rbse - flashback chemistry€¦ · 1 rbse - flashback chemistry solid state rbse 2012 q.1 give...
TRANSCRIPT
1
RBSE - FLASHBACK CHEMISTRY
Solid State
RBSE 2012
Q.1 Give packing efficiency and coordination number of the following crystal structures-
(i) Body centered cubic (ii) Cubic close packing
(1 out of 56)
Q.2 Ferromagnetic substances make permanent magnet. Give reason.
(1 out of 56) Q.3 Glass is considered as a super-cooled liquid. Give reason.
(1 out of 56)
RBSE 2013
Q.1 Which of the following are polar molecular solids?
Solid sulphur dioxide, solid ammonia, iodine crystals, graphite, carbon tetrachloride. (1 out of 56)
Q.2 Explain insulator and semiconductor on the basis of conduction of electricity. (2 out of 56)
RBSE 2014
Q.1 (a) Assuming that atoms are touching each other, calculate the efficiency of packing in case of a
crystal for simple cubic metal.
(b) On the basis of nature of ionic solids compare Frenkel defect with Schottky defect. (3 out of 56)
RBSE 2015
Q.1 (a) Write the values of axial angles for hexagonal crystal system.
(b) Which type of semiconductor is obtained by doping boron with silicon.Explain
(3 out of 56)
Solution
RBSE 2012
Q.1 Define reverse osmosis. Give one use of it.
(1 out of 56)
Q.2 Calculate the molality of H2SO4 solution if the density of 10% (w/w) aqueous solution of
H2SO4 is 1.84 gm cm-3
. (Molar mass of H2SO4=98 gm cm-3
)
(1 out of 56)
Q.3 Arrange the following solution in increasing order of Vant Hoff factor :
0.1M CaCl2, 0.1M Al2(SO4)3, 0.1M C12H22O11
(1 out of 56)
RBSE 2013
Q.1 5g of NaOH are dissolved in 500 ml water. Find the molarity of the solution. (1 out of 56)
Q.2 0.2 L of aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such
solution at 300 K is found to be 2.57×10-3
bar. Calculate the molar mass of the protein.
(R = 0.083 L bar mol-1
K-1
) (2 out of 56)
RBSE 2014
Q.1 (a) (i) What happens to vapour pressure of water if a tablespoon of sugar is added to it?
(ii) Which colligative property is preferred for the molar mass determination of macromolecules?
(b) Will the elevation in boiling point be same if 0.1 mole of sodium chloride or 0.1 mole of sugar is
dissolved in 1 L of water?
(c) Can we separate the compounds of azeotropic mixture by fractional distillation? Explain.
(3 out of 56)
2
RBSE 2015
Q.1 Write the formula to calculate the molality.
(1 out of 56)
Q.2 Osmotic pressure of a solution is 0.0821 atm at a temperature of 400k. Calculate the concentration of
solution in mol/litre.
[R=0.821 L atm K-1
mol-1
]
(2 out of 56)
Electrochemistry
RBSE 2012
Q.1 A solution of AgNO3 is electrolysed for 30 Minutes through a current of 2.0 amperes.
What will be the mass of silver deposited at the cathode?[Molar mass of Ag = 108 gm
mol-1
; F = 96500 coulomb mol-1
]
(2 out of 56)
Q.2 0
m for KCI, HCI and CH3COOK are 149.8 S cm2mol
-1, 425.9 S cm
2 mol
-1 and 114.4 S cm
2 mol
-1
respectively. Calculate the value of 0
m for CH3COOH.
(2 out of 56)
RBSE 2013
Q.1 Write Nernst equation for Daniel Cell at equilibrium condition and derive relationship between
E0
(cell) and equilibrium constant (Kc). (2 out of 56)
Q.2 (a) Draw the labelled diagram of Daniell cell.
(b) Write the half-reactions of oxidation and reduction taking place on electrodes. (2 out of 56)
RBSE 2014
Q.1 (a) Can we store CuSO4 solution in an iron vessel? Explain.
(b) Write Kohlrausch law and give one application of it. (2 out of 56)
Q.2 (a) Write the complete chemical reaction of rusting of iron.
(b) Give overall reaction of Daniel cell and write mathematical expression of Nernst equation for it.
(2 out of 56)
RBSE 2015
Q.1 Write the SI unit of Resistivity.
(1 out of 56)
Q.2 Limiting molar conductivity of 2Mg and Cl ions are 106.0 Scm2 mol
-1 and 76.35 Scm
2 mol
-1
respectively. Calculate the limiting molar conductivity of MgCl2.
(1 out of 56) Q.3 (a) Write Faraday‘s second law of electrolysis.
(b) Draw labelled diagram of standard Hydrogen electrode.
(2 out of 56)
Chemical Kinetics
RBSE 2012
Q.1 (a) Rate constant of chemical reaction rises to double by 10oC increase in temperature.
Explain the statement with labelled distribution curve.
(b)The rate constant of a first order reaction increases four times when the temperature
changes from 350K to 400K Calculate the energy of activation of the reaction assuming that it
does not change with temperature. (R =8.314 JK-1
mol-1
, log 4 = 0.6021)
3
(4 out of 56)
RBSE 2013
Q.1 Explain pseudo first order reaction by taking the example of hydrolysis of ethyl acetate. (1 out of 56)
Q.2 Write the units of rate constants for zero order and first order reactions. (1 out of 56)
Q.3 What is half-life of a reaction? Derive formula for finding out half-life from first order rate reaction.
(2 out of 56)
RBSE 2014
Q.1 Give the order of the following reactions:
(a) Artificial radioactive decay
(b) Decomposition of gaseous ammonia on a hot Pt-surface at high pressure
(c) Hydrogenation of ethane
(d) Decomposition of N2O5. (2 out of 56)
Q.2 What is half-life period of reaction? Justify from the rate equation for first order reactions that the half
life period for such reaction is independent of the initial concentrations of the reactants. (2 out of 56)
RBSE 2015
Q.1 For a reaction the unit of rate constant is sec-1
. What will be the order of reaction ?
(1 out of 56)
Q,2 Write differential rate equation for the reaction, 2A B Product .
(1 out of 56)
Q.3 Half life period for first order reaction is 10sec. Calculate the rate constant for the reaction.
(2 out of 56)
Surface Chemistry
RBSE 2012
Q.1 Colloidal solution show Tyndall effect. Give two reasons.
(1 out of 56)
Q.2 Hydrophobic sol is easily coagulated. Give reason.
(1 out of 56)
Q.3 Give specific term to show the effect of following processes-
(a) Ferric hydroxide sol is mixed with arsenic sulphide sol.
(b) Ferric chloride solution is mixed with freshly prepared precipitate of ferric hydroxide.
(c) H2S gas is passed through arsenic oxide solution.
(d) A beam of light is passed through a colloidal solution.
(2 out of 56)
RBSE 2013
Q.1 (a) Write two differences between physical adsorption and chemical adsorption.
(b) Define homogeneous and heterogeneous catalysis. Give an example of each.
(c) Draw labelled diagram of eletro-dialysis.
OR
(a) Write two differences between lyophilic and lyophobic colloides.
(b) Give the manes of two enzymes present in yeast. Also write equations of the enzyme catalysed
reactions.
4
(c) Draw labelled diagram of Bredig‘s Arc method. (4 out of 56)
RBSE 2014
Q.1 (a) Describe the method for the preparation of gold colloidal sol with diagram.
(b) What is the difference between a sol and a gel?
(c) What is coagulation? Explain.
OR
(a) Describe the method for the purification of colloidal sol containing excess amount of electrolytes
with diagram.
(b) What is the difference between aerosol and foam?
(c) What is peptization? Explain. (4 out of 56)
RBSE 2015
Q.1 (a) What is enzyme catalysis ? Write an example
(b) Which type of emulsion is milk ? Explain.
(c) Explain electrophoresis with labelled diagram.
OR
(a) What is selectivity of catalyst ? Write example.
(b) What happens when hydrated ferric oxide sol is mixed with arsenius sulphide sol ?
(c) Explain Tyndall effect with labelled diagram.
(4 out of 56)
General Principles & Processes of Isolation of Elements
RBSE 2012
Q.1 Give the name and role of collector and froth stabilizer in froth floatation process.
(2 out of 56)
RBSE 2013
Q.1 Write the names of any two impurities present in bauxite ore? (1 out of 56)
Q.2 Write the chemical reactions related to Mond process for refining nickel metal. (1 out of 56)
RBSE 2014
Q.1 Draw labeled diagram of electrolytic cell for the extraction of aluminium and the overall reaction
taking place in it.
