Question

The sequence of bases along the DNA and RNA chain establishes its primary structure which controls the specific properties of the nucleic acid. An RNA molecule is usually a single chain of ribose-containing nucleotide. On the basis of X-ray analysis of DNA, J.D., Watson and F.H.C. CYST (shared noble prize in 1962) proposed a three dimensional secondary structure for DNA. DNA molecule is a long and highly complex, spirally twisted, double helix, ladder like structure. The two polynucleotide chains or strands are linked up by hydrogen bonding between the nitrogenous base molecules of their nucleotide monomers. Adenine (purine) always links with thymine (pyrimidine) with the help of two hydrogen bonds and guanine (purine) with cytosine (pyrimidine) with the help of three hydrogen bonds. Hence, the two strands extend in opposite directions, i.e., are antiparallel and complimentary. In these questions (Q. No. i-iv), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. a) Assertion and reason both are correct statements and reason is correct explanation for assertion. b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. c) Assertion is correct statement but reason is wrong statement d) Assertion is wrong statement but reason is correct statement. i) Assertion - DNA molecules and RNA molecules are found in the nucleus of a cell. Reason : There are two types of nitrogenous bases, purines and pyrimidines. Adenine (A) and guanine (G)are substituted purines; cytokine (C), thymine (T) arid uracil (U) are substituted pyrimidines. ii) Assertion .- In both DNA and RNA, heterocyclic base and phosphate ester linkages are at C- 1’ and C-5’respectively of the sugar molecule. Reason : Nucleotides and nucleosides mainly differ from each other in presence of phosphate units. iii)Assertion .- The backbone of RNA molecule is a linear chain consisting of an alternating units of heterocylic base, D-ribose and a phosphate. Reason : The segment of RNA which acts as the instruction manual for the synthesis of protein is ribose. iv) Assertion.- In DNA, the complementary bases are, adenine and guanine; thymine and cytosine. Reason : The phenomenon of mutation is chemical change in DNA molecule.

An organic compound A having molecular formula \(C_6\)\(H_6\)O turn blue litmus solution into red but does not react with sodium bicarbonate, but when treated with bromine water then form a white ppt of compound B. when compound A react with chloroform in presence of aqueous caustic soda solution at 340K then form two compound C and D. When compound A treated with caustic soda then form compound E compound E when treated with methyl halide then form compound F. Read the above passage carefully and answer the following questions: (i) The name of compound is: (a) 2-methyl propene-2-ol (b) 2-methyl phenol (c) 2,4,6-tribromophenol (d) Butane 1-ol (ii) Which are isomers of each other (a) A and C (b) B and C (c) C and D (d) D and E (iii) The IUPAC name of compound F is (a) Anisole (b) Methoxybenzene (c) Salicylaldehyde (d) 2-methyl propene-2-ol (iv) When compound E treated with ethyl iodide then ...... form. (a) Ehoxybenzene (b) Ethoxy hexane (c) Propoxypropane (d) Benzaldehyde (v) On oxidation with sodium dichromate and conc \(H_2\)\(SO_4\) phenol gives (a) Benzaldehyde (b) p-Benzoquinone (c) o-Benzoquinone (d) m-benzoquinone

