Grass-G/jee-main-questions huggingface dataset:
Subject: Chemistry — Physical Chemistry — Mole Concept
Difficulty: Easy

Question: Sulphur exist in different allotropic forms like S2, S6 and S8 etc.
If equal moles of these three forms are taken in separate containers,
then the ratio of number of atoms present in them respectively is:, S2, S6 S8 
 -, 
 -
(1) 1: 3: 4 (2) 1: 1: 1 (3) 12: 4: 3 (4) 4: 3: 1

2) 1: 1: 1
3) 12: 4: 3
4) 4: 3: 1

Correct option: 1

Solution:
S_2 S_6 S_8
Mole () n n n
Molecule() n × N_A n × N_A n × N_A
Atoms () n × N_A × 2 n × N_A × 6 n × N_A × 8
[IMAGE] Ratio of atoms of S (S ) = 2: 6:
8 = 1: 3
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Tough

Question: What will be the product of chlorine water and acetylene

1) Dichloro propanol
2) Dichloroethanal
3) Propanol
4) 2-Chloroethanal

Correct option: 2

Solution:
Subject: Chemistry —  — 

Question: Which of the followings represent enantiomer



Solution:
[IMAGE] Enantiomer (Every wedge converted into dash & Vice-Versa)
Subject: Chemistry — Inorganic Chemistry — P-block nitrogen
Difficulty: Moderate

Question: Which of the following chemical reactions depicts the oxidizing
behaviour of H2SO4?
iznf'kZr djrh

1) 2HI + H2SO4 → I2 + SO2 + 2H2O
2) Ca(OH)2 + H2SO4 → CaSO4 + 2H2O
3) NaCl + H2SO4 → NaHSO4 + HCl
4) 2PCl5 + H2SO4 → 2POCl3 + 2HCl + SO2Cl2

Correct option: 1

Solution:
Subject: Chemistry —  — 

Question: The conversion of oxygen to ozone occurs to the extent of 15 only.
The mass of ozone
that can be prepared from 67.2 L of oxygen at STP will be



Solution:
[IMAGE] 3 mol
3 - 3 × 15 moles of O_3 = 3 × 15 × 2
[IMAGE] mass of O3 [IMAGE] = 14.4 gm
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Easy

Question: pH of 0.1 M Na_2HPO_4 and 0.2 M NaH_2PO_4 are respectively:
(pK_a for H_3PO_4 are 2.2, 7.2 and 12)

1) 4.7, 9.6
2) 9.6, 4.7
3) 9.3, 4.4
4) 4.4, 93

Correct option: 2

Solution:
For Na_2HPO_4, pH =
$\frac{{pK}_{a_{2}} + {pK}_{a_{3}}}{2} = \frac{7.28 + 12}{2} = 9.6$.
For Na_2HPO_4, pH =
$pH = \frac{{pK}_{a_{1}} + {pK}_{a_{2}}}{2} = \frac{2.2 + 7.2}{2} = 4.7$.
Subject: Chemistry — IUPAC Nomenclature — IUPAC -Nomenclature of chain terminating Functional groups
Difficulty: Easy

Question: The correct IUPAC name of following compound is

1) Methyl -2-ethylpropanoate
2) Methyl butane-2-carboxylate
3) Methyl- 2-methylbutanoate
4) Methoxypentanone

Correct option: 3

Solution:
[IMAGE] Methyl 2 -methylbutanoate
Subject: Chemistry — Chemical Bonding — Lewis acids, coordinate bond, electrondeficient bond
Difficulty: Moderate

Question: Which of the following statement is false for trisilylamine?

1) Three sp2 orbitals are used for s bonding, giving a plane
triangular structure.
2) The lone pair of electrons occupy a p-orbital at right angles to
the plane triangle and this overlaps with empty p-orbitals on each of
the three silicon atoms resulting in p bonding.
3) The N-Si bond length is shorter than the expected N-Si bond
length.
4) It is a weaker Lewis base than trimethyl amine.

