Question-1 Arrange the following in the order of increasing rnass (atomic mass; O = 16, Cu = 63, N = 14)
(I) One atom of oxygen
(II) One atom of nitrogen
(III) 1 x 10-10 mole of oxygen
(IV) 1 x 10-10 mole of copper
A) II < I < III < IV
B) I < II < III < IV
C) IV < II < III < I
D) II < IV < I < III
Answer: A) II < I < III < IV
Explanation
1. Mass of one atorn of oxygen
=\frac{16}{6.022\times10^{23}}=2.66\times10^{-23}2. Mass of one atorn of nitrogen
=\frac{14}{6.022\times10^{23}}=2.32\times10^{-23}\;g3. Mass of 1 × 10-10 mole of oxygen = 16×10-10g
4. Mass of 1 × 10-10 mole of copper = 63×10-10
Hence. mas of atoms in irl,creasing order
II < I < III < IV
Question-2 In the N2 molecule, the rumber of electron pairs shared by the nitrogen atoms is
A) 1
B) 2
C) 3
D) 5
Answer: C) 3
Explanation
In the N2 molecule, 3 electron pairs are shared by the nitrogen atoms, as triple bond is present in this molecule.
Question-3 The correct statement about the product of the following reaction is
A) configuration of the product is 100% inverted.
B) configuration of the product is 100% retention.
C) of inverted product is less than retention product.
D) the product will be a racemic mixture.
Answer: D) the product will be a racemic mixture.
Explanation
Reaction proceed via SN1 mechanism and in this reaction carbocation is formed as intermediate which is planar.
On this intermediate, nucleophile can attack from both sides due to which racemic mixture is formed.
Question-4 Lead (IV) oxide is:
A) only basic.
B) only acidic.
C) neutral.
D) amphoteric.
Answer: D) amphoteric.
Explanation
The lead (IV) oxide (PbO2) as acidic and bcsic oxides, hence it is amphoteric in nature.
Whan lead (IV) oxide (PbO2) is dissolved in a strong base [Pb(OH)6]2- is obtained. It is an acidic property.
When lead (IV) oxidePb02 dissolved in HCI, PbCl2 and Cl2 are forrned. It is a basic property.
Thus, lead (IV) oxide (PbO2) behaves as amphoteric oxide
Question-5 A 5% solution (w/w) of cane sugar (molar = 342 g mol-1) has freezing point 271K. What will be the freezing point of 5% glucose (molar mass = 180 g mol-1) in water, if freezing point of pure water is 273.15K?
A) 273.07 K
B) 269.07 K
C) 273.15 K
D) 260.09 K
Answer: B) 269.07 K
Explanation
The expræsion for the depression in freezing point for determining the molar mas of solute is as follows:
\triangle T_f=\frac{K_f\times W_B}{M_B\times W_A}For cane sugar solution,
2.15K=\frac{K_f\times5}{342\times0.095}(95g\;of\;water\;=\;0.095kg)For glucose solution,
\triangle T_f=\frac{K_f\times5}{180\times0.095} \frac{\triangle T_f}{2.15}=\frac{K_f\times5}{180\times0.095}\times\frac{342\times0.095}{K_f\times5} \triangle T_f=\frac{342}{180}\times2.15=4.085KFreezing of glucose solution = 273.15 – 4.085 — 269.07 K
Question-6 In the steady state approximation, if I is the intermediate formed, then
A)
\lbrack I\rbrack=0B)
\left[I\right]\neq0C)
\frac{d\left[I\right]}{dt}=0D) None of the above.
Answer: C)
Explanation
In steady state approxirnation. we asurne that concentration of intermediate is constant during the course of reaction due to which \frac{d\left[Intermediate\right]}{dt}=0
Question-7 Arrange the following in correct order of basicity
A) II > I > III
B) III > II > I
C) I > III > II
D) None
Answer: A) II > I > III
Explanation
This problern includes conceptual of basic hybridization of nitrogen atorn and extent of conjugation.
While solving such problem students are advised to draw the structure and mark the type of hybridization on N-atom, then answer the question by using combined concept of hybridization and conjugation.
Hybridization of N-atom in below cornpounds are sp2, sp3 and sp2 respectively.
