Grass-G/jee-main-questions huggingface dataset:
Subject: Chemistry — Organic Chemistry — Chemistry in everyday life
Difficulty: Easy

Question: Which one of the following is known as broad spectrum
antibiotics?

1) Streptomycin
2) Ampicillin
3) Chloramphenicol
4) Penicillin G

Correct option: 3

Solution:
Chloramphenicol is broad spectrum antibiotic used in the treatment
of typhoid, dysentery,
acute fever.
Subject: Chemistry — Physical Chemistry — Redox Reaction
Difficulty: Easy

Question: In XeO_3 and XeF_6 the oxidation state of Xe is


Answer: 6

Solution:
The oxidation state of Xe in both XeO_3 and XeF_6 is +6
[IMAGE] x = +6 x = +6
Subject: Chemistry — Chemical Kinetics — Temperature dependence of rate
Difficulty: Tough

Question: Find out the percentage of the reactant molecules crossing over
the activation energy barrier at 325 K, given that DH325 = 0.12 kcal,
Ea(b) = + 0.02 kcal.


Answer: 80.65

Solution:
Given that, $\Delta H = 0.12 \times 10^{3}cal$
$E_{a(b)} = 0.02 \times 10^{3}cal$ $(E_{a}Never\ negative)$
$\because\Delta H = E_{a(f)} - E_{a(b)}$
$\therefore E_{a(f)} = 0.12 \times 10^{3} + 0.02 \times 10^{3}cal = 0.14 \times 10^{3}cal$ of molecule crossing over the barrier
$$= 100 \times e^{- E_{a}(f)/RT} = 100 \times \times 325} = 100 \times 0.8065 = 80.65\ $$
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Tough

Question: Calculate in a 0.2 M solution of dichloroacetic acid
(K_a = 5 × 10^-2) that also contains 0.05 M H_2SO_4. Report your
answer after multiplying it by 400.

1) Expecting α < < 1, 5×10^-2 = $\frac{0.2\alpha \times 0.1}{0.2}$
$\therefore$ a = 0.5 (not negligible)
So, solve quadratic

Answer: 60

Solution:
$\begin{matrix}
\ HA\ \ \ \ \ \ \ \ \ \ \ \ \rightleftharpoons \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A^{-}\ \ \ \ + \ \ \ \ H^{+}
\end{matrix}$
t = 0 0.2 0.1
t = eq 0.2(1 - α) 0.2 α 0.1 + 0.2 α
$K_{a} = \frac{\left\lbrack H^{+} \right\rbrack\left\lbrack A^{-} \right\rbrack}{\lbrack HA\rbrack} = \frac{0.2\alpha(0.1 + 0.2\alpha)}{0.2(1 - \alpha)}$....(1)
Expecting α < < 1, 5×10^-2 = $\frac{0.2\alpha \times 0.1}{0.2}$
$\therefore$ a = 0.5 (not negligible)
So, solve quadratic (1) & get α = 0.25
$\therefore$ = 0.1 + 0.2 × 0.25
= 0.15 M
0.15 × 400 = 60
Subject: Chemistry —  — 

Question: The element having second highest ionization enthalpy in 3d-series.

1) Cr
2) Mn
3) Fe
4) Ti


Solution:
After removal one electron of Cr, it achieve stable half filled
configuration (3d5) which is more stable than other electronic
configuration, hence its second ionisation energy value will be high
other than.
Subject: Chemistry — Organic Chemistry — REACTION MECHANISM
Difficulty: Tough

Question: The major product obtained when $Br_{2}/Fe$ is treated with


Correct option: 1

Solution:
The phenyl ring having H-N< group is activated while another
one is deactivated due to
[IMAGE], so electrophilic aromatic bromination0
will occur at para position with respect to H - N < group inactivated
ring.
Subject: Chemistry — Physical Chemistry — 
Difficulty: Moderate

Question: An aqueous solution of a weak monobasic acid containing 0.1 g
in 21.7g of water freezes at 272.813 K. If the value of K_f for water
is 1.86 k/m, what is the molecular mass of the monobasic acid.