Correct option: 2

Solution:
There is $p\pi - d\pi$ delocalization of lone pair of electron
on nitrogen atom and empty d-orbital of silicon but not$p\pi - d\pi$,
one of the 2s-orbital electrons jumps to the last Pz orbital and does
not participate in sp2hybridisation.
Subject: Chemistry — Physical Chemistry — Redox Reaction
Difficulty: Tough

Question: How many moles of [IMAGE] can be reduced by 1 mole

4) 1

Correct option: 1

Solution:
[IMAGE] It is clear from this equation that 3 moles of
[IMAGE] reduce one mole of[IMAGE], hence
1mol. of [IMAGE] will reduce 1/3 moles
Subject: Chemistry — Organic Chemistry — Alcohols, Phenols and Ethers
Difficulty: Moderate

Question: With oxalic acid, glycerol at 260º C gives

1) Allyl alcohol
2) Glyceryl mono-array oxalate
3) Formic acid
4) Glyceraldehyde

Correct option: 1

Solution:
Subject: Chemistry — Inorganic Chemistry — Environmental Chemistry
Difficulty: Easy

Question: Choose the factor that is responsible for ozone layer depletion.

1) CFCs
2) CBrCs
3) CH_2Br_2
4) all are correct

Correct option: 1

Solution:
Explanation: Chlorofluorocarbons (CFCs) that are produced by
air conditioners, refrigerators etc. are responsible for ozone layer
depletion. It is also known as Freon and its life in the atmosphere is
about 20 - 100 years.
Subject: Chemistry — Organic Chemistry — Organic Chemistry
Difficulty: Moderate

Question: In the following benzyl/allyl system R - CH = CH2 and
[IMAGE] (R is alkyl group) Then decreasing order
of inductive effect is

1) (CH3)3 C-- > (CH3)2CH-- > CH3CH2
2) CH3CH2-- > (CH3)2CH-- > (CH3)3C
3) (CH3)2CH-- > CH3CH2-- > (CH3)3C
4) (CH3)3C-- > CH3CH2-- > (CH3)2CH

Correct option: 1

Solution:
CH3 group has + I effect, as number of - CH3 group increases
the inductive effect increases.
Subject: Chemistry — Physical Chemistry — Redox Reaction
Difficulty: Moderate

Question: SnCl_2 a gives a precipitate with a solution of HgCl_2. In this
process HgCl_2 is

1) Reduced
2) Oxidized
3) Converted into a complex compound containing both Sn and Hg
4) Converted into a chloro complex of Hg

Correct option: 1

Solution:
Subject: Chemistry — Physical Chemistry — Chemical & Ionic Eqilibrium
Difficulty: Easy

Question: Vant hoff factor of BaCl_2 of conc. 0.01M is 1.98. Percentage
dissociation of BaCl_2 on this conc. Will be


Answer: 49

Solution:
[IMAGE] Initially 1 0 0
After dissociation [IMAGE] Total [IMAGE] For a mole [IMAGE] For 0.01 mole [IMAGE]
Subject: Chemistry — Organic Chemistry — Organic Chemistry
Difficulty: Moderate

Question: Which of the following resonating structures of 1-methoxy-1, 3
-butadiene is least stable


Correct option: 3

Solution:
The octets of all atoms are complete in structures (1) and (2).
In structure (4) electron deficiency of positively charged carbon is
duly compensated by lone pair electrons of adjacent oxygen atom while
such neighbour group support is not available in structure (3).
Subject: Chemistry — Physical Chemistry — Some Basic Concepts of Chemistry
Difficulty: Moderate

Question: BaCl_2 + Na_3PO_4[IMAGE] Ba_3(PO_4)_2 + NaCl
Which of the following statement (s) are correct, when 2 mol each of
Na_3PO_4 and BaCl_2 reacts ?
(I) 6 mol NaCl is formed.
(II) 2/3 mol of Ba3(PO_4)_2 is formed.
(III) Na_3PO_4 is Limiting reagent
(IV) 2/3 mol of Na_3PO_4 is left.
(V) BaCl_2 is Limiting reagent.