Greater the s-character more will be electronegativity of N-atom and lesser will be its basicity on this basic I is less basic than II.
Conjugation If lone pairs of electrons of N is involved in conjugation causes decrease in basicity of compound due to leser availability of lone pair for donation to show basic nature
Lone involved in formation of aromatic sextet of 6\pi -electron (least basic).
Question-8 Match the following
| Column-I | Column-II |
| (A) Co | (i) Wilkinson catalyst |
| (B) Zn | (ii) Chlorophyll |
| (C) Rh | (iii) Vitamin B-12 |
| (D) Mg | (iv) Carbonic anhydrase |
A) (A —iii), (B —iv), (C —i), (D — ii)
B) (A — i), (B — ii), (C —iii), (D —iv)
C) (A — ii), (B —i), (C —iv), (D — iii)
D) (A — iv), (B —iii), (C — i), (D — ii)
Answer: A) (A —iii), (B —iv), (C —i), (D — ii)
Explanation
Carmnic anhydrase
Chlorophyll is a green pigment found in plants. Plants use chlorophyll and light to food. People sometimes use chlorophyll as medicine Comrnon source of chlorophyll used for medicine include alfalfa, algae and silkworm dropping.
Question-9 Which one of the following types of drugs reduces fever?
A) Tranquiliser
B) Antibiotic
C) Antipyretic
D) Analgesic
Answer: C) Antipyretic
Explanation
Antipyretic drugs reduce fever. Analgesic relieves in pain, antibiotics act against bacterial infections while tranquilizers are used
against mental disorders.
Question-10 Follow the provided scheme and identify X and Y.
| X | Y | P | Q | |
| i) | MgO | Mg(OH)2 | Mg(OH)2 | N2 |
| ii) | MgO | Mg3N2 | Mg(OH)2 | NH3 |
| iii) | MgO | Mg3N2 | Mg(OH)2 | N2 |
| iv) | MgO | MgCO3 | Mg(OH)2 | CO2 |
A) (i)
B) (ii)
C) (iii)
D) (iv)
Answer: B) (ii)
Explanation
Magnæium reacts with air to form oxide and nitride. On reaction with water the oxide gives hydroxide and nitride gives hydroxide and ammonia.
2Mg+O_2\rightarrow\underset{(X)}{2MgO} 3Mg+N_2\rightarrow\underset{(Y)}{Mg_3N_2} MgO+H_2O\rightarrow\underset{(P)}{Mg{\left(OH\right)}_2} Mg_3N_2+H_2O\rightarrow\underset{(P)}{3Mg{\left(OH\right)}_2}+\underset{(Q)}{2NH_3}Question-11 The geometry of ClO3¯ ion according to Valence Shell Electron Pair Repulsion (VSEPR) theory will be
A) Planar triangular
B) Pyramidal
C) Tetrahedral
D) Square planar
Answer: B) Pyramidal
Explanation
Cl has 3 sigma bond pairs and 1 lone pair in ClO3¯. So, it has pyrarnidal structure
Question-12 Assume each reaction is carried out in an open container. For which reaction, \triangleH = \triangleE?
A)
H_2(g)+Br_2(g)\rightarrow2HBr(g)B)
C(s)+2H_2O(g)\rightarrow2H_2(g)+CO_2(g)C)
PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)D)
2CO(g)+O_2(g)\rightarrow2CO_2(g)Answer: A)
Explanation
\triangleH = \triangleE + P\triangleV
\triangleH = \triangleE + \triangleng RT
If undefined=O for reactions which is carried out in an container, therefore \triangleH = \triangleE
So, for the reaction (a). \triangle ng = 2-2=0
Hence, \triangleH = \triangleE
Question-13 The elements with maximum and minimum melting in the second transition series respectively are
A) Cr and Zn.
B) Mn and Cd.
C) Cr and Hg.
D) Mo and Cd.
Answer: D) Mo and Cd.
Explanation
Melting and boiling of the d— block is generally higher than the s—block and p—blocki this is due to stronger metallic bonds formed by unpaired d—electrons.
The strength of the rnetallic bonds is on the nurnber of ungx)ired electrons in the outermost shell of the given atom.