Answer: 60

Solution:
Subject: Chemistry — Equivalent Concept — Equivalent Concept for Acid Base Titration and Precipitation
Difficulty: Easy

Question: How many millilitres of 0.1N H2SO4 solution will be required for
complete reaction with a solution containing 0.125 g of pure Na2CO3

1) 23.6 mL
2) 25.6 mL
3) 26.3 mL
4) 32.6 mL

Correct option: 1

Solution:
$M.\ eq.\ ofH_{2}{SO}_{4} = m.\ eq.\ of{Na}_{2}{CO}_{3}$
$0.1 \times \frac{V}{1000} = \frac{0.125}{106} \times 2\ $
$V = 23.6mL$
Subject: Chemistry — Organic Chemistry — Some Basic Principles and Techniques
Difficulty: Moderate

Question: In which of the following resonance is not possible ?


Correct option: 4

Solution:
Delocalization is possible in the cases where atoms with lone
pair & vacant p or d-orbital are adjacent.
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: Chlorination of toluene in the presence of light and heat
followed by treatment with aqueous NaOH gives
tyh; NaOH ds lkFk mipkj ds ckn izdk"k

1) o-cresol
2) p-cresol
3) 2, 4-dihydroxy toluene
4) Benzyl alcohol

Correct option: 4

Solution:
Subject: Chemistry — Chemical Bonding — Multicentered species
Difficulty: Moderate

Question: Which is correct about the cyclic silicate

1) The value of n is 12
2) each Si atom is bonded with three oxygen atoms
3) each oxygen atom is bonded with two Si atoms
4) all the above are correct.

Correct option: 1

Solution:
[IMAGE] General formula of cyclic silicates is [Sin O3n]2n- ¼
Subject: Chemistry — Inorganic Chemistry — Metallurgy
Difficulty: Easy

Question: After partial roasting, the sulphide of copper is reduced by

1) Reduction by carbon
2) Electrolysis
3) Self-reduction
4) Cyanide process

Correct option: 3

Solution:
Self reduction: - Reduction of oxide ore of a metal by its own
sulphide
2Cu_2O + Cu_2 S → 6Cu + SO_2
Subject: Chemistry — Physical Chemistry — Thermodynamics
Difficulty: Moderate

Question: An engine operating between 150ºC and 25ºC takes 500 J heat
from a higher temperature reservoir if there are no frictional losses,
then work done by engine is

1) 147.7 J
2) 157.75 J
3) 165.85 J
4) 169.95 J

Correct option: 1

Solution:
$T_{2} = 150 + 273 = 423K$
$T_{1} = 25 + 273 = 298K$
$Q = 500K$
$\frac{W}{Q} = \frac{T_{2} - T_{1}}{T_{2}};W = 500\left( \frac{423 - 298}{423} \right) = 147.7J$
Subject: Chemistry —  — 

Question: In an atom an electron is moving with a speed equal to 600 m/s with
an accuracy upto 0.001 with what certainity its position can be
determined.

1) 9.75 × 10^-3 m
2) 5.2 × 10^-4 m
3) 2 × 10^-3 m
4) 0.8 ×
10^-3 m


Solution:
v = 600 m/sec
Subject: Chemistry — Inorganic Chemistry — Metallurgy
Difficulty: Moderate

Question: Which method of purification is represented by the equations ?

