Grass-G/jee-main-questions huggingface dataset:
Subject: Chemistry —  — 

Question: [IMAGE] Reagent for the above reaction will be

1) Hot acidic [IMAGE]
4) dil


Solution:
Mild oxidizing agent like PCC can oxidize only alcohols into
carbonyl compounds.
Subject: Chemistry — Physical Chemistry — 
Difficulty: Tough

Question: Ferrous oxide has a cubic structure and each edge of the unit
cell is 5.0 Assuming density of the oxide as 4.0 g - cm^-3, then
the number of Fe^2+ and O^2- ions present in each unit cell will be

1) Four Fe^2+ and four O^2
2) Two Fe^2+ and four O^2
3) Four Fe^2+ and two O^2
4) Three Fe^2+ and three O^2

Correct option: 1

Solution:
Let the units of ferrous oxide in a unit cell = n, molecular
weight of ferrous oxide
$\left( F_{e}O \right) = 56 + 16 = 72gmol^{- 1}$
 weight of n units $= \frac{72 \times n}{6.023 \times 10^{23}}$
 Volume of one unit = (length of corner )^3
$= (5A)^{3} = 125 \times 10^{- 24}{cm}^{3}$
$\ Density\ = \frac{\ wt.\ of\ cell\ }{\ volume\ }$
$4.09 = \frac{72 \times n}{6.023 \times 10^{23} \times 125 \times 10^{- 24}}$
$n = \frac{3079.2 \times 10^{- 1}}{72} = 42.7 \times 10^{- 1} = 4.27 \approx 4$
Subject: Chemistry — Inorganic Chemistry — Metallurgy
Difficulty: Easy

Question: The carbon-based reduction method is NOT used for the
extraction of

1) tin from SnO2
2) iron from Fe2O3
3) aluminium from Al2O3
4) magnesium from MgCO3, CaCO3
èkkrqvks ds fu"d"kZ.k cZu vkèkkfjr vip;d fofèk

Correct option: 3

Solution:
SnO2 and Fe2O3 are reduced by C reduction method.
Al2O3 andMgCO
Subject: Chemistry — Inorganic Chemistry — P-block Halogen
Difficulty: Moderate

Question: The total number of electrons present in 4th shell of Astatatine
(85At) are:,LVsjhu (_85At) ds 4^th dks'k es mifLFkr bysDVªkWu


Answer: 32

Solution:
Electronic configuration of
[IMAGE] _85At
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Moderate

Question: The figure show the change in concentration of species A and B as
a functional of time. The equilibrium constant K_C for the reaction
$A(g) \rightleftharpoons 2B(g)$ is

1) K_c > 1
2) K < 1
3) K = 1
4) Data insufficie

Correct option: 1

Solution:
$K_{c} = \frac{\lbrack B\rbrack^{2}}{\lbrack A\rbrack} = \frac{(0.4)^{2}}{0.1} = 1.6$
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: [IMAGE] ([X] gks ldrk

4) None of these

Correct option: 1

Solution:
Subject: Chemistry — Physical Chemistry — Thermodynamics
Difficulty: Easy

Question: Joule-Thomson expansion is

1) Isobaric
2) Isenthalpic
3) Isothermal
4) None of these

Correct option: 2

Solution:
Joule-Thomson expansion is isenthalpic.
Subject: Chemistry — Physical Chemistry — 
Difficulty: Moderate

Question: 0.6 g of a solute is dissolved in 0.1 litre of a solvent,
which develops an osmotic pressure of 1.23 atm at 27ºC. The
molecular mass of the substance is


Answer: 120

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

Question: Acetylene may be prepared by electrolysis of

1) potassium oxalate
2) potassium acetate
3) potassium maleate
4) potassium succinate,lhVyhu

Correct option: 3

Solution:
Subject: Chemistry — Inorganic Chemistry — P-block Halogen
Difficulty: Moderate

