NCERT Solutions for Class 12 Chemistry Chapter 5: Coordination Compounds

Answer

(i) [Co(NH₃)₄(H₂O)₂]Cl₃ (ii) K₂[Ni(CN)₄] (iii) [Cr(en)₃]Cl₃ (iv) [Pt(NH₃)BrCl(NO₂)]⁻ (v) [PtCl₂(en)₂](NO₃)₂ (vi) Fe₄[Fe(CN)₆]₃

Answer

(i) Hexaamminecobalt(III) chloride (ii) Pentaamminechloridocobalt(III) chloride (iii) Potassium hexacyanidoferrate(III) (iv) Potassium trioxalatoferrate(III) (v) Potassium tetrachloridopalladate(II) (vi) Diamminechlorido(methanamine)platinum(II) chloride

Answer

(i) Geometrical isomerism: in the octahedral complex the two H₂O ligands may be cis (adjacent) or trans (opposite). The cis isomer has no plane of symmetry, so it also shows optical isomerism (a pair of non-superimposable mirror images, d and l). The trans isomer is optically inactive.

(ii) Optical isomerism: the three bidentate en ligands wrap around Co like a propeller in either a right-handed or left-handed sense, giving d and l enantiomers that are mirror images of each other.

(iii) Linkage isomerism: NO₂⁻ can bind through N (nitrito-N, [Co(NH₃)₅(NO₂)]²⁺) or through O (nitrito-O, [Co(NH₃)₅(ONO)]²⁺). Ionisation isomerism: the nitrite inside and a nitrate outside can exchange, e.g. [Co(NH₃)₅(NO₃)](NO₂)(NO₃). Combinations of these give several isomers.

(iv) Geometrical isomerism (square planar): cis isomer with the two Cl atoms adjacent, and trans isomer with the two Cl atoms opposite (then NH₃ and H₂O are opposite each other).

Answer

They have the same formula but give different ions in solution:

[Co(NH₃)₅Cl]SO₄
⇒ [Co(NH₃)₅Cl]²⁺ + SO₄²⁻: its solution gives a white precipitate of BaSO₄ with BaCl₂, but no precipitate with AgNO₃.

[Co(NH₃)₅(SO₄)]Cl
⇒ [Co(NH₃)₅(SO₄)]⁺ + Cl⁻: its solution gives a white precipitate of AgCl with AgNO₃, but no precipitate with BaCl₂.

Different ions in solution confirm that they are ionisation isomers.

Answer

In both, Ni is +2: Ni²⁺ = [Ar]3d⁸ with two unpaired electrons.

[Ni(CN)₄]²⁻: CN⁻ is a strong field ligand and pairs the two unpaired 3d electrons, leaving one 3d orbital empty. One 3d, one 4s and two 4p orbitals undergo dsp² hybridisation, giving a square planar complex with no unpaired electrons
⇒ diamagnetic.

[NiCl₄]²⁻: Cl⁻ is a weak field ligand and cannot pair the electrons. The 4s and three 4p orbitals undergo sp³ hybridisation, giving a tetrahedral complex with two unpaired electrons
⇒ paramagnetic.

Answer

In [NiCl₄]²⁻, Ni is +2 (3d⁸) and the weak field Cl⁻ cannot pair the two unpaired d electrons; sp³ hybridisation leaves 2 unpaired electrons
⇒ paramagnetic. In [Ni(CO)₄], Ni is in the 0 state (3d⁸4s²); the strong field CO ligand causes the 4s electrons to pair into the 3d orbitals giving 3d¹⁰, and sp³ hybridisation then leaves no unpaired electron
⇒ diamagnetic.

Answer

Fe³⁺ is 3d⁵. With the strong field ligand CN⁻, the 3d electrons pair up leaving only one unpaired electron; two empty 3d orbitals give d²sp³ hybridisation (inner orbital complex)
⇒ weakly paramagnetic. With the weak field ligand H₂O, the electrons do not pair, outer 4d orbitals are used in sp³d² hybridisation (outer orbital complex) with five unpaired electrons
⇒ strongly paramagnetic.