1) I and V
2) II and III
3) II, III and IV
4) II, IV and V
BaCl_2 + Na_3PO_4[IMAGE] Ba_3(PO_4)_2 + NaCl

Correct option: 4

Solution:
3BaCl_2 + 2Na_3PO_4[IMAGE] Ba_3(PO_4)_2 +
6NaCl
$\because$ 3 mol of Basl_2 reacts with 2 mol Na_3PO_4
$\therefore$ 2 mol of BaCl_2 reacts with
$= \frac{2}{3} \times 2 = \frac{4}{3}$ mol Na_3 PO_4
Reacted mol of Na_3PO_4$= \frac{4}{3}$ mol
Subject: Chemistry —  — 

Question: The correct order of bond length (C - O) is

1) CO_2 < CO < CO_3 ^2
2) CO_3 ^2- < CO < CO_2
3) CO < CO_2 < CO_3 ^2
4) CO < CO_3 ^2- < CO_2


Solution:
[IMAGE] Bond Order = 2
[IMAGE] Bond order = 3
Subject: Chemistry — Physical Chemistry — Electrochemistry
Difficulty: Easy

Question: The electrolytic cells, one containing acidified ferrous chloride
and another acidified ferric chloride are connected in series. The ratio
of iron deposited at cathodes in the two cells when electricity is
passed through the cells will be

1) 3: 1
2) 2: 1
3) 1: 1
4) 3: 2

Correct option: 4

Solution:
At Ratio of weight of Fe liberated[IMAGE]
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Moderate

Question: Consider the reaction equilibrium
2SO_2 (g) + O_2 (g)[IMAGE] 2SO_3 (g); ΔHº = - 198 kJ.
On the basis of Le Chatelier's principle, the condition favourable for
the forward reaction is

1) lowering of temperature as well as pressure
2) increasing temperature as well as pressure
3) lowering the temperature and increasing the pressure
4) any value of temperature and pressure.
fuEu vfHkfØ ij fopkj djks
2SO_2 (g) + O_2 (g)[IMAGE] 2SO_3 (g); ΔHº = - 198 kJ.
gksxh

Correct option: 3

Solution:
The conversion of SO2 to SO3 is an exothermic reaction, hence
there is no need to increase the temperature. There is also a decrease
in volume or moles in product side. Thus the reaction is favoured by low
temperature and high pressure. (Le-Chatelier's principle).
c<+kus sbZ ugha gSaA pw¡fd mijksDr
vksj ¼nk¡;h vksj½ ksa gks jgh gSA vr yhµ'krkfy, vuqlkj fuEu rki,oe~ mPp nkc bl vfHkfØ;k
Subject: Chemistry — Physical Chemistry — 
Difficulty: Tough

Question: 1.22 g of benzoic acid is dissolved in 100 g of acetone and 100
g of benzene separately. Boiling point of the solution in acetone
increases by 0.17ºC, while that of, in the benzene increases by 0.13ºC;
Kb for acetone and benzene is 1.7 K kg mol-1 and 2.6 K kg mol-1. Find
molecular weight of benzoic acid in two cases and justify your answer.
1.22 g csatksbd vEy dks i`Fkd:i ls 100 g,lhVksu rFkk 100 g csathu esa
tcfd csathu esa 0.13ºC rd DoFkukad c<+rk gSA,lhVksu rFkk csathu ds fy,
Kb Øe'k 1.7 K kg mol-1 rFkk 2.6 K kg mol-1 gSA nksuksa csatkWbd vEy dk vkf.od Hkkj Kkr djrs gq,] vius mÙkj