OR
Draw labeled diagram of Zone refining process. In which this process is useful? (2 out of 56)
RBSE 2015
Q.1 Explain calcination and roasting with example.
(2 out of 56)
The p-block Elements
RBSE 2012
Q.1 (a) Complete the following reactions-
(i) Cl2 + NaOH (cold & dil.) → …….
(ii) C + conc. HNO3 → ………….
(b) Explain the following –
(i) F2 is a strong oxidizing agent in 17th
group.
(ii) Oxygen is a gas but sulphur is a solid
(a) Draw structures of the following-
5
(i) N2O5 (ii) H3PO4
(4 out of 56)
RBSE 2013
Q.1 (a) Write the name and electronic configuration of the third element from the top in group 15.
(b) Draw the structure of ammonia molecule.
(c) Why does NH3 act as Lewis base?
(d) Give equations for the reactions of dilute and concentrated HNO3 with Zn.
OR
(a) Write the name and electronic configuration of metal element of group 15.
(b) Explain the reason of bleaching action of chlorine gas.
(c) Write equations of brown ring test.
(d) Draw the structure of PCl5 molecule. (4 out of 56)
RBSE 2014
Q.1 (a) Write the chemical formulae of four oxy acids of chlorine.
(b) Write the general electronic configuration of noble gas group. Which element of this group is
useful in magnetic resonance imaging (MRI)?
(c) Write the chemical reactions of C2H5OH with PCl3 and PCl5 separately.
OR
(a) Write the chemical formulae of four oxy acids of sulphur.
(b) Write the general electronic configuration of chalcogen group. Write the chemical formula of
Epsom salt.
(c) Ammonia act as a Lewis base. Explain. (4 out of 56)
RBSE 2015
Q.1 (a) What is Lanthanoid contraction
(b) What are Interstitial compounds ? Give an example.
(c) Calculate the ‗spin only‘ magnetic moment of M2+
(aq.) ion. (z = 29)
(3 out of 56)
Q.2 What is the basicity of H3PO3?
(1 out of 56) Q.3 (a) Write an equation of the reaction of chlorine with cold and dilute NaOH solution.
(b) Explain the reducing nature of H3PO2.
OR
(a) Write an equation of reaction of chlorine with hot and concentrated NaOH solution
(b) Why are all the five bonds of PCl5 not equivalent ? Explain
(2 out of 56)
The d and f block Elements
RBSE 2012
Q.1 Calculate the magnetic moment of Ni2+
ion.
(1 out of 56)
Q.2 Give reasons-
(a) Mn shows maximum oxidation states in 3d series of transition elements.
(b) Both Cr2+
and Mn3+
have d4 configuration but Cr
2+ has reducing while Mn
3+ has
oxidizing properties.
(2 out of 56)
RBSE 2013
Q.1 Zn, Cd and Hg are not regarded as transition elements. Give reason. (1 out of 56)
Q.2 What is the shape of chromate ion ? Draw its structure. (1 out of 56)
6
Q.3 Ti4+
ion is colourless. Give reason. (1 out of 56)
RBSE 2014
Q.1 Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only? (1 out of 56)
Q.2 (a) Silver atom has completely filled d-orbitals (4d10
) in its ground state, even then it is a transition
element. How?
(b) Explain actinoid contraction. (2 out of 56)
RBSE 2015
Q.1 (a) What is Lanthanoid contraction
(b) What are Interstitial compounds ? Give an example.
(c) Calculate the ‗spin only‘ magnetic moment of M2+
(aq.) ion. (z = 29)
(3 out of 56
Coordination Compounds
RBSE 2012
Q.1 [NiCl4]2-
ion is paramagnetic but [Ni(CN)4]2-
ion is diamagnetic Explain with the help of
valence bond theory.
(2 out of 56)
Q.2 Write I.U.P.A.C. names of the following complex compounds-
(a) [CoCl2 (en)2] Cl [b] K3 [Fe(CN)6]
(2 out of 56)
RBSE 2013
Q.1 Taking an example of an ambidentate ligand, explain why it is called as ambidentate ligand.
(1 out of 56)
Q.2 Give the oxidation number and coordination number of the central metal atom in the complex
compound K3[Fe(C2O4)3]. (1 out of 56)
Q.3 Give the structures of cis-[CoCl2(en)2] and facial-[Co(NH3)3(NO2)3] isomers. (2 out of 56)
Q.4 Explain on the basis of valence bond theory that [Ni(CN)4]2-
is a low spin complex ion.
(2 out of 56)
RBSE 2014
Q.1 (a) Explain the applications of coordination compounds in the field of metal purification giving an
example.
(b) Write the IUPAC name of [Ag(NH3)2][Ag(CN)2] coordination compound. (2 out of 56)
Q.2 Explain the difference between a double-salt and a complex, giving an example in each case.
(2 out of 56)
RBSE 2015
Q.1 Write the formula of ionisation isomer of 2 24Cr H O Br Cl .
(1 out of 56)
Q.2 Write the formula of coordination compound mercury tetrathiocyanato-cobaltate (III).
(1 out of 56)
7
Q.3 Explain on the basis of valence bond theory that [Ni(CN)4]2-
is a low spin complex ion.
(2 out of 56)
Haloalkanes and Haloarenes
RBSE 2012
Q.1 Which compound in each of the following pairs will react faster by SN1 reaction in aqueous KOH?
Give reason-
(a) CH3 - CH2 - Br or CH3 - CH2 – Cl ;
(b) CH3 – CH2 – CH2 – CH2 – X or (CH3)3 CX.
OR
Write the mechanism of the following reactions-
(a) CH3 – Cl + aq. KOH → CH3 – OH + KCI
(b) (CH3)3CCl + aq. KOH → (CH3)3COH+KCI
(2 out of 56)
Q.2 Write the equation for the following reactions-
(a) Fittig reaction (b) Finkelstein reaction.
(2 out of 56)
RBSE 2013
Q.1 Write the structural formulae of benzylic chloride and vinylic chloride. Also give the state of
hybridisation of the carbon atoms in these two compounds linked to chlorine atoms. (2 out of 56)
Q.2 Write the equation for the following reactions :
(a) Wurtz reaction
(b) Wurtz-fitting recation (2 out of 56)
RBSE 2014
Q.1 Predict the products of the following reactions:
3 2 (aq. alcohol)(a)CH CH Br AgCN
3 2 2 (aq. alcohol)(b)CH CH Br KNO
(c)
Cl
+ Nadry ether
dry ether
3(d)CH I Mg (2 out of 56)
Q.2 Describe any two differences in substitution nucleophilic bimolecular (SN2) and substitution
nucleophilic unimolecular (SN1) reaction mechanisms. (2 out of 56)
RBSE 2015
Q.1 Complete the following reaction and explain its mechanism :
3 3CH C Br OH aq.
(2 out of 56)
Q.2 Complete the following chemical reactions.
(a)
8
(b) 3 2CH CH Cl KCN
(c) 3 2CH CH CH HI
(d) 3 2 2 4CH CH OH NaBr H SO
(2 out of 56)
Alcohols, Phenols and Ethers
RBSE 2012
Q.1 Give reasons for the following-
(a) Phenol is more acidic than Ethanol.
(b) Boiling Point of propanol is greater than that of butane.
(2 out of 56)
RBSE 2013
Q.1 (a) Write the IUPAC name and structural formula of secondary-Butyl alcohol.
(b) Give equations for preparation of phenol from the following compounds:
(i) Benzene
(ii) Aniline
(c) Write the resonance structures of phenol.
OR
(a) Write the IUPAC name and structural formula of methyl n-propyl ether.
(b) Give equations for preparation of ether by the following reactions :
(i) Dehydration of alcohols
(ii) Williamson synthesis
(c) Write the resonance structures of alkoxy benzene. (4 out of 56)
RBSE 2014
Q.1 How will you convert phenol to benzene? (1 out of 56)
Q.2 What is the major product when ethanol is treated with conc. H2SO4 at 413 K? (1 out of 56)
Q.3 What happens when the 1o and 2
o alcohols are oxidized by anhydrous chromium trioxide (Cr O3)?
Write chemical equations. (2 out of 56)
RBSE 2015
Q.2 (a) Write IUPAC name of the following compounds :
(i) 2 2CH CH CH OH
(ii) 2 2CH CH CH
| | |OH OH OH
(b) Write equations of the following chemical reactions :
(i) Phenol reacts with Br2 in presence of CS2
(ii) Ethanol is heated at 573 k in presence of Cu.
OR
(a) Write IUPAC name of the following compounds :
(i) 3 3
3
CH O CH CH|CH
(ii) 3 2
3
CH CH CH OH|CH
(b) Write equation of the following chemical reactions.