Therefore. greater is the number of unpaired electrons, stronger is the metallic bonding.
Cd \rightarrow [Kr] 4d105s2 has zero unpaired electrons.
Mo\rightarrow [Kr] 4d55s1 has 6 unpaired electrons.
Question-14 The detergent which has hydrogen bonding group at the soluble ends of the chain is
A) Anionic detergent
B) Cationic detergent
C) Non-ionic detergent
D) Sodium lauryl sulphate detergent
Answer: C) Non-ionic detergent
Explanation
Non-ionic detergent has hydrogen group at the soluble ends of the chain.
Example :Pentaerythrityl stearate
Question-15 The empirical formula of a compound is CH2O and its vapour density is 30. The molecular formula of the compound is:
A)
C_3H_6O_3B)
C_2H_4O_2C)
C_2H_4OD)
CH_4OAnswer: B)
Explanation
Empirical forrnula= CH2O
Empirical formula mass= 12 + 2 + 16 = 30
Molecular mass= 2 x V. D. = 2 x 30 = 60
n=\frac{Molecular\;mass}{Empirical\;mass}Molecular formula= (Elillmcal formula)n
= (CH2O)2 =C2H4O2
Question-16 Quantitative of C, H, and extra elements (eg., N , S, P, and halogens) is carried out by LiebigS combustion , Carob Dumas, and Kjeldahl’s method.
Carius method is used for the quantitative estimation of:
A) C and H
B) Halogens, S and P
C) N
D) All
Answer: B) Halogens, S and P
Explanation
Carius method:
For this a long sealed hard glass tube called carius tube is used, which is used for the estirnation of Sulptwr and phosphorus.
\rightarrow Halogen is estirnated by the formed silver halide
1 mole of AgX cmtains 1 gm atom of X (X = Cl, Br or I) hence % of halogens in the is:
\frac{At.\;mass\;of\;ha\log en}{108+At.\;mass\;of\;ha\log en}\times\frac{mass\;of\;silver\;halide\;formed}{Mass\;of\;subs\tan ce\;taken}\times100\rightarrow Sulphur is estimated by formed Barium sulphate that is 1 mole BaS04 (233 gm) give 1 gm atom of sulphur (32 gm).
Hence, % of S estimated as follows:
\frac{32}{233}\times\frac{mass\;of\;BaSO_4}{Mass\;of\;subs\tan ce\;taken}\times100\rightarrow Phosporus is estimated by forrned Mg2 P2O7 that is 1 rnole Mg2 P2O7 gives 2 gm atom of P (62 gm).
\rightarrow Hence % of P estimated as follows:
\frac{62}{222}\times\frac{mass\;of\;Mg_2P_2O_7}{Mass\;of\;subs\tan ce\;taken}\times100Question-17 Acetyl Bromide on reaction with an excess of CH3MgI followed by treatment with a saturated solution of NH4Cl gives:
A) 2-methyl-2-propanol
B) Acetamide
C) Acetone
D) Acetyl iodide
Answer: A) 2-methyl-2-propanol
Explanation
Nucleophilic substitution followed by mcleophilic addtion reaction takes place with methyl rnagnesium iodide and acetyl brornide.
Question-18 Number of isomeric alcohols of molecular formula C6H14O , which gives positive Iodoform test is:
A) Two
B) Three
C) Four
D) Five
Answer: C) Four
Explanation
For an alcohol to show haloform reaction, it should be of the type
Here, R can any alkyl group or hydrogen. This type of alcohol, oxidises to give a compound containing a methyl ketone group and thus show positive Iodoform.
C6H14O have 17 isorneric alcohols, out of which four alcohols can show iodoform test. One can think of these isomers by varying the —R group in the structural requirement as described above
The isomers are:
Question-19 Calculate the freezing point of a solution containing 18 g glucose, C6H12O6 and 68.4 g sucrose, C12H22O11 in 200 g of water. The freezing point of pure water is 273 K and Kf for water is 1.86 K kg mol-1.
