1. (i) Chromium hydroxide is converted into soluble yellow sodium chromate.
3 (ii) Zinc dissolves in caustic potash solution evolving hydrogen
↑
(iii) Deuteromethane is evolved.
4
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Chromium hydroxide is oxidised by H2O2 in presence of NaOH into sodium chromate.
[H2O2 ⎯⎯→ H2O + [O]] × 3 2Cr(OH)3 + 4NaOH + 3[O] ⎯⎯→ 2Na2CrO4 + 4H2O
---2Cr(OH)3 + 4NaOH + 3H2O2 ⎯⎯→ 2Na2CrO4 + 8H2O H2O2 as reductant.
Potassium ferricyanide is reduced to ferrocyanide in presence of KOH by H2O2. 2K3[Fe(CN)6] + 2KOH ⎯⎯→ 2K4 [Fe(CN)6] + H2O + [O]
NH3 gives brown precipitate with Nessler’s reagent.
∴ D is Nessler’s reagent (K2HgI4)
∴ E formed is O Hg
NH I2 Hg
oxydimercuric ammonium iodide
NH3 + 3NaOH + K2HgI4→ O Hg
NH I2 Hg
+ 4KI + 3NaI + 2H2O.
AakashIIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 5. Gas B is chlorine (Cl2)
Ethyl alcohol gives anaesthetic CHCl3
∴ E = CHCl3
The milky precipitate C is CaCO3
∴ A is a compound of Ca and Cl CaCO3 + H2O + CO2→ Ca(HCO3)2 D = Ca(HCO3)2
The inert gas at room temperature by oxidation of NH3 is N2
∴ F = N2
∴ A also has O
A is a compound of Ca, O, Cl
∴ A is CaOCl2 i.e. bleaching powder CaOCl2 + H2O → Ca(OH)2 + Cl2
3CaOCl2 + 2NH3→ 3CaCl2 + 3H2O + N2
Cl2 + H2O → 2HCl + O; O + CH3CH2OH → CH3CHO + H2O
CH3CHO + 3Cl2→ CCl3CHO + 3HCl; 2CCl3CHO + Ca(OH)2→ 2CHCl3 + Ca(HCOO)2 6. 6NaOH + 4S + 5H2O → Na2S2O3·5H2O + 2Na2S + 3H2O
∴ B is Na2S2O3·5H2O Na2S +
) excess
( 4S → Na2S5
∴ C is (Na2S5) A + S → Na2S2O3
∴ A is Na2SO3 D is Na2S2O3
Na2S2O3 + H2SO4→ Na2SO4 + SO2 +
) yellow
( S + H2O
∴ From description F is SO2 E is sulphur
2Na2S2O3 + I2→ NaI + Na2S4O6
∴ G is Na2S4O6 (sodium tetrathionate).
7. C + K2CrO4→ D (yellow ppt.)
∴ D is BaCrO4
C is a carbonate
∴ C is BaCO3
∴ A is Ba(NO3)2 B is BaCl2
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∴ E is Ba(OH)2 F is CO2
6Ba(OH)2 + 6Cl2→
. ppt 3)2
ClO (
Ba + 5BaCl2 + 6H2O
∴ G is Ba(ClO3)2. 8. D = AgI (from property)
B = KI (from 3rd reaction) C = CuI2
∴ A = CuSO4.
9. The first part is bead test using microcosmic salt
∴ C is NaPO3 and A is Na(NH4)HPO4
Na(NH4)HPO4 ⎯⎯→Δ NaPO3 + H2O + NH3 Na2HPO4 + NH4Cl → Na(NH4)HPO4 + NaCl
∴ B is Na2HPO4
E is Na2SiO3 (present in glass) So, D is a oxide of Si
D is SiO2
Na2SiO3 + 6HF → Na2SiF6 + 3H2O
∴ F is Na2SiF6.
10. Oxide B is amphoteric as it dissolves in NaOH. BeO is the only amphoteric oxide
∴ B = BeO A = Be
BeO + 2NaOH → Na2BeO2 + H2O
∴ C = Na2BeO2
BeO + C + Cl2→ BeCl2 + CO
∴ D = BeCl2 E = CO BeCl2 +
) air of (H2O
2 → Be(OH)2 +
) fumes (2HCl F = HCl
BeO + 4NH4F → (NH4)2BeF4 + 2NH3 + H2O (NH4)2BeF4 ⎯⎯→Δ BeF2 + 2NH4F
AakashIIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 11. Red ppt. is of HgI2
∴ D is HgI2
A is iodide salt Only KI dissolves free I2
∴ KI + I2→ KI3
12. Ellingham diagram can be used to predict which metal/non-metal can be used to extract any other metal from its oxide by using smelting process.