(i) Phenol reacts with conc. HNO3
9
(ii) Phenol reacts with Zinc dust
(4 out of 56)
Aldehydes, Ketones and Carboxylic Acids
RBSE 2012
Q.1 Give the structure of cyclopropanon – 2, 4-dinitrophenyl hydrazine
(1 out of 56)
Q.2 (a) Write the equations for the following reactions –
(i) Cannizzaro reaction (ii) Aldol condensation.
(b) Arrange the following compounds in increasing order of their boiling points-
3 2 2 3 2 2 2
3 2 3 3 2 2 3
, ,
,
||
O
CH CH CH CHO CH CH CH CH OH
CH CH C CH CH CH O CH CH
OR
(b) Write the equations for the following reactions-
(i) Rosenmund (ii) Gattermann-Koch reaction.
(b) Arrange the following compounds in increasing order of their acidity-
(CH3)2 – CH – COOH, H – COOH, CH3 – COOH, (CH3)3 – C – COOH
(c) How will you distinguish between carboxylic acid and phenol by sodium carbonate?
(4 out of 56)
RBSE 2013
Q.1 An organic compound having molecular formula C3H6O, gives orange-red precipitate with 2, 4-
dinitrophenyl hydrazine, but does not reduce Tollen‘s reagent. Give IUPAC name and sturucture
formula of the compound. (1 out of 56)
Q.2 Give the structural formulae and IUPAC names of the following compounds :
(a) Formaldehyde
(b) Acetone
OR
Give the structural formulae and IUPAC names of the following compounds :
(a) Malonic acid
(b) Succinic acid (2 out of 56)
Q.3 (a) How carboxylate ion get stablilized by resonance? Explain by the structures.
(b) Carboxylic acids are more acidic than phenols. Explain. (2 out of 56)
RBSE 2014
Q.1 How will you convert sodium acetate to methane in one step? (1 out of 56)
Q.2 Explain the following with suitable example:
(a) Rosenmund reduction
(b) Gatterman-Koch reaction.
OR
Explain the following with suitable example:
(a) Clemmensen reduction
(b) Stephen reaction. (4 out of 56)
10
RBSE 2015
Q.1 Arrange benzoic acid, 4-methyl benzoic acid and 4-nitro-benzoic acid in increasing order of their
acidic strength. (1 out of 56)
Q.2 (a) Write a note on Rosenmund reduction.
(b) Why is fluroacetic acid more acidic than chloroacetic acid ?
(c) Draw the resonating structure of carboxylate ion.
OR
(a) Write a note on Wolff-Krishner reduction.
(b) Boiling point of carboxylic acids are higher than aldehyde and ketone of comparable
molecular masses. Why ?
(c) Draw the dimer structure of ethanoic acid in vapour state.
(4 out of 56)
Amines
RBSE 2012
Q.1 Identify A, B, C and D in the following reaction sequences –
(a) 2 2
6 5 2
H O Br
WaterC H N Cl A B
(b) 2 4 2 4. .
2 2 413
dil H SO conc H SO
KCH CH C D
(2 out of 56)
Q.2 (a) Give one chemical test to distinguish between aniline and N-methyl aniline.
(b) Dimethyl amine is a stronger base than methyl amine. Give reason.
(c) How will you obtain from aniline (Give equations only)-
(i) 2, 4, 6-tribromoaniline ;
(ii) Benzene diazonium chloride?
(3 out of 56)
RBSE 2013
Q.1 Write the chemical name and formula of Hisberg‘s reagent. (1 out of 56)
Q.2 (a) Identify X and Y in the following reaction sequence and also write the names of the two
reactions involved :
32
.CHCl / KOH alc.Br NaOH
2R CONH X Y
(b) Alkanamine is more basic than ammonia. Give reasons. (3 out of 56)
RBSE 2014
Q.1 Write the coupling reaction of benzene diazonium chloride with phenol. (1 out of 56)
Q.2 Write the reactions of (i) aliphatic and (ii) aromatic primary amines with nitrous acid. (1 out of 56)
Q.3 Write Hofmann‘s bromamide reaction. (1 out of 56)
RBSE 2015
Q.1 Write IUPAC name of 2 5 2 2 2 3
2 5
C H N CH CH CH CH|C H
(1 out of 56)
11
Q.2 What happens when benzene diazonium chloride reacts with potassium iodide ? (write only equation
of the reaction).
(1 out of 56)
Q.3 Arrange the following compounds in increasing order of their basic strength.
Explain reason :
2 5 2 3 6 5 2C H NH , NH ,C H NH
(2 out of 56)
Biomolecules
RBSE 2012
Q.1 (a) Draw a labeled diagram of double strand helix structure for DNA.
(b) Vitamin C cannot be stored in our body. Give reason.
(c) How will you obtain from glucose (give equations only)-
(i) Gluconic acid, (ii) n-hexane.
(3 out of 56)
RBSE 2013
Q.1 (a) Write the Haworth structure of sucrose.
(b) Sucrose is a non-reducing sugar. Why? (2 out of 56)
RBSE 2014
Q.1 Write the chemical reaction which confirm the presence of carbonyl group in glucose. (1 out of 56)
Q.2 Classify Vitamins A, B, C and D depending upon their solubility in water and fat and compare them.
(1 out of 56)
Q.3 What are monosaccharides? (1 out of 56)
RBSE 2015
Q.1 Explain denaturation of protein with an example.
(2 out of 56)
Polymers
RBSE 2012
Q.1 Give the names of monomers used for obtaining Nylon – 6, 6.
(1 out of 56)
Q.2 Give the structural formula of the initiator and its utility in free radical additional
Polymerization.
(1 out of 56)
Q.3 Write two differences between thermosetting and thermoplastic polymers.
(1 out of 56)
RBSE 2013
Q.1 (a) Explain the addition polymer by an example.
(b) Write the equation of preparation of synthetic rubber.
(c) Give the names of monomers used for obtaining Dacron.
OR
(a) Explain the condensation polymer by an example.
(b) Give one example each of thermoplastic and thermosetting polymers.
(c) Give the names of monomers used for obtaining Buna-N. (3 out of 56)
RBSE 2014
Q.1 What are thermosetting polymers? (1 out of 56)
Q.2 Write the names of the monomers of PHBV polymer. (1 out of 56)
12
Q.3 How is Nylon 6, 6 obtained? Give reaction. (1 out of 56)
RBSE 2015
Q.1 Explain that vulcanized rubber is an elastomer. (1 out of 56)
Q.2 Write any two differences between addition and condensation polymerisation
(2 out of 56)
Chemistry in Everyday life
RBSE 2012
Q.1 (a) Give two difference between soap and detergent.
(b) Classify the following into artificial sweeteners, preservatives, sop, detergents-
Sodium palmitate, sucralose, salts of sorbic acid, cetyl terimethyl ammonium bromide.
OR
(a) Give two differences between antiseptic and disinfectant.
(b) Classify the following into antihistamine, anti-acid, tranquillizer, antibiotic drug-Penicillin,
meprobamate, terfenadine, ranitidine.
(3 out of 56)
RBSE 2013
Q.1 (a) Why do diabetic persons require artificial sweetening agents?
(b) Write name of a food preservative. How does it preserve food?
(c) Give the name and formula of cationic detergents. (3 out of 56)
RBSE 2014
Q.1 Explain the following:
(a) Antiseptics
(b) Artificial sweetening
(c) Antacids.
OR
Explain the following:
(a) Disinfectants
(b) Food preservatives
(c) Tranquilizers. (3 out of 56)
RBSE 2015
Q.1 (a) What are non-narcotic analgesics ? Write an example.
(b) Why is artificial sweetening agent, Aspartame used only in soft drinks and cold foods ?
(c) Write any one difference between antiseptics and disenfectants.
OR
(a) What is anionic detergent ? Write an example.
(b) Why are chemicals added to food ? Write any two reasons.
(c) Write any one difference between broad spectrum and narrow spectrum antibiotics.
(3 out of 56)
ANSWERS
Solid State
RBSE 2012
Q.1 . Packing Efficiency Coordination
13
(i) Body centered cubic 68% 8
(ii) Cubic close packing 74% 12
Q.2 Ferromagnetic substances make permanent magnets as when placed in magnetic field the
domains aligned in the orientation of magnetic field.
Q.3 Glass is considered as a super-cooled liquids because it shows tendency to flow.
RBSE 2013
Q.1 Solid sulphur dioxide and solid ammonia are polar molecular solid.
Q.2 Insulators are substances having extremely low conductivities in range of 10-10
and 10-20
(Ωm) -1
while semiconductors having intermediate conductivities ranging between 10-6
and 104 (Ωm)
-1.