Question: The set with correct order of acidity is:
vEyh;rk

1) HClO < HClO2 < HClO3 < HClO4
2) HClO4 < HClO3 < HClO2 < HClO
3) HClO < HClO4 < HClO3 < HClO2
4) HClO4 < HClO2 < HClO3 < HClO

Correct option: 1

Solution:
As charge dispersion increases, the stability of conjugated
base increases and thus acidity increases.
vkos'k
Subject: Chemistry — Physical Chemistry — Chemical and Ionic Equilibrium
Difficulty: Moderate

Question: For preparing a buffer solution of pH 6 by mixing sodium
acetate and acetic acid, the ratio of the concentration of salt and acid
should be (K_a = 10^-5).

1) 1: 10
2) 10: 1
3) 100: 1
4) 1: 100

Correct option: 2

Solution:
CaOCl_2 has two anions Cl^- and OCl^- along with
Ca^2+ ions.
Subject: Chemistry — Physical Chemistry — 
Difficulty: Moderate

Question: What volume of 98 sulphuric acid (in ml) should be mixed with
water to obtain 200 mL of 15 solution of sulphuric acid by weight ?
Given density of H2O = 1.00 g cm^-3, sulphuric acid (98 ) = 1.88 g
cm-3 and sulphuric acid (15 ) = 1.12 g cm-3.


Answer: 18

Solution:
It is a case of dilution, simplest way is to determine
normality of the 98 and 15 H2SO4.
(i) For 98 H2SO4: 100 g H2SO4 solution $= \frac{100}{1.88}{cm}^{3}$
solution has H2SO4 = 98 g
$N_{1}(98\%) = \frac{1000w_{1}}{E_{1}V} = \frac{1000 \times 98}{49 \times \frac{100}{1.88}} = 37.6N$
where w1 is the weight of solute of equivalent weight E1 in V mL
solution
(ii) For 15 H2SO4: 100 g H2SO4 solution $= \frac{100}{1.12}{cm}^{3}$
has H2SO4 = 15 g
$N_{2}(15\%) = \frac{1000w_{1}}{E_{1}V} = \frac{1000 \times 15}{49 \times \frac{100}{1.12}} = 3.43$
Using $N_{1}V_{1} = N_{2}V_{2}$
$37.6 \times V_{1} = 3.43 \times 200$
$V_{1} = 18.2mL$
Subject: Chemistry —  — 

Question: The passage of current liberates [IMAGE] at cathode
and [IMAGE] at anode. The solution is

1) Copper chloride in water
2) [IMAGE] in water
3) Mercuric chloride in water
4) [IMAGE] in water


Solution:
Subject: Chemistry — Physical Chemistry — 
Difficulty: Easy

Question: How many molecules are there in the unit cell of sodium
chloride?


Answer: 4

Solution:
In NaCl (rock salt): Number of Na^+ ions = 12
(at edge centers) $x\frac{1}{4} + 1$ (at body centre) × 1 =
Subject: Chemistry — Chemical Bonding — Oxy Acids
Difficulty: Moderate

Question: How many of the following oxides are acidic?


Answer: 5

Solution:
${Mn}_{2}O_{7},{SO}_{2},{NO}_{2},{CrO}_{3},{SiO}_{2}$ = acidic
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Moderate

Question: Which of the following is most soluble in water ? Assume no
reaction of cation/anion.
dksbZ vfHkfØ;k u ekusaA

1) MnS (Ksp = 2.5 × 10-13 )
2) ZnS (Ksp = 1.6 × 10-24)
3) Bi2S3 (Ksp =1.6 × 10-72)
4) Ag2S (Ksp = 10-51)

Correct option: 1

Solution:
Find solubility for each separately by s2 = Ksp for MnS and
ZnS, 108s5 = Ksp for Bi2S3 and 4s3 = Ksp for Ag2S.
MnS rFkk ZnS ds fy, s2 = Ksp, Bi2S3 ds fy, 108s5 = Ksp rFkk Ag2S ds
fy, 4s3 = Ksp
Subject: Chemistry —  — 

Question: A following mechanism has been proposed for a reaction
The rate law expression for the reaction is



Solution:
Slowest step is rate determining.
Hence r =
Subject: Chemistry — Inorganic Chemistry — Chemical Bonding
Difficulty: Moderate

Question: Among the following groupings which represents the collection of
isoelectronic species ?