Answer: 122

Solution:
(i) In first case, çFke voLFkk esa]
$\Delta T_{b} = K_{0} \times m = K_{0} \times \frac{W_{t}\.\ of\ solute\ }{\ Mol.\ wt.\ of\ solute\ } \times \frac{1000}{\ wt.\ of\ solvent\ }$
or $0.17 = 1.7 \times \frac{1.22}{M \times 100 \times 10^{- 3}}$
or M = 122 g/mole
Thus the benzoic acid exists as a monomer in acetone
vr csUtkWbd vEy],lhVksu esa,dyd ds:i aA
(ii) In second case, f}rh; voLFkk esa]
$\Delta T_{b} = K_{0} \times \frac{W\ t.of\ solute\ }{\ Mol.wt.of\ solute\ } \times \frac{1000}{\ wt.\ of\ solvent\ }$
or $0.13 = 2.6 \times \frac{1.22}{M \times 100 \times 10^{- 3}}$
or M` = 224
Double molecular weight of benzoic acid (244) in acetone solution
indicates that benzoic acid exists as a dimer in acetone.
lhVksu csUtkWbd vEy dk nqxquk v.kqHkkj (244) fd csUtksbd vEy,lhVksu
Subject: Chemistry — Coordination Compounds — Valence Bond Theory and Crystal Field Theory (Part I)
Difficulty: Moderate

Question: The crystal field splitting energy for octahedral complex
$\left( \Delta_{0} \right)$ and that for tetrahedral complex
$\left( \Delta_{t} \right)$ are related as:
$\left( \Delta_{t} \right)$ ds fy, fØLVy {ks=k foikVu fuEu izdkj
lEcfU/kr

1) $\Delta_{t} = \frac{4}{9}\Delta_{0}$
2) $\Delta_{t} = 0.5\Delta_{0}$
3) $\Delta_{t} = 0.33\Delta_{0}$
4) $\Delta_{t} = \frac{9}{4}\Delta_{0}$

Correct option: 1

Solution:
(A) This may attributes to the following two reasons.
(i)There are only four ligands instead of six, so the ligand field is
only two thirds the size; as the ligand field splitting is also the
two thirds the size and (ii) the direction of the orbitals does not
concide with the direction of the ligands. This reduces the crystal
field splitting by roughly further two third.
So$\Delta_{1} = \frac{2}{3} \times \frac{2}{3} = \frac{4}{9}\Delta_{0}$
gy uEu nks dkj.kksa ls,slk gks ldrk gSaA
(i) i) vkdkj
 oy nks frgkbZ
frgkbZ gSaA (ii) fyxs.M lkFk lEikrh ugha
blfy,
$\Delta_{1} = \frac{2}{3} \times \frac{2}{3} = \frac{4}{9}\Delta_{0}$
Subject: Chemistry — Periodic table — Ionisation energy
Difficulty: Moderate

Question: Which one of the following statements is incorrect in relation
to ionisation enthalpy ?

1) Ionization enthalpy increases for each successive electron.
2) The greatest increase in ionization enthalpy is experienced on
removal of electron from core of noble gas configuration.
3) End of valence electrons is marked by a big jump in ionization
enthalpy.
4) Removal of electron from orbitals bearing lower n value is
easier than from orbitals having higher n value.

Correct option: 4

Solution:
Orbitals bearing lower value of n will be more closer to the
nucleus and thus electrons will experience greater attraction from
nucleus and so its removal will be difficult, not easier.
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: But-2-ene can be obtained by electrolysis of an aqueous solution
of

1) 2, 3-Dimethyl maleic acid
2) 2, 2-Dimethyl butanedioic acid
4) 2,3-Dimethylbutanedioic acid
C;wV-2-bZu

Correct option: 4

Solution:
Subject: Chemistry — P-block — Miscellaneous (Silicones, Silicates, Zeolites & Alums)
Difficulty: Moderate

