Time: Two Hours Maximum Marks : 100
- i) There are TWELVE questions in this paper. Attempt ALL
- ii) Answer each question starting on a new page. The corresponding question number must be written in the left margin. Answer all parts of a question at one place only.
iii) Use of Logarithmic tables is permitted.
- iv) Use of calculator is NOT PERMITTED
Gas Constant R = 8.314 J mol–1K–1
= 0.0821 lit atm mol–1 K–1
= 2 Cal mol–1
Avogadro’s Number N = 6.023 ´ 1023
Planck’s constant h = 6.625 ´ 10–34 J sec.
Velocity of light c = 3 ´ 108 m sec–1
1 electron volt ev = 1.6 ´ 10–19 J
F = 96500 C
Atomic Masses Ag = 108, Mn = 55, Cr = 52, Ca = 40, C = 12, O = 16, K = 39,
Cl = 35.5, N = 14, S = 32, Na = 23, H = 1, P = 31, I = 127, As = 75,
Fe = 56, Ag = 108
Enrolment No. :
- Provide plausible mechanism for each of the following:
|b)||[4 + 4]|
- An organic compound A (MF – C12H13N) does not decolourise Br2/CCl4 readily though it can be nitrated and sulphonated. Upon boiling with dil. acid it is hydrolysed with evolution of NH3 gas. The organic product of hydrolysis upon permanganate oxidation gives a dicarboxylic acid B (MF – C8H6O4) of which only one mononitroisomer is possible and which upon decarboxylation gives benzene. The same compound (hydrolysis product) upon treatment with NBS and subsequent treatment with hot alc. KOH gives a compound C
(MF – C12H12O2). The compound C upon reductive ozonolysis with only one equivalent of O3 gives a compound D(MF – C12H12O4). In the presence of dil. NaOH the compound D undergoes intramolecular aldol condensation. Identify A, B, C, D and two probable aldol condensation products of D which are expected to be formed after acidification and
subsequent heating. 
- a) Given at 25°C:
Cu(s) ¾® Cu2+(aq) + 2e, DG0 = + 65.04 kJ
Ag(s) ¾® Ag+(aq) + e, DG0 = +77.200 kJ
What would be the concentration of Ag+ ion in a solution containing 0.06 M of Cu2+ ion such that both metals can be deposited together? Assume that Ag and Cu do not dissolve in one another.
- b) Write the IUPAC nomenclature of the following:
- i) K3[Fe(NC)5NO]×2H2O
- ii) [Pt(NH3)2Cl2] [ZnCl4] [5 + 5]
- a) An organic liquid, A , immiscible with water, when boiled together with water, has a boiling point of 90°C at which the partial vapour pressure of water is 526 mm Hg. The superincumbent pressure is 736 mm Hg. If mass ratio of liquid and water collected is
2.5 : 1, what is the molecular mass of the liquid?
- b) Photoelectrons are emitted when 400 nm radiation is incident on a surface of work function 1.9 eV. These photoelectrons pass through a region containing a-particles. A maximum energy electron combines with an a-particle to form a He+ ion, emitting a single photon in this process. He+ ions thus formed are in their fourth excited state. Find the energies in eV of the photons, lying in the 2 to 4 eV range that are likely to be emitted during and after the combination (h = 4.14 × 10–15 eVs). [5 + 5]
- a) 100 ml of water containing soluble Ca(HCO3)2 required 30 ml. N/50 HCl solution for the titration using methyl orange indicator. Calculate the temporary hardness of water in ppm unit.
[Density of water is 1 gm/cc]
- b) Calculate the heat of oxidation of ethyl alcohol to CO2 when the heats of formation of CO2 and H2O(l) are given as – 94.05 and – 68,32 kcal and that of liquid ethyl alcohol is
– 66.4 kcal.
- c) Calculate the no. of moles of electrons in 18 ml of water water at 4°C temperature.
[4 + 4 + 2]
- Explain the followings:
- a) The bleaching action of Cl2 is permanent while that of SO2 is temporary.