1) $NO,{CN}^{-},N_{2},O_{2}^{-}$
2) ${NO}^{+},C_{2}^{2 -},O_{2} - CO$
3) $N_{2},C_{2}^{2 -},CO,NO$
4) $CO,{NO}^{+},{CN}^{-},C_{2}^{2 -}$

Correct option: 4

Solution:
Substance CO NO^+ CN^- $C_{2}^{2 -}$ N_2 NO $O_{1}^{-}$
No. of electrons 14 14 14 14 14 14 15 17
Subject: Chemistry — Inorganic Chemistry — Hydrogen And Its Compound
Difficulty: Easy

Question: $1ml\text{ of }H_{2}O_{2}$solution gives$10ml\text{ of }O_{2}$
at NTP. It is


Answer: 10

Solution:
10 volume of $H_{2}O_{2}$ means 10 mlof $O_{2}$ is obtained from 1
ml of $H_{2}O_{2}$.
Subject: Chemistry — Alcohol, Phenol and Ethers — Properties of alcohol, Phenol and Ethers
Difficulty: Moderate

Question: What amount of bromine will be required to convert 2 g of phenol
into 2, 4, 6 -tribromophenol
fQuksy dh 2 xzke dks esa cnyus


Answer: 10.22

Solution:
[IMAGE] 94 grams of phenol reacts with 480 gms. of$Br_{2}$.
$2\ gm.\ of\ phenol\ - \frac{480}{94} \times 2 = 10.22gms$
Qsuksy dk 94 xzke $Br_{2}$ ds 480 xzke ds lkFk
Qsuksy dk 2 xzke & $\frac{480}{94} \times 2 = 10.22$ xzke
Subject: Chemistry — Physical Chemistry — Chemical Bonding
Difficulty: Easy

Question: The total number of lone pairs in chlorate ion is


Answer: 08

Solution:
Subject: Chemistry — Physical Chemistry — Thermodynamics
Difficulty: Moderate

Question: Molar heat of vaporization of a liquid is $6\ k{mol}^{- 1}$, If
the entropy change is $16{Jmol}^{- 1}K^{- 1}$, the boiling point of the
liquid is _______________ K


Answer: 375

Solution:
$S = 16\ J\ {mole}^{- 1}K^{- 1}$
$T_{b}, = \frac{\Delta H_{vapour\ }}{\Delta S_{vapour\ }} = \frac{6 \times 1000}{16} = 375K$
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Moderate

Question: The solubility product of BaCrO4 is 2.4 × 10-10 M2. The maximum
concentration of Ba(NO3)2 possible without precipitation in a 6 × 10-4
M K2CrO4 solution is....... × 10-7:
BaCrO4 xq.kuQy 2.4 × 10-10 M2 gSA K2CrO4


Answer: 4

Solution:
Ksp (BaCrO4) = 2.4 × 10-10 M2
 4-2] = 2.4 ×
10-10
$\Rightarrow$ × 6 × 10-4 = 2.4 × 10-10
 = 4 × 10-7 M
Subject: Chemistry — Organic Chemistry — Biomolecule & Polymers
Difficulty: Moderate

Question: Which one of the following bases is not present in DNA ?