Answer

In [Co(NH₃)₆]³⁺, Co³⁺ is 3d⁶. NH₃ pairs these electrons into three 3d orbitals, leaving two 3d orbitals empty, so d²sp³ hybridisation (using inner 3d orbitals) occurs – an inner orbital, diamagnetic complex.

In [Ni(NH₃)₆]²⁺, Ni²⁺ is 3d⁸. Even after pairing, only one 3d orbital could be emptied, which is not enough for d²sp³; so the outer 4d orbitals are used in sp³d² hybridisation – an outer orbital complex with two unpaired electrons.

Answer

Pt²⁺ is 5d⁸. For a square planar structure, dsp² hybridisation is needed, so the electrons pair up to leave one 5d orbital empty. Hence the number of unpaired electrons is zero (diamagnetic).

Answer

Mn²⁺ is 3d⁵. In an octahedral field the d orbitals split into t₂g (lower) and e_g (higher).

H₂O is a weak field ligand: Δo < pairing energy P, so electrons occupy all five orbitals singly – t₂g³e_g² with 5 unpaired electrons (high spin).

CN⁻ is a strong field ligand: Δo > P, so electrons pair in the t₂g set – t₂g⁵e_g⁰ with 1 unpaired electron (low spin).

Answer

(i) In coordination compounds metals show two types of valencies – primary and secondary.

(ii) Primary valencies are ionisable and are satisfied by negative ions; they correspond to the oxidation state of the metal.

(iii) Secondary valencies are non-ionisable and are satisfied by neutral molecules or negative ions (ligands). The secondary valence equals the coordination number and is fixed for a metal.

(iv) The ions/groups bound by secondary valencies have a characteristic spatial arrangement corresponding to definite geometries (octahedral, tetrahedral, square planar), called coordination polyhedra.

Example: in CoCl₃·6NH₃ ([Co(NH₃)₆]Cl₃), the primary valence is 3 (three ionisable Cl⁻) and the secondary valence is 6 (six NH₃), arranged octahedrally. In CoCl₃·5NH₃ ([Co(NH₃)₅Cl]Cl₂), only two Cl⁻ are precipitated by AgNO₃, since one Cl satisfies both a primary and a secondary valence.

Answer

FeSO₄ and (NH₄)₂SO₄ form a double salt, Mohr's salt FeSO₄·(NH₄)₂SO₄·6H₂O, which dissociates completely in water into Fe²⁺, NH₄⁺ and SO₄²⁻ ions, so Fe²⁺ tests are positive. CuSO₄ and NH₃ (1:4) form a complex, [Cu(NH₃)₄]SO₄, in which Cu²⁺ is bound in the stable complex ion [Cu(NH₃)₄]²⁺ and is not released as free Cu²⁺, so it does not give the test of Cu²⁺.

Answer

Coordination entity: a central metal atom/ion bonded to a fixed number of ions or molecules, e.g. [CoCl₃(NH₃)₃], [Fe(CN)₆]⁴⁻.

Ligand: an ion or molecule that donates an electron pair to the central atom, e.g. Cl⁻, NH₃, H₂O, CN⁻.

Coordination number: number of ligand donor atoms directly bonded to the metal, e.g. 6 in [PtCl₆]²⁻, 4 in [Ni(NH₃)₄]²⁺; in [Fe(C₂O₄)₃]³⁻ it is 6 (three didentate ligands).

Coordination polyhedron: the spatial arrangement of donor atoms around the metal, e.g. octahedral [Co(NH₃)₆]³⁺, tetrahedral [Ni(CO)₄], square planar [PtCl₄]²⁻.

Homoleptic: metal bound to only one kind of donor group, e.g. [Co(NH₃)₆]³⁺, [Fe(CN)₆]⁴⁻.

Heteroleptic: metal bound to more than one kind of donor group, e.g. [Co(NH₃)₄Cl₂]⁺, [Co(NH₃)₅Br]²⁺.

Answer

Unidentate: bind through a single donor atom, e.g. NH₃, Cl⁻, H₂O.