A) 270.21 K
B) 270 K
C) 273 K
D) 279 K
Answer: A) 270.21 K
Explanation
Molar mass of C6H12O6 = 12 x 6 + 1 x 12 + 16 x 6 = n + 12 + 96 = 180
Molar mass of C12H22O11 = 12 x 12+ 1 x 22 + 16 x 11 = 144 + 22 + +176 = 342
Total moles of solutes,
n_B=\frac{w_1}{M_1}+\frac{w_2}{M_2}=\frac{18}{180}+\frac{68.4}{342}= 0.1 + 0.2 = 0.3 mol
\triangle T_f=\frac{K_f\times n_B\times1000}{W_A} \triangle T_f=\frac{1.86\times0.3\times1000}{200} \triangle T_f=1.86\times0.3\times5=2.79\;K \triangle T_f=T_f^0-T_fWhere Tf0 is the freezing point of solvent and Tf is the freezing point when non-volatile solute is dissolved in it.
Hence, Freezing point of aqueous solution = 273 — 2.79 = 270.21 K.
Question-20 The correct order of nucleophilicity:
| A |  |
| B | CH3O¯ |
| C | CN¯ |
| D |  |
A) A>B>C>D
B) D>C>B>A
C) C>B>A>D
D) None
Answer: C) C>B>A>D
Explanation
Nucleophiles are electron rich cornpounds which possess a negative charge or a lone These are rucleus loving species. The strength of a nucleophile is known as nucleophilicity.
Electronegative atorns pull the electron density and stabilises the negative charge.
More the electronegativity of a donor atorn, les will be its nucleophilicity.
Higher the tendency to dorvate electrons or lone pair, higher is the nucleophilicity.
Nucleophilicity\propto\frac1{Electronegativity\;of\;donor\;atom}CN¯ is the strongest rucleophile amongst all. As carbon and nitrogen are both less electronegative atoms than oxygen, the negative charge on the carbon or nitrogen atom rnakes them highly reactive towards
rucleophilic reactions. The negative charge in cornpound I and IV are resonance stabiliæd, as negative charge on oxygen ion gets delocalised and thus decreases the nucleophilicity.
Question-21 Schottky defect in crystals is observed, when :
A) Density of crystal is increased
B) An ion leaves its normal site and occupies an interstitial site
C) Equal of cations and anions are missing from the lattice
D) Unequal number of cations and anions are missing from the lattice
Answer: C) Equal of cations and anions are missing from the lattice
Explanation
Schottky defect is due to missing of equal number of cations and anions from their respective position leaving behind a pair of holes.
Question-22 The number of molecules of chlorine liberated at the from molten sodium chloride in one minute by a current of 300 milliamperes is
A)
5.618\times10^{18}B)
5.618\times10^{19}C)
2.4\times10^{20}D)
2.4\times10^{18}Answer: B)
Explanation
Molten NaCI has Na+ as the cation and CI¯ as the anion. Electrolysis of molten NaCI results in deposition of Na metal at the cathode and of Chlorine gas at the anode.
Using Faradays law of electrolysis,
n=\frac{i\times t}{x\times F}n is the nurnber of moles of metal deposited. x is valency factor.
n_{Cl_2}=\frac{i\times t}{2\times F}=\frac{300\times10^{-3}A\times60s}{2\times96500}=9.33\times10^{-5}Number of Cl2 molecules = 9.33 × 10-5 × NA
=9.33 × 10-5 × 6.022 x 1023 = 5.618 x 1019
Question-23 What is the formula of carbolic acid ?
A)
HCOOHB)
CH_3COOHC)
C_6H_5COOHD)
C_6H_5OHAnswer: D)
Explanation
Phernol (C6H5 OH) is carbolic acid.
Question-24 Arnong the following, the reaction that proceeds through an electrophilic substitution, is:
A) 
B) 
C) 
D) 
Answer: B)
Explanation
Option B proceeds through an electrophilic substitution reaction.
Cl2 reacts with AlCl3 to form Cl+ and [AlCl4]¯. This Cl+ attaches the molecule making it electrophilic substitution.
Option A is nucleophilic addition.
Option C is an addition reaction.
Option D is rucleophilic substitution.
