak field}) \] \[t_{2g}^{6}\, e_g^{0} \] \[t_{2g}^{4}\, e_g^{2} \]Step $\displaystyle 2$: Color is set by the size of \(\displaystyle \Delta_o \), because that is the energy an electron absorbs to jump from \(\displaystyle t_{2g} \) to \(\displaystyle e_g \).A colored complex absorbs visible light of energy \(\displaystyle E = h\nu = \Delta_o \) to promote an electron from the \(\displaystyle t_{2g} \) level to the \(\displaystyle e_g \) level; the color we see is the complementary color of the light absorbed. Since:\[\Delta_o \, [\text{Fe(CN)}_6]^{4-} \;\; (\text{strong field, CN}^-) \;\; \gg \;\; \Delta_o \, [\text{Fe(H}_2\text{O})_6]^{2+} \;\; (\text{weak field, H}_2\text{O}) \]the two complexes absorb light of very different energy (hence very different wavelength) for their \(\displaystyle d\text{-}d \) transitions. Different absorbed wavelength means different transmitted/complementary wavelength — so the two complexes show different colors, even though the metal ion and its electron count are identical.The trap here is assuming color is a fixed property of the metal ion. It is not — it is a property of the metal–ligand combination, because \(\displaystyle \Delta_o \), not the raw \(\displaystyle d^n \) count, sets the absorbed energy.Answer: Both ions contain \(\displaystyle \text{Fe}^{2+} \) (\(\displaystyle d^6 \)), but \(\displaystyle \text{CN}^- \) is a strong-field ligand giving a large \(\displaystyle \Delta_o \) and a low-spin \(\displaystyle t_{2g}^6 e_g^0 \) arrangement, while \(\displaystyle \text{H}_2\text{O} \) is a weak-field ligand giving a small \(\displaystyle \Delta_o \) and a high-spin \(\displaystyle t_{2g}^4 e_g^2 \) arrangement. Since the color of a complex is decided by the energy \(\displaystyle \Delta_o \) of the \(\displaystyle d\text{-}d \) transition (the light absorbed), and \(\displaystyle \Delta_o \) differs greatly between the two complexes, \(\displaystyle [\text{Fe(CN)}_6]^{4-} \) and \(\displaystyle [\text{Fe(H}_2\text{O})_6]^{2+} \) absorb light of different wavelengths and therefore show different colors.

The M–C bond in a metal carbonyl is not a one-way dative bond from carbon to the metal — it is synergic, built from two bonds pointing in opposite directions that reinforce each other.1. The σ bond (ligand → metal). Carbon monoxide is \(\displaystyle :C\!\equiv\!O: \). The carbon atom carries a lone pair sitting in a hybrid orbital that points away from the oxygen. This lone pair is donated into a vacant hybrid orbital on the metal atom, forming a σ bond: \[M \;\leftarrow\; C\!\equiv\!O \] Taken alone, this is exactly the kind of donor–acceptor (coordinate) bond you see in any Werner complex, with CO acting as a two-electron donor through carbon.2. The π bond (metal → ligand), i.e. back bonding. Metal carbonyls form with the metal in a very low, often zero, oxidation state (as in \(\displaystyle \mathrm{Ni(CO)_4} \), \(\displaystyle \mathrm{Fe(CO)_5} \), \(\displaystyle \mathrm{Cr(CO)_6} \)), so the metal atom is electron-rich and holds filled \(\displaystyle d \)-orbitals of \(\displaystyle t_{2g} \) symmetry. These d-orbitals are correctly oriented to overlap sideways with the empty antibonding \(\displaystyle \pi^{*} \) orbitals of CO. Electron density flows the other way, from metal to ligand: \[M \;\rightarrow\; CO(\pi^{*}) \] This donation from a filled metal d-orbital into an empty ligand π\ orbital is called π back bonding, and it can only happen because CO has a low-lying empty π\ level able to accept the density.Why "synergic," not just "two bonds." The σ-donation from CO piles extra negative charge onto the metal. That build-up makes the metal even more willing to push density back out through π back-donation — and because draining density out of CO's π\* orbital relieves the electron crowding that back-donation itself would otherwise cause, each bond makes the other more favorable. Together they give far more stabilization than either would alone. Students often picture only the σ-donation step and stop there — that misses the whole reason CO is such an effective ligand specifically toward metals in low oxidation states, where there are plenty of d-electrons available to back-donate.The experimental signature. Populating an antibonding orbital always weakens the bond it belongs to, so the more the metal back-donates into CO's π\, the weaker and longer the C–O bond becomes, while the M–C bond simultaneously gains partial double-bond character and becomes shorter and stronger. This shows up directly in IR spectroscopy: free gaseous CO stretches at about \(\displaystyle 2143\ \mathrm{cm^{-1}} \), while the C–O stretch in a neutral metal carbonyl such as \(\displaystyle \mathrm{Ni(CO)_4} \) is lowered to around \(\displaystyle 2060\ \mathrm{cm^{-1}} \) — a direct measure of how much π\ population (back donation) has occurred. The more electron-rich the metal (lower oxidation state, more electropositive metal), the greater the back donation, the lower the C–O stretching frequency, and the shorter/stronger the M–C bond becomes.