The solubility of gases increases with increase of pressure. William Henry made a systematic investigation of the solubility of a gas in a liquid. According to Henry’s law “the mass of a gas dissolved per unit volume of the solvent at constant temperature is directly proportional to the pressure of the gas in equilibrium with the solution”. Dalton during the same period also concluded independently that the solubility of a gas in a liquid solution depends upon the partial pressure of the gas. If we use the mole fraction of gas in the solution as a measure of its solubility, then Henry’s law can be modified as “the partial pressure of the gas in the vapour phase is directly proportional to the mole fraction of the gas in the solution”. Answer the following MCQs by choosing the most appropriate options: (i) Henry’s law constant for the solubility of methane in benzene at 298 K is 4.27 × \(10^5\) mm Hg. The solubility of methane in benzene at 298 K under 760 mm Hg is (a) 4.27 × \(10^{-5}\) (b) 1.78 × \(10^{-3}\) (c) 4.27 × \(10^{-3}\) (d) 1.78 × \(10^{-3}\) (ii) The partial pressure of ethane over a saturated solution containing 6.56 × \(10^{-2}\) g of ethane is 1 bar. If the solution contains 5.00 × \(10^{-2}\) g of ethane then what will be the partial pressure (in bar) of the gas? (a) 0.762 (b) 1.312 (c) 3.81 (d) 5.0 (iii) KH (K bar) values for Ar(g), \(CO_2\)(g), HCHO(g) and \(CH_4\)(g) are 40.39, 1.67, 1.83 × \(10^{-5}\) and 0.413 respectively. Arrange these gases in the order of their increasing solubility. (a) HCHO < \(CH_4\) < \(CO_2\) < Ar (b) HCHO < \(CO_2\) < \(CH_4\) < Ar (c) Ar < \(CO_2\) < \(CH_4\) < HCHO (d) Ar < \(CH_4\) < \(CO_2\) < HCHO (iv) When a gas is bubbled through water at 298 K, a very dilute solution of the gas is obtained. Henry’s law constant for the gas at 298 K is 150 K bar. If the gas exerts a partial pressure of 2 bar, the number of millimoles of the gas dissolved in 1 L of water is (a) 0.55 (b) 0.87 (c) 0.37 (d) 0.66

The properties of dilute or ideal solutions which depend only upon the concentration of the solute in the solution and no other characteristics are known as colligative properties. There are in all four such properties i.e. relative lowering in vapour pressure, osmotic pressure, elevation in boiling point temperature and depression in freezing point temperature. All of them help in calculating the observed molar mass of the solute which is inversely proportional to the colligative property involved. Out of these, osmotic pressure may be regarded as the best for the determination of molecular mass of the solute. According to Van’t Hoff theory of dilute solution, π = CRT, where ‘π’ is the osmotic pressure while ‘C’ is the molar concentration of the solution. (i) When liquids A and B are mixed, hydrogen bonding occurs. The solutions will show: a) Positive deviation from Raoult’s law b) Negative deviation from Raoult’s law c) No deviation from Raoult’s law d) Slightly increase in volume (ii) The azeotropic mixture of water and HCl boils at \(108.5^\circ\)C when the mixture is distilled. It is possible to obtain: a) Pure HCl b) Pure water c) Pure water as well as pure HCl d) Neither HCl nor water in their pure states. (iii) On freezing an aqueous solution of sugar, the solid which starts separating out is: a. Sugar b. Ice c. Solution with the same composition d. Solution with different composition (iv) The value of osmotic pressure does not depend upon: a) Concentration of the solution b) Temperature of the solution c) Number of the particles of the solute present d) Structure of the solute particles (v) Effect of adding a non-volatile solute to a solvent is : a) to lower the vapour pressure b) to increase the freezing point c) to decrease the boiling point d) to decrease the osmotic pressure

Read the passage given below and answer the following questions: The transition metals when exposed to oxygen at low and intermediate temperatures form thin, protective oxide films of up to some thousands of Angstroms in thickness. Transition metal oxides lie between the extremes of ionic and covalent binary compounds formed by elements from the left or right side of the periodic table. They range from metallic to semiconducting and deviate by both large and small degrees from stoichiometry. Since d electron bonding levels are involved, the cations exist in various valence states and hence give rise to a large number of oxides. The crystal structures are often classified by considering a cubic or hexagonal close-packed lattice of one set of ions with the other set of ions filling the octahedral or tetrahedral interstices. The actual oxide structures, however, generally show departures from such regular arrays due in part to distortions caused by packing of ions of different size and to ligand field effects. These distortions depend not only on the number of d-electrons but also on the valence and the position of the transition metal in a period or group. In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices on the basis of the above passage. A. Assertion and reason both are correct statements and reason is correct explanation for assertion. B. Assertion and reason both are correct statements but reason is not correct explanation for assertion. C. Assertion is correct statement but reason is wrong statement. D. Assertion is wrong statement but reason is correct statement. 1) Assertion: Cations of transition elements occur in various valence states Reason: Large number of oxides of transition elements are possible. 2) Assertion: Crystal structure of oxides of transition metals often show defects. Reason: Ligand field effect cause distortions in crystal structures. 3) Assertion : Transition metals form protective oxide films. Reason: Oxides of transition metals are always stoichiometric. 4) Assertion: CrO crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by chromium ions. Reason: Transition metal oxide may be hexagonal close-packed lattice of oxide ions with metal ions filling the octahedral voids.