1) Quinoline
2) Adenine
3) Cytosine
4) Thymine

Correct option: 1

Solution:
Quinoline is an alkaloid, it is not present in DNA, DNA has
four nitrogenous bases in adenine, guanine, cytosine & thymine
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Moderate

Question: For the reaction, CO(g) + Cl2 (g) [IMAGE] COCl2
(g) then Kp / Kc is equal to:
CO(g) + Cl2 (g) [IMAGE] COCl2 (g) vfHkfØ fy,] Kp /
Kc fuEu

2) 1.0
3) $\sqrt{RT}$
4) RT

Correct option: 1

Solution:
CO(g) + Cl2 (g) [IMAGE] COCl2 (g)
Δn = 1 - 2 = - 1; K_p = K_c (RT)^Δn $\therefore$
$\frac{K_{p}}{K_{c}} = (RT)^{- 1} = \frac{1}{RT}$
Subject: Chemistry — Organic Chemistry — Organic Chemistry
Difficulty: Moderate

Question: Choose the chain terminating step
(a)$H_{2} \rightarrow H^{*} + H^{*}$
(b)$Br_{2} \rightarrow Br^{*} + Br^{*}$
(c)$Br^{*} + HBr \rightarrow H^{*} + Br_{2}$
(d)$H^{*} + Br_{2} \rightarrow HBr + Br^{*}$
(e) $Br^{*} + Br^{*} \rightarrow Br_{2}$

1) a
2) c
3) d
4) e

Correct option: 4

Solution:
In chain termination step free radical present in reactant
side.
Subject: Chemistry — Mole concept — Limiting reagent, % Excess, % Yield / Efficiency
Difficulty: Tough

Question: 0.05 mole of LiAlH4 in ether solution was placed in a flask
containing 74g (1 mole) of t-butyl alcohol. The product LiAlHC12H27O3
weighed 12.7 g. If Li atoms are conserved, the percentage yield is: (Li
= 7, Al = 27, H = 1, C = 12, O = 16).


Answer: 100

Solution:
$Li{AlH}_{4} + t - butyl\ alcohol\overset{Ether}{\longrightarrow}{LiAlHC}_{12}H_{27}O_{3}(M \cdot W. = 254)$
0.05 mole 12.7 gram
[IMAGE] Li atom remain conserved so
No. of mole of LiAlH4 = No. of mole of LiAlHC12H27O3
So No. of mole of LiAlHC12H27O3 = 0.05 yield [IMAGE] = 100
Subject: Chemistry — Isomerism — Structural Isomerism
Difficulty: Easy

Question: The general formula [IMAGE] could be for open chain

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

Correct option: 2

Solution:
Diketones:[IMAGE], Carboxylic acid
Diols:[IMAGE], Dialdehydes
Subject: Chemistry — Halogen — Introduction of Halogen containing compounds
Difficulty: Easy

Question: In${CH}_{3}{CH}_{2}Br,\%$ of Br is ${CH}_{3}{CH}_{2}Br$


Answer: 75

Solution:
$\%\ of\ Br = \frac{\ Mass\ of\ Br}{\ Mass\ of\ organic\ compound\ } \times 100$
$= \frac{80}{109} \times 100 = 73.39\%$or approx. $75\%$
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Moderate

Question: When different types of salts have nearly same solubility
product constants Ksp, but less than one, the most soluble salt is that

1) Which produces maximum number of ions per formula unit
2) Which produces minimum number of ions per formula unit
3) Which produces ions with maximum charge
4) Which produces ions with miniumum charge
tc fofHkUu çdkj ds yo.kksa xq.kuQy Ksp yxHkx cjkcj gaS]
ijUrq 1 ls de tks

Correct option: 1

Solution:
For different salts such as AB, AB2, AB3,..........etc.
Ksp = s2, 4s3, 27s4............ respectively. If Ksp is same for
different salts, then s is more for the salt in which produces more
number of ions.
AB, AB2, AB3,..........bR;kfn tSls fofHkUu yo.kksa ds fy, Ksp =
s2, 4s3, 27s4............ gksrs gSaA;fn fHkUu yo.kksa ds fy, Ksp
leku gksrk gS] rc s ml yo.k
Subject: Chemistry — Physical Chemistry — Redox Reaction
Difficulty: Easy

Question: In the following reaction
Which element is reduced?