- b) PbCl4 is available in the form of unstable liquid and gradually decomposes to PbCl2 and Cl2 while PbBr4 and PbI4 couldn’t be prepared.
- c) F2O2 is more stable than H2O2. [3 + 3 + 3]
- A metal sulphate ‘A’ gives sky blue colour in oxidising flame and opaque red in reducing flame during borax bead test.When ‘A’ reacts with BaCl2 it gives a white ppt. ‘B’ which is insoluble in HCl. Into the solution of ‘A’ when excess KI is added two compound ‘C’ and ‘D’ are produced and the gas liberated when passed through starch solution it becomes violet. When potassium ferrocyanide is added to a solution of ‘A’ a chocolate coloured ppt. E appeared along with formation of ‘D’. Identify A to E. 
- Write down the products in the following
[3 + 3]
- Find out the missing alphabets in the following sequence
CH3 – – CO2Et
- The compound Co(en)2(NO2)2Cl has been prepared in a number of isomeric forms. One form undergoes no reaction with either AgNO3 or ethylene diamine. A second form reacts with AgNO3 but not with ethylenediamine and the third form reacts with both AgNO3 and ethylenediamine. Identify each of the these forms by their IUPAC names and discuss the above reaction in light of Werner’s theory. 
- After solid SrCO3 was equilibrated with a pH = 8.60 buffer, the solution was found to have [Sr2+] = 2.2 ´ 10–4. Determine the solubility product of SrCO3? K2 for H2CO3 = 4.7 ´ 10–11.
- A proportional counter is filled in succession with the same amount of 3 samples of CO2 at a fixed pressure and the following counts were noted.
- a) Dead CO2 from coal : 10,800 counts in 900 minutes.
- b) CO2 from contemporary wood : 33,000 counts in 200 minutes.
- CO2 from test sample : 18,800 counts in 400 minutes
If half life of 14C = 5570 year, then estimate the age of sample? 
- a) DG0 = –
So, 65040 = – 2 ´ 96500 ´
\ = – 0.337V
Similarly, 77200 = – 1 ´ 96500 ´ 
\ = – 0.80V
According to Nernst equation
= 0.337 + log(0.06) = 0.301
= 0.80 + log[Ag+] 
Two metals will be deposited together when their electrode potentials are equal
0.301 = 0.80 + 0.059 log[Ag+]
[Ag+] = 10–8.47 = 0.348 ´ 10–8M 
- b) i) Potassium pentacyano(N) nitrosyl ferrate (II)dihydrate
- ii) Diamminedichloroplatinum(IV) tetrachlorozincate(II) [2.5 + 2.5]
- a) At boiling point Pmix = 736 mm
Thus at boiling point,
= 526 mm
= 210 mm
Also, × mole fraction in vapour phase 
Let ‘a’ g of liquid and water is collected or this is the amount of vapours at equilibrium
Thus, Mass of liquid vapours =
Mass of water vapours = 
Now for liquid
210 = 736 ´ …(1) 
Now for H2O
526 = 736 ´ …(2) 
From (1) and (2)
Þ m = 112.7 
- b) The energy of the incident photon is
E1 = = 3.1 eV
The maximum kinetic energy of the emitted electrons is
Emax = E1 – W = 3.1 – 1.9 eV = 1.2 eV 
It is given that
emitted electrons of maximum energy + 2He2+ ¾® + Photon
The 4th excited state implies that the electron enters in the n = 5 electronic state. In this state the energy is
E5 = = – 2.18 eV
The energy of the emitted photon in the above combination reaction is
E = Emax + (–E5) = 1.2 + 2.1 = 3.3 eV 
After recombination reaction, the electron may undergo transitions from a higher level to a lower level thereby emitting photons.
The energies in the lower electronic levels of He+ are
E4 = = – 3.4eV
E3 = = – 6.04 eV
E2 = = – 13.6 eV
The possible transitions are
n = 5 ¾® n = 4 DE = E4 – E5 = –3.4 –(–2.1) = –1.3 eV 
n = 5 ¾® n = 3 DE = E3 – E5 = –6.04–(–2.1) = –3.94 eV 
n = 4 ¾® n = 3 DE = E3 – E4 = –6.04 – (–3.4)= –2.64 eV 
Hence the photons that are likely to be emitted in the range of 2eV to 4 eV are 3.3 eV, 3.94 eV and 2.64 eV.