Question: Select incorrect statement

1) Red lead is Pb3O4
2) (Me)2SiCl2 on hydrolysis and then on subsequent intermolecular
condensation gives cross linked silicones.
3) SiO44- on hydrolysis with water or acid produces Si2O76
4) None

Correct option: 2

Solution:
When a compound like CH3SiCl3 undergoes hydrolysis, a complex
cross-linked polymer is obtained as chain can grow in three places as
Subject: Chemistry — Inorganic Chemistry — Metallurgy
Difficulty: Easy

Question: Heating mixture of Cu_2O and Cu_2S will give

1) Cu + SO_2
2) Cu + SO_3
3) CuO + CuS
4) Cu_2 SO_3

Correct option: 1

Solution:
Cu_2 S + 2Cu_2 O → 6Cu + SO_2  (Auto-reduction).
This reaction occurs in reverberatory furnace to get metallic copper.
Subject: Chemistry — Electrochemistry — Completely miscible liquids : Raoult's law
Difficulty: Moderate

Question: An ideal solution contains two volatile liquids A (pº = 100
torr) and B (pº = 200 torr). If mixture contain 1 mole of A and 4 mole
of B then total vapour pressure of the distillate is


Answer: 188.88

Solution:
[IMAGE] yA (composition of A in vapour phase) [IMAGE] For condensation
$y_{A} = x_{A}^{1};y_{B} = x_{B}^{1}$
$P_{bal} = P_{A}^{\circ}x_{A}^{1} + P_{B}^{\circ}x_{B}^{1}$
$P_{wal} = 100 \times \frac{1}{9} + \frac{8}{9} \times 200\ \Rightarrow 100 \times \frac{1}{5} + 200 \times \frac{4}{5}$
Subject: Chemistry — Physical Chemistry — Chemical equilibrium
Difficulty: Tough

Question: When the pressure is applied over system ice $\rightleftharpoons$
water what will happen?

1) More water will form
2) More ice will form
3) There will be no effect over equilibrium
4) Water will decompose in H_2 and O_2

Correct option: 1

Solution:
$\begin{matrix}
Ice \\
more\ volume
\end{matrix} \rightleftharpoons \begin{matrix}
Water \\
less\ volume
\end{matrix}$
On increasing pressure, equilibrium shifts forward.
Subject: Chemistry — Physical Chemistry — Electrochemistry
Difficulty: Moderate

Question: The molar conductance of at infinite dilution are 126.45,426.16 and
[IMAGE] respectively. The molar conductance of
[IMAGE] at infinite dilution is


Correct option: 2

Solution:
Subject: Chemistry —  — 

Question: The electrode potentials for [IMAGE] and
[IMAGE] are are respectively. The value of [IMAGE] will be



Solution:
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Tough

Question: How many of the following relations are correct for the
solubility product (Ksp) & solubility (s g/litre) of sparingly soluble
salt A3B2 (producing A2+ & B3- ions; mol. wt. M) in water ?
(Assume no hydrolysis of any ion).
ty esa vYi cukrk M)
lgh gSa (ekuk fd

1) K_sp = 108s^5
2) $K_{sp} = \left\lbrack \frac{3s}{M} \right\rbrack^{3}\left\lbrack \frac{2s}{M} \right\rbrack^{2}$
3) $K_{sp} = \left( 3\left\lbrack A^{2 +} \right\rbrack \right)^{3}\left( 2\left\lbrack B^{3 -} \right\rbrack \right)^{2}$
4) $\left\lbrack B^{3 -} \right\rbrack = \frac{2s}{M}$

Answer: 4

Solution:
A3B2 [IMAGE] 3A2+ + 2B3+
KSP =
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: A metallic carbide on treatment with water gives a colourless
gas which burns readily in air and gives a precipitate with ammonical
silver nitrate solution. The gas evolved is:
tks fd rsth ls gok

1) C_5 H_5
2) C_6 H_6
3) C_2 H_4
4) C_2 H_2

Correct option: 4

Solution:
Subject: Chemistry — Periodic table — Shielding Effect &Z_eff
Difficulty: Moderate