RBSE 2014
Q.1 (a) If Edge length of unit cell = a
Radius of each atom = r
Here, a = 2r
Efficiency = No. of atoms ×Vol. of each atom
100Vol. of unit cell
=
3
3
41 r
3 100a
3
3
41 r
3 100 52.4%2r
(b)
FRENKEL DEFECT SCHOTTKY DEFECT
(i) Arises in ionic crystals in which huge
difference in size of anion and cation.
(ii) Cation dislocates from its position into
interstitial site of crystal.
(iii)Does not changes density of crystal.
(i) Sizes of anion and cation are almost same.
(ii) Equal number of anion and cation leaves
crystal.
(iii)Density of crystal decreases.
RBSE 2015
Q.1 (a) For hexagonal crystal system, Axial angles,
o o90 ; r 120
(b) When Boron is doped with silicon is n-type of semiconductor due to presence of a free
electron.
Solutions
RBSE 2012
Q.1 Reverse osmosis- When pressure more than osmotic pressure is applied to solution
movement of solvent particles occurs from its lower concentration to its higher
concentration through a semipermeable membrane resulting in process called reverse
osmosis. Reverse osmosis is used in desalination of sea water.
Q.2 Wt. of H2SO4 = 10g
Wt. of water = 90g Molar mass = 98gmol-1
14
soluten
. ( )Molality
Wt of solvent inkg
3
1
10 / 98 10
90
1.23
m
m mol kg
Q.3 Increasing order of vant Hoff factor :
0.1 M C12 H22 O11 < 0.1 M CaCl2 < 0.1 M Al2 (SO4)3
RBSE 2013
Q.1 Weight of NaOH = 5 gm.
Volume of water = 500 ml = 0.5 lts.
Molar mass of NaOH = 40 gmol-1
Moles of NaOH = 5
0.12540
Molarity =No. of moles of solute
Vol. of solution (in lts.)
M = -10.1250.25mol L
0.5
Q.2 Volume of solution = 0.2 L
Weight of protein = 1.26 gm
Temperature = 300 K
Osmotic pressure (π) = 2.57 × 10-3
bar
R = 0.083 L bar mol-1
K-1
solute
solute
w R T
M Vol.(in lts.)
Msolute 3
1.26 0.083 300
2.57 10 0.2
= 13137461,038gmol
0.514
RBSE 2014
Q.1 (a) (i) If a tablespoon of sugar is added to water, vapour pressure of water will decrease.
(ii) Osmotic pressure is preferred for the determination of molar mass of macro molecules.
(b) No, the elevation in boiling point will not be same rather the value will be higher for 0.1 mole of
sodium chloride.
(c) No, we cannot separate the compounds of azotropic mixtures by fractional distillation because
there should be some difference in boiling point of components to be separated by technique of
fractional distillation but the boiling point of components of azeotropes is same.
RBSE 2015
Q.1 Formula to calculate molality is: solutenm
wt.of solvent (in kg)
Q.2 = 0.0821 atm, T = 400 k.
R = 0.821 L atm k–1
mol–1
= MRT
15
0.0821
MRT 0.821 400
M = 2.5 × 10–4
mol L–1
Electrochemistry
RBSE 2012
Q.1 t = 30 min I = 2A
= 30 × 60 = 1800 Sec
Q = It
= 2 × 1800 = 3600 C
Ag+ +e
- → Ag
96500 C is depositing 108 g of Ag
3600 C will deposit 108
360096500
= 4.02g of Ag.
Q.2
2 -1
2 -1
2 -1
3
- - +
3 3
+ -
Kcl = 149.8 Scm mol
Hcl = 425.9 Scm mol
CH COOK = 114.4 Scm mol
CH COOH = Cl + CH COO + k
K - Cl
o
o
o
o o o o o
o o
H
2 -1
= 425.9 + 114.4 - 149.8
= 390.5 Scm mol
RBSE 2013
Q.1 Nernst equation for Daniel cell equilibrium:
0
c
RT0 E ln K
nF
Relationship between 0
cellE and equilibrium constant (Kc)
nFE0 = RTlnKc
lnKc = onFE
RT
or 0
10 c
nFE2.303log K
RT
Q.2 (a) Daniel cell:
16
(b) At anode: 2Zn Zn 2e
At cathode: 2Cu 2e Cu
Overall reaction: 2 2Zn Cu Zn Cu
RBSE 2014
Q.1 (a) No, we cannot store CuSO4 solution in an iron vessel because iron is a better reducing agent.
(b) Kohlrausch‘s Law of the independent Migration of ions which states that limiting molar
conductivity of an electrolyte can be represented as the sum of the individual contributions of the
anion and cation of the electrolyte.
Application: Limiting molar conductivity of weak electrolyte can be calculated using Kohlrausch
law.
RBSE 2015
Q.1 S.I. Unit of Resistivity is ' m' (ohm meter)
Q.2 o 2
mMg 106.0 Scm2 mol
–1
o
mCl 76.3 Scm2 mol
–1
2
2
o o o
MgCl Mg Cl2
= 106.0 + 2 × 76.3
= 258.6 Scm2 mol
–1
Q.3 (a) According to second law of electrolysis, ―The amount (in weight) of different substances
liberated by the same quantity of electricity passing through the electrolytic solution is
proportional to their chemical equivalent weights‖.
(b)
Chemical Kinetics
RBSE 2012
Q.1 (a) All the molecules in the reacting species do not possess the same kinetic energy. Boltzmann and
Clark Maxwell tried to explain the behavior of large number of molecules by plotting the fraction of
molecules possessing certain kinetic energy E vs kinetic energy, i.e. distribution curve.
The peak of the curve corresponds to the most probable kinetic energy or the energy of maximum
number of molecules with energies lower of higher than this go on decreasing on either side. When
temperature rises, the maximum of the curve moves towards the higher energy value and the curve
spreads to the right such that there is a probability must be one, the area under the curve must be
17
constant. Position of Ea is marked on the curve. On increasing the temperature the fraction of
molecules that collide with energies equal to and greater than Ea is increased.
From the curve it is evident that the area showing the fraction of such molecules at temperature (t +
10) gets doubled, as a result the rate of reaction is doubled (a) Rate increases to 4 times, so k2 = 4k : T1 = 350 K ; T2 = 400 K
a2 1 2 2
1 1 2 2 1 1
Ek 2.303 R TT k1 1log or Ea = . log
k 2.303 T T T - T kR
a
2.303 8.314 350 400 E =
400 350
2.303 8.314 350 400log 4 = 0.6021 =
50
32.28kJ mol
-1
RBSE 2013
Q.1 In some reactions the rate of a reaction is proportional to the first power of the concentration of two
of the reactions, than the reaction is said to be first-order with respect to both the reactants hence
called, Pseudo first order reaction.
H
3 2 5 2 3 2 5CH COOC H H O CH COOH C H OH
t = 0 0.01 mole 10 mole 0 mole 0 mole
t = 0 mole 9.9 mole 0.01 mole 0.01 mole
The concentration of water, does not get altered much, so in rate equation
Rate = 3 2 5 2K [CH COOC H ][H O]
The term [H2O] can be taken as constant. The equation, thus becomes,
Rate = K [CH3COOC2H5]
where, K = 2K [H O]
and the reaction behaves as first order.
Q.2 Units of rate constant (K)
(i) Zero order mol L-1
S-1
(ii) First order S-1
Q.3 Half life of a reaction is the time in which the concentration of a reactant is reduced to one half of its
initial concentration.
For First order, R P
0
R2.303K log
t R
at 1/2 0
1t , R R
2
18
0
1/20
R2.303K log
1tR
2
1/2
2.303t 0.301
K
1/2
0.693t
K
RBSE 2014
Q.1 The order of the following reactions:
(a) Artificial radioactivity decay takes places by first order kinetics.
(b) The decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction at high
pressure. (c) Hydrogenation of ethane is a first order reaction. (d) Decomposition of N2O5 is first order reactions.
Q.2 Half life of a reaction is the time in which the concentration of a reactant is reduced to one half of its
initial concentration.
For First order, R P
0
R2.303K log
t R
at 1/2 0
1t , R R
2
0
1/20
R2.303K log
1tR
2
1/2
2.303t 0.301
K
1/2
0.693t
K
Hence, this relation is independent of initial concentration of reactant.
RBSE 2015
Q.1 For the unit of rate constant is sec–1
the order of reaction is Fist order.
Q.2 2A B Product
Differential rate equation, 2 1d[R]K[A] [B]
dt
Q.3 1/ 2t 10sec.
