LEVEL – I
- K2[CO(CN)5] Potassium pentocyanocobaltate(II)
Co3+: d6 configuration
As CN– is a strong ligand hence it forces the electrons to pair up. There are no unpaired electrons left so diamagnetic
But in aq. solution the ligand is H2O which is a weak one in comparison to CN–
There are 4 unpaired electrons and hence paramagnetic
- Hydrazinium tris – hydrazine carboxylato nickellate (II), or, hydrazinium tris-carbazato nickelate (II)
EAN = 28– 2 + 6 × 2 = 38
μs = = 2.8 B.M
- Cu is very slowly oxidised on the surface in moist air giving a green coating, which is basic copper carborate CuCO3.Cu(OH)2.
2Cu + H2O + CO2 + O2 ⎯→
- This is because a part of the oxygen produced from KMnO4 + HCl is used is oxidizing HCl to Cl2
4HCl + 2(O) → 2H2O + 2Cl2
- No structural isomers because co ordination number of orbital is 6 but two geometrical isomers. viz Cl2 Cis and trans.
- On addition of chloride ion, the pale red coloured octahedral complex [Co(H2O)6]2+ changes into tetrahedral [CoCl4]2– ion. In Tetrahedral complexes, the d–d absorption bands are considerably more intense than in octahedral complexes.
- The lanthanoids which will most likely react with this strong reducing agent are those with stable +2 oxidation states to which they can be reduced –Sm, Eu and Yb.
- The larger the number of ions and the larger the charge on each, the larger the conductivity. The compounds from lowest conductivity to highest conductivity are:
b < a < d < c
- EDTA coordinates lead ion in the body, in which form it is passed out of the body without harmful effects. The calcium salt is used so that any excess EDTA will not remove calcium ions from the body.
LEVEL – II
- i) (NH4)3[ZrF7]
- ii) [FeCl2(H2O)4]+
iii) [Ag(NH3)2]4 [Fe(CN)6]
- iv) [Cr(en)4Cl2]2[PdCl4]
- v) [Al(H2O)2(OH)4]2−
- vi) [Co(en)2(CN)2]ClO3
- ix) K2[NiF6]
- x) [Pt (NH3)3Br]NO2
- xi) Na2[Co(H2O)2(OH)4]
- xv) Na4[Fe(CN)5(NO)S]
- xx) K[MnO4]
- i) Bromoaquatetraamminecobalt(III)nitrate
- ii) Sodiumdicyanoaurate(I)
- iv) Tris(ethylenediamine)cobalt(III)chloride
- v) μ-hydroxobis(pentamminechromium(III)) chloride
- vi) chlorotriphenylphosphinepalladium(II)-μ dichlorochlorotriphenylphosphinepalladium (II)
- ix) Bis(cyclopentadienyl)iron(II)
- x) Sodiumpentacarbonylmanganate(–I)
- xi) Dichlorobis(dimethylglyoxime)cobalt(II)
xiii) Hexaamminechromium (III) ion
xiv) Hexaaquamanganese (II) ion
- xv) Hexacyanoferrate (II) ion
xvi) Hexaamminenickel (II) chloride
xvii) hexacyanocobaltate (III) ion
xviii) Calciumhexacyanoferrate (II)
xix) Hexammine cobalt (III) chloride
- xx) Tetraaquadichlorochromium (II) nitrate
xxi) Tetraaminechlorornitrocobalt (III) nitrate
xxii) Potassium amminetrichloro platinate (II)
xxiii) Tetrapyridine platinium (II) tetrachloroplatinate (II)
xxiv) Sodium bisthiosulphatoargentate(I)
xxv) Hexaamminecobalt(III) chloride
xxix) Tris(ethylenediamine)chromium(III) chloride.
xxx) Potassium hexacyano-N-ferrate(II)
xxxi) Pentaamminenitritocobalt(III) chloride
xxxii) Potassium hexacyano-C-chromate(III)
xxxiii) Pentaamminenitrocobalt(III) nitrate
xxxiv) Diamminebromochloroethylenediaminecobalt(III) nitrate
LEVEL – III
- For complex [Mn(CN)6]3–, the number of unpaired electrons is calculated as,
2.8 = ⇒ n = 2
[Mn(CN)6]3– has two unpaired electrons. Hence the geometry is octahedral with d2sp3 hybridisation. For complex [MnBr4]2–, the number of unpaired electrons is calculated as
5.9 = ⇒ n = 5
[MnBr4]2–, has 5 unpaired electrons. Hence the geometry is tetrahedral with sp3 hybridisation.
- One form does not undergo reaction with AgNO3 indicating absence of free Cl– i.e. Cl– in the outer sphere and with ethylene diamine indicating that the Cl– and NO2– are anti to each other. Had they been cis, they should have been replaced by ethylene diamine which is a bidentate ligand.
Second form reacts with AgNO3 but not with ethylene diamine which indicates availability of free Cl–. The third compound reacts with AgNO3 and ethylene diamine indicating Cl– in the outer sphere and presence of NO2– in the cis-position.