- a) Ca(HCO3)2 + 2HCl = CaCl2 + 2H2O + 2CO2
No. of ml eq. of HCl used = 30 ´ = no. of ml eq. of CaCl2 produced
= no. of ml eq. of CaCO3 present in 100 ml of water.
\no. of ml eq. of CaCO3 in 106 ml or 106 gm of water = 30 ´ ´ 104 
= ´ 10 gm eq. of CaCO3 = ´ 10 ´ 50 gm of CaCO3
= 300 gm of CaCO3
\ Temporary hardness of water is 300 ppm. 
- b) C2H5OH(l) + 3O2 = 2CO2 + 3H2O(l)
DHf0 – 66.4 0 2 ´(–94.05) 3 ´ (–68.32)
DH for the reaction =
= 2 ´ (–94.05) + 3 ´ (–68.32) – (–66.4)
= – 326.6 kcal
\ 326.6 kcal heat will be evolved in the oxidation of ethanol. 
- c) 18 ml of water = 18 gm water = 1 mole water
as density of water is 1 gm/cc at 4°C temperature
\ in 18 gm of water, no. of molecules of water = 6.023 ´ 1023
In one molecule of water no. of moles of electron is 8
\ Total moles of electron in 18 gm of water
= 8 ´ 6.023 ´ 1023 = 4.818 ´ 1024 
- a) Chlorine bleaches substance by oxidation while SO2 bleaches by reduction. So reduced bleached product after sometimes undergoes oxidation in presence of air or oxygen and SO2 bleaching is only a temporary bleaching. 
- b) ‘Pb’ is the bottom element of group IV A with electronic configuration…….. 4f145d106s26p2. Due to presence of so many ‘d’ and ‘f’ electrons the effective nuclear charge of Pb is very high and two electrons of ‘6s’ orbital remain inert and don’t behave as valence cell electrons (inert pair effect). So (+4) oxidation state of ‘Pb’ is not stable.
Now Pb+4 being hard cation (having high effective nuclear charge) a considerable percentage of covalency is incorporated in PbCl4 and it is liquid and Pb+4 is hard, hence it is to some extent stable with hard base like Cl–. But due to inert pair effect it dissociates into PbCl2 & Cl2. Br– and I–‑ being soft base and Pb+4 being hard and unstable PbBr4 and PbI4 is not at all stable. 
- c) H2O2 and F2O2 both have peroxy linkage which is not so much stable due to lone pair – lone pair repulsion
Due to higher electronegativity of ‘F´atom the electron density on oxygen atoms decreases and interelectronic repulsion in peroxy linkage also decreases. 
- A = CuSO4, B = BaSO4, C = Cu2I2, D = K2SO4, E = Cu2[Fe(CN)6]. [5 ´ 1]
- 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)2]Cl
C = [Co(en)2(NO2)2]Cl
- Ksp = [Sr2+] [CO32–]
CO32– + H2O H CO3– + OH–
Kh = 
[H3O+] = antilog (– 8.6) = 2.51 ´ 10–9
\ [OH–] =
The carbonate ion which dissolves forms HCO3– in a 1 : 1 ratio or remains unreacted. Thus
[Sr2+] = [HCO3–] + [CO32–]
= 53.4 [CO32–] + [CO32-–]
= 54.4 [CO32–]
\ [CO32–] = \ Ksp = 2.2 ´ 10–4 = 8.9 ´ 10–10 
- Counting rate of dead CO2 = = 12 counts/min.
This gives the background counting rate counting rate of contemporary wood
= = 165 cpm (counts/min)
\ Actual counting rate = 165 – 12 = 153 cpm
Counting rate of test sample = cpm
True counting rate of test sample = 47 – 12 = 35 cpm 
\ Age =
= = 11858.19 years