For I order,
1/ 2
0.693t
k
0.693
k10
K = 0.0693 sec–1
19
Surface Chemistry
RBSE 2012
Q.1 Colloidal solution shows Tyndall effect because
(i) Diameter of dispersed particle is between 1 to 100 nm. Therefore the light gets &
Scattered.
(ii) The refractive index of the dispersed phase and the dispersion medium differ in
magnitude
Q.2 Hydrophobic sol is easily coagulated because these are highly unstable due to water
repelling nature.
Q.3 (a) Coagulation
(b) Peptisation
(c) Metathesis
(d) Tyndall effect
RBSE 2013
Q.1 (a) Two differences between physical and chemical adsorption are as follows:
Physical Adsorption
(i) It arises because of vander waal‘s forces.
(ii) It is not specific in nature.
Chemical Adsorption
(i) It is caused by chemical bond formation.
(ii) It is highly specific in nature.
(b) Homogeneous Catalysis:-If the catalyst is present in the same phase as the reactant, the catalysis
is called as homogeneous catalysis.
H aq.
2
O O|| ||
R C OR H O R C OH R OHEster
Heterogeneous Catalysis: If the catalyst is present in a different phase than the reactants the catalysis
is called as heterogeneous catalysis.
Oxidation of sulphur dioxide into sulphur trioxide in the presence of Pt.
Pt s
2 3SO g 2SO g
(c) Electro-dialysis:
OR
(a) Two differences between lyophilibic and lyophobic colloides.
20
Lyophobic Solution
(i) Made by special method
(ii) Irreversible in nature
Lyophilic Solution
(i) Made by simple method.
(ii) Reversible in nature.
(b) Two enzymes of yeast are : Invertase and Zymase.
Invertase
Zymase
Sucrose Fructose
Glucose Ethyl alcohol and carbon dioxide
(c) Bredig‘s Arc Method:
RBSE 2014
Q.1 (a) Gold solution is made by Bredig‘s are method.
(b) The difference between solution and a gel is that in solution dispersed phase is solid while
dispersion medium is liquid while in gel dispersed phase is liquid while dispersion medium is solid.
(c) Coagulation: The process of aggregation of colloidal particles into an insoluble precipitate in
presence of a suitable electrolyte is known as coagulation.
OR
(a) Electro dialysis is the method of purification of colloidal solution containing excess amount of
electrolyte.
21
In this method a bag made up of a semipermiable membrane is filled with the colloidal solution and
is the suspended in fresh water. The electrolyte particles pass out leaving behind the colloidal
solution. Movement of ion is made faster by applying electric field.
(b) In Aerosol the dispersion medium is gas and dispersed phase can be solid or liquid but in foam
dispersion medium is liquid and dispersed phase is solid.
(c) Peptization: It is defined as the process of converting a precipitate into colloidal solution by
shaking it with dispersion medium in the presence of a small amount of electrolyte called peptizing
agent.
RBSE 2015
Q.1 (a) Numerous reactions that occurs in the bodies of animals and plants to maintain the life
process
are catalysed by enzymes and the phenomenon is known as enzymes catalysis.
Example: Inversion of cane sugar Invertase
12 22 11 2 6 12 6C H O (ag.) H O( ) C H O (aq.) Glucose C6H12O6(aq.) Fructose
(b) Milk is oil in water type emulsion where the dispersion medium is water and dispersed phase
is oil.
(c)
Fig.: Electrophoresis
OR
(a) The selectivity of a catalyst is its ability to direct a reaction to yield a particular product.
Example: (i) Ni
2 4 2CO(g) 3H (g) CH (g) H O(g)
(ii) 2 3Cu / ZnO Cr O
2 3CO(g) 2H (g) CH OH(g)
(iii) Cu
2CO(g) H (g) HCHO(g)
22
(b) When hydrated ferric Oxide sol is mixed with arsenius sulphide sol coagulation occurs.
(c)
Fig.: Tyndall effect
General Principles & Processes of Isolation of Elements
RBSE 2012
Q.1 Collector used in froth floatation process is pine oil, or fatty acids, xanthales, etc. Role of
collector is to increase the non-wettability of the mineral particles.
Stabilizer used in the process is aniline or cresols. Its role is to stabilize the froth.
RBSE 2013
Q.1 Two impurities present in bauxite ore are – SiO2 and TiO2
Q.2 The chemical reactions related to Mond‘s process of Nickel are as follows:
350K
4Ni 4CO Ni CO
450K
4Ni CO Ni 4CO
RBSE 2014
Q.1 Diagram of electrolytic cell for extraction of Aluminium.
The overall reaction in the cell is :
2Al2O3 + 3C 4Al + 3CO2
23
P
HO OH
OH
O||
OR
Diagram of Zone Refining process.
This process is useful for refining of metals of semiconductor of very high purity i.e., Ge, Si, B, Ga
etc.
RBSE 2015
Q.1 Calcination – It is a process in which an ore is heated in the absence or limited supply of air without
fusion (melting) in order to remove moisture and volatile impurities. The ore become s porous.
3 2ZnCO ZnO CO
2 3 2 2 3 2Fe O .xH O(s) Fe O (s) xH O(g)
Roasting – Roasting is the process in which concentrated ore is heated in a furnace at a temperature
below its melting point in the presence of excess of air. The sulphide ore is converted into metallic
oxide and the volatile impurities are removed.
2 22ZnS 3O 2ZnO 2SO
2 2 2 22Cu S 3O 2Cu O 2SO
The p-Block Elements
RBSE 2012
Q.1 (a) (i) 2NaOH + Cl2 → NaCl + NaOCl + H2O
(ii) C + 4 HNO3 → CO2 + 4NO2 + 2H2O
(b) (i) F2 + 2Cl- → 2F
- + Cl2
F can Oxidize Cl2 easily, due to its high electronegativity
(ii) S2 is a solid because at octatomic configuration & Large magnitude of Vanderwal
forces
(c) (i) (ii)
RBSE 2013
Q.1 (a) Group 15
3rd
element – Arsenic
Electronic Configuraton = [Ar]18 3d10
4s2 4p
3
N O N
24
(b) Structure of ammonia
(c) NH3 acts as a Lewis Base due to presence of lone pair of electron.
(d) Reactions
4Zn + 10HNO3 (dil) 4Zn(NO3)2 + 5H2O + N2O
Zn + 4HNO3 (Conc.) Zn(NO3)2 + 2H2O
OR
(a) Group 15.
Metal element – Bismuth
Configuration – [Xe]4f14
5d10
6s2 6p
3
(b) Bleaching action of chlorine gas is due to release of Nascent Oxygen.
Cl2 + H2O 2HCl + [O].
(c) Equations 2 3
3 2NO 3Fe 4H NO 3Fe 2H O
2 2
2 2 26 5Fe H O NO Fe H O NO H O
.
(d) Structure of PCl5
RBSE 2014
Q.1 (a) Chemical formula of oxyacids of chlorine HClO; HClO2; HClO3; HClO4
(b) General electronic configuration of noble gas group is – ns2 np
6
(c) Reactions
3 2 5 2 5 3 3PCl 3C H OH 3C H Cl H PO
5 2 5 2 5 3PCl C H OH C H Cl POCl HCl
OR
25
(a) Oxyacids of sulphur
H2SO3; H2SO4; H2S2O7; H2S2O8
(b) Configuration of chalcogen group – ns2 np
4
Epsom salt = MgSO4. 7H2O
(c) Ammonia acts as a Lewis Base due to lone pair of electron.
RBSE 2015
Q.1 3 2 2Ba(N ) Ba 3N
Q.2 It is dibasic acid due to the present of two replaceable hydrogen therefore its basicity is 3.
O
P
||
H
OHOH|
Q.3 (a) 2 22Cl NaOH NaCl NaClO H O
(cold & dil)
(b) 3 2H Po
It is good reducing agent, it is monobasic acid due the presence of one replaceable hydrogen.
OR
(a) 2 3 23 6 5 3Cl NaOH NaCl NaClO H O
(Hot, Con) (Sodium Chlorate)
(b) Axial chlorine in 5PCl have longer bond length than equatorial chlorine Because axial bond
pairs suffer more repulsion than equatorial bond pairs.
The d and f – Block Elements
RBSE 2012
Q.1 Ni2+
in aqueous solution has 3d8 configuration, i.e., it has 2 unpaired electrons. Therefore
its magnetic moment is-
( 2) 2(2 2) 8 2.84 .n n BM
Q.2 Nickel (28) has configuration [Ar] 4s2 3d
8, so Ni
2- is [Ar] 4s
0 3d
8,
[Ni(CN)4]2-
-
26
In [Ni(CN)4]2-
- Nickel is in +2 oxidation state. There is dsp2 hybridization each CN
- ion (strong
ligand) donates a pair of electrons and all electrons are paired up, no unpaired electron. Hence
[Ni(CN)4]2-
has square planar structure and is diamagnetic.