1) Cupellation
2) Poling
3) Van Arkel
4) Zone refining

Correct option: 3

Solution:
Purification of Ti and Zr are performed by Van Arkel method as
given in the question.
Subject: Chemistry — Inorganic Chemistry — P-block
Difficulty: Moderate

Question: Which of the following reactions lead to chemical inertness

1) Lead with dilute H2SO4
2) Lead with conc. HCl
3) Aluminium with oxygen
4) All of above reactions

Correct option: 4

Solution:
(1)
$Pb + H_{2}{SO}_{4} \longrightarrow {PbSO}_{4}\ (layer) + H_{2} \uparrow$
(2)
$Pb + Conc.HCl \longrightarrow {PbCl}_{2}(\ Coating) + H_{2} \uparrow$
(3) $2Al + 3/2O_{2} \longrightarrow {Al}_{2}O_{3}(s)$
Subject: Chemistry — Organic Chemistry — Alcohols, Phenols and Ethers
Difficulty: Moderate

Question: The reaction of C_2H_5OH with H_2SO_4 does not
give

1) Ethylene
2) Diethyl ether
3) Acetylene
4) Ethyl hydrogen

Correct option: 3

Solution:
Subject: Chemistry —  — 

Question: What is the final product of the following reaction?



Solution:
Subject: Chemistry — Eudiometry — Eudiometry
Difficulty: Tough

Question: 20 ml of a gaseous hydrocarbon requires 100 ml of O_2 for
complete combustion and produces 60 ml of CO_2 gas. Determine the
molecular formula of the hydrocarbon.


Correct option: 3

Solution:
Let the molecular formula of the hydrocarbon is
[IMAGE] From ques. Vol. of CO_2 formed = 60 ml
20 x=60
x=3
vol. of O_2 used = 100 ml
Hence the hydrocarbon is [IMAGE]
Subject: Chemistry — Inorganic Chemistry — Quantum number
Difficulty: Easy

Question: Which set of quantum numbers are not possible from the following

1) $n = 3,l = 2,m = 0,s = - \frac{1}{2}$
2) $n = 3,l = 2,m = - 2,s = - \frac{1}{2}$
3) $n = 3,l = 3,m = - 3,s = - \frac{1}{2}$
4) $n = 3,l = 0,m = 0,s = - \frac{1}{2}$

Correct option: 3

Solution:
If $n = 3$ then, $l = 0$ to $n = - 1$&$m = - l\ to\ + l$
Subject: Chemistry —  — 

Question: [IMAGE] of the solution in the anode compartment of
the following cell at [IMAGE] is x when
[IMAGE] Find out x

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


Solution:
Subject: Chemistry — Organic Chemistry — HALOALKANES & HALOARENES
Difficulty: Moderate

Question: Which compound does not form iodine form with alkali and iodine?

1) Acetone
2) Ethanol
3) Diethyl ketone
4) Isopropyl alcohol

Correct option: 3

Solution:
[IMAGE] Yellow ppt
Subject: Chemistry —  — 

Question: For a reaction: 2A(g) [IMAGE] 5B(g), if the rate
constant for disappearance of is 4 × 10-2 M sec-1 then calculate
rate of formation of B after 10 seconds from start of the reaction. If
initially 10 M of A is taken.

1) 4 × 10-2 Msec-1
2) 0
3) 0.1 Msec-1
4) 2 × 10-1 Msec-1


Solution:
2A (g) → 5B (g)
Subject: Chemistry — Organic Chemistry — Organic Reaction mechanisms-I
Difficulty: Moderate

Question: Rate of the reaction is fastest when Z is

1) Cl
2) ${OCOCH}_{3}$
3) ${OC}_{2}H_{5}$
4) ${NH}_{2}$

Correct option: 1

Solution:
Among the given option Cl- is the best leaving group hence the
rate of reaction will be fastest in case of RCOCl.
Subject: Chemistry — Physical Chemistry — Redox Reaction
Difficulty: Easy

Question: Which of the following is the most powerful oxidizing agent

1) F_2
2) Cl2
3) Br_2
4) I_2

Correct option: 1

Solution:
Fluorine is a most powerful oxidizing agent because it consist
of E^0 = + 2.5 volt.
Subject: Chemistry — Real Gas — Vander waal equation and virial equation of state
Difficulty: Easy