1) Cr
2) H
3) O
4) I

Correct option: 1

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

Question: Which of the following will not lose asymmetry on reduction with
LiAlH_4


Correct option: 2

Solution:
On reduction of C=O group it will give COH.
Subject: Chemistry — Electrochemistry — Nernst equation & its Applications (including concentration
Difficulty: Tough

Question: Consider the cell [IMAGE] The measured EMF of the
cell is 1.0 V. What is the value of x ? [IMAGE]


Answer: 2

Solution:
Subject: Chemistry — Physical Chemistry — Ionic equilibrium
Difficulty: Moderate

Question: When pure water is saturated with both CaCO3 and CaC2O4, the
concentration of calcium ion in the solution under equilibrium is 8 ×
10-5 M. If the ratio of the solubility product of CaCO3 to that of
CaC2O4 is 3, what is the solubility product of CaCO3 in pure water ?
Report your answer after multiplying by 109 and rounding it off to
nearest whole number.


Answer: 5

Solution:
${CaCO}_{3} \rightleftharpoons {Ca}^{+ 2} + {CO}_{3}^{- 2}$
${CaC}_{2}O_{4} \ \ \rightleftharpoons \ {Ca}^{+ 2} + \ C_{2}O_{4}^{- 2}$
x x y y
(x+y) x (x+y) y
$\frac{K_{SP}\left( {CaCO}_{3} \right)}{K_{SP}\left( {CaC}_{2}O_{4} \right)} = \frac{x}{y} = 3 \Rightarrow x = 3y$
$(x + y) = 8 \times 10^{- 5}$
$(3y + y) = 8 \times 10^{- 5}$
$\Rightarrow y = 2 \times 10^{- 5}x = 6 \times 10^{- 5}$
$K_{SP}\left( {CaCO}_{3} \right) = \left\lbrack {Ca}^{+ 2} \right\rbrack\left\lbrack {CO}_{3}^{- 2} \right\rbrack$
$= 8 \times 10^{- 5} \times 6 \times 10^{- 5}$
$= 48 \times 10^{- 10}$
$K_{SP}\left( {CaCO}_{3} \right) = 4.8 \times 10^{- 9}$
Subject: Chemistry — Physical Chemistry — Redox Reaction
Difficulty: Easy

Question: Oxidation number of Fe in [IMAGE] is


Answer: 3

Solution:
Subject: Chemistry —  — 

Question: The Vander Waals constant for the three gases A, B & C are as shown.
[IMAGE] Which of the following gases will behave most ideally around 300 K.?

1) B
2) C
3) A
4) All the gases


Solution:
Subject: Chemistry — Physical Chemistry — Equivalent concept of Titration
Difficulty: Moderate

Question: The volume of 0.1 N dibasic acid sufficient to neutralize 1 g of
a base that furnishes 0.04 mole of ${OH}^{-}$ in aqueous solution is

1) 200 mL
2) 400 mL
3) 600 mL
4) 800 mL

Correct option: 2

Solution:
The law of equivalence states that the equivalence of acid
isequal to the equivalence of base. $s = ut + \frac{1}{2}{at}^{2}$
Therefore, the volume required is calculated as,
$s = (0)t + \frac{1}{2}{at}^{2}$
$s = \frac{1}{2}{at}^{2}$
$t = \sqrt{\frac{2s}{a}}$
Subject: Chemistry — Inorganic Chemistry — Periodic Properties
Difficulty: Moderate

Question: What is the correct order of second ionisation energy for the
following ?