Didentate: bind through two donor atoms, e.g. ethane-1,2-diamine (H₂NCH₂CH₂NH₂, en) and oxalate (C₂O₄²⁻).

Ambidentate: have two different donor atoms but bind through either one at a time, e.g. NO₂⁻ (via N as nitrito-N or via O as nitrito-O) and SCN⁻ (via S as thiocyanato-S or via N as thiocyanato-N/isothiocyanato).

Answer

(i) x + 0 + (−1) + 0 = +2
⇒ x = +3 (ii) x + 2(−1) = +1
⇒ x = +3 (iii) x + 4(−1) = −2
⇒ x = +2 (iv) [Fe(CN)₆]³⁻: x − 6 = −3
⇒ x = +3 (v) x + 3(−1) = 0
⇒ x = +3

Answer

(i) [Zn(OH)₄]²⁻ (ii) K₂[PdCl₄] (iii) [Pt(NH₃)₂Cl₂] (iv) K₂[Ni(CN)₄] (v) [Co(NH₃)₅(ONO)]²⁺ (vi) [Co(NH₃)₆]₂(SO₄)₃ (vii) K₃[Cr(C₂O₄)₃] (viii) [Pt(NH₃)₆]⁴⁺ (ix) [CuBr₄]²⁻ (x) [Co(NH₃)₅(NO₂)]²⁺

Answer

(i) Hexaamminecobalt(III) chloride

(ii) Diamminechlorido(methanamine)platinum(II) chloride

(iii) Hexaaquatitanium(III) ion

(iv) Tetraamminechloridonitrito-N-cobalt(III) chloride

(v) Hexaaquamanganese(II) ion

(vi) Tetrachloridonickelate(II) ion

Answer

A. Stereoisomerism (same bonds, different spatial arrangement):

(i) Geometrical isomerism – e.g. cis- and trans-[Pt(NH₃)₂Cl₂]; fac- and mer-[Co(NH₃)₃(NO₂)₃].

(ii) Optical isomerism – e.g. d- and l-[Co(en)₃]³⁺; cis-[PtCl₂(en)₂]²⁺.

B. Structural isomerism (different bonds):

(i) Linkage isomerism – [Co(NH₃)₅(NO₂)]Cl₂ and [Co(NH₃)₅(ONO)]Cl₂.

(ii) Coordination isomerism – [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆].

(iii) Ionisation isomerism – [Co(NH₃)₅(SO₄)]Br and [Co(NH₃)₅Br]SO₄.

(iv) Solvate (hydrate) isomerism – [Cr(H₂O)₆]Cl₃ (violet) and [Cr(H₂O)₅Cl]Cl₂·H₂O (grey-green).

Answer

(i) None – all three ligands are the same symmetrical didentate ligand, so no geometrical isomer exists (it shows only optical isomerism).

(ii) Two – facial (fac, three Cl on one face of the octahedron) and meridional (mer, three Cl in one plane around the meridian).

Answer

(i) [Cr(C₂O₄)₃]³⁻: Cr at the centre of an octahedron with three oxalate rings spanning cis positions, arranged as a left-handed or right-handed propeller. The two forms (d and l) are non-superimposable mirror images.

(ii) [PtCl₂(en)₂]²⁺: only the cis isomer is optically active. In it the two Cl ligands are adjacent and the two en rings occupy the remaining four positions; the d and l forms are mirror images of each other. (The trans isomer has a plane of symmetry and is inactive.)

(iii) [Cr(NH₃)₂Cl₂(en)]⁺: the isomer in which the two NH₃ are cis to each other and the two Cl are also cis (with en spanning two cis positions) has no plane of symmetry, so it exists as a pair of enantiomers (d and l), which are mirror images.

Answer

(i) [CoCl₂(en)₂]⁺: two geometrical isomers – trans (Cl opposite, optically inactive) and cis (Cl adjacent). The cis form exists as d and l enantiomers. Total 3 isomers.