Question-25 In an aqueous solution of B2+ (1M) at 25ºC, when a metal rod A is dipped, then (electrode potential of A2+ || A = —O. 76 V; B2+ || B = +0.34 V)
A) A will gradually dissolve
B) B will gradually dissolve
C) No reaction will occur
D) Water will decompose into Hydrogen and Oxygen gas
Answer: A) A will gradually dissolve
Explanation
Feasibility of any reaction can be determined by the value of Gibbs free energy change of the reaction.
| Gibbs Free Energy Change | Feasibility of the Reaction |
| \triangleG < 0 | Feasible |
| \triangleG > 0 | Not feasible |
| \triangleG = 0 | Equilibriurn |
For an electrochemical cell:-
\triangleGº = – nFE°cell
Where, \triangleGº = Gibbs free energy
n = of moles of electrons in the reaction
F = Faraday’s constant ~ 96500 C mol-1
E°cell = EMF of Cell
If E°cell > 0 then \triangleGº will be les than 0 and reaction will be feasible.
For the reaction. A + B2+ \rightarrow A2+ + B
A|A2+ will act as an anode and B2+ |B will act as a cathode.
For an electrochernical cell:-
E°cell = E°cathode – E°anode
= o. 34 — (—0.76) = 1.1 v
Since E°cell > 0, hence, the reaction is feasible and A will gradually dissolve in solution of B2+.
Question-26 An element of 3d—transition series shows two oxidation states x and y differing by two units. Then compounds in oxidation state
A) x are ionic if x > y
B) x are ionic if x < y
C) y are covalent if x = y
D) y are covalent if y < x
Answer: B) x are ionic if x < y
Explanation
As Fajan’s rule,
Covalent nature \propto Polarisation \propto Charge on metal ion.
Therefore. order of covalent nature is M+x < M+y
Usually in +2 and +3 oxidation states, the bonds formed are mostly ionic, while in higher oxidation states the bonds formed are covalent.
Question-27 Calculate the half-life of first-order reactions from their rate constants given below:
200 s-1, 2 min-1, 4 years-1
A)
3.46\times10^{-3}s,\;3.46\times10^{-1}\;min,\;1.73\times10^{-1}\;yearsB)
2.46\times10^{-3}s,\;2.46\times10^{-1}\;min,\;0.73\times10^{-1}\;yearsC)
1.46\times10^{-3}s,\;1.46\times10^{-1}\;min,\;2.73\times10^{-1}\;yearsD) None of these
Answer: A)
Explanation
Half-life period for a first-order reaction:
t_{1/2}=\frac{0.693}k i)\;t_{1/2}=\frac{0.693}{200s^{-1}}=0.346\times10^{-2}s=3.46\times10^{-3}s ii)\;t_{1/2}=\frac{0.693}{2min^{-1}}=0.346\;min=3.46\times10^{-1}min ii)\;t_{1/2}=\frac{0.693}{4\;yr^{-1}}=0.173\;years=1.73\times10^{-1\;years}Question-28 Which of the following reaction and name of the reaction is incorrectly matched?
A) Williamson synthesis
B) Kolbés reaction
C) Wolff -Kishner reduction
D) Gattermann-Koch reaction
Answer: C)
Explanation
The reagents used in the different reactions:
Williarnson synthesis: Sodium alkoxide or Sodium phenoxide
Kolbe’s reaction: Carbon dioxide in basic medium
Wolff-Kishner reduction: Hydrazine in the presence of ethylene glycol and base
Gattermarm-Koch reaction: CO + HCI / Cu2 Cl2, \triangle
Question-29 Which group is called buffer group of the periodic table
A) I
B) VII
C) VIII
D) Zero
Answer: D) Zero
Explanation
Zero group is called a buffer group because it lies between highly electronegative halogens and highly electropositive alkali metal elernent.
Question-30 When the sample of copper with zinc impurity is to be purified by electrolysis, the appropriate electrodes cathode and anode are respectively:
A) Pure zinc ,pure copper
B) Impure sample, pure copper
C) Impure zinc, impure sample
D) Pure copper, impure sample
Answer: D) Pure copper, impure sample
Explanation
When the sample of with zinc impurity is to be purified by electrolysis impure is made the anode and pure copper acts as the cathode.
From impure sample Zn goes into the solution as Zn ion and Cu ion deposited as pure copper at cathode after reduction.