1) F < N < C < O
2) N < C < F < O
3) C < N < F < O
4) O < C < F < N

Correct option: 3

Solution:
After removal of one electron:
C^+: 1S^2 2S^2 2p^1; N^+: 1S^2 2S^2 2p^2
O^+: 1S^2 2S^2 2p^3; F^+: 1S^2 2S^2 2p^4
Subject: Chemistry — Physical Chemistry — Atomic Structure
Difficulty: Moderate

Question: In an atom, the total number of electrons having quantum numbers
n = 4, $\left| m_{\mathcal{l}} \right| = 1$ and $m_{s} = - 1/2$ is


Answer: 4

Solution:
n = 4 $m_{\mathcal{l}} = 1$
Hence$\ \mathcal{l}$ can be = 3,2,1
i.e. Hf; 2 orbitals
Hd; 2 orbitals
Hp; 2 orbitals
Hence total of 6 orbitals, and we want $ms\ = \ - \frac{1}{2}$, that
is only one kind of spin. So, 6 electrons.
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: Write the major product of following reactions.


Correct option: 1

Solution:
[IMAGE] (iii) $Ph - Ph + {CH}_{3} - {CH}_{3} + Ph - {CH}_{3}$
Subject: Chemistry — Organic Chemistry — IUPAC & Structural Isomerism
Difficulty: Tough

Question: The number of isomers of dibromobiphenyl (Biphenyl - C6H5) is


Answer: 12

Solution:
Overall 12 isomers.
Subject: Chemistry — Amines — Diazonium Salts
Difficulty: Moderate

Question: Which of the following compounds will not undergo azo coupling
reaction with benzene diazoniumchloride?
A Aniline
B Phenol
C Anisole
D Nitrobenzene


Correct option: 3

Solution:
Diazonium cation is weak electrophile and hence it reacts with
electron rich compounds containing electron donatinggroups such as
$- OH, - {NH}_{2}\ and{\ OCH}_{3}$ groups and not with compounds
containing electron withdrawing groups such as$- {NH}_{2}$etc.
Subject: Chemistry — Inorganic Chemistry — Chemical Bonding
Difficulty: Tough

Question: The number of molecules having two or more $p\pi –d\pi$ bond
amongs the following is:
$H_{3}{PO}_{4},P_{4}O_{10},{SO}_{3},{SO}_{2},{HClO}_{4},{HClO}_{3},{HClO}_{2},N_{2}O_{5}$

1) 4
2) 5
3) 6
4) 7

Correct option: 1

Solution:
H3PO4, HClO2: Only $1\pi$-bond
N_2O_5: No d-orbital's
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Moderate

Question: The dissociation constant of acetic acid at a given temperature
is 1.69 × 10^-5. The degree of dissociation of 0.01 M acetic acid in
the presence of 0.01 M HCl is

1) 1.69 × 10^-7
2) 1.69 × 10^-5
3) 1.69 × 10^-3
4) 2.9 × 10^-2

Correct option: 3

Solution:
${CH}_{3}COOH(aq) \rightleftharpoons H^{+}(aq) + {CH}_{3}{COO}^{-}(aq)$
t = 0 0.01
t = eq 0.01 - x x x
[H^+] = x + 0.01[IMAGE] 0.01 M
$\therefore$
$K_{a} = \frac{\left\lbrack H^{+} \right\rbrack\left\lbrack CH_{3}COO^{-} \right\rbrack}{\left\lbrack CH_{3}COOH \right.\ }$
$\Rightarrow$ 1.69 ×10^-5 =
$\frac{0.01 \times \left\lbrack {CH}_{3}COO^{-} \right\rbrack}{0.01}$
$\therefore$ CH_3COO-] = 1.69 × 10^-5 M
So, degree of dissociation of CH_3COOH
$= \frac{1.69 \times 10^{- 5}}{0.01} = 1.69 \times 10^{- 3}$
Subject: Chemistry — Physical Chemistry — Chemical equilibrium
Difficulty: Moderate