(ii) [Co(NH₃)Cl(en)₂]²⁺: trans (NH₃ opposite Cl, optically inactive) and cis (NH₃ adjacent to Cl). The cis form exists as d and l enantiomers. Total 3 isomers.

(iii) [Co(NH₃)₂Cl₂(en)]⁺: three geometrical isomers – (a) NH₃ groups trans to each other, Cl groups cis (inactive); (b) Cl groups trans, NH₃ groups cis (inactive); (c) both NH₃ cis and both Cl cis, which is chiral and exists as d and l forms. Total 4 isomers.

Answer

Square planar MABCD type gives three geometrical isomers, decided by which ligand is trans to NH₃:

(I) NH₃ trans to Br (Cl trans to py); (II) NH₃ trans to Cl (Br trans to py); (III) NH₃ trans to py (Br trans to Cl).

None of them shows optical isomerism, because a square planar complex lies in a plane, which is a plane of symmetry.

Answer

Aqueous CuSO₄ contains the blue complex [Cu(H₂O)₄]²⁺. Its water ligands are replaced by stronger/available ligands:

(i) [Cu(H₂O)₄]²⁺ + 4F⁻
⇒ [CuF₄]²⁻ + 4H₂O; the fluoride complex forms a green precipitate.

(ii) [Cu(H₂O)₄]²⁺ + 4Cl⁻
⇒ [CuCl₄]²⁻ + 4H₂O; tetrachloridocuprate(II) is soluble and gives a bright green solution.

Answer

Excess CN⁻ replaces the water ligands and forms a very stable cyanide complex. CN⁻ reduces Cu²⁺ to Cu⁺, and the entity formed is the tetracyanidocuprate(I) ion, [Cu(CN)₄]³⁻ (often written at this level as [Cu(H₂O)₄]²⁺ + 4CN⁻
⇒ [Cu(CN)₄]²⁻ + 4H₂O).

Because the cyanide complex is very stable (very high stability constant), it hardly dissociates, so the concentration of free copper ions is too low for the ionic product of copper sulphide to exceed its Ksp. Hence no CuS precipitate forms with H₂S.

Answer

(i) [Fe(CN)₆]⁴⁻: Fe²⁺ is 3d⁶. Strong field CN⁻ pairs the electrons into three 3d orbitals, leaving two empty; d²sp³ hybridisation
⇒ octahedral, inner orbital, low spin, diamagnetic (0 unpaired).

(ii) [FeF₆]³⁻: Fe³⁺ is 3d⁵. Weak field F⁻ causes no pairing; 4s, 4p and 4d orbitals give sp³d² hybridisation
⇒ octahedral, outer orbital, high spin, 5 unpaired electrons, paramagnetic.

(iii) [Co(C₂O₄)₃]³⁻: Co³⁺ is 3d⁶. Oxalate causes pairing here; two empty 3d orbitals give d²sp³ hybridisation
⇒ octahedral, inner orbital, low spin, diamagnetic.

(iv) [CoF₆]³⁻: Co³⁺ is 3d⁶. Weak field F⁻ causes no pairing; sp³d² hybridisation
⇒ octahedral, outer orbital, high spin, 4 unpaired electrons, paramagnetic.

Answer

In a free metal ion the five d orbitals are degenerate. When six ligands approach along the x, y and z axes, the average energy rises (spherical field), and then the orbitals split into two sets:

e_g set (d(x²−y²) and d(z²)): lobes point directly at the ligands, so they are repelled more and rise by +0.6Δo above the average energy.

t₂g set (d(xy), d(yz), d(xz)): lobes point between the axes, so they are lowered by −0.4Δo.

The energy gap between e_g (upper level) and t₂g (lower level) is the crystal field splitting energy Δo. In the diagram: one level for the free ion, a slightly higher level for the spherical field, then splitting into two upper e_g lines and three lower t₂g lines.

Answer

The spectrochemical series is the experimentally determined order of ligands according to increasing field strength (increasing Δo they produce):

I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < edta⁴⁻ < NH₃ < en < CN⁻ < CO

Weak field ligands produce small splitting (Δo < P); electrons enter e_g before pairing, giving high spin complexes (e.g. F⁻, Cl⁻, H₂O). Strong field ligands produce large splitting (Δo > P); electrons pair in t₂g first, giving low spin complexes (e.g. CN⁻, CO, en).