Question-31 As per IUPAC nomenclature, the name of the complex [Co(H2O)4(NH3)2]Cl3 is
A) Tetraaquadiaminecobalt (Ill) chloride
B) TetraaquaDiamminecobal (Ill) chloride
C) DiminTetraaqua Cobalt (Ill) chloride
D) Diamminetetraaquacobalt (Ill) chloride
Answer: D) Diamminetetraaquacobalt (Ill) chloride
Explanation
[Co(H2O)4(NH3)2]Cl3 is Diamminetetraaquacobalt (Ill) chloride
IUPAC Rules
(i) To name a coordination compoundi no matter whether the complex ion is the cation or the anion. always the cation before the anim. (This is just like naming an ionic compound.)
(ii) In naming the complex ion:
1. Name the ligands first, in order, then the metal atom or ion. Note: The rnetal atom or ion is before the ligands in the chernical formula.
For anionic ligands end in “-o” for anions end in “-ide”(e.g. chloride). “-ate” (e.g. sulfate. nitrate). and “-ite” (e.g. nirite), change the endings as follows: -ide -ido; -ate -ato; -ite -ito.
For neutral ligands. the common name of the rnolecule is used e.g. H2 NCH2 CH2 NH2 (ethylenediamine).
Important exceptions: water is called ‘aqua’. ammonia is ‘ammine’, carbon monoxide is called ‘cabonyl’and the N2 and 02 are called ‘dinitrogen’ and dioxygen’.
Question-32 The first compound of noble gas prepared by Bartlett was
A)
XeOF_4B)
Xe^+{\lbrack Pt\;F_6\rbrack}^-C)
XeF_4D)
XeF_6Answer: B)
Explanation
The first ever compound of noble gas, Xe^+{\lbrack Pt\;F_6\rbrack}^- was by Neil Bartlett in 1962.
Question-33 According to Faradays first law
A)
w=\frac{96500\times E}{I\times t}B)
w=\frac{E\times I\times t}{96500}C)
E=\frac{I\times t\times96500}wD)
E=\frac{I\times w}{t\times96500}Answer: B)
Explanation
According to Faraday’s first law. ITIass of substance deposited on electrode is directly proportional to the electricity or passed through it.
w=\frac{EIt}{96500}Where,
I = Current
T = Time
E = Equivalent weight
W = Weight
Question-34 The heats of combustion of carbon and carbon monoxide are —393.5 and — 283.5 kJ mol-1, respectively. The heat of formation (in kJ) of carbon monoxide per mole is:
A) -676.5
B) -110
C) 110.5
D) 676.5
Answer: B) -110
Explanation
Given: Heat of combustion of carbon and carbon monoxide are —393.5 and — 283.5 kJ mol-1, respectively.
i)\;C+O_2\rightarrow CO_2;\;\;\triangle H_1=-393.5kJ\;mol^{-1} i)\;CO+\frac12O_2\rightarrow CO_2;\;\;\triangle H_2=-283.5kJ\;mol^{-1}To find the heat of forrnation of CO per mole.
C+\frac12O_2\rightarrow CO_2\triangleHf of CO
Frorn equations (i) and (ii),
=-393.5- (-283.5)
=—110 kJ mol-1
Question-35 The monorner of neoprene rubber is
A) 
B) 
C) 
D) 
Answer: A)
Explanation
Neoprene or polychlorogyene is part of a family of rubbers that are produced by polyrnerization of chloroprene.
Polychloroprene is the main component of Neoprene and characterised by its heat and chemical resistance, as well as tenperature performance.
Chloroprane is a of monorner of neoprene.
Question-36 Match the following
| Column I | Column II |
| A. Na | p. Can form peroxides |
| B. Ba | q. Imparts colour to Bunsen flame |
| C. Li | r. Can form nitrides |
| D. Mg | s. Forms bicarbonates only in solution |
A)
A\rightarrow r,q,\;\;B\rightarrow s,p,\;\;C\rightarrow q,s,\;\;D\rightarrow rB)
A\rightarrow p,q,\;\;B\rightarrow r,p,\;\;C\rightarrow r,q,\;\;D\rightarrow sC)
A\rightarrow p,q,\;\;B\rightarrow p,q,r,s,\;\;C\rightarrow q,r,s\;\;\;D\rightarrow r,sD)
A\rightarrow q,r,\;\;B\rightarrow s,p,\;\;C\rightarrow r,q,\;\;\;D\rightarrow qAnswer: C)
Explanation
A \rightarrow p,q
Sodium can form peroxide and oxides . Due to low ionisation potential, sodium imparts colour to the flame.