In [Ni(CN)4]
2- Nickel is in 0 oxidation state. There is sp
3 hybridization, each Cl
- ion (weak ligand)
donates a pair of electrons and there are two unpaired electrons in d-orbital‘s. Hence, [Ni(CN)4]2-
has
tetrahedral structure and is paramagnetic.
Q.3 (a) Manganese (Z=25) exhibits the largest number of oxidation states, as its atom has the
maximum number of unpaired electrons 5 in the 3d-orbitals which can be available for
bond formation so both 3d & 4s e– can participate in bonding.
(b) Cr2+
is strongly reducing because it changes to Cr3+
which is more stable and Cr3+
has
higher hydration energy where as Mn3+
is strongly oxidizing because it changes to
Mn2+
which is more stable due to half filled d-orbital‘s.
RBSE 2013
Q.1 Zn, Cd and Hg are not transition elements since they have completely filled d-orbitals
(d10
configuration) in ground state as well as excited state.
Q.2 Structure of chromate ion
Q.3 Ti+4
is colourless due to no electron in d-orbitals (do configuration)
RBSE 2014
Q.1 The highest oxidation state of a metal is exhibited in its oxide or fluoride-
- The ability of fluorine is due to release of high lattice energy or high bond enthalpy
- The ability of oxygen is due to tendency to form multiple bonds with metals.
Q.2 (a) Silver has d10
configuration still it is a transition metal as silver has incompletely filled orbitals
in the excited state.
(b) Actinoid Contraction - The gradual decrease in size of actinoids across the series because of poor
shielding by 5f electrons.
RBSE 2015
Q.1 (a) There is a decrease in the atomic and conic radii of lanthanoids as the atomic number
increase. It is because for every additional protein in the nucleus the corresponding e
goes into the 4 f shell which is too diffused to screen the nucleus as affective more
localized inner shell. Hence the attraction of the nucleus for the outer most e increases
steadily with the atomic number of lanthanoids.This contraction in size quite regular
and is called as lanthanoids contraction.
27
(b) Transition elements trap small size atoms such as H,C,N etc, in the voids of the crystal lattice
such compounds are called interstitial compounds.
eg-3Fe H ,TiC,
4Mn N etc
(c) The electronic configuration 29M is
1 10[ ]4 3Ar S d , here M is cu.
The outer configuration of 2cu is 9[ ]3Ar d
Thus unpaired electrons presentation one
( 2)n n 1(1 2)
Coordination Compounds
RBSE 2012
Q.1 (a) dichloridobis (ethane-1, 2-diamine) cobalt (III) chloride.
(b) Potassium hexacyanoferrate (III)
RBSE 2013
Q.1 Ambidentate ligand - Ligand which can ligate through two different atoms present in it. For eg 2NO
ion can coordinate through either nitrogen or oxygen atom to the central metal atom/ion.
Q.2 Complex K3[Fe(C2O4)3]
Oxidation Number of Central element = +3.
Coordination Number = 6
(Since C2O4 is a bidentate ligand)
Q.3 Structures :
Cis-[CoCl2(en)2]
Facial – [Co(NH3)3(NO2)3]
the fac-and mer – isomers of [Co(NO2)3(NH3)3]
Q.4 The oxidation state of Ni in [Ni(CN)4]2–
ion is +2. In Ni2+
ion, the co-ordination number is 4.
Electronic configuration of Ni is
4s 4p
Electronic configuration Ni
2+ ion is
4s 4p3d
According to valence bond theory, due to the presence of strong liganed [CN
–], the unpaired
electrons present in d-orbitals are paired resulting one empty d-orbital.
28
CN
CN
4s 4p
dsp2
Therefore a low spin complex, dsp2 hybridisation takes place.
In this way four empty four dsp2 hybrid orbitals of Ni form four co-ordinate bonds with ion
RBSE 2014
Q.1 (a) Application of coordination compounds in metal purification, an example:
- Impure nickel is converted to [Ni(CO)4]; which is decomposed to yield pure nickel.
(b) IUPAC name of [Ag(NH3)2] [Ag(CN)2]
Diammine silver (I) dicyanoargentate (I)
Q.2 Difference between:-
Double Salts: Double salts are the compounds formed by combination of two or more stable
compounds; which can dissociate into simple ions completely dissolved in water.
Ex. – Carnalite KCl.MgCl2.6H2O
Complexes: are coordination compounds where metal atoms are bound to a number of anions or
neutral molecules which do not dissociate into individual ions.
Ex. – K4[Fe(CN)6]
RBSE 2015
Q.1 [Cr(H2O)4BrCl]Br
Q.2 Hg[Co(CNS)4]
Q.3 The oxidation state of Ni in [Ni(CN)4]2–
ion is +2. In Ni2+
ion, the co-ordination number is 4.
Electronic configuration of Ni is
4s 4p
Electronic configuration Ni
2+ ion is
4s 4p3d
According to valence bond theory, due to the presence of strong liganed [CN
–], the unpaired
electrons present in d-orbitals are paired resulting one empty d-orbital.
4s 4p
dsp2
Therefore a low spin complex, dsp2 hybridisation takes place.
In this way four empty four dsp2 hybrid orbitals of Ni form four co-ordinate bonds with ion
29
Haloalkanes and Haloarenes
RBSE 2012
Q.1 (a) CH3 CH2 Br will react faster as it‘s a better leaving group.
(b) (CH3)3 C-X will react faster as 3o carbonation in more stable.
OR
(b) CH3Cl + aq. KOH → CH3OH + KCI
Mechanism followed : SN2.
(c) (CH3)3C - Cl + aq. KOH → (CH3)3C – OH + KCI
Mechanism Followed : SN1
Q.2 (a) Fitting reaction : -
(b) Finkelstein reaction : -
3 3CH Cl + NaI CH l + NaCldry ether
RBSE-2013
Q.1 Benzylic Chloride
Hybridisation of carbon is Sp
3
Vinylic Chloride
C=CCl
Hybridisation of carbon is Sp2
Q.2 (a) Wurtz reaction :-
dryether2R X 2Na R R 2NaX
(b) Wurtz-fittig reaction :-
RBSE – 2014
Q.1 (a) aq.alcohol
3 2 3 2CH CH Br AgCN CH CH NC
(b) 3 2 2 3 2CH CH Br KNO CH CH O N O
(c)
(d) dry
3 3etherCH I Mg CH Mg I
30
Q.2 Two differences in substitution nucleophilic biomolecular (SN2) and substitution nucleophilic
unimolecular (SN1)
SN2
(i) Molecularity of reaction is two
(ii) It‘s a single step mechanism
SN1
(i) Molecularity of reaction is One.
(ii) Its two step mechanism
RBSE 2015
Q.1
+ KI
N2Cl+ –
I
Q.2 3 3 3 3(CH ) C Br OH(ag.) (CH ) C OH
Mechanism:
Step I: Formation of carbocation:
CH3—C—Br
|CH3
|
CH3
polar protic
solventCH3
|CH3
CH3CH3
+ Br–
Step II : Attack of Nucleophile:
C
|CH3
CH3CH3
+ OH (CH3)3 C—OH
Q.3
(a)
C
+ CH3—C Anhydrous AlCl3
C
CH3 (b) 3 2 3 2CH CH Cl KCN CH CH CN
CH3CH = CH2 + HI CH3—CH—CH3|
I
(c)
(d) 3 2 2 4 3 2CH CH OH NaBr H SO CH CH Br
31
3 2 3 3
|
CH CH CH HI CH CH CH
H
3 2 2 4 3 2 2 4 2CH CH OH NaBr H SO CH CH Br Na SO H O
Q.4 (a)
minor major
1-chloro-2-methyl 1-chloro-4-methyl
benzene benzene
(b) 3 2 3 2CH CH Cl KCN CH CH CN KCl
(c)
(d)
Alcohols, Phenols and Ethers
RBSE 2012
Q.1 (a) A is phenol and B is 2, 4, 6- tribromophenol
2
6 5 2 6 5
Phenol
H O Br
WaterC H N Cl C H OH
(b) C is Ethyl alcohol and D is Diethyl ether.
2 4 2 4. .
2 2 3 2 413
Ehtyl alcohol
dil H SO conc H SO
KCH CH CH CH OH
2 5 2 5C H O C H
Diethyl ether
Q.2 (a) In phenol lone pair of oxygen are in resonance with benzene there by making OH
bond polar hence release of proton is easier.