Question: A gas obeys the equation of state P(V - b) = RT (The parameter
b is a constant). The slope for an isochore will be

1) Negative
2) Zero
3) R/(V - b)

Correct option: 3

Solution:
$P(V - b) = RT;P = \frac{RT}{(V - b)}$
$P = (\frac{R}{(V - b)})T + 0$
Subject: Chemistry — Physical Chemistry — Atomic Structure
Difficulty: Moderate

Question: Given that in H-atom, the transition energy for n = 1 to n = 2
is 10.2 eV, the energy for the same transition in Be^3+ is

1) 20.4 eV
2) 30.6 eV
3) 40.8 eV
4) None of these

Correct option: 4

Solution:
$\Delta$E = z^2 x 10.2 ev
=16 x 10.2 ev
=162.3 ev
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Moderate

Question: (a)$\ {CuSO}_{4}.5H2O(s) \rightleftharpoons {CuSO}_{4} \cdot 3H_{2}O(s) + 2H_{2}O(g)$
K_P = 4 × 10^-4 atm^2
(b)$\ {Na}_{2}{SO}_{4} \cdot 10H_{2}O(s) \rightleftharpoons {Na}_{2}{SO}_{4.5}H_{2}O(s) + 5H_{2}O(g)$
K_P = 2.43 × 10^-8 atm^5
(c)
${Na}_{2}S_{2}O_{3} \cdot 5H_{2}O(s) \rightleftharpoons {Na}_{2}S_{2}O_{3} \cdot 2H_{2}O(s) + 3H_{2}O(g)$K_P
= 6.4 × 10^-5 atm^3
What is order of pressure of water vapours at equilibrium and relative
humidity respectively.

1) c > b > a V.P.
c > b > a R.H.
2) c < b < a V.P.
c > b > a R.H.
3) a > c > b V.P.
a > c > b R.H.
4) a > c > b V.P.
a < c < b R.H.

Correct option: 1

Solution:
[IMAGE] = 2 × 10^-2
[IMAGE] = 243 × 10-10 = [IMAGE] = 3 × 10^-2
[IMAGE] = 6.4 × 10^-5 = [IMAGE] = 4 × 10^-2
Order of V.P c > b > a
and same as c > b > a
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: Consider the following statements: A hydrocarbon of molecular
formula C5H10 is a
1. monosubstituted alkene
2. disubstituted alkene
3. trisubstituted alkene
How many statement(s) is (are) correct?


Answer: 3

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

Question: When ethyl chloride and n-propyl chloride undergoes wurtz
reaction which is not obtained

1) n-butane
2) n-pentane
3) n-hexane
4) isobutene
tks izkIr ugha gksrk

Correct option: 4

Solution:
[IMAGE] (a) [IMAGE] (b) [IMAGE] (c) [IMAGE] Isobutane is not obtained.
Subject: Chemistry — Organic Chemistry — Hydrocarbons
Difficulty: Moderate

Question: ${CH}_{3}C \equiv {CCH}_{3}\overset{Na,{NH}_{3}(I)}{\longrightarrow}Q\frac{dil\ Cold\ }{{KMO}_{4}} \longrightarrow R$
If Q is an unsaturated hydrocarbon then products R is/are

1) i and ii
2) ii and iii
3) Only v
4) Only iv

Correct option: 2

Solution:
${CH}_{3}C = {CCH}_{3}\overset{Na,{NH}_{3}(l)}{\rightarrow}$
Subject: Chemistry — Inorganic Chemistry — D & F -Block Elements
Difficulty: Moderate

Question: When calomel reacts with [IMAGE] we get


Correct option: 1

Solution:
Subject: Chemistry —  — 

Question: A reaction follows the given concentration reaction (M) v/s time
graph. The instantaneous rate for this reaction at 20 seconds will be



Solution:
Subject: Chemistry — Mole concept — Empirical Formula, % Composition of a given compound by mass,
Difficulty: Moderate