Question: From the given set of species, point out the species from each
set having highest Z_eff
(a) O2-, F-, Na+ (b) Li, Be, Na (c) He, Li+, H-
a b c

1) Na+ Be Li+
2) O2- Li H
3) F- Na He
4) Na+ Be He

Correct option: 1

Solution:
For isoelectronic species, as Z increases, Z_eff increases
(and vice versa).
Subject: Chemistry — Organic Chemistry — Chemistry in Everyday life
Difficulty: Moderate

Question: Which of the following is a bactericidal antibiotic ?

1) Erythromycin
2) Tetracycline
3) Chloramphenicol
4) Ofloxacin

Correct option: 4

Solution:
Ofloxacin is a bactericidal antibiotic. On the other
hand,erythromycin, tetracycline and chloramphenicol arebacteriostatic
antibiotic.
Subject: Chemistry — Organic Chemistry — Biomolecules
Difficulty: Moderate

Question: Which of the following pentoses will be optically active

1) All
2) II and III
3) I
4) II

Correct option: 1

Solution:
All are optically active.
Subject: Chemistry — Surface Chemistry — Catalysis
Difficulty: Easy

Question: A catalyst increases rate of reaction by

1) Decreasing enthalpy
2) Decreasing internal energy
3) Decreasing activation energy
4) Increasing activation energy

Correct option: 3

Solution:
Catalyst provides new path to the chemical reaction which has
lower value of activation energy. Reactant and product will not be
affected, so there will not be any change in state parameter like
enthalpy and internal energy.
Subject: Chemistry — Inorganic Chemistry — Gaseous state
Difficulty: Moderate

Question: At 400 K, the root mean square (rms) speed of a gas X (molecular
weight = 40) is equal to the most probable speed of gas Y at 60 K. The
molecular weight of the gas Y is



Solution:
$\sqrt{\frac{3RT_{1}}{M_{1}}} = \sqrt{\frac{2RT_{2}}{M_{2}}}$
$\sqrt{\frac{3 \times 400}{40}} = \sqrt{\frac{2 \times 60}{x}}$
$30 = \frac{120}{x}$
x = 4 g/mol
Subject: Chemistry — Organic Chemistry — Organic Chemistry
Difficulty: Moderate

Question: The oxidation state of Cr in [IMAGE] is


Answer: 3

Solution:
Subject: Chemistry — IUPAC and Structural Isomerism — IUPAC -Nomenclature of chain terminating Functional groups
Difficulty: Moderate

Question: The correct IUPAC name of following compound is

1) 4-Aminomethyl-3-hydroxycyclohex-5-ene-1-carboxylic acid
2) 2-Aminomethyl-5-carboxycyclohex-3-en-1-ol
3) 4-Aminomethyl-5-hydroxycyclohex-2-ene-1-carboxylic acid
4) 3-Hydroxy-4-aminomethylcyclohex-5-en-1-oic acid

Correct option: 3

Solution:
[IMAGE] 4-Aminomethyl-5-hydroxycyclohex-2-ene-1-carboxylic acid
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: The major organic compound formed by the reaction of 1,1,1-
trichloroethane with silver powder is ____-Butyne.
cM+h ek=kk


Answer: 2

Solution:
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Tough

Question: Match the reactants in column I with product in column II:
Column I Column II
dkWye I dkWye II
(A) CaC_2 + H_2O (p) CH_3 - C $\equiv$ C - CH_3
(B) Mg_2C_3 + H_2O (q) CH_3 - C $\equiv$ CH
(C) [IMAGE] (r) [IMAGE] (D) [IMAGE] (s) [IMAGE] (E) [IMAGE] (t) CH_3 - CH_3
(F) [IMAGE] (u) I_2CH - CHI_2
(G) [IMAGE] (v)

1) (A) - (s) (C) - q (E) - s 
(G) - s
2) (A) - (s) (C) - s (E) - s 
(G) - p
3) (A) - (s) (C) - p (E) - s 
(G) - s
4) (A) - (s) (C) - s (E) - s 
(G) - p

Correct option: 4

Solution:
(A) CaC_2 + H_2O [IMAGE] Ca(OH)_2 + CH
(B) Mg_2C_3 + H_2O[IMAGE] Mg (OH)_2 + CH_3
Subject: Chemistry —  — 

Question: Which of the following compounds was starting material for the
oxidation shown below?