(b) Propanol has higher boiling point as it has inter molecular H-bonding.
RBSE 2013
Q.1 (a) Secondary-Butyl alcohol
|
3 2 3
OH
CH CH CH CH
IUPAC name: Butan – 2-ol.
(b) Preparation of phenol :-
(i) From Benzene :
32
(ii) From aniline :
(c) Resonating structures of phenol :-
OR
(a) Methyl n-propyl ether
CH3OCH2CH2CH3
IUPAC name :- Methoxy propane
(b) (i) Dehydration of alcohols :-
(ii) Williamson‘s Ether synthesis
R X R O Na R O R NaX
(c) Resonating structures of alkoxy benzene
RBSE 2014
Q.1 Convert phenol to benzene :
Q.2 2 4
413K
3 2 3 2 2 3H SOCH CH OH CH CH OCH CH
Q.4
33
33 2 3
o
CrOCH CH OH CH CHO
2 alcohol
3
3 3 3 3
O O| ||
CrOCH CH CH CH C CH
(2° alcohol)
RBSE 2015
Q.1 (a) IUPAC names of following
(i) CH2 = CH—CH2—OH
Prop-2-enol.
(ii)
CH2—CH—CH2
|OH
|OH
|OH
Propane-1,2,3-triol.
(b) (i)
OH
+ Br2
CS2
OH
Br
+
OH
Br
573K
3 2 3Cu(ii)CH CH OH CH CHO
OR
(a) IUPAC names of following
(i) CH3—O—CH—CH3|
CH3
2-Methoxy propane
(ii) CH3—CH—CH2—OH|
CH3
2-Methyl propanol
(b) (i)
OH
Conc. HNO3
OH
NO2
NO2
O2N
34
(ii)
OH
Zn (dust)
Aldehydes, Ketones and Carboxylic acids
RBSE 2012
Q.1 Structure of cyclopropanon -2, 4-dinitro phenyl hydrazone :-
Q.2 (a) (i) Cannizaro‘s Reaction- Aldehydes having no a-Hydrogen ato like HCHO, CCI3CHO (chloral),
(CH3)3 C-CHO (Pivaldehyde(, C6H5CHO (Benzaldehyde), etc. When heated with concentrated
alkali, two molecules of aldehydes are used. One molecules is oxidized into while another molecules
is reduced into alcohol.
3
6 5 6 5 2 6 5
2HCHO + NaOH CH OH + HCOONa
conc. Methanol Sodium formate
2C H CHO + NaOH C H CH OH + C H COONa
c
onc. Benzyl alcohol Sodium Benzoate
When two different aldehydes are taken for Cannizaro reaction it is known as cross Cannizaro
reaction. Easily oxidisable molecule is oxidized into acid while othe molecule is reduced into
alcohol.
(ii) Aldol condensation- Whane teo molecules of an aldehyde are heated in the presence of dilute
NaOH, a condensation product known as aldol is obtained. This process is called aldol condensation.
For example,
3 2 3 2
O OH
|| |
CH -C-H + HCH CHO CH -CH-CH -CHO
NaOH
Acetaldol
When two difference aldehydes each having a-hydrogen atom condense in the presence of NaOH,
four product are obtained, this process is called cross-aldol condensation. For example,propanal and
butanal on condensation give four products, two simple and two crossed.
(b)
3 2 2 3 3 2 2
3 2 3 3 2 2 2
CH -CH -O-CH -CH < CH -CH -CH -CHO
< CH -CH -C-CH < CH -CH -CH -CH -OH
||
O
35
RBSE 2013
Q.1
Structural formula:
3 3
O||
CH C CH
IUPAC name: Propan-2-one.
Q.2 (a) Formaldehyde
Structural formula :
O||
H C H
IUPAC name : Methanal
(b) Acetone
Structural formula :
3 3
O||
CH C CH
IUPAC name : Propan-2-one
OR
(a) Malonic acid
Structural formula :
2
O O|| ||
H C CH C H
IUPAC name : Propane -1, 3-dioicacid.
(b) Succinic acid :
Structural formula :
2 2HOOC CH CH COOH
IUPAC name: Butane-1, 4-dioic acid
Q.3 (a) Carboxylate ion:
Resonating Structure:
36
Carboxylate ion has two equivalent resonating structures in which negative charge is delocalised
over more electronegative oxygen atom has it is highly stablised due to resonance.
(b) Carboxylic acids are more acidic than phenol because in carboxylic acid, the carboxylate ion has
two equivalent resonating structures in which negative charge is delocalised over more
electronegative oxygen atom while in phenol, the phenoxide ion has five non-equivalent resonating
structure in which negative charge is delocalised over less electronegative carbon atom hence
resonance is carboxylic acid is more-stablised hence more acidic.
RBSE 2014
Q.1 3 4 2 3
CaOCH COONa NaOH CH Na COheat
Sodium acetate
Q.2 (a) Rosen mund reduction :-
Acyl Chloride is hydrogenated over Catalyst, palladium on barium sulphate. This reaction is called
Rosen mund reduction.
(b) Gatterman-Koch reaction :-
When benzene or its derivative is treated with carbon monoxide and hydrogen chloride in the
presence of anhydrous aluminium chloride, it gives benzaldehyde. This reaction is called as
Gatterman – Koch reaction.
OR
(a) Clemmensen reduction :
The Carbonyl group of aldehyde and Ketone are reduced to CH2 group on treatment with Zn
amalgam and Conc. HCl called Clem mensen‘s reduction.
(b) Stephen‘s reaction :
3H O
2R CN SnCl HCl RCH NH RCHO
Nitrile are reduced to corresponding imine with stannous chloride in the presence of hydrochloric
acid, which on hydrolysis give corresponding aldehyde.
37
RBSE 2015
Q.1 (a) Rosenmund reduction:
CC H2
Pd - BaSO4
||O
CHO
Benzoyl chloride Benzaldehyde Acyl chloride (acid chloride) is hydrogenated over catalyst, palladium on barium sulphate.
This reaction is called rosenmund reduction.
(b) Fluroacetic acid is more acid than chloroacetic acid as Fluorine is more electronegative than
chlorine.
(c) Resonating structure of carboxylation:
R—C
O
O
R—C
O
O
_
OR
(a) Wolff-Krishner reduction:
C = ONH2—NH2
—H2OC = N NH2
KOH/ethyleneglycol
heatCH2 + N2
When carbonyl group of aldehyde and ketone is reduced with hydrazine followed by heated
with sodium or potassium hydroxide in high soiling solvent such as ethylene glycol.
(b) Boiling point of carboxylic acids are higher than aldehyde and ketone of comparable
molecular masses because intermolecular H-bonding present in carboxylic acid
(c) Dimer structure of ethanoic acid :
CH3—C C—CH3
OH
O HO
O
Q.2 Order of increasing basic strength is as follows:
4-Methyl benzoic acid < Benzoic acid < 4-Nitrobenzoic acid
38
Amines
RBSE 2012
Q.1 (a) Aniline and N-methyl aniline can be distinguished using isocyanide test
3CHCl KOH
3CH Cl KOH
No reaction
(b) Dimethyl amine is more basic due to pressure of two allylic groups which increase
electron density on nitrogen. Hence tendency to accept electron increases.
(c) (i) Aniline to 2, 4, 6 – tribromo aniline
2
Br Water
(ii) Aniline to Benzene diazonium chloride
2NaNO HCl
RBSE 2013
Q.1 Hisberg‘s reagent
Formula :
Name : Benzene sulphonyl chloride.
(a) 2 2
Br ,NaOH2R CONH R NH
(X)
CHCl3/KOH(alc.)
R-NC
(Y)
(b) Alkanamine is more basic than ammonia due to +I effect of alkyl group in alkanamine which
increases electron density on nitrogen these by making it more basic.
39
RBSE 2014
Q.1 Coupling reaction :-
Q.2 (i) 2 2NaNO H O
2 2 2 2HCl ColdR NH NNO R N Cl R OH N
(ii)
Q.3 Hoffmann‘s bromamide reaction :-
2 2 2 2 3 2
O||
R C NH Br 4NaOH R NH Na CO 2NaBr 2H O
RBSE 2015
Q.1 IUPAC name of
C2H5—N—CH2—CH2—CH2CH3
|
C2H5 M, N- diethyl butanamine.
Q.2 Inerseasing order of basic strength is as follows:-
C6H5NH2 < NH3 < C2H5NH2
Reasons: Aniline is least basic as lone pair of nitrogen is in resonance with benzene hence are less
available to accept proton, while in Ethanamine, due to +I offset of alkyl group electron density on
nitrogen is increased, making it more basic. While no such effects are observed in ammonia.
