SECTION- A
1.Give one chemical test to distinguish betweena. Pentan-2-one and Pentan-3-one
b. Benzaldehyde and Acetophenone
Ans. a. Pentan-2-one and Pentan-3-one can be distinguish by haloform test.
Pentan-2-one have methyl ketone so it gives haloform test but Pentan-3-one does not have methyl ketone so it does not give this test.
CH3CH2CH2-CO-CH3 + 3NaOX → CH3CH2CH2COONa + CHX3 + 2NaOH
CH3CH2-CO-CH2CH3 + 3NaOX → No ppt. of haloform.
b. Benzaldehyde reduces Tollen's reagent to give a red-brown precipitate of Cu2O, but acetophenone does not give this test.
Acetophenone undergoes oxidation by NaOX to give a yellow ppt. of haloform, but benzaldehyde does not respond to this test.
2. What pressure of H2 would be required to make the e.m.f. of the hydrogen electrode zero in pure water at 25°C?
Ans.
(a) What is Fehling’s solution?
(b) To what oxidation state ethanal converts Fehling’s solution?
Ans. (a) Fehling’s solution is alkaline solution of CuSO4 along with some Rochelle salt.
(b) Ethanol converts Cu(II) of Fehling’s solution to Cu(I) i.e., + 1 oxidation state.
SECTION- B
4.(a) On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0 > P.(b) [NiCl4]2− is paramagnetic, while [Ni(CO)4] is diamagnetic, though both are tetrahedral. Why? (Atomic number of Ni = 28).
(c) What is meant by the chelate effect? Give an example.
Ans. (a) If Δ0 > P, it becomes more energetically favourable for the fourth electron to occupy a t2g orbital with configuration t2g4 eg.
(b) Though both [NiCl4]2− and [Ni(CO)4] are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligands. Cl− is a weak field ligand and it does not cause the pairing of unpaired 3d electrons. Hence, [NiCl4]2− is paramagnetic. On the other hand, CO is a strong field ligand and it causes the pairing of unpaired 3d electrons. Hence, [Ni(CO)4] is dimagnetic.
(c) When a polydentate ligand attaches to the metal ion in a manner that forms a ring, then the metal-ligand association is found to be more stable. In other words, we can say that complexes containing chelate rings are more stable than complexes without rings. This is known as the chelate effect.
OR
4.(a) What are the oxidation state of Ni and Fe in Ni(CO)4 and Fe(CO)5.
(b) NH3 acts as a ligand but NH4+ does not.
(c) CN– is an ambidentate ligand.
Ans. (a) In Ni(CO)4 and Fe(CO)5 the oxidation state of Ni and Fe is zero oxidation state.
(b) NH3 acts as a ligand because it has lone pair of electrons, whereas NH4+ does not have lone pair of electrons.
(c) CN– is an ambidentate ligand because it can coordinate through either the nitrogen or the carbon atom to central metal ion.
5. Account for the following:
(a) Europium(II) is more stable than Cerium(II).
(b) Transition elements from interstitial compounds.
(c) Separation of Zr and Hf from a mixture is difficult.
Ans.(a) Electronic configuration Eu2+ is [Xe] 4f 7 5d0 while Ce2+ is [Xe] 4f2 5d0. Hence, Eu(II) has stable configuration as d-orbital is half filled whereas Ce2+ configuration has no such extra stability.
(b) In transition metals small size atoms like carbon, boron, nitrogen etc., occupy the interstices or holes present in the metal lattice. e.g., TiC, Fe3H. These compounds are more malleable, have high melting point and are chemically inert.
(c) Separation of Zr and Hf from a mixture is difficult due to their similar atomic radii they have common physical properties due to which their separation is difficult.
6. An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.
Ans. The given molecular formula C9H10O forms 2,4-DNP derivative and also reduces Tollens reagent. So, the given compound must be an aldehyde. Again, the compound undergoes Cannizzaro reaction and on oxidation, it gives 1,2-benzenedicarboxylic acid. So, the -CHO group is directly attached to the benzene ring and this benzaldehyde is ortho- di-substituted.Hence, the given organic compound is 2-ethylbenzaldehyde.
7. (a) Explain what is observed when electric current is passed through a colloidal sol.
(b) Why is the ester hydrolysis slow in the begining and becomes faster after sometime?
(c) Why does physisorption decreases with rise in temperature?
Ans. (a) On passing electric current through a sol, colloidal particles start moving towards oppositely charged electrode where they lose their charge and get coagulated. This process is known as electrophoresis.
(b) In simple ester hydrolysis reaction that can be written as:
Ester + Water → Acid + Alcohol
The acid i.e. product of the hydrolysis will release H+ ions in solution which acts as catalyst (autocatalysis) for the reaction. Therefore, the slow process of hydrolysis will become faster.
(c) Physissorption is exothermic in nature (ΔH is negative) which means that the rise in temperature favours the reverse process i.e., desorption. Therefore, physisorption decreases with rise in temperature.
OR
7. Explain what is observed when:
(a) Silver nitrate solution is added to potassium iodide solution.
(b) The size of the finest gold particles increases in a gold sol.
(c) Two oppositely charged sols are mixed in almost equal proportions.
Ans. (a) AgNO3(aq) + KI(aq) —→ AgI(s) + KNO3(aq)
A yellow precipitate or coagulated silver iodide is formed.
(b) On dissolution, a large number of gold atoms agitate. Thus, the size of the finest gold sol particles increases in the gold. A large number of atoms gold aggregate together to form species having size in the colloidal range (1–1000 nm).
(c) When two oppositely charged sols are mixed in almost equal proportions, their charges are neutralized resulting in coagulation. This type of coagulation is called mutual coagulation or material coagulation.
8. What happens when
a. Acetic acid is heated with P2O5
b. Two moles of acetone are condensed in presence of Ba(OH)2
c. Aniline reacts with chloroform in presence of alcoholic KOH.
Ans.a.
b.
c. When aniline reacts with chloroform in presence of alcoholic KOH it gives phenyl isocyanide.
9. Depict the galvanic cell in which the reaction:
Zn(s)+ 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place.
Further show:
(a) Which electrode is negatively charged?
(b) The carriers of the current in the cell.
(c) Individual reaction at each electrode.
Ans. The galvanic cell in which the given reaction takes place is depicted as:
Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)
(a) Zn electrode (anode) is negative charged.
(b) Ions are carriers of current in the cell and in the external circuit, current will flow from silver to zinc.Ions are Zn2+ and Ag+.
(c) The reaction taking place at the anode is given by
Zn(s) → Zn2+(aq) + 2e–
The reaction taking place at the cathode is given by,
Ag+(aq) + e– → Ag(s)
10. Answer the following questions:
(a) Why is KMnO4 solution used to clean surgical instruments in hospitals?
(b) Out of cobalt and zinc salts, which is attracted in a magnetic field. Explain with reasons.
(c) Name a transition element which does not exhibit variable oxidation states.
Ans. (a) KMnO4 solution is used to clean surgical instruments because of the germicidal action of KMnO4.
(b) In cobalt salts, Co has d7 electrons in the outer shell and hence it has three unpaired electrons. Therefore it will be attracted in a magnetic field. But in case of Zr, it has got d10 configuration, and therefore, will not be attracted by magnetic field.
(c) Scandium.
11.(a) What is Hinsberg Test ?
(b) How can you achieve aniline to benzonitrile?
(c) Why ethylamine is soluble in water but aniline is insoluble?
Ans. (a) Hinsberg test is used to distinguish between 1°, 2° and 3° amines. Here amines are treated with benzene sulphonyl chloride and excess of KOH.
(b)
SECTION- C
12. Read the passage given below and answer the questions that follow:All chemical reactions proceed through one or more transition-state intermediates whose content of free energy is greater than that of either the reactants or the products. For the simple reaction R (reactants)-------> P (products),
The free energy of activation ΔG is equal to the difference in free energy between the transition-state intermediate S and the reactant R. Because ΔG+ generally has a very large positive value, only a small fraction of the reactant molecules will at any one time have acquired this free energy, and the overall rate of the reaction will be limited by the rate of formation of S. (a) What is the main difference between a photosensitizer and a catalyst?
(b) Give reason, why the rate of a reaction generally increases with rise in temperature.
(c) A reaction is found to be zero order, will its molecularity be zero?
(d) A reaction is second order with respect to a reactant. How will the rate of reaction be affected if the concentration of this reactant is:
(i) doubled and (ii) reduced to half.
OR
(d) Why order of a reaction cannot be determined by looking at the balanced chemical reaction?
Ans.(a) A catalyst can only change the speed of the reaction while a photosensitizer only initiates the reaction.
(b) Increase in temperature causes a total increase in the energy of the reacting species. Therefore, more and more reacting species are able to cross the activation energy required to form the product. Hence overall rate of reaction increases.
(c) No, molecularity of a reaction can not be zero.
(d) Rate = k[A]2.
(i) When concentration of ‘A’ is doubled, the rate becomes 4 times.
(ii) When concentration of ‘A’ is reduced to half, the rate becomes ¼ times.
OR
(d) The sum of the stoichiometric coefficients of the reactants in a balanced chemical reaction displays the total number of moles involved in the reacting species but may or may not depict the correct order of the reaction. If the reaction is not an elementary reaction then only the slowest step reactants decides the order of the reaction.