Answer

Crystal field splitting energy (Δo for octahedral) is the energy difference between the two sets of d orbitals (e_g and t₂g) produced when ligands approach the metal ion.

For d¹–d³ the electrons simply occupy t₂g. For d⁴ onwards, the fourth electron can either pair in t₂g (costing pairing energy P) or enter e_g (costing Δo):

If Δo < P (weak field): the electron goes to e_g
⇒ t₂g³e_g¹, high spin.

If Δo > P (strong field): the electron pairs in t₂g
⇒ t₂g⁴e_g⁰, low spin.

Answer

In [Cr(NH₃)₆]³⁺, Cr³⁺ is 3d³. The three electrons occupy three different 3d (t₂g) orbitals singly, and two empty 3d orbitals are used in d²sp³ hybridisation; three unpaired electrons remain, so it is paramagnetic (μ ≈ 3.87 BM).

In [Ni(CN)₄]²⁻, Ni²⁺ is 3d⁸. The strong field CN⁻ pairs the two unpaired electrons, freeing one 3d orbital for dsp² hybridisation (square planar). No unpaired electron remains, so it is diamagnetic.

Answer

In [Ni(H₂O)₆]²⁺, Ni²⁺ (3d⁸) with weak field water has two unpaired electrons in e_g; d–d transitions absorb light in the red region of the visible spectrum and the complementary green colour is seen.

In [Ni(CN)₄]²⁻, strong field CN⁻ gives a large splitting and all electrons are paired in the square planar arrangement; no d–d transition occurs by absorbing visible light, so the solution is colourless.

Answer

Both contain Fe²⁺ (3d⁶), but the ligands differ. CN⁻ is a strong field ligand giving a large Δo (low spin t₂g⁶), while H₂O is a weak field ligand giving a small Δo (high spin t₂g⁴e_g²). Since the colour depends on the wavelength of light absorbed for the d–d transition, which depends on Δo, the two complexes absorb different wavelengths and show different colours.

Answer

The metal–carbon bond in metal carbonyls has both σ and π character:

(i) σ bond: the lone pair on the carbon atom of CO is donated into a vacant orbital of the metal (M ← CO).

(ii) π back bond: a filled d orbital of the metal donates electron density into the vacant antibonding π* orbital of CO (M
⇒ CO).

This synergic bonding strengthens the M–C bond and weakens the C≡O bond, and it stabilises the metal in low (often zero) oxidation states, e.g. Ni(CO)₄ (tetrahedral), Fe(CO)₅ (trigonal bipyramidal), Cr(CO)₆ (octahedral).

Answer

(i) Co: +3; 3d⁶ (low spin t₂g⁶e_g⁰); CN = 6

(ii) Cr: +3; 3d³ (t₂g³); CN = 6

(iii) Co: +2; 3d⁷ (tetrahedral, e⁴t₂³, 3 unpaired); CN = 4

(iv) Mn: +2; 3d⁵ (high spin t₂g³e_g²); CN = 6

Answer

(i) Potassium diaquadioxalatochromate(III) trihydrate; Cr +3; 3d³ (t₂g³); CN 6; octahedral (shows cis/trans isomers, cis is optically active); 3 unpaired, μ = √15 = 3.87 BM.

(ii) Pentaamminechloridocobalt(III) chloride; Co +3; 3d⁶ (low spin t₂g⁶); CN 6; octahedral; 0 unpaired, μ = 0 (diamagnetic).

(iii) Trichloridotripyridinechromium(III); Cr +3; 3d³ (t₂g³); CN 6; octahedral (fac and mer isomers); μ = 3.87 BM.

(iv) Caesium tetrachloridoferrate(III); Fe +3; 3d⁵ (e²t₂³); CN 4; tetrahedral (sp³); 5 unpaired, μ = √35 = 5.92 BM.