Question: A compound possess 8 sulphur by mass. The least molecular mass
is

1) 300
(B) 400
(C) 155
(D) 355

Correct option: 2

Solution:
8 sulphur by mass means - 8 g sulphur is present in 100 g
solid.
[IMAGE] 32 g sulphur (1 mole atom) will be present in
 Q compound must be having at least one atom of
[IMAGE] min. mol. mass = 400 g.
Subject: Chemistry — Electrochemistry — Electrochemical series & its Applications
Difficulty: Easy

Question: KCl can be used in salt bridge as electrolyte in which of the
following cells?

1) Zn | ZnCl2 || AgNO3 | Ag
2) Pb | Pb(NO3)2 || Cu(NO3)2 | Cu
3) Cu | CuSO4 || AuCl3 | Au
4) Fe | FeSO4 || Pb(NO3)2 | Pb

Correct option: 3

Solution:
KCl can make precipitate with AgNO3, Pb(NO3)2 so can't be used
along these electrolyte.
Subject: Chemistry — Inorganic Chemistry — Coordination Compounds
Difficulty: Moderate

Question: The number of geometric isomers that can exist for square planar
[Pt (Cl) (py)
$\left. \ \left( {NH}_{3} \right)\left( {NH}_{2}OH \right) \right\rbrack^{+}$
is (py = pyridine )


Answer: 3

Solution:
The complex is of the type 
M = metal
a, b, c, d = Monodentate ligands.
Subject: Chemistry — Organic Chemistry — ALKANE
Difficulty: Easy

Question: 2-Methylbutane on reacting with bromine in the presence of
sunlight gives mainly

1) 1 -bromo-2-methylbutane
2) 2 -bromo-2-methylbutane
3) 2 -bromo-3-methylbutane
4) 1 -bromo-3-methylbutane

Correct option: 2

Solution:
Subject: Chemistry — Organic Chemistry — Alcohols, Phenols and Ethers
Difficulty: Moderate

Question: When Phenol is heated with phthalic anhydride in concentrated
sulphuric acid and the hot reaction mixture is poured into a dilute
solution of sodium hydroxide, the product formed is

1) Alizarin
2) Methyl orange
3) Fluorescein
4) Phenolphthalein

Correct option: 4

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

Question: Assertion: Among the carbon allotropes, diamond is an insulator,
whereas, graphite is a good conductor of electricity. Reason:
Hybridization of carbon in diamond and graphite aresp^3 and sp^2,
respectively.

1) Both assertion and reason are correct, and the reason is the
correct explanation for the assertion.
2) Both assertion and reason are correct, but the reason is not the
correct explanation for the assertion.
3) Assertion is incorrect statement, but the reason is correct.
4) Both assertion and reason are incorrect.

Correct option: 1

Solution:
Diamond has carbon atoms that are bonded via a covalent bond to
four other carbon atoms in a tetrahedral manner.
The$a = \frac{a_{0}}{\sqrt{2}} = \frac{bt}{m}$
hybridization of carbon atom in diamond is
$\frac{a_{0}}{\sqrt{2}} = \frac{bt}{m}$. In
$= \frac{10^{- 2}t}{0.1} = \frac{t}{10}$
diamond, carbon utilizes its unpaired electrons in the formation of the
bond. Therefore, it is a poor conductor of electricity.
A trigonal geometry is obtained when a carbon atom in graphite is bonded
via a covalent bond to three other carbon atoms.
The$\frac{a_{0}}{\sqrt{2}} = a_{0}e^{\frac{t}{10}}$
 hybridization of carbon atom in graphite
is $\frac{1}{\sqrt{2}} = e^{- \frac{1}{10}}$. In 
$ln\sqrt{2} = \frac{t}{10}$
t = 3.5 sec
graphite, the three unpaired electrons of carbon is utilized in the
formation of the bond. Therefore, it is a good conductor of electricity.
Subject: Chemistry — Inorganic Chemistry — Hydrogen And Its Compound
Difficulty: Tough