**Answer: Bonding in a metal carbonyl is synergic — a σ bond in which carbon's lone pair is donated into an empty metal orbital, reinforced by a π bond in which filled metal \(\displaystyle t_{2g} \) d-orbitals back-donate into the empty π\* orbitals of CO. This back donation strengthens the M–C bond (partial double-bond character, shorter bond) and weakens the C–O bond (lower bond order, longer bond, lower IR stretching frequency than free CO's \(\displaystyle 2143\ \mathrm{cm^{-1}}\)).**

The oxidation state of the central metal comes from charge balance: whatever ion sits outside the square bracket fixes the charge on the complex ion, and inside the bracket, (oxidation state of metal) + (sum of the ligand charges) must equal that complex-ion charge.(i) \(\displaystyle K_3[Co(C_2O_4)_3]\)Three \(\displaystyle K^+\) ions sit outside the bracket, contributing \(\displaystyle +3\) total. For the compound to be neutral, the complex ion \(\displaystyle [Co(C_2O_4)_3]^{3-}\) must carry \(\displaystyle -3\).Oxalate, \(\displaystyle C_2O_4^{2-}\), binds through two oxygen atoms (bidentate) and carries a charge of \(\displaystyle -2\). With \(\displaystyle x\) as the oxidation state of cobalt: \[x + 3(-2) = -3 \implies x = +3 \]Neutral cobalt (\(\displaystyle Z=27\)) is \(\displaystyle [Ar]3d^{7}4s^{2}\). When a metal forms a cation, electrons leave the 4s orbital before any leave 3d — the reverse of the order they were filled in. Removing the two 4s electrons and one 3d electron gives \(\displaystyle Co^{3+}: [Ar]3d^{6}\).Because oxalate is bidentate, each of the three ligands occupies two coordination positions, so the coordination number is \(\displaystyle 3 \times 2 = 6\).\(\displaystyle [Co(C_2O_4)_3]^{3-}\): oxidation state \(\displaystyle +3\), configuration \(\displaystyle 3d^{6}\), coordination number \(\displaystyle 6\).(ii) cis-\(\displaystyle [CrCl_2(en)_2]Cl\)Only one \(\displaystyle Cl^-\) sits outside the bracket, contributing \(\displaystyle -1\), so the complex ion \(\displaystyle [CrCl_2(en)_2]^{+}\) must carry \(\displaystyle +1\).Inside the bracket, each chloride ligand carries \(\displaystyle -1\), and ethylenediamine (en) is a neutral bidentate ligand (charge \(\displaystyle 0\)). With \(\displaystyle x\) as the oxidation state of chromium: \[x + 2(-1) + 2(0) = +1 \implies x = +3 \]Neutral chromium (\(\displaystyle Z=24\)) has the configuration \(\displaystyle [Ar]3d^{5}4s^{1}\) (a half-filled 3d is more stable than \(\displaystyle 3d^44s^2\), so chromium "borrows" a 4s electron even in the neutral atom). Removing the one 4s electron and two 3d electrons gives \(\displaystyle Cr^{3+}: [Ar]3d^{3}\).For coordination number: the $\displaystyle 2$ chloride ligands (monodentate) contribute \(\displaystyle 2\), and the $\displaystyle 2$ en ligands (bidentate, two N donors each) contribute \(\displaystyle 2 \times 2 = 4\). Total \(\displaystyle = 2 + 4 = 6\).\(\displaystyle [CrCl_2(en)_2]^{+}\): oxidation state \(\displaystyle +3\), configuration \(\displaystyle 3d^{3}\), coordination number \(\displaystyle 6\).(iii) \(\displaystyle (NH_4)_2[CoF_4]\)Two \(\displaystyle NH_4^+\) ions outside the bracket contribute \(\displaystyle +2\), so the complex ion \(\displaystyle [CoF_4]^{2-}\) must carry \(\displaystyle -2\).Fluoride is monodentate, charge \(\displaystyle -1\). With \(\displaystyle x\) as the oxidation state of cobalt: \[x + 4(-1) = -2 \implies x = +2 \]From \(\displaystyle Co: [Ar]3d^{7}4s^{2}\), forming \(\displaystyle \mathrm{Co^{2+}}\) removes only the two 4s electrons — no 3d electron is touched this time, since two electrons are all that need to leave. So \(\displaystyle Co^{2+}: [Ar]3d^{7}\).Four fluoride ligands, each monodentate, give a coordination number of \(\displaystyle 4\).