B \rightarrow p,q,r,s
Barium exhibit all as mentioned.
C \rightarrow q,r,s
Lithium exhibit all properties except peroxide forrnation as size of lithium is very small . So due to size, holding capacity of lithium is lees, so it cant hold oxygen atom into peroxide bond. But it can directly with oxygen and form its oxides.
D \rightarrow r,s
Magnesium and doesn’t irngx:rts colour to the flame nor form peroxides due to high nuclear charge.
Question-37 After the reaction is over between adsorbed reactants, it is important to create space for the other reactant molecules to approach the surface and react. The process responsible for this is known as
A) sorption
B) desorption
C) physisorption
D) chemisorptim
Answer: B) desorption
Explanation
Desorption is a whereby a substance is released from or through a surface. The process is the opposite of sorption.
This occurs in a system being in the state of sorption equilibrium between bulk phase
and an adsorbing surface.
Whern the concentration of substance in the bulk phase is lowered, sorne of the sorbed substance changes to the bulk state.
Question-38 The of rnoles of BaCO3 contains 1.5 moles of oxygen atoms is –
A) 0.5
B) 1
C) 3
D)
6.02\times10^{23}Answer: A) 0.5
Explanation
Ans = 0.5 mol
Question-39 Arrange the following compounds in increasing order of their boiling points.
A) II < I < III
B) I < II < III
C) III < I < II
D) III < II < I
Answer: C) III < I < II
Explanation
Boiling point decreases with increase in branching. Compound (Ill) has two branches, compound (l) has one branch and compound (II) is a normal alkyl halide with no branch.
So, the boiling point is minimum for compound (Ill) and maximum for compound (II).
Question-40 Zn converts from molten state to its solid state in hcp structure. What is the of nearest atoms?
A) 4
B) 6
C) 8
D) 12
Answer: D) 12
Explanation
Coordination nurnber is the of nearest neighbor atoms or ions surrounding an atom or ion.
In hcp, unit cell is hexagonal and coordination rumber is 12. A hcp unit cell contains six atoms unit cell.
Question-41 The deliquescent among the given compounds is
A)
ZnCl_2B)
Hg_2Cl_2C)
HgCl_2D)
CdCl_2Answer: A)
Explanation
ZnCl2 is deliquescent as it absorbs moisture from atmosphere and gets dissolved in it.
Question-42 In Lassaignds test, the sulphur present in the organic compound, on fusion with sodium, is converted into
A)
Na_2SB)
C_4H_4SC)
Na_2S_2O_3D)
CH_3SHAnswer: A)
Explanation
During the preparation of Lassaignes extract, sulphur from organic cornpound with sodium to form sodium sulphide.
It gives a purple colour with sodiurn nitroprusside due to the forrnation of sodium thionitroprusside.
2Na+S \rightarrow Na2S.
Question-43 Which of the following nitrogenous compounds does not give blue colour in the usual Lassaignds test for the detection of nitrogen?
A) Glycine
B) Urea
C) Aniline
D) Hydrazine
Answer: D) Hydrazine
Explanation
Lassaignes test is based upon the formation of NaCN. Since hydrazine (NH2 NH2) does not contain carbon, it cannot form NaCN during Na fusion and hence does not give blue colour.
All the remaining three compounds (aniline, glycine and urea) contain both C and N needed for the formation of NaCN and thus give blue colour.
Question-44 Which of the following is the correct form of alanine at given pH conditions?
A) II and III
B) I and II
C) III and IV
D) None of these
Answer: A) II and III
Explanation
If the pH is lower (in acidic conditions) than the isoelectric point then the amino acid acts as a base and accepts a proton at the
amino group.
This gives it a positive charge. If the pH is higher (in basic conditions) than the isoelectric point then the amino acid
acts as an acid.