Molecular Nitrogen \(N_2\) comprises about 78% by volume of Earth’s atmosphere. It occurs as Sodium nitrate, \(NaNO_3\)(chile saltpeter) & Potassium nitrate, \(KNO_3\)(Indian altpeter) in earth’s crust. Since nitrate are very soluble in water so these are not wide spread in the earth’s crust. Nitrogen is also an important constituent of amino acids, protein & nucleic acids in plants & animals. Nitrogen shows anomalous behavior from rest of the elements due to following reasons; Smaller size, high ionization enthalpy, high electronegativity & absence of d-orbital. It has unique ability to form p∏-p∏ multiple bonds with itself & with small size atoms like C & O as they have small size & high electronegativity. Heavier elements of this group do not form p∏-p∏ bonds as their atomic orbitals are so large & diffuse that they can’t have effective overlapping. Thus Nitrogen exists as diatomic molecules \(N_2\) with a triple bond. Consequently, its bond enthalpy (941.4 KJ \(mol^{-1}\)) is very high. P, As & Sb form only single bonds as P-P, As-As & Sb-Sb. Due to much bond enthalpy N is much less reactive than P. Single N-N bond is weaker than single P-P bond due to high interelectronic repulsion of the non bonding electrons, owing to small bond length. As a result, the catenation tendency is weaker in nitrogen. Hence nitrogen exists as gas while phosphorus exists as solid. Nitrogen can’t form d∏- d∏ bond due to absence of d- orbitals so it can’t expand its covalency beyond four as heavier members can. The following questions are multiple choice questions. choose the most appropriate answer. 1) Among group 15 elements which exists as gas at room temperature a) Arsenic b) Bismuth c) Nitrogen d) Phosphorous 2) The stability of +5 oxidation state decreases and that of +3 state increases down the group in group 15 elements due to a) inert pair effect b) decrease in ionisation enthalpy c) increase in size d) shielding effect 3) Nitrogen is restricted to a maximum covalency of 4 because of a) absence of d-orbitals b) presence of d-orbitals c) absence of s and p-orbitals d) none of the above 4) Extra pure \(N_2\) can be obtained by heating a) \(NH_3\) with CuO b) \(NH_4NO_3\) c) \(\left(NH_4\right)_2Cr_2O_7\) d) \(Ba\left(N_3\right)_2\) 5) Catenation tendency is weaker in nitrogen, because of a) single N–N bond is weaker b) single N–N bond is stronger c) ability to form pi bonds by N atoms d) none of the above

Read the passage given below and answer the following questions: All real structures are three-dimensional structures. They can be obtained by stacking two dimensional layers one above the other while placing the second square close packed layer above the first we follow the same rule that was followed when one row was placed adjacent to the other. The second layer is placed over the first layer such that the spheres of the upper layer are exactly above there of the first layer. In his arrangement spheres of both the layers are perfectly aligned horizontally as well as vertically. A metallic element crystallise into a lattice having a ABC ABC pattern and packing of spheres leaves out voids in the lattice. 1) What type of structure is formed by this arrangement? (A) ccp (B) hcp (C) ccp/fcc (D) none of the above 2) Name the non-stoichiometric point defect responsible for colour in alkali metal halides. (A) Frenkel defect (B) Interstitial defect (C) Schottky defect (D) F-centres 3) What is the total volume of atoms in a face centred cubic unit cell of a metal? (r is atomic radius). (A) 16/3 \(πr^3\) (B) \(πr^3\) (C) 24/3 \(πr^3\) (D) 12/3 \(πr^3\) 4) Which of the following statements not true for the amorphous and crystalline solids? (A) Amorphous solids are isotropic and crystalline solids are anisotropic. (B) Amorphous solids are short range order and crystalline solids are long range order. (C) Amorphous solids melt at characteristic temperature while crystalline solids melt over a range of temperature. (D) Amorphous solids have irregular shape and crystalline solids have a geometrical shape.