Question: 20 volume $H_{2}O$ solution has a strength of about in gm/ml


Answer: 6.71

Solution:
$\because 22.4\text{ litre }O_{2}$at N.T.P. obtained by 68
gmof$H_{2}O_{2}$
$\therefore 1\text{ litre }O_{2}$at N.T.P. obtained by
$\frac{68}{22.4} \times 20gm\text{ of }H_{2}O_{2} = 60.71gm\text{ of }H_{2}O_{2}$
$\therefore 1000mlO_{2}$ at N.T.P. obtained by $= 60.71\ gm\ of\ H_{2}O$
$\therefore 100mlO_{2}$ N.T.P. obtained by
$= \frac{60.71}{1000} \times 100 = 6.71\%$
Subject: Chemistry — Organic Chemistry — Mesomeric Effect
Difficulty: Moderate

Question: Which of the following groups (attached with benzene ring)
show + M effect?


Answer: 6

Solution:
[IMAGE] have + M group.
Subject: Chemistry — Inorganic Chemistry — P-block Halogen
Difficulty: Moderate

Question: In case of nitrogen, NCl3 is possible but no NCl5 while in case
of phosphorus, PCl3 as well as PCl5 are possible. It is due to

1) Availability of vacant d-orbital in P but not in N.
2) Lower electronegativity of P then N.
3) Lower tendency of H bond formation in P than N.
4) Occurrence of P in solid while N in gaseous state at room
temperature.
ukbVªkstu dh fLFkfr esa] NCl3 lEHko fdUrq NCl5 ugha] tcfd QkWLQksjl
ds fLFkfr esa PCl3 rFkk PCl5 nksuks lEHko gSA;g bl dkj.k gksrk

Correct option: 1

Solution:
In phosphorous the vacant 3d-orbitals are available. So it can
increase its covalence beyond three.
Subject: Chemistry — Organic Chemistry — Organic Chemistry
Difficulty: Moderate

Question: Among the following compounds, the most acidic is

1) p-nitrophenol
2) p-hydroxybenzoic acid
3) o-hydroxybenzoic acid
4) p-toluic acid

Correct option: 1

Solution:
[IMAGE] Due to intramolecular hydrogen bonding in conjugate base of
o-Hydroxybenzoic acid, it is strongest acid.
Subject: Chemistry — Physical Chemistry — Mole Concept
Difficulty: Tough

Question: Fluoxymesterone, C_20H_29FO_3, is an anabolic steroid. A solution
is prepared by dissolving 10.0 mg of the steroid in 500 mL of water. How
many moles of Fluoxymesterone are present in 1 mL of solutions.
 C_20H_29FO_3 500 mL 
10.0 mg 1 mL

1) 1.16 × 10-10
2) 1.19 × 10-17 (3) 5.95 × 10-8
4) 2.38 ×
10-11

Answer: 3

Solution:
Number of moles in 500 mL [IMAGE] 500 mL [IMAGE] Number of moles in 1 mL [IMAGE] =5.95 × 10-8.
1 mL [IMAGE] 5.95 × 10-8.
Subject: Chemistry — Alcohol, Phenol and Ethers — Preparation of alcohol, Phenol and Ethers
Difficulty: Moderate

Question: Ethanol is prepared industrially by

1) Hydration of ethylene
2) Fermentation of sugars
3) Both the above
4) None of these,FkukWy vkS|ksfxd:i