\(\displaystyle [CoF_4]^{2-}\): oxidation state \(\displaystyle +2\), configuration \(\displaystyle 3d^{7}\), coordination number \(\displaystyle 4\).(iv) \(\displaystyle [Mn(H_2O)_6]SO_4\)Sulfate, \(\displaystyle SO_4^{2-}\), sits outside the bracket carrying \(\displaystyle -2\), so the complex ion \(\displaystyle [Mn(H_2O)_6]^{2+}\) must carry \(\displaystyle +2\).Water is a neutral ligand, so it contributes no charge and the entire \(\displaystyle +2\) sits on manganese: \(\displaystyle x = +2\).Neutral manganese (\(\displaystyle Z=25\)) is \(\displaystyle [Ar]3d^{5}4s^{2}\). Removing the two 4s electrons gives \(\displaystyle Mn^{2+}: [Ar]3d^{5}\).Six water molecules, each monodentate, give a coordination number of \(\displaystyle 6\).\(\displaystyle [Mn(H_2O)_6]^{2+}\): oxidation state \(\displaystyle +2\), configuration \(\displaystyle 3d^{5}\), coordination number \(\displaystyle 6\).Answer: (i) \(\displaystyle \mathrm{Co^{3+}}\), \(\displaystyle 3d^{6}\), CN = $\displaystyle 6$ (ii) \(\displaystyle \mathrm{Cr^{3+}}\), \(\displaystyle 3d^{3}\), CN = $\displaystyle 6$ (iii) \(\displaystyle \mathrm{Co^{2+}}\), \(\displaystyle 3d^{7}\), CN = $\displaystyle 4$ (iv) \(\displaystyle \mathrm{Mn^{2+}}\), \(\displaystyle 3d^{5}\), CN = $\displaystyle 6$

Every part of this question reduces to one balance: the sum of the oxidation states and ligand charges inside the square brackets must equal the charge carried by the complex ion, which is fixed by whatever ions sit outside it.For a formula (counter-ion)\(\displaystyle _a\)[M(ligand)\(\displaystyle _n\)](counter-ion)\(\displaystyle _b\), the whole compound is neutral, so the complex ion's charge is exactly opposite to the total charge of the counter-ions. Once you know the metal's oxidation state, remove that many electrons from the free-atom configuration — the \(\displaystyle ns\) electrons first, then \(\displaystyle (n-1)d\) — to get the \(\displaystyle d\)-electron count. The number of unpaired electrons in that count then gives the magnetic moment through the spin-only formula \[\mu = \sqrt{n(n+2)} \ \text{BM} \] where \(\displaystyle n\) is the number of unpaired electrons.The step people get wrong: whether those \(\displaystyle d\)-electrons pair up depends on the ligand's field strength. Strong-field ligands (\(\displaystyle NH_3\), \(\displaystyle CN^{-}\)) force pairing into the lower \(\displaystyle t_{2g}\) set before any electron reaches \(\displaystyle e_g\) (low spin); weak-to-medium field ligands (\(\displaystyle H_2O\), \(\displaystyle \mathrm{Cl^{-}}\), oxalato, pyridine) generally don't (high spin). For \(\displaystyle d^3\) it makes no difference — three electrons singly fill three \(\displaystyle t_{2g}\) orbitals either way — but for \(\displaystyle d^5\) and \(\displaystyle d^6\) it changes the magnetic moment completely.(i) \(\displaystyle K[Cr(H_2O)_2(C_2O_4)_2]\cdot 3H_2O\)The complex ion \(\displaystyle [Cr(H_2O)_2(C_2O_4)_2]^{-}\) must carry charge \(\displaystyle -1\) to balance the one \(\displaystyle K^+\) outside it (the water of crystallisation carries no charge). Aqua is neutral; oxalato (\(\displaystyle C_2O_4^{2-}\)) is bidentate and contributes \(\displaystyle -2\) each: \[x + 2(0) + 2(-2) = -1 \ \Rightarrow\ x = +3 \] Chromium is in the +$\displaystyle 3$ oxidation state.Cr (Z = $\displaystyle 24$) has ground-state configuration \(\displaystyle [Ar]3d^54s^1\). Removing $\displaystyle 3$ electrons for \(\displaystyle \mathrm{Cr^{3+}}\) (the \(\displaystyle 4s\) electron first, then two from \(\displaystyle 3d\)) gives \(\displaystyle [Ar]3d^3\), i.e. \(\displaystyle t_{2g}^3e_g^0\).Coordination number: $\displaystyle 2$ aqua ligands (monodentate, $\displaystyle 1$ donor atom each) + $\displaystyle 2$ oxalato ligands (bidentate, $\displaystyle 2$ donor atoms each) \(\displaystyle = 2+4 = 6\). Stereochemistry: octahedral.Three electrons occupy three separate \(\displaystyle t_{2g}\) orbitals (Hund's rule) whatever the field strength, so \(\displaystyle n=3\): \[\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ \text{BM} \] — paramagnetic.IUPAC name: potassium diaquadioxalatochromate(III) trihydrate.(ii) \(\displaystyle [Co(NH_3)_5Cl]Cl_2\)The complex ion \(\displaystyle [Co(NH_3)_5Cl]^{2+}\) must be \(\displaystyle +2\) to balance the two chloride ions outside: \[x + 5(0) + (-1) = +2 \ \Rightarrow\ x = +3 \] Cobalt is in the +$\displaystyle 3$ oxidation state.Co (Z = $\displaystyle 27$): \(\displaystyle [Ar]3d^74s^2\). Removing $\displaystyle 3$ electrons for \(\displaystyle \mathrm{Co^{3+}}\) gives \(\displaystyle [Ar]3d^6\).Coordination number: \(\displaystyle 5(NH_3) + 1(Cl) = 6\), octahedral.