Solution:
Subject: Chemistry — Organic Chemistry — Structure Identification & POC
Difficulty: Moderate

Question: How many isomeric alkynes on catalytic hydrogenation gives
3-Ethyl-4-methylheptane ?


Answer: 3

Solution:
Subject: Chemistry — Physical Chemistry — Chemical Bonding
Difficulty: Moderate

Question: The correct order of increasing covalent character is

1) LiCl, NaCl, BeCl_2
2) BeCl_2, NaCl, LiCl
(3) NaCl, LiCl, BeCl_2
4) BeCl_2, LiCl, NaCl

Answer: 3

Solution:
According to Fajans' rule,
Covalent character (when anion is same)
[IMAGE] charge of cation
Na and Li have +1 charge while Be has +2 charge. Moreover, the order
of size is Be < Li < Na
Thus, the order of covalent character is BeCl2 > LiCl > NaCl,
 
 [IMAGE] 
Na Li +1 Be +2 
 Be < Li < Na,:
BeCl_2 > LiCl > NaCl
Subject: Chemistry — Physical Chemistry — Redox Reaction
Difficulty: Easy

Question: In acid solution, the reaction [IMAGE] involves

1) Oxidation by 3 electrons
2) Reduction by 3 electrons
3) Oxidation by 5 electrons
4) Reduction by 5 electrons

Correct option: 4

Solution:
[IMAGE] In this reaction [IMAGE] are needed for the reduction of [IMAGE] as
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Moderate

Question: Phosphorus pentachloride dissociates as follows in a closed
reaction vessel.
PCl_5(g) [IMAGE] PCl_3(g) + Cl_2(g)
If total pressure at equilibrium of the reaction mixture is P and
degree of dissociation of PCl_5 is x, the partial pressure of PCl3
will be:
gksrh gS
PCl_5(g) [IMAGE] PCl_3(g) + Cl_2(g)
fn vfHkfØ;k feJ.k ij dqy nkc P gS rFkk PCl_5 sfV x

1) $\left( \frac{x}{x + 1} \right)P$
2) $\left( \frac{2x}{1 - x} \right)P$
3) $\left( \frac{x}{x + 1} \right)P$
4) $\left( \frac{x}{1 - x} \right)P$

Correct option: 1

Solution:
Given: PCI5 (g)[IMAGE] PCI3(g) + CI2(g)
t = 0 1 0 0
teq 1-x x x
Total number of moles = 1 - x + x + x = 1 + x
Thus partial pressure of PCI3 = $\left( \frac{x}{1 + x} \right)$ P.
fn;k gqvk gS PCI5 (g)[IMAGE] PCI3(g) + CI2(g)
t = 0 1 0 0
teq 1-x x x
dqy = 1 - x + x + x = 1 + x
vr PCI3 = $\left( \frac{x}{1 + x} \right)$ P.
Subject: Chemistry — Physical Chemistry — Ionic equilibrium
Difficulty: Moderate

Question: A solution of Na2CO3 is added drop by drop to one litre of a
solution containing 10-4 mole of Ba2+ and 10-5 mole of Ag+. If Ksp for
BaCO3 is 8.1 × 10-9 and Ksp for Ag2CO3 is 6.9 × 10-12, then which is
not true ? Assume no hydrolysis of CO32- ion.