A = [Co(en)2 (NO2) Cl]NO2
B = [Co(en)2(NO2)]Cl
C = [Co(en)2(NO2)2]Cl
C = Tetraammineaquahydroxocobalt(III)sulfate
D = Di-μ-hydroxobis[tetraamminecobalt(III)]sulfate
- i) CuII is reduced to CuI by I– but not by Cl–.
- ii) The oxidation is aided by the complexation of the product gold (III) as
iii) The stability of ion is so great that the E value for oxidation is reduced to a point where H2O can oxidise the copper.
- iv) When zinc is exposed to moist air, the surface is affected with the formation of a film of basic zinc carbonate on it. Due to this zinc becomes dull.
4Zn + 3H2O + CO2 + 2O2 ⎯→ ZnCO3⋅3Zn(OH)2
- i) Ferrous sulfate is a salt of a weak base and a strong acid. Thus its hydrolysis occurs when it is dissolved in water and solutioin becomes turbid due to formation of ferrous hydroxide
FeSO4 + 2H2O Fe(OH)2 + H2SO4
Addition of a small amount of acid shifts the equilibrium towards left and prevents hydrolysis.
- ii) HgCl2 is an oxidizing agent while SnCl2 is a reducing agent. When both are present, a redox reaction occurs forming Hg and stannic chloride as final products.
SnCl2 + 2HgCl2 ⎯→ Hg2Cl2 + SnCl4
Hg2Cl2 + SnCl2 ⎯→ 2Hg + SnCl4
iii) A ferrous salt turns brown in air due to oxidation to ferric salt
- iv) Cu(OH)2 dissolves in NH4OH by forming a complex.
Cu(OH)2 + 4NH2OH ⎯→ [Cu(NH3)4](OH)2 + 4H2O
Cu(OH)2 is insoluble in NaOH as no such complex is formed
- In aq. solution (0.001 M) the complexes will dissociate to give the ions.
[Pt(NH3)6]Cl4 [Pt(NH3)6]4+ + 4Cl– (5 ions)
[Cr(NH3)6]Cl3 [Cr(NH3)6]3+ + 3Cl– (4 ions)
Co(NH3)4⋅Cl3 [Co(NH3)4Cl2]+ + Cl– (2 ions)
K2[PtCl6] [PtCl6]2– + 2K+ (3 ions)
Thus, their conductivities (in solution) will increase with increasing number of ions liberated
Co(NH3)4⋅Cl3 < K2PtCl6 < Cr(NH3)6Cl3 < Pt(NH3)6Cl4
(least conducting) (most conducting)
- i) The isomers can be distinguished by using AgNO3 reagent. One gives curdy precipitate of AgCl soluble in ammonia while the other will form light yellow precipitate of AgBr partly soluble in ammonia.
- ii) The isomers can be distinguished by passing their aq. solutions through a cation exchanger. In the isomer [Co(NH3)6] [Cr(NO2)6] the cation [Co(NH3)6]3+ will be replaced by H+ ions from the resin. The resulting solution will thus contain
H3[Cr(NO2)6]. In the other case, the resulting solution after passage through the exchanger will contain H3[Co(NO2)6]. On adding KCl solution, [Co(NO2)6]3– ion will give an yellow precipitate of potassium cobaltinitrate, K3[Co(NO2)6]. Hence the resulting solutions obtained from the two isomers can be distinguished.
- The compound which conducts electricity means it gives ions in aq. solution. Keeping in view the six co-ordination number of chromium and assuming that the H2O molecules enter into the co-ordination sphere, its possible structures can be one of the following.
- i) [Cr(NH3)3(NO2)2H2O]NO2
- ii) [Cr(NH3)3NO2(H2O)2](NO2)2
The compound which does not conduct electricity means it does not give ions in solution. Hence it is neutral (or) uncharged complex. Its structure may be written as
- 143.45g of AgCl contains 35.45 g Cl– ions
∴ 2.87g of AgCl will contain = 0.709 g Cl– ions
266.35 g CrCl3⋅6H2O contains n × 35.45 g of ionisable Cl– ions (where n = no. of Cl– ions outside the coordination sphere).
Thus, 2.665g CrCl3⋅6H2O will contain g
Also = 0.709 ⇒ n ≈ 2
Keeping in view the octahedral geometry of the complex, its structure may be written as
- i) Fe3+ is more paramagnetic than Fe2+ as Fe3+ consists five unpaired electrons while Fe2+ possesses four unpaired electrons.
- ii) Any ion of transition elements which possesses unpaired d-electrons, i.e. d – d transition is possible shows a characteristic colour (n –1)d0 (or) (n–1)d10 configuration does not involve d–d transition and hence, is colourless
Fe2+, Mn2+ and Cr3+ are coloured, while Cu+, Se3+ and Ti4+ are colourless.
iii) Chromium and copper. Chromium attains 3d54s1 configuration in which all the d-orbitals are unpaired in order to get extra stability. Copper attains 3d104s1 configuration in which all the d-orbitals are paired in order to get extra-stability.