Correct option: 3

Solution:
Hydration of ethylene,Fkyhu 
${CH}_{2} = {CH}_{2} + \overset{+}{HH}\overset{-}{{SO}_{4}} \rightarrow {CH}_{3} - {CH}_{2} - {HSO}_{4}$
${CH}_{3} - {CH}_{2}{HSO}_{4}\frac{H_{2}O}{\ Boil\ } \rightarrow {CH}_{3} - {CH}_{2} - OH + H_{2}{SO}_{4}$
Fermentation of sugars
"kdZjk dk fd.ou
$C_{12}H_{22}O_{11} + H_{2}O\overset{Intertaes}{\rightarrow}\begin{matrix}
C_{6}H_{12}O_{6} \\
\ Glucose\ 
\end{matrix} + \begin{matrix}
C_{6}H_{12}O_{6} \\
\ Fructose\ 
\end{matrix}$
$\begin{matrix}
C_{6}H_{12}O_{6} \\
\ Glucose\ or\ Fructose\ 
\end{matrix}\overset{Zymase}{\rightarrow}2C_{2}H_{5}OH + 2{CO}_{2}$
Subject: Chemistry — Organic Chemistry — Aldehydes, Ketones & Carboxylic Acids
Difficulty: Moderate

Question: [IMAGE] product in the reaction is

4) None of these

Correct option: 1

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

Question: [IMAGE] Compound (A) is


Correct option: 4

Solution:
Subject: Chemistry — Physical Chemistry — Gaseous state
Difficulty: Moderate

Question: An unknown gaseous hydrocarbon and oxygen gas are mixed in volume
ratio 1: 7 and exploded. The resulting mixture upon cooling occupied a
volume of V mL, half of which got absorbed in aq. KOH (absorbs $CO_{2}$)
and the remaining half was absorbed in alkalline pyrogallol solution
(absorbs $O_{2}$). Assuming all volumes to be measured under identical
conditions

1) The molecular formula of hydrocarbon can be $C_{2}H_{6}$
2) The resulting mixture obtained after cooling contains 50 $O_{2}$
and 50 CO by mole.
3) The molecular formula of hydrocarbon can be $C_{3}H_{4}$
4) Molecular mass of hydrocarbon is definitely less than 28 u

Correct option: 3

Solution:
$CxHy + \left( x + \frac{y}{4} \right)O_{2} \longrightarrow {xCO}_{2} + \frac{y}{2}H_{2}O\mathcal{(l)}$
$i$ $a$ $\text{7a}$ 0
$f$ 0 $7a - \left( x + \frac{y}{4} \right)a$ xa
$\text{Given: }7a - \left( x + \frac{y}{4} \right)a = xa$
$\begin{aligned}
\therefore & 8x + y = 28
\end{aligned}$
Only $C_{3}H_{4}$ hydrocarbon satisfies the above relation.
Subject: Chemistry —  — 

Question: The decreasing order of the rate of nitration of the following
compounds is:
(i) Benzene [IMAGE] (ii) Hexadeutero benzen [IMAGE] (iii) Nitrobenzene
(iv) Chlorobenzene

1) i > ii > iii > iv
2) i > ii > iv > iii
3) i = ii > iv > iii
4) i = ii > iii > iv


Solution:
Rate of Nitration [IMAGE] Rate of E.S.R
Subject: Chemistry —  — 

Question: has four choices

1) ,
2) ,
3) ,
4) out of which ONLY ONE is correct]

Answer: 3

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

Question: C_6 H_12 (P) has two types of alkenes that can be reduced to
one type of C_6H_14 (Q). Q is:
C_6 H_12 (P) nks rjg dh,Ydhu tks fd,d rjg,Ydsu C_6H_14
(Q)


Correct option: 2

Solution:
Subject: Chemistry — Inorganic Chemistry — p-Block (Nitrogen and Oxygen family)
Difficulty: Moderate

Question: Hydrogen peroxide in its reaction with KIO4 and NH2OH
respectively, is acting as a

1) reducing agent, oxidising agent
2) reducing agent, reducing agent
3) oxidising agent, oxidising agent
4) oxidising agent, reducing agent