Question: Ammonia gas at 15 atm is introduced in a rigid vessel at 300 K.
At equilibrium the total pressure of the vessel is found to be 40.11 atm
at 300ºC. The degree of dissociation of NH3 will be


Answer: 0.4

Solution:
$P_{1} = 15\ atm\;T_{1} = 300K$
Equilibrium temperature is 300ºC that is 573 K.
So first of all we have to calculate pressure of NH3 at 573 K.
$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} = \frac{15}{300} = \frac{P_{2}}{573}$
$P_{2} = 28.65atm\ at\ 300^{\circ}C$
${NH}_{3}(g) \rightleftharpoons \frac{1}{2}N_{2}(g) + \frac{3}{2}H_{2}(g)$
t=0 28.65 atm 0 0
$t = t_{eq}$ $\lbrack 28.65 - x\rbrack$ $\frac{x}{2}\ atm\ $
$\frac{3}{2}x$
But according to question.
$P_{total\ } = 28.65 - x + \frac{x}{2} + \frac{3}{2}x$ or
$28.65 + x = 40.11$
$x = 11.46$
Degree of dissociation of ${NH}_{3} = \frac{11.46}{28.65} = 0.4$
Subject: Chemistry — Physical Chemistry — Redox Reaction
Difficulty: Easy

Question: Which of the following is redox reaction?

2) In atmosphere, [IMAGE] from [IMAGE] by lightning
3) Evaporation of [IMAGE]
4) Nitrogen oxides form nitrogen and oxygen by lightning

Correct option: 4

Solution:
[IMAGE] Here O.N. of N increases from O in [IMAGE] to +2 in
[IMAGE] and that of decreased from O in
[IMAGE] to -2 in O, therefore, it is a redox reaction.
Subject: Chemistry — Inorganic Chemistry — Periodic Properties
Difficulty: Moderate

Question: Select equation having exothermic step

1) S^- (g) + e^- (g) [IMAGE] S^2- (g)
2) Ne (g) + e^- (g) [IMAGE] Ne^- (g)
3) N (g) + e^- (g) [IMAGE] N^- (g)
4) Al^+ (g)[IMAGE] Al (g) - e^- (g)

Correct option: 4

Solution:
[IMAGE] H_e.g.= (+) ve because of electrostatic repulsion.
(2) Ne (g) + e^-(g)[IMAGE] Ne^-(g) H_e.g.= (+) vebecause of stable completely
filled electron configuration
(+) ve because of stable half filled electron configuration.
= (-) ve because of the addition of electron to cation.
Subject: Chemistry — Physical Chemistry — Atomic Structure
Difficulty: Easy

Question: The energy required to melt 1g ice is 33J. The number of quanta
of radiation of frequency 4.67 ×10^13 sec^-1, that must be absorbed
to melt 10 g ice is

1) 1.065 ×10^22
2) 3.205 ×10^23
3) 9.076 ×10^20
4) None of these

Correct option: 1

Solution:
ΔE = n.h.(ν) 
33 x 10 J= n. $6.62 \times 10⁻³⁴\ J \cdot s\ $`<!- ->`4.67
×10^13 sec^-1
n = 1.065 ×10^22
Subject: Chemistry — Amines — Physical Properties of Amines
Difficulty: Moderate

Question: Which of the following explode on heating?

1) Azides
2) Sulphates
3) Chlorides
4) Phosphides

Correct option: 1

Solution:
some experiments describe on calcium and barium azid and
tetrazene, which explode while they are solid. The explosion of the
molten azides is due to self heating of the liquid. Explosion is
facilitated by the presence of an inert gas above the decomposing liquid
self heating of the liquid azides is due to the retention of heat
products of reaction near the surface.
Subject: Chemistry — Isomerism — Geometrical isomerism
Difficulty: Moderate

Question: Which of the following will form only one oxime on reaction with
NH2OH solution?

1) Only A is correct
2) Only B & C correct
3) Only C & D correct
4) Only A & D correct

Correct option: 2

Solution:
I [IMAGE] +
Subject: Chemistry — Physical Chemistry — Ionic equilibrium
Difficulty: Moderate

Question: A certain buffer solution contains equal concentration of
$X^{-}$and HX. The $K_{b}$ for $X^{-}$ is$10^{- 10}$. The pH of the
buffer is


Answer: 4

Solution:
$X^{-} + H_{2}O = OH^{-} + HX$
$K_{b} = \frac{\lbrack OH^{-}\rbrack\lbrack HX\rbrack}{\lbrack X^{-}\rbrack}$
$HX = H^{+} + X^{-}$
$K_{a} = \frac{\lbrack H^{+}\rbrack\lbrack X^{-}\rbrack}{\lbrack HX\rbrack}$
$\therefore K_{a} \times K_{b} = \lbrack H^{+}\rbrack\lbrack OH^{-}\rbrack = K_{v} = 10^{- 14}$
Hence, $K_{a} = 10^{- 4}$
Now as $\lbrack X^{-}\rbrack = \lbrack HX\rbrack,pH = pK_{a} = 4$
Subject: Chemistry —  — 

Question: [IMAGE] Product is



Solution:
[This section contains
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Easy

Question: Molar concentration of 96 g of O_2 contained in a 2 liter vessel
is


Correct option: 2

Solution:
Molar
conc. $= \frac{\ no.\ of\ moles\ of\ O_{2}}{\ volume\ (in\ litre\ )} = \frac{96}{32} \times \frac{1}{2}$
= 1.5M
Subject: Chemistry — Equivalent Concept — Equivalent Concept for Redox reactions, KMnO4 / K2Cr2O7 v/s
Difficulty: Moderate

Question: 25 mL of a solution of Fe2+ ions was titrated with a solution of
the oxidizing agent Cr2O72-. 50 mL of 0.01 M K2Cr2O7 solution was
required. What is the molarity of the Fe2+ solution?


Answer: 0.12

Solution:
[IMAGE] No. of m.eq. of Fe+2 = No. of m.eq. of Cr2O7-2
$\lbrack M \times 25\rbrack \times 1 = \lbrack 0.01 \times 50\rbrack \times 6$
$M = \frac{0.01 \times 50 \times 6}{25}$
$M = 0.02 \times 6 = 0.12$
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Tough

Question: An amount of solid NH_4HS is placed in a flask already
containing ammonia gas at a certain temperature at 0.50 atm pressure.
Ammonium hydrogen sulphide decomposes to yield NH_3 and H_2S gases in
the flask. When the decomposition reaction reaches equilibrium, the
total pressure in the flask rises to 0.84 atm? The equilibrium constant
for NH_4HS decomposition at this temperature is:
Bksl NH_4HS dh,d ek=kk dks ¶ykLd ftl 
rFkk 0.50 atm nkc ij vksj igq¡prh gS rks ¶ykLd esa dqy nkc 0.84 atm c<+ A blh
rki ij NH_4HS fo?kVu


Answer: 0.11

Solution:
NH4HS(s) [IMAGE] NH3(g) + H2S(g)
Initial presens 0 0.5 0
At equi. 0 0.5+x x
Total pressure = 0.5 + 2x = 0.84 $\therefore\ $ x = 0.17 atm
Kp = [IMAGE] = 0.11 atm
Subject: Chemistry — Coordination Compounds — Nomenclature of coordination compounds
Difficulty: Moderate

Question: The correct IUPAC name for the compound [Co(NH3)4CI(ONO)]Cl is

1) Tetraamminechloridonitrito-N-cobalt (III) chloride
2) Chloridonitrito-O-tetraamminecobalt (II) chloride
3) Dichloridonitrito-O-tetraamminecobalt (III)
4) Tetraamminechloridonitrito-O-cobalt (III) chloride

Correct option: 4

Solution:
Cation is named first followed