\(\displaystyle NH_3\) is a strong-field ligand, so all six \(\displaystyle d\) electrons pack into the lower set before touching \(\displaystyle e_g\): \(\displaystyle t_{2g}^6e_g^0\), leaving \(\displaystyle n=0\) unpaired electrons. \[\mu = \sqrt{0(0+2)} = 0 \ \text{BM} \] — diamagnetic. (This is exactly the case where field strength changes the outcome: the same \(\displaystyle d^6\) count with weak-field ligands would give $\displaystyle 4$ unpaired electrons instead of 0.)IUPAC name: pentaamminechloridocobalt(III) chloride.(iii) \(\displaystyle [CrCl_3(py)_3]\)This complex carries no ions outside it, so it is neutral overall: \[x + 3(-1) + 3(0) = 0 \ \Rightarrow\ x = +3 \] Chromium is in the +$\displaystyle 3$ oxidation state — the same as part (i), so the electronic configuration is again \(\displaystyle [Ar]3d^3\) (\(\displaystyle t_{2g}^3e_g^0\)).Coordination number: \(\displaystyle 3(Cl) + 3(py) = 6\), octahedral.As in (i), \(\displaystyle d^3\) gives $\displaystyle 3$ unpaired electrons regardless of field strength: \[\mu = \sqrt{3\times5} = \sqrt{15} \approx 3.87 \ \text{BM} \] — paramagnetic.IUPAC name: trichloridotripyridinechromium(III).(iv) \(\displaystyle Cs[FeCl_4]\)The complex ion \(\displaystyle [FeCl_4]^{-}\) balances the single \(\displaystyle Cs^+\): \[x + 4(-1) = -1 \ \Rightarrow\ x = +3 \] Iron is in the +$\displaystyle 3$ oxidation state.Fe (Z = $\displaystyle 26$): \(\displaystyle [Ar]3d^64s^2\). Removing $\displaystyle 3$ electrons for \(\displaystyle \mathrm{Fe^{3+}}\) gives \(\displaystyle [Ar]3d^5\).Coordination number: $\displaystyle 4$ — only four chloride ions are bound. With a coordination number this low around a first-row \(\displaystyle M^{3+}\) ion, the geometry is tetrahedral, not octahedral (the ligand-field splitting in a tetrahedral field is too small to favour anything else with a weak-field halide).Tetrahedral splitting is small, so \(\displaystyle \mathrm{Cl^{-}}\) (already weak-field) gives high spin: all five \(\displaystyle d\) orbitals singly occupied, \(\displaystyle n=5\): \[\mu = \sqrt{5\times7} = \sqrt{35} \approx 5.92 \ \text{BM} \] — strongly paramagnetic.IUPAC name: caesium tetrachloridoferrate(III).(v) \(\displaystyle K_4[Mn(CN)_6]\)The complex ion \(\displaystyle [Mn(CN)_6]^{4-}\) balances four \(\displaystyle K^+\) ions: \[x + 6(-1) = -4 \ \Rightarrow\ x = +2 \] Manganese is in the +$\displaystyle 2$ oxidation state.Mn (Z = $\displaystyle 25$): \(\displaystyle [Ar]3d^54s^2\). Removing $\displaystyle 2$ electrons for \(\displaystyle \mathrm{Mn^{2+}}\) (both from \(\displaystyle 4s\)) gives \(\displaystyle [Ar]3d^5\) — the same \(\displaystyle d^5\) count as iron in part (iv), but here the ligand is different.Coordination number: $\displaystyle 6$, octahedral.\(\displaystyle \mathrm{CN^{-}}\) is a strong-field ligand, so all five electrons pack into the three \(\displaystyle t_{2g}\) orbitals before any reach \(\displaystyle e_g\): \(\displaystyle t_{2g}^5e_g^0\), leaving only \(\displaystyle n=1\) unpaired electron. (Compare this directly with the same \(\displaystyle d^5\) count in part (iv), which was high-spin with \(\displaystyle n=5\) because \(\displaystyle Cl^-\) is weak-field and the geometry was tetrahedral — same electron count, opposite spin state.) \[\mu = \sqrt{1\times3} = \sqrt{3} \approx 1.73 \ \text{BM} \] — weakly paramagnetic.IUPAC name: potassium hexacyanidomanganate(II).Answer: (i) potassium diaquadioxalatochromate(III) trihydrate — Cr(+$\displaystyle 3$), \(\displaystyle 3d^3\), CN $\displaystyle 6$, octahedral, \(\displaystyle \mu\approx3.87\) BM; (ii) pentaamminechloridocobalt(III) chloride — Co(+$\displaystyle 3$), \(\displaystyle 3d^6\) low spin, CN $\displaystyle 6$, octahedral, \(\displaystyle \mu=0\) BM; (iii) trichloridotripyridinechromium(III) — Cr(+$\displaystyle 3$), \(\displaystyle 3d^3\), CN $\displaystyle 6$, octahedral, \(\displaystyle \mu\approx3.87\) BM; (iv) caesium tetrachloridoferrate(III) — Fe(+$\displaystyle 3$), \(\displaystyle 3d^5\) high spin, CN $\displaystyle 4$, tetrahedral, \(\displaystyle \mu\approx5.92\) BM; (v) potassium hexacyanidomanganate(II) — Mn(+$\displaystyle 2$), \(\displaystyle 3d^5\) low spin, CN $\displaystyle 6$, octahedral, \(\displaystyle \mu\approx1.73\) BM.