Correct option: 1

Solution:
${KIO}_{4} + H_{2}O_{2} \rightarrow {KIO}_{3} + H_{2}O + O_{2}$
$H_{2}O_{2}$ acts as a reductant 
$2{NH}_{2}OH + H_{2}O_{2} \rightarrow N_{2} + 4H_{2}O$
$H_{2}O_{2}$ acts as an oxidant.
Subject: Chemistry — Carbonyl Compound (Aldehyde and Ketone) — Aldol condensation
Difficulty: Moderate

Question: In the scheme given below, the total number of intermolecular
aldol condensation products formed from is


Answer: 1

Solution:
Subject: Chemistry — Atomic Structure — Bohr Model
Difficulty: Moderate

Question: A hydrogen sample is prepared in a particular excited state.
Photons of energy 2.55 eV get absorbed into the sample to take some of
the electrons to a further excited state B. Find orbit numbers of the
states A and B. Given the allowed energies of hydrogen atom:
$E_{1} = - 13.6eV,E_{2} = - 3.4eV,E_{3} = - 1.5eV,E_{4} = - 0.85eV,E_{5} = - 0.54eV$,d gkbMªkstu uewus dks,d fuf'pr mÙksftr voLFkk esa
2.55 eV ds QksVkWu dh vo'kksf"kr djds uewus ds dqN bysDVª mÙksftr voLFkk B esa vk tkrs gSA voLFkk A rFkk B 
Kkr djksA gkbMªkstu ijek.kq $E_{1} = - 13.6eV,E_{2} = - 3.4eV,E_{3} = - 1.5eV,E_{4} = - 0.85eV,E_{5} = - 0.54eV$

1) A =2, B = 4
2) A =4, B = 2
3) A =2, B = 2
4) A =4, B = 4

Correct option: 1

Solution:
Clearly,
$E_{4} - E_{2} = ( - 0.85) - ( - 3.4) = 2.55eV \Rightarrow A = 2,B = 4$
gy Li"V:i 
$E_{4} - E_{2} = ( - 0.85) - ( - 3.4) = 2.55eV \Rightarrow A = 2,B = 4$
Subject: Chemistry — IUPAC and Structural Isomerism — Chain Isomerism
Difficulty: Easy

Question: The general formula CnH2nO2 could be for open chain

1) diketones
2) carboxylic acids
3) diols
4) dialdehydes.

Correct option: 2

Solution:
Diketones: CnH2n -2O2, Carboxylic acid: CnH2nO2,
Diols: CnH2n+2O2, Dialdehydes: CnH2n-2O2
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Moderate

Question: A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some
of the CO2 is converted into CO on the addition of graphite. If the
total pressure at equilibrium is 0.8 atm, the value of K is:
dqN CO2, CO esa ifjofrZr gks tkrh gSA ij dqy nkc 0.8 atm gks
rks K


Answer: 1.8

Solution:
CO2(g) + C(s)[IMAGE] 2CO (g)
0.5 atm
0.5-p 2p
Total pressure ¼dqy nkc½ = 0.5 - P + 2P = 0.8
P = 0.3
$K_{p} = \frac{P_{CO}^{2}}{P_{{CO}_{2}}} = \frac{(2P)^{2}}{(0.5 - P)} = \frac{(0.6)^{2}}{(0.5 - 0.3)}$
Kp = 1.8
Subject: Chemistry — Atomic Structure — Quantum numbers & Electronic configuration
Difficulty: Moderate

Question: For an electron, with n = 3 has only one radial node. The
orbital angular momentum of the electron will be

1) 0
2) $\sqrt{6}\frac{h}{2\pi}$
3) $\sqrt{2}\frac{h}{2\pi}$
4) $3(\frac{h}{2\pi})$

Correct option: 3

Solution:
Number of radial nodes
$= n\mathcal{- l -}1 = 1,n = 3.\therefore\mathcal{l} = 1$
Orbital angular momentum
$= \sqrt{\mathcal{l}(\mathcal{l} + 1)}\frac{h}{2\pi} = \sqrt{2}\frac{h}{2\pi}$