Biomolecules
RBSE 2012
Q.1 (a) The double helical structure of a common DNA is shown in the adjoining figure.
(b) Vitamin C is soluble in water therefore gets excreted in urine. So it cannot be
stored in our body.
(c) (On reaching it Br2 water)
4 2 4
CHO COOH
| |
(CHOH) +Br Water (CHOH)
|
2 2
|
CH OH CH OH
D-Glucose
(ii) On boiling glucose with HI for long time n-hexane is obtained.
40
4 3 2 2 2 2 3
2
CHO
|
(CHOH)
| n-Hexane
CH OH
HI CH CH CH CH CH CH
RBSE 2013
Q.1 (a) Haworth structure of sucrose
OH
HO HO
H HO
H
H
CH2OH
4
5
6
3 2
1
O
HOH
H HO
HOH2C
CH2OH
2 5
43 6
1
–D–Fructose
H
O
H
–D–Glucose
Glycosidiclinkage
Sucrose
(b) Sucrose is a non-reducing sugar since the reducing groups of glucose and fructose are involved in
glycosidic bond formation in sucrose.
RBSE 2014
Q.1 Reaction to confirm the presence of carbonyl group in glucose:
Glucose reacts with hydroxylamine to give monoxime.
Q.2 Vitamins classified based on solubility:
(a) Fat Soluble: These are oily and not soluble in water.
Ex. – Vitamin A and D
(b) Water Soluble: These readily dissolve in water.
Ex. – Vitamin B and C
Q.3 Monosaccharides – These are carbohydrates which cannot be hydrolysed to simpler units.
Ex.- Glucose
RBSE 2015
Q.1 When a protein lose its biological activity due to change H pH or temp. is knows as denaturation
eg. Curdling of milk, Coagulation of egg.
Polymers
RBSE 2012
Q.1 Monomers of nylon-6, 6 is
[ - NH(CH2)6-NH-CO-(CH2)4-CO-]
Which is derived from adipic acid and hexamethyl diamine
COOH ( CH2 )4COOH NH2 ( CH2 )6NH2
41
Q.2 Mostly Benzoyl peroxide is used as the common initiator in free radical additional
polymerization.
6 5 6 5
O O
|| ||
C H - C - O - O - C - C H
Benzoyl peroxide
2 C6H5
Free Radical
This acts as initiator and starts the reaction by forming a free radical phenyl.
Q.3 A thermoplastic polymer can be repeatedly softened on heating and hardened on cooling.
Hence it can be used again and again. e.g., polythene, polypropylene. A thermosetting
polymer and sets during moudling process and cannot be softened again. For examp Bakelite
and melamine.
RBSE 2013
Q.1 (a) Addition Polymer: These are formed by addition reactions.
Ex.
(b) Synthetic Rubber:
(c) Monomers of Dacron: Ethylene glycol and terepthalic acid.
OR
Q.1 (a) Condensation Polymers – These are formed by condensation between monomeric units by
elimination of small molecules like water, ammonia etc.
Ex. – Nylon 6,6
NH2 (CH2)6 NH2 + HOOC – (CH2)4–COOH [NH–(CH2)6 NH–CO(–CH2)4–CO]n (b) Thermoplastic Polymers – Polymers which can be easily moulded on heating.
Ex. – Polythene
Thermosetting Polymer – Polymers which on heating convert into hard mass.
Ex. – Bakelite
(c) Monomers of Buna-N – 1,3 – Butadiene and acrylonitrile.
RBSE 2014
Q.1 Thermosetting Polymers - Polymers which on heating convert into hard mass.
Ex. – Bakelite
42
Q.2 Monomers of PHBV
3 – hydroxybutanoic acid and
3 – hydroxypentanoic acid
Q.3 Nylon – 6,6
RBSE 2015
Q.1 (1) The chains of elastomers are some what coiled get open on stress, so they become elongated.
(2) Their elasticity can be increased by introducing few cross links in between different chains.
(3) The elasticity of natural rubber is poor therefore it is heated with sulphur to increase cross-links
bonds in form of sulphur bridge. Therefore vulcanized rubber is an elastomer.
Q.2 Addition Polymerisation Condensation Polymerisation
1. These are formed by adding moments 1. Monomers combine together with loss of
to a growing polymer chain without small molecules like 2H O , 3NH , 2CO ,
loss of any molecules. 3CH OH etc
2. These are formed from unsaturated 2. It does not involve chain reaction.
compounds.
3. It involves chain reaction 3. monomers have di or polyfunctional groups
eg - PVC Teflon eg- Nylon-66, Bakelite etc
Chemistry in Everyday Life
RBSE 2012
Q.1 (a) 1. Soap are salts of weak acid and strong base, whereas, detergents are the salts of
strong acid and strong base.
2. Aqueous solution of soap is basic, but aqueous solution of detergent is neutral.
(b) (1) Sodium palmitate is a soap.
(2) Sucralose is artificial sweetener
(3) Salts of sorbic acid is a preservative
(4) Cetyl trimethyl ammonium bromide is detergent.
OR
(a) The chemicals which either kill or prevent the growth of harmful micro organisms but
Are not harmful to living tissues are called antiseptic. Whereas, the chemicals which
either kill or prevent the growth of harmful microorganisms but are not safe to living
tissues, are called disinfectants.
Antiseptics are applied to cuts, wounds or ulcers on skin . But disinfectants cannot be
applied on the skin. These are commonly applied to non living substance such as
floors, toilets, surgical instruments, clothes and utensils etc.
Some example of antiseptics are Dettol (a mixture of chloroxylenol and tereneol),
soframicine, acryflavine, bithional, boric acid and loidine etc. Phenol, chlorine,
43
sulphur dioxide, hydrogen peroxide, naphthalene balls etc are important disinfectants.
(b) (1) Penicillin is an antibiotic drug
(2) Meprobamate is a tranquillizer
(3) Terferandin is antihistamine
(4) Ranitidine is an anti-acid
RBSE 2013
Q.1 (a) Diabetic persons require artificial sweeting agent to reduce calorie intake.
(b) Food preservative prevents spoilage of food by retarding microbial growth.
Ex.– Salt, Sugar, Oil
(c) Catinoic detergent
RBSE 2014
Q.1 (a) Antiseptics – Chemicals which prevent growth of microbes on living tissues such as wounds,
ulcers etc. Ex. – dettol, 0.2% Phenol
(b) Artificial Sweetners – They are chemical food additives which reduce calorie intake.
Ex.– Saccharin
(c) Antacids – Chemicals which reduce excess acid in stomach.
Ex.ss– Magnesium hydroxide
OR
(a) Disinfectants – Chemicals applied to non – living objects to prevent microbial growth.
Ex. – 1% Phenol
(b) Food Preservatives – Food preservative prevents spoilage of food by retarding microbial growth.
Ex. – Salt, Sugar, Oil
(c) Tranquilizers – Chemicals used for treatment of stress or mental diseases.
Ex. – Equanal
RBSE 2015
Chemistry in everyday life
Q.1 (a) Non narcotie (Non addictive) Analgesics
Analgesics reduces pain with out causing impairment of consciousness confusion or paralysis
etc Non Narcotic analgesics are those analgesics which are non addictive.
(b) Aspartame is used only in soft drinks and cold foods, because aspartame decomposes at
cooking temperature and hence can be used as a sweetning agent only in cold foods and
drinks.
44
(c)
Antiseptics Disinfectants
These are the chemicals which check the
growth of the microorganisms or kill them
but are not harmful to the living tissues.
These are the chemicals which kill the
microbes but are dangerous or poisonous for
living tissues
These are the chemicals which check the
growth of the microorganisms or kill them
but are not harmful to the living tissues.
These are the chemicals which kill the
microbes but are dangerous or poisonous for
living tissues.
can not be applied on wounds cuts scars or
on skin surfaces.
eg- detol Furacin etc
can not be ingested orally
can not be applied on living tissues nor can
be ingested orally.
can be applied to inanimate object like floor
drainage etc
eg : Phenol
OR
(i) Anionic detergents are the sodium salts of sulfonated long chain alcohols or hydrocarbons.
in anionic detergents anion part of the molecule is involved in cleansing action.
eg- Sodium lauryl sulphates.
(ii) Chemicals are added to food to prevent spoilage due to microbial growth
eg- sodium benzoate and also sometimes to provides sweetness without increasing calories
eg- aspartame
(iii) Antibiotics which kill or inhibit a wide range of harmful or disease causing bacteria are called
broad spectrum antibiotics.
eg- ampicillin where as narrow spectrum antibiotics are specific against single type of
microbe are called narrow or limited spectrum antibiotics.
eg- pincillin G.