Question

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

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

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

Boiling point or freezing point of liquid solution would be affected by the dissolved solids in the liquid phase. A soluble solid in solution has the effect of raising its boiling point and depressing its freezing point. The addition of non-volatile substances to a solvent decreases the vapor pressure and the added solute particles affect the formation of pure solvent crystals. According to many researches the decrease in freezing point directly correlated to the concentration of solutes dissolved in the solvent. This phenomenon is expressed as freezing point depression and it is useful for several applications such as freeze concentration of liquid food and to find the molar mass of an unknown solute in the solution. Freeze concentration is a high quality liquid food concentration method where water is removed by forming ice crystals. This is done by cooling the liquid food below the freezing point of the solution. The freezing point depression is referred as a colligative property and it is proportional to the molar concentration of the solution (m), along with vapor pressure lowering, boiling point elevation, and osmotic pressure. These are physical characteristics of solutions that depend only on the identity of the solvent and the concentration of the solute. The characters are not depending on the solute’s identity. Answer the following MCQs by choosing the most appropriate options: (i) When a non volatile solid is added to pure water it will- (a) boil above \(100^\circ\)C and freeze above \(0^\circ\)C. (b) b) boil below \(100^\circ\)C and freeze above \(0^\circ\)C. (c) boil above \(100^\circ\)C and freeze below \(0^\circ\)C. (d) boil below \(100^\circ\)C and freeze below \(0^\circ\)C. (ii) Colligative properties are (a) dependent only on the concentration of the solute and independent of the solvent’s and solute’s identity. (b) dependent only on the identity of the solute and the concentration of the solute and independent of the solvent’s identity. (c) dependent on the identity of the solvent and solute and thus on the concentration of the solute. (d) dependent only on the identity of the solvent and the concentration of the solute and independent of the solute’s identity. (iii) Assume three samples of juices A, B and C have glucose as the only sugar present in them. The concentration of sample A, B and C are 0.1 M, 0.5 M and 0.2 M respectively. Freezing point will be highest for the fruit juice (a) A (b) B (c) C (d) all have same freezing point (iv) Identify which of the following is a colligative property? (a) Freezing point (b) Boiling point (c) Osmotic pressure (d) All 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.

Group 18 elements are called noble gases and not inert gases because compounds of Kr, Xe and Rn have been prepared. Their general electronic configuration is \(ns^2\)\(np^6\) except He(\(1s^2\) ). They have highest ionisation enthalpy and positive electron gain enthalpy due to stable electronic configuration. Helium is found in sun and stars. Noble gases have low boiling points due to weak van der Waals’ forces of attraction. Xenon forms \(XeF_2\), \(XeF_4\), \(XeF_6\), \(XeOF_4\), \(XeO_3\), \(XeO_2\)\(F_2\), their structures can be drawn on bases of VSEPR theory. Helium is mixed with oxygen by deep sea divers to avoid pain. Neon is used in coloured advertising lights. Argon is used in bulbs as inert gas. Kr and Xe are used in high efficiency lamps, head light of cars. Radon is radioactive formed by a-decay of Radium 226 88Ra Argon is most abundant (0.9%) noble gas in atmosphere. The following questions are multiple choice questions. Choose the most appropriate answer. 1) What are the elements in group 18 (the far right) of the periodic table called? a) Alkali metals b) Alkaline earth metals c) Halogens d) Noble gases 2) Out of (i) \(XeO_3\) (ii) \(XeOF_4\) and (iii) \(XeF_6\) , the molecules having the same number of lone pairs on Xe are - a) (i) and (ii) only b) (i) and (iii) only c) (ii) and (iii) only d) (i) , (ii) and (iii) 3) Which one has linear shape? a) \(XeF_2\) b) \(XeF_4\) c) \(XeF_6\) d) \(XeO_3\) 4) Which of the outer electronic configuration represent Argon? a) \(ns^2\)\(np^4\) b) \(ns^2\)\(np^3\) c) \(ns^2\)\(np^6\) d) \(ns^1\)\(np^6\) 5) Which of the following statement is false? a) Radon is obtained from the decay of radium b) Helium is an inert gas c) Xenon is the most reactive among the rare gases d) The most abundant rare gas found in the atmosphere is helium

Read the passage given below and answer the following questions: Phenol contains -OH group directly attached to carbon atoms of an aromatic system \(C_6\)\(H_5\)OH in phenol the group is attached to \(sp^2\) hybridised carbon of aromatic ring. The carbon oxygen bond length is 1:36 pm in phenol is slightly less than that in in methanol this is due to first point partial double bond character on account of the conjugation of unshared electron pair of oxygen with the aromatic ring s point \(sp^3\) hybridised state of carbon to which oxygen is attached it can be prepared by various means or methods. Some important methods are alkali fusion of sulphonates, hydrolysis of diazonium salts decarboxylation of salicylic acid and from Grignard reagent, it is prepared from Dow's process and from cumene. Aerial oxidation of human produce cumene peroxide which on hydrolysis produce phenol and acetone. 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. (a) Both assertion and reason are true and the reason is the correct explanation of assertion. (b) Both assertion and reason are true but the reason is not the correct explanation of assertion. (c) Assertion is true but reason is false. (d) Assertion is false but reason is true. (i) Assertion: C-O bond length in phenol is less than C-O bond length in ethanol. Reason: In phenol carbon atom is \(sp^2\) hybridize while in ethanol carbon atom is \(sp^3\) hybridised. (ii) Assertion: p-nitrophenol is more acidic than phenol. Reason: Nitro group helps in the stabilisation of the phenoxide ion by dispersal of negative charge due to resonance. (iii) Assertion: Phenol is prepared by the reaction of chlorobenzene with caustic soda at 623K and 300 atm pressure. Reason: The substitution of chorine atom from chlorobenzene is difficult due to resonance. (iv) Assertion: Methanol is less acidic than phenol. Reason: Due to resonance phenate ion become more stable. (v) Assertion: On nitration phenol forms 2-nitrophenol and 4- nitrophenol. Reason: The presence of –OH group in phenols activates the aromatic ring towards electrophilic substitution and directs the incoming group to ortho and para positions.