Crystal field splitting means the complex absorbs one colour of light and we see its complementary colour, not "no colour."Titanium has the configuration \(\displaystyle [Ar]\,3d^{2}4s^{2}\). To form \(\displaystyle \mathrm{Ti^{3+}}\), three electrons are removed — the two \(\displaystyle 4s\) electrons first, then one \(\displaystyle 3d\) electron — leaving \(\displaystyle \mathrm{Ti^{3+}}\) as \(\displaystyle [Ar]\,3d^{1}\): a single electron in the d subshell.In the free \(\displaystyle \mathrm{Ti^{3+}}\) ion, all five 3d orbitals \(\displaystyle (d_{xy}, d_{yz}, d_{xz}, d_{x^{2}-y^{2}}, d_{z^{2}})\) are degenerate (identical energy) because there is nothing around the ion to disturb them.In \(\displaystyle [Ti(H_2O)_6]^{3+}\), six \(\displaystyle H_2O\) ligands sit along the \(\displaystyle +x,-x,+y,-y,+z,-z\) axes (octahedral geometry). Their electron pairs repel electrons in the metal d orbitals, but not equally: Aside — this is the step people gloss over: it isn't the presence of ligands that splits the orbitals, it is their direction of approach. A perfectly spherical field of the same charge would lower all five orbitals equally and split nothing.The energy gap between the two sets is the crystal field splitting energy, \(\displaystyle \Delta_o\) (also written \(\displaystyle 10Dq\)). In the ground state the lone 3d electron goes into the lower-energy set, giving the configuration \(\displaystyle t_{2g}^{1}e_g^{0}\).When white light (which contains every visible wavelength at once) falls on an aqueous solution of \(\displaystyle [Ti(H_2O)_6]^{3+}\), the complex can absorb exactly the photon energy that matches \(\displaystyle \Delta_o\), promoting the electron from \(\displaystyle t_{2g}\) to \(\displaystyle e_g\):\[t_{2g}^{1}e_g^{0} \;\xrightarrow{\;h\nu = \Delta_o\;}\; t_{2g}^{0}e_g^{1} \]This single-electron jump between crystal-field-split d orbitals is called a d–d transition. For \(\displaystyle [Ti(H_2O)_6]^{3+}\) the absorbed energy corresponds to a wavenumber of about \(\displaystyle 20{,}300\ \text{cm}^{-1}\), which lies in the yellow-green region of the visible spectrum (roughly $\displaystyle 490$–$\displaystyle 500$ nm).Aside — the other easy mistake: the solution does not simply "lose" the absorbed light and go dark. Every wavelength except yellow-green passes through untouched; what reaches the eye is white light with the yellow-green component removed.Removing yellow-green from white light leaves behind its complementary colour on the colour wheel, which is violet. That transmitted violet light is what the eye detects, so the aqueous \(\displaystyle [Ti(H_2O)_6]^{3+}\) ion looks violet.(A further consistency check: because \(\displaystyle \mathrm{Ti^{3+}}\) has only one d electron, only one d–d transition — \(\displaystyle t_{2g}\to e_g\) — is possible, so the absorption spectrum of this ion shows a single broad band, matching what is observed experimentally.)Answer: The single 3d electron of \(\displaystyle \mathrm{Ti^{3+}}\) sits in the lower \(\displaystyle t_{2g}\) set; it absorbs yellow-green light (\(\displaystyle \Delta_o \approx 20{,}300\ \text{cm}^{-1}\)) to jump to the higher \(\displaystyle e_g\) set (a d–d transition). Since yellow-green is removed from the incident white light, the transmitted light is its complementary colour, violet — which is the colour observed for \(\displaystyle [Ti(H_2O)_6]^{3+}\).

A chelating ligand does not just sit at one point on the metal — it wraps around and grips the metal ion from two or more places at once, closing a ring. That ring is what makes the complex unusually stable, and this extra stability is exactly what "chelate effect" names.Start with denticity. A ligand is called monodentate if it has only one atom that can donate an electron pair to the metal — for example \(\displaystyle \text{NH}_3 \) or \(\displaystyle \text{Cl}^- \), each attaching through a single donor atom. A ligand is called polydentate (or chelating) if it has two or more donor atoms that can simultaneously bind to the same metal ion. Ethane-$\displaystyle 1,2$-diamine, \(\displaystyle \text{H}_2\text{N–CH}_2\text{–CH}_2\text{–NH}_2 \) (abbreviated "en"), is the standard example: it has two \(\displaystyle -\text{NH}_2 \) groups, and both nitrogen atoms can coordinate to one metal centre at the same time.When a polydentate ligand attaches through two or more donor atoms to a single metal ion, it closes a ring that includes the metal atom as one of the ring members. Such a ring is called a chelate ring, and the complex is called a chelate (from the Greek chele, meaning claw). With ethylenediamine, the ring formed is\[\text{M} \; \longleftarrow \; \overset{\displaystyle H_2N}{\underset{\displaystyle H_2C}{\Big\backslash}}\!-\!\text{CH}_2\!-\!\overset{\displaystyle NH_2}{}\; \longrightarrow \; \text{M} \]— a five-membered ring: M, N, C, C, N joined back to M.The chelate effect is the observed fact that a complex formed by polydentate (chelating) ligands is thermodynamically far more stable than a comparable complex formed by an equal number of donor bonds supplied through separate monodentate ligands. In other words, tying the donor atoms together into one ligand molecule, rather than leaving them as independent ligands, increases the stability constant of the complex.Example. Compare two nickel(II) complexes that both use six Ni–N bonds:\[[\text{Ni}(\text{NH}_3)_6]^{2+} \quad \text{(six separate, monodentate } \text{NH}_3 \text{ ligands)} \] \[[\text{Ni}(\text{en})_3]^{2+} \quad \text{(three ethylenediamine molecules, each chelating through its two N atoms)} \]Both complexes have the same donor atom (N) and the same number of metal–nitrogen bonds (six), yet \(\displaystyle [\text{Ni}(\text{en})_3]^{2+} \) is very much more stable than \(\displaystyle [\text{Ni}(\text{NH}_3)_6]^{2+} \) — its overall stability (formation) constant is several orders of magnitude larger. The only structural difference is that in \(\displaystyle [\text{Ni}(\text{en})_3]^{2+} \) the nitrogen atoms are held together in pairs by the ethylene bridge, so each "en" closes a five-membered chelate ring around the Ni(II) ion, whereas the six \(\displaystyle \text{NH}_3 \) molecules in the other complex are completely independent of one another.A short aside on why this happens: it is not about the Ni–N bond itself being stronger in one case than the other — bond strength per Ni–N bond is comparable. The gain comes mainly from entropy. Replacing six separate \(\displaystyle \text{NH}_3 \) molecules with three "en" molecules means three particles are combining with the metal ion instead of six, so more free ligand molecules are released into solution when \(\displaystyle \text{NH}_3 \) is displaced by "en" than are consumed — an increase in the number of free species (higher entropy) that favours the chelated product. Five- and six-membered chelate rings are the most stable ring sizes, which is why "en" (giving $\displaystyle 5$-membered rings) is such an effective chelating agent.Another everyday example of the same effect is EDTA, a hexadentate ligand that wraps around a metal ion (e.g., \(\displaystyle \text{Ca}^{2+} \)) through six donor atoms at once, forming several fused five-membered rings and giving an extremely stable complex — which is why EDTA is used to sequester metal ions so tightly in water-softening and in treating heavy-metal poisoning.Answer: The chelate effect is the extra thermodynamic stability a complex gains when it is formed from a polydentate (chelating) ligand — one that binds the metal through two or more donor atoms at once, closing a ring — compared with an otherwise similar complex built from monodentate ligands. Example: \(\displaystyle [\text{Ni}(\text{en})_3]^{2+}\), in which ethylenediamine chelates Ni(II) through pairs of N atoms forming five-membered rings, is far more stable than \(\displaystyle [\text{Ni}(\text{NH}_3)_6]^{2+}\), even though both have six Ni–N bonds, because "en" ties the donor atoms together into rings while \(\displaystyle \text{NH}_3\) ligands remain independent.