This gives it a negative charge. Alanine is a neutral amino acid, hence, at pH = 1 it is a cation and at pH=13 it is an
anion.
Question-45 A tripeptide (X) on partial hydrolysis gave two dipeptides Cys-Gly and Glu-Cys, i.e.,
Identify the tripeptide.
A) Glu-Cys-Gly
B) Gly-Glu-Cys
C) Cys-Gly-Glu
D) Cys-Glu-Gly
Answer: A) Glu-Cys-Gly
Explanation
Since the tripeptide on hydrolysis gave two dipeptides Glu-Cys and Cys-Gly. Hence, cysteine must be in between glutamic acid and glyciæ as given below.
Question-46 The total præsure of a mixture of H2 and 02 is 1.00 bar. Water, which is formed due to reaction between H2 and 02 , is completely removed in the form of liquid to leave pure stains of H2 at a presure of 0.35 bar. Assuming ideal gas behavior and all pressure rneasurements were made
under the same conditions of and volume, what is the mole fraction of H2 in the original mixture?
A) 0.78
B) 0.28
C) 0.22
D) 0.72
Answer: A) 0.78
Explanation
Let initial pressure of O2 be x.
Hence, Presure of H2 = 1 — x
H2 remaining = 1 — 3x = 0.35
x = 0.65/3
XO2 = 0.217 = 0.22
XH2 = 0.78
Question-47 Which is not true in respect of beryllium chemistry?
A) Beryllium is amphoteric
B) It forms unusual carbide Be2C
C) Be(OH)2 is amphoteric
D) Beryllium halides are electron deficient
Answer: A) Beryllium is amphoteric
Explanation
Be(OH)2 is amphoteric as it with both acids and bases.
Be reacts with alkali & not acids to liberate H2 due to the lowest oxidation potential.
Be+2NaOH+2H_2O\rightarrow Na_2BeO_2\cdot2H_2O+H_2\uparrow Be{\left(OH\right)}_2+2HCl\rightarrow BeCl_2+2H_2O Be{\left(OH\right)}_2+2NaOH\rightarrow\underset{(Sodium\;berylate)}{Na_2BeO_2}+2H_2OBe2C evolves CH4 on hydrolysis as comr:xred to acetylene which other alkaline earth metal carbides evolve.
BeC+4H_2O\rightarrow2Be{(OH)}_2+CH_4\uparrowBeCl2 is covalent. Lewis acid.
Question-48 Adrenaline and ephedrine contain
A) 1o amino group
B) 2o amino group
C) 3o amino group
D) Quaternary amino group
Answer: B) 2o amino group
Explanation
A prirnary amine has one alkyl or aryl group on the nitrogen atom, a secondary amine has two alkyl or aryl groups and a tertiary amine has three alkyl or aryl groups on the nitrogen atorn.
The structure of Adrenaline is
The structure of Ephedrine is
In both the compounds nitrogen is surrounded by two alkyl groups. Therefore, these two are 20 amines.
Question-49 A colloidal system having a solid substance as a dispersed phase and a liquid as a dispersion medium is dassified as
A) Solid sol
B) Gel
C) Ernulsion
D) Sol
Answer: D) Sol
Explanation
A colloidal system having a solid substance as a dispersed phase and a liquid as a dispersion medium is classified as sol.
Salt dissolved in water is an exarnple of sol.
Question-50 A flask of capacity one litre contains NH3 at 1 atm & 25 ºC . A spark is passed through until all the NH3 is decomposed into N2 & H2 . Calculate the pressure of gases left at 25ºC .
A) 2 atm
B) 0.5 atm
C) 1.5 atm
D) 1 atm
Answer: A) 2 atm
Explanation
Arnrnonia decomposes into N2 and H2 cornpletely when a spark is through it.
| NH3(g) | \rightarrow | 1/2 N2(g) | + | 3/2 H2(g) | |
| Initid presure in atm | 1 | 0 | 0 | ||
| Final pressure | 0 | 1/2 | 3/2 |
The total pressure of N2 and H2 after complete decomposition is given by
PT = PN2 + PH2 = 0.5 + 1.2 = 2 atm