(v) Potassium hexacyanidomanganate(II); Mn +2; 3d⁵ (low spin t₂g⁵); CN 6; octahedral (d²sp³); 1 unpaired, μ = √3 = 1.73 BM.

Answer

Ti³⁺ is 3d¹; in the octahedral field the single electron occupies the t₂g level (t₂g¹e_g⁰). On absorbing visible light of wavelength about 498 nm (blue-green region), the electron is promoted to the e_g level (d–d transition, t₂g¹e_g⁰
⇒ t₂g⁰e_g¹). The transmitted light, complementary to the absorbed blue-green, makes the complex appear violet. In the absence of ligands there would be no splitting and the ion would be colourless.

Answer

When a didentate or polydentate ligand binds a metal through two or more donor atoms, it forms a ring (chelate). Such chelate complexes are more stable than similar complexes of unidentate ligands; this extra stability is the chelate effect. Example: [Ni(en)₃]²⁺ is more stable than [Ni(NH₃)₆]²⁺; EDTA forms very stable chelates with Ca²⁺ and Mg²⁺.

Answer

(i) Biological systems: chlorophyll is a magnesium complex (photosynthesis), haemoglobin is an iron complex (oxygen carrier), vitamin B₁₂ (cyanocobalamin) is a cobalt complex.

(ii) Medicinal chemistry: cis-platin, [Pt(NH₃)₂Cl₂], is used against tumours; EDTA is used to treat lead poisoning; D-penicillamine and desferrioxamine B remove excess copper and iron (chelate therapy).

(iii) Analytical chemistry: Ni²⁺ is detected and estimated with dimethylglyoxime (red complex); hardness of water is estimated by titration with Na₂EDTA, which forms stable complexes with Ca²⁺ and Mg²⁺.

(iv) Metallurgy: gold and silver are extracted as soluble cyanide complexes, e.g. [Au(CN)₂]⁻, and then recovered with zinc; nickel is purified by the Mond process via volatile Ni(CO)₄.

Answer

(iii) 3. [Co(NH₃)₆]Cl₂
⇒ [Co(NH₃)₆]²⁺ + 2Cl⁻, giving three ions.

Answer

(ii) [Fe(H₂O)₆]²⁺. Unpaired electrons: Cr³⁺ (3d³) 3; Fe²⁺ (3d⁶, high spin with water) 4; Zn²⁺ (3d¹⁰) 0. So Fe²⁺ has the highest μ ≈ 4.90 BM.

Answer

(iii) [Fe(C₂O₄)₃]³⁻. Oxalate is a didentate (chelating) ligand forming three five-membered rings around Fe³⁺, so the chelate effect makes it the most stable.

Answer

Field strength: NO₂⁻ > NH₃ > H₂O, so Δo decreases in that order. Since energy absorbed E = hc/λ, a larger Δo means a shorter wavelength.

Order of wavelength absorbed: [Ni(NO₂)₆]⁴⁻ < [Ni(NH₃)₆]²⁺ < [Ni(H₂O)₆]²⁺

Answer

The splitting in a tetrahedral field is small, Δt ≈ (4/9)Δo, so it is usually less than the pairing energy. Electrons therefore occupy the higher orbitals rather than pair, giving high spin complexes.

Answer

Tetraammineaquachloridocobalt(III) chloride; cobalt is in the +3 state.

Answer

For M + 4L ⇌ ML₄, the overall stability constant β₄ = [ML₄]/([M][L]⁴). A high value means the complex is formed to a large extent and is very stable in solution, i.e. the metal–ligand interaction is strong.

Answer

(i) It cannot explain the colour of complexes or give quantitative interpretation of magnetic data. (ii) It does not distinguish between weak and strong ligands and cannot predict exactly whether a 4-coordinate complex will be tetrahedral or square planar; it also gives no account of thermodynamic or kinetic stability.

Answer

A double salt (e.g. Mohr's salt, potash alum) dissociates completely into simple ions in water and exists only in the solid state. A complex (e.g. K₄[Fe(CN)₆]) keeps its complex ion intact in solution, so the central metal ion is not released as a free ion.