Coordination compounds are not lab curiosities — a metal ion held by donor atoms in a fixed geometry is exactly what living cells, drugs, analytical tests, and metal extraction all exploit. Four examples, one for each area asked.(i) Biological systemsLife runs several of its core machines as metal complexes. (ii) Medicinal chemistryCoordination compounds are used both to treat poisoning and to treat disease. (iii) Analytical chemistryCoordination compounds give sharply coloured, insoluble, or otherwise distinctive complexes that are used to detect and estimate ions. (iv) Extraction and metallurgy of metalsComplex formation is the working step in extracting certain metals from their ores. \[4\,\text{Au(s)} + 8\,\text{CN}^-(\text{aq}) + 2\,\text{H}_2\text{O(aq)} + \text{O}_2(\text{g}) \rightarrow 4\,[\text{Au(CN)}_2]^-(\text{aq}) + 4\,\text{OH}^-(\text{aq}) \] The metal, which would otherwise not dissolve, is pulled into solution only because it forms this stable dicyanoaurate(I) complex. The pure metal is then recovered by displacing it from the complex with a more reactive metal: \[2\,[\text{Au(CN)}_2]^-(\text{aq}) + \text{Zn(s)} \rightarrow [\text{Zn(CN)}_4]^{2-}(\text{aq}) + 2\,\text{Au(s)} \] The same cyanide-complexation, zinc-displacement sequence is used for silver. Without complex formation, these noble metals would simply stay behind in the insoluble ore.Answer: (i) O₂-transport in haemoglobin (Fe–porphyrin) and photosynthesis in chlorophyll (Mg–porphyrin) — both metal-coordination complexes; (ii) EDTA chelates and removes Pb²⁺ in lead-poisoning treatment, and cis-platin (a Pt coordination compound) is used in cancer chemotherapy; (iii) Ni²⁺ is estimated via its red DMG complex, and water hardness (Ca²⁺, Mg²⁺) is estimated by EDTA titration; (iv) Ag and Au are extracted from their ores as soluble \(\displaystyle [\text{M(CN)}_2]^-\) cyanide complexes, then displaced as the free metal by zinc.

Only ligands and counter-ions outside the square brackets ionize — those inside the coordination sphere travel together as one unit.The formula \(\displaystyle \left[\mathrm{Co(NH_3)_6}\right]\mathrm{Cl_2}\) has two distinct parts: Because the compound as a whole is neutral and the two \(\displaystyle \mathrm{Cl^-}\) ions carry a total charge of \(\displaystyle -2\), the complex ion left behind, \(\displaystyle \left[\mathrm{Co(NH_3)_6}\right]^{2+}\), must carry a charge of \(\displaystyle +2\) (this fixes the oxidation state of cobalt at \(\displaystyle +2\), with a coordination number of $\displaystyle 6$ — a side fact, not something you need for the ion count).Dissociation in water proceeds as: \[\left[\mathrm{Co(NH_3)_6}\right]\mathrm{Cl_2}\ (aq) \longrightarrow \left[\mathrm{Co(NH_3)_6}\right]^{2+}(aq) + 2\,\mathrm{Cl^-}(aq) \]The mistake to avoid: it is tempting to count every \(\displaystyle \mathrm{NH_3}\) as a separate particle, as if the formula behaved like a simple salt with $\displaystyle 9$ pieces. It does not — the six \(\displaystyle \mathrm{NH_3}\) groups are coordinately bonded to \(\displaystyle \mathrm{Co}\) and move as a single complex ion, not as free ammonia molecules.Counting the particles actually produced: \[1 \times \left[\mathrm{Co(NH_3)_6}\right]^{2+} \;+\; 2 \times \mathrm{Cl^-} \;=\; 3 \text{ ions total} \]Answer: (iii) $\displaystyle 3$ — one \(\displaystyle \left[\mathrm{Co(NH_3)_6}\right]^{2+}\) ion and two \(\displaystyle \mathrm{Cl^-}\) ions.

Magnetic moment depends on the number of UNPAIRED electrons in the metal ion, and that number depends on the ion's d-electron count plus whether the ligand forces pairing.The spin-only magnetic moment formula is \[\mu = \sqrt{n(n+2)}\ \text{BM} \] where \(\displaystyle n\) is the number of unpaired electrons and BM stands for Bohr Magneton. To use it, first find the d-electron configuration of each metal ion, then decide how those electrons are arranged in the octahedral field of six \(\displaystyle H_2O\) ligands.Step $\displaystyle 1$ — find the d-electron count of each ionElectrons are removed from the 4s orbital before the 3d orbital when a transition metal atom is ionised — that is the step people get wrong, since it looks like the 3d electrons should go first because 3d is written after 4s. Step $\displaystyle 2$ — decide high-spin or low-spin\(\displaystyle H_2O\) is a weak-field ligand, so it does not supply enough crystal field splitting energy to force electrons into pairing against Hund's rule. All three complexes are therefore high-spin, and the d electrons occupy the five d orbitals singly for as long as empty orbitals remain. Step $\displaystyle 3$ — compute \(\displaystyle \mu\) for each ion\[\mu\big[Cr(H_2O)_6\big]^{2+} = \sqrt{4(4+2)} = \sqrt{24} = 4.90\ \text{BM} \]\[\mu\big[Fe(H_2O)_6\big]^{3+} = \sqrt{5(5+2)} = \sqrt{35} = 5.92\ \text{BM} \]\[\mu\big[Zn(H_2O)_6\big]^{2+} = \sqrt{0(0+2)} = 0\ \text{BM (diamagnetic)} \]Step $\displaystyle 4$ — compare\[5.92\ \text{BM} > 4.90\ \text{BM} > 0\ \text{BM} \]\(\displaystyle \mathrm{Fe^{3+}}\) has the highest magnetic moment because its half-filled \(\displaystyle 3d^{5}\) configuration gives it the maximum possible number of unpaired electrons ($\displaystyle 5$) among these three ions — more unpaired electrons always means a larger magnetic moment, since \(\displaystyle \mu\) grows with \(\displaystyle n\) in \(\displaystyle \sqrt{n(n+2)}\).Answer: \(\displaystyle [Fe(H_2O)_6]^{3+}\) has the highest magnetic moment, \(\displaystyle \mu \approx 5.92\) BM ($\displaystyle 5$ unpaired electrons in its high-spin \(\displaystyle 3d^5\) configuration), compared to \(\displaystyle 4.90\) BM for \(\displaystyle [Cr(H_2O)_6]^{2+}\) and \(\displaystyle 0\) BM for \(\displaystyle [Zn(H_2O)_6]^{2+}\).

Stability here isn't decided by crystal-field splitting — it's decided by the chelate effect, because every metal ion in this list is high-spin \(\displaystyle d^5\) and has zero CFSE.Step $\displaystyle 1$ — find the oxidation state of iron in each complex. Balance the overall charge against the ligand charges.\(\displaystyle [\text{Fe}(\text{H}_2\text{O})_6]^{3+} \): \(\displaystyle \text{H}_2\text{O}\) is neutral, so \(\displaystyle x + 0 = +3 \Rightarrow \text{Fe}^{3+}\).\(\displaystyle [\text{Fe}(\text{NH}_3)_6]^{3+} \): \(\displaystyle \text{NH}_3\) is neutral, so \(\displaystyle x + 0 = +3 \Rightarrow \text{Fe}^{3+}\).\(\displaystyle [\text{Fe}(\text{C}_2\text{O}_4)_3]^{3-} \): each oxalate carries \(\displaystyle -2\), three of them give \(\displaystyle -6\); \(\displaystyle x + (-6) = -3 \Rightarrow x = +3 \Rightarrow \text{Fe}^{3+}\).\(\displaystyle [\text{FeCl}_6]^{3-} \): six \(\displaystyle \text{Cl}^-\) give \(\displaystyle -6\); \(\displaystyle x + (-6) = -3 \Rightarrow x = +3 \Rightarrow \text{Fe}^{3+}\).So all four have the same central ion, \(\displaystyle \text{Fe}^{3+}\), which is \(\displaystyle 3d^5\). The only thing that differs between them is the ligand.Step $\displaystyle 2$ — check whether crystal field stabilisation energy (CFSE) can distinguish them. This is the step people try first and it's a dead end here. None of \(\displaystyle \text{H}_2\text{O}\), \(\displaystyle \text{NH}_3\), \(\displaystyle \text{C}_2\text{O}_4^{2-}\) or \(\displaystyle \text{Cl}^-\) is a strong enough field to pair up a \(\displaystyle d^5\) ion (that takes a very strong field ligand like \(\displaystyle \text{CN}^-\)), so in every one of these four complexes iron stays high spin: \(\displaystyle t_{2g}^{3}e_g^{2}\) — three unpaired-orbital electrons in \(\displaystyle t_{2g}\), two in \(\displaystyle e_g\).CFSE for an octahedral field is \[\text{CFSE} = n(t_{2g})(-0.4\,\Delta_o) + n(e_g)(+0.6\,\Delta_o) \] For \(\displaystyle t_{2g}^{3}e_g^{2}\): \[\text{CFSE} = 3(-0.4\,\Delta_o) + 2(+0.6\,\Delta_o) = -1.2\,\Delta_o + 1.2\,\Delta_o = 0 \] This is zero regardless of how large \(\displaystyle \Delta_o\) is — it cancels for every high-spin \(\displaystyle d^5\) ion. So no matter where each ligand sits on the spectrochemical series, the CFSE contributed is zero in all four complexes. CFSE cannot be the reason any one of them is more stable than the others — the discriminator has to be something else.Step $\displaystyle 3$ — the real discriminator: denticity (the chelate effect). Look at how each ligand binds: Forming a chelate ring releases extra small molecules/ions into solution (for example, water molecules displaced from the hydrated metal ion) for every bidentate ligand that goes on, compared to putting on the same number of donor atoms one monodentate ligand at a time. That raises the number of free particles in solution, so \(\displaystyle \Delta S\) for complex formation is more positive, and \[\Delta G = \Delta H - T\Delta S \] becomes more negative — the formation constant goes up sharply. This entropy-driven boost from ring closure is the chelate effect, and it has nothing to do with \(\displaystyle \Delta_o\) or CFSE: it applies even when, as here, CFSE is zero for every option.Step $\displaystyle 4$ — apply it to the four complexes. \(\displaystyle [\text{Fe}(\text{H}_2\text{O})_6]^{3+} \), \(\displaystyle [\text{Fe}(\text{NH}_3)_6]^{3+} \) and \(\displaystyle [\text{FeCl}_6]^{3-} \) are all built from monodentate ligands, so none of them gets a chelate-effect boost — their stability differences (small, and irrelevant to CFSE since that's zero for all) come only from how good a \(\displaystyle \sigma\)-donor the ligand is. \(\displaystyle [\text{Fe}(\text{C}_2\text{O}_4)_3]^{3-} \) is the only one built from a bidentate, ring-closing ligand, so it is the only one that gets the chelate-effect stability boost on top of ordinary donor strength.That makes \(\displaystyle [\text{Fe}(\text{C}_2\text{O}_4)_3]^{3-} \) the most stable of the four.Answer: (iii) \(\displaystyle [\text{Fe}(\text{C}_2\text{O}_4)_3]^{3-} \) is the most stable complex. All four contain high-spin \(\displaystyle \text{Fe}^{3+}\) (\(\displaystyle 3d^5\), \(\displaystyle t_{2g}^3e_g^2\)), for which CFSE is exactly zero, so crystal-field stabilisation cannot separate them. The oxalate ion is bidentate and closes three five-membered chelate rings around the iron, and this chelate effect makes \(\displaystyle [\text{Fe}(\text{C}_2\text{O}_4)_3]^{3-} \) far more thermodynamically stable than the complexes built from the monodentate ligands \(\displaystyle \text{H}_2\text{O}\), \(\displaystyle \text{NH}_3\), and \(\displaystyle \text{Cl}^-\).

The color a coordination complex absorbs comes from a d–d transition, and its energy is fixed by how strongly the ligand splits the metal's d-orbitals — the spectrochemical series ranks that strength directly, and wavelength runs opposite to it.For an octahedral \(\displaystyle Ni^{2+} \) complex, the six ligands split the five d-orbitals into a lower \(\displaystyle t_{2g} \) set and a higher \(\displaystyle e_g \) set separated by the crystal-field splitting energy \(\displaystyle \Delta_o \). Absorbing a photon promotes an electron from \(\displaystyle t_{2g} \) to \(\displaystyle e_g \), so the photon's energy must equal \(\displaystyle \Delta_o \):\[\Delta_o = h\nu = \frac{hc}{\lambda} \]Here \(\displaystyle h\) is Planck's constant, \(\displaystyle \nu\) is the frequency of the light absorbed, \(\displaystyle c\) is the speed of light, and \(\displaystyle \lambda\) is its wavelength. Because \(\displaystyle \Delta_o\) sits in the numerator and \(\displaystyle \lambda\) in the denominator, they move in opposite directions: a ligand that causes a bigger splitting makes the complex absorb a shorter wavelength, not a longer one — that inversion is the exact trap this question is checking.The metal ion is the same, \(\displaystyle \mathrm{Ni^{2+}}\), in all three complexes, so only the ligand changes: \(\displaystyle NO_2^-\), \(\displaystyle NH_3\), and \(\displaystyle H_2O\). The spectrochemical series ranks ligands by the field strength — and hence the \(\displaystyle \Delta_o\) — they produce, and for this trio it reads, in increasing strength,\[H_2O \;<\; NH_3 \;<\; NO_2^- \]\(\displaystyle H_2O\) is a weak-field ligand, \(\displaystyle NH_3\) is a moderately strong-field ligand, and \(\displaystyle NO_2^-\) (bonded to the metal through nitrogen) is one of the strongest-field ligands in the whole series, well above \(\displaystyle NH_3\) — that ordering is the one detail worth memorizing here, since it is easy to assume \(\displaystyle NH_3\) is stronger.Ligand field strength and \(\displaystyle \Delta_o\) track together, so the same order carries over:\[\Delta_o\big([Ni(H_2O)_6]^{2+}\big) \;<\; \Delta_o\big([Ni(NH_3)_6]^{2+}\big) \;<\; \Delta_o\big([Ni(NO_2)_6]^{4-}\big) \]and since \(\displaystyle \lambda \propto 1/\Delta_o\), the wavelength order flips relative to it:\[\lambda\big([Ni(NO_2)_6]^{4-}\big) \;<\; \lambda\big([Ni(NH_3)_6]^{2+}\big) \;<\; \lambda\big([Ni(H_2O)_6]^{2+}\big) \]The complex with the strongest-field ligand, \(\displaystyle [Ni(NO_2)_6]^{4-}\), needs the most energetic photon and so absorbs the shortest wavelength; the weakest-field complex, \(\displaystyle [Ni(H_2O)_6]^{2+}\), gets by on the least energy and absorbs the longest wavelength.Answer: In increasing order of the wavelength absorbed, \(\displaystyle [Ni(NO_2)_6]^{4-} < [Ni(NH_3)_6]^{2+} < [Ni(H_2O)_6]^{2+}\).