Any metal having its metal-metal oxide line above any other metal can be extracted by other metals having their metal-metal oxide line below in Ellingham diagram. Ex. Al can be used to extract Cr, Fe etc. from their oxide.
14. In Ellingham diagram the line of Al – Al2O3 is below metal-metal oxide line of other metals. Thus coupled reaction of its oxide with other metal will have positive Gibb’s free energy change value. Which doesn’t favour its reduction using smelting.
15. (i) AlCl3, 6H2O exists as [Al(H2O)6]Cl3. Upon heating AlCl3 undergoes hydrolysis and forms Al2O3. (ii) AlCl3 lacks back-bonding as in BCl3 because of large size of Aluminium.
Aluminium metal forms complete octet by coordinate bridges by chlorine atoms between two Al atoms.
(iii) On heating, borax first swells up due to elimination of water molecules. On further heating, it melts to a liquid which then solidifies to a transparent glassy mass.
4
When the hot bead is touched with a coloured salt, B2O3 displaces the volatile oxides and combines with basic oxides to form metaborates.
e.g. CuSO + B O CuO · B O + SO4 2 3 2 3 3
Cu(BO ) 2 2
Blue
Colour of metaborates : Cu
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Acidic character decreases down the group.
(ii) NF3 > NCl3 > NBr3
Stability decreases due to increasing size of halogen atom.
(iii) HNO3 > H3PO4 > H3AsO4 > H3SbO4
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A weighed quantity of Bleaching powder is suspended in water and treated with excess of acetic acid and KI. The liberated iodine is estimated by titrating it with a standard solution of hypo using starch as indicator.
(ii) Chlorides when heated with K2Cr2O7 and the H2SO4 evolve chromyl chloride (orange vapours) which when passed through lead acetate gives yellow ppt. of PbCrO4
COOH
(iii) (fluoride salt)
6
AakashIIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 21. It disproportionates in solution giving purple colored permanganate and brown colored MnO2.
4MnO42– + 4H+→ 3MnO4– + MnO2 + 2H2O
acidic medion is obtained due to dissociation of H2CO3 formed.
22. [VO ]43– pH12 [VO3·OH]2– pH10 [V O2 6·OH]3– pH·9 [V O ]3 93–
colorless colorless colorless orange
[V O ]5 143–
pH6.5
V O2 5·(H O)2 n
pH2.2
[V O ]10 286–
[VO ]2+ pH1
pH·7
brown ppt. red
23. (i) Ni + 22+
CH C = NOH3 CH C = NOH3
+ 2NH OH4
CH C = N3 CH C = N3
Ni2+
N = C – CH3 N = C – CH3 OH ···O
O ···HO ·
bis(dimethyl glyoximato) nickel (II) (ii) Ni is in +2 oxidation state and the complex is square due to dsp2 hybridization.
(iii) The complex is diamagnetic due to absence of unpaired electron.
24. (i) SnCl2 reduces HgCl2 to Hg2Cl2 first and then to Hg SnCl2 + 2HgCl2→ Hg2Cl2 + SnCl4
Hg2Cl2 + SnCl2→ 2Hg + SnCl4
(ii) In the solution, the following equilibria exists : Cr2O72– + H2O 2CrO42– + 2H+
In acidic medium (pH < 7), it exists as Cr2O72– ions and has orange color while in basic medium (pH > 7), it exists as CrO42– ions and has yellow color.
25. Ksp of CuS is less than Ksp of ZnS. On passing H2S in acidic medium, the dissociation of H2S is suppressed due to common ion effect and its provides a limited conc. of S2– ions enough of exceed Ksp of CuS but not ZnS. Thus only CuS gets precipitated.
26. (i) 82 (vi) 34
(ii) 35 (vii) 33
(iii) 30 (viii) 86
(iv) 38 (ix) 54
(v) 53 (x) 86
27. (i) Tris (ethylenediammine) cobalt (III) chloride (ii) Tris-oxalatocobaltate (III) ion
(iii) Decaammine-µ-peroxodicobalt (III) ion
(iv) Hexaammine chromium (III) hexa cyano cobaltate (III) (v) Bis (dimethylglyoximato) nickel (II)
27(a). Answer (3) IIT-JEE-2008
IUPAC name is tetraamminenickel(II)-tetrachloronickelate(II).
AakashIIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 28. (i) [NH4]2[FeF5(H2O)] (vi) [Ni(NH3)4] (ClO4)2
(ii) [Co(H2O)2(en)2]2 (SO4)3 (vii) [Ni(NH3)6]3 [Co(NO2)6]2 (iii) [Zn(NCS)4]2– (viii) [Ni(CO)2(PPh3)2]
(iv) Na3[Co(NO2)6] (ix) (en) Co Co(en)2 2 NH2
OH
(SO )4 2
(v) [Fe(en)3] [Fe(CN)4] (x) (en) Co Co(en)2 2 NH2
OH
(SO )4 2
29. (i) d 2sp3 (ii) sp 3d 2 (iii) dsp2 (iv) sp3
30. Au 53+ d8
5d 6s 6p
[AuCl ]¯4
5d 6s 6p
dsp2 hybridisation
Ga3+
3d10 4s 4p
[GaCl ]¯4
3d 4s 4p
sp3 hybridisation
31. A = [Cr(H2O)6]Cl3 ⎯⎯H2SO⎯⎯4→ No reaction
(∴ All H2O molecule are present in co-ordination sphere) B = [CrCl(H2O)5]Cl2⋅H2O⎯⎯H2SO⎯⎯4→ one mole of H2O is lost Molecular weight of complex = 266.5
% loss = 100 6.75% 5
. 266
18 × =
C = [CrCl2(H2O)4]Cl2⋅2H2O⎯⎯H2SO⎯⎯4→ 2 moles of H2O is removed
∴ % loss in weight = 100 13.50% 5
. 266
18 2× × =
AakashIIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 32. [Pt(NH3)4] [PtCl4] ; Tetraammineplatinum (II) tetrachloroplatinate (II)
[Pt(NH3)4] [NO3]2 ; Tetraammineplatinum (II) nitrate Ag2[PtCl4]; Silvertetrachloroplatinate (II)
33. (A) = H2S2O8, (B) = H2SO4, (C) = H2O2 and (D) = BaSO4
34. (i) Hg2(NO3)2 + 2KI Hg2I2 + 2KNO3 (ii) Hg2I2 + 2KI K2HgI4 + Hg
(iii) NiSO + 2H C — C == NOH 4 3 H C — C == NOH3
H C — C == N3 OH
H C — C == N3 O
Ni
N == C — CH3
HO O
N == C — CH3
+ (NH ) SO + 2H O4 2 4 2 + 2NH OH4
35. A = Hg2(NO3)2, B = Hg2Cl2, C = HgCl2, D = K2HgI4 E = Hg2, F = FeSO4 · NO 36. A = PH4I,
B = PH3, C = KI, D = P2O5 E = Cu2I2 37. A = K2MnO4,
B = KMnO4, C = KIO3, D = Mn2O7, E = MnO2
38. (i) Turns lime water milky, thus (A) is either CO2 or SO2 gas
(ii) (Y) gives alkaline solution and its solution forms white ppt (Z) with BaCl2 and (Z) on heating with acid gives effervescences of CO2, so (Z) is BaCO3 and (Y) is metal carbonate
(iii) Since (Y) and (A) are formed from (X) and thus, (X) is metal bicarbonate and (A) is CO2
AakashIIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 (v) The above data reveal that
2MHCO3→ CO2 + H2O + M2CO3 Hence metal is Na, i.e.,
)
39. White phosphorus ⎯⎯ →Air⎯ waxy crystalline solid having garlic smell (A).
(A) + Hot water →
PH3 thus smell,rottenfish having
(B)
Gas + Acid (C)
CuSO4 solution + gas (B) → Black ppt of cupric phosphide (D) Reactions involved
:-P4 + 3O2→
40. Let us summarise the reaction
(A) NaCl conc.H SO (B) yellow solution
AakashIIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 Na2CrO4 + 2AgNO3→ Ag2CrO4↓ + 2NaNO3
Red ppt
2Na2CrO4 + H2SO4⎯⎯→ Na2Cr2O7 + Na2SO4 + H2O Cr2O72– + 14H+ + 6 I– ⎯⎯→ 2Cr3+ + 7H2O + 3I2 Quantitative analysis
CrO2Cl2≡ Na2CrO4≡ 2AgNO3
155 g 2 mol
155 g of CrO2Cl2 requires 2 moles of AgNO3
⇒ 0.155 g of CrO2Cl2 requires 0.002 mole (2 m.mole) Similarly. 2 CrO2Cl2≡ 2 CrO4
2–≡ Cr2O72–≡ 3I2
⇒ 2 × 155 g
0.155 g will liberate 10 3 2
3 −
× moles = 1.5 milli mole