1) No precipitate of BaCO3 will appear until
$\left\lbrack {CO}_{3}^{2 -} \right\rbrack$ reaches 8.1 × 10-5 mol
per litre.
2) A precipitate of Ag2CO3 will appear when
$\left\lbrack {CO}_{3}^{2 -} \right\rbrack$ reaches 6.9 × 10-5 mol
per litre.
3) No precipitate of Ag2CO3 will appear until
$\left\lbrack {CO}_{3}^{2 -} \right\rbrack$ reaches 6.9 × 10-2 mole
per litre.
4) BaCO3 will be precipitated first.

Correct option: 2

Solution:
For precipitation of Ag2CO3
$\left\lbrack {CO}_{3}^{2 -} \right\rbrack = \frac{K_{SP}}{\left\lbrack {Ag}^{+} \right\rbrack^{2}} = \frac{6.9 \times 10^{- 12}}{\left\lbrack 10^{- 5} \right\rbrack^{2}} = 6.9 \times 10^{- 2}$
and for precipitation of BaCO
Subject: Chemistry — Halogen — Preparation of Halogen containing compounds
Difficulty: Moderate

Question: Best method of preparing alkyl chloride is

1) $ROH + SOCl_{2} \rightarrow$
2) $ROH + PCl_{2} \rightarrow$
3) $ROH + PCl_{3} \rightarrow$
4) $ROH + HCl\overset{anhy\ ZnCl_{2}}{\rightarrow}$

Correct option: 1

Solution:
The chlorination of alcohol by $SOCl_{2}$ (thionyl chloride) is
the best method for the preparation of alkyl halides as in this method
all the other product are gaseous and thus halides are obtained on quite
pure state
gy $SOCl_{2}$ ¼ fFk;ksukby DyksjkbM½ n~okjk vYdksgy bl lHkh
gSykbM~l dkQh "kqn~/k voLFkk esa izkIr gksrs gSaA
$R - OH + SOCl_{2}\overset{\mathrm{\Delta}}{\rightarrow}R - Cl + HCl \uparrow + SO_{2} \uparrow$
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: The products A and B in the reaction
 are given by the set

1) CH3--CH2--CH(OH)--CH3 and CH3--CH2--CHCH2
2) CH3--CHCH--CH3 and CH3--CH2--CH2--CH2OH
3) CH3--CHCH--CH3 and CH3--CH2--CHCH2
4) CH3--CH2--CH2--CH2OH and CH3--CH2--CH CH2

Correct option: 3

Solution:
[IMAGE] alc. KOH gives elimination reaction hence the product A and B are
alc. KOH mUewyu
Subject: Chemistry — Coordination Compounds — Werner's Theory
Difficulty: Moderate

Question: $1\ g$of complex $\lbrack Cr(H2O)5Cl\rbrack Cl2.H2O$ was passed
through a cation exchanger to produce $HCl$. The acid liberated was
diluted to 1 litre. What will be the molarity of acid solution
 ?
ls ladqy $\lbrack Cr(H2O)5Cl\rbrack Cl2.H2O$


Answer: 0.0075

Solution:
1 mole of complex will give two moles of $Cl–$ ion, i.e. 2 mole
$HCl$
ladqy ds 1 2 eksy $Cl–$ vFkkZr~ $HCl$ ds 2
eksy
$\left\lbrack Cr\left( H_{2}O \right)_{5}Cl \right\rbrack{Cl}_{2} \cdot H_{2}O \rightleftharpoons \left\lbrack Cr\left( H_{2}O \right)_{5}Cl \right\rbrack^{+ 2} + 2{Cl}^{-}$
1 g
mole of HCl$= \frac{2 \times 1}{266.5}$ (for HCl, V.f. = 1 $\therefore$
N = M)
HCl $= \frac{2 \times 1}{266.5}$ (HCl ds fy, la;kstdrk xq.kkad =
1 ∴ N = M)
$\therefore N = \frac{2/266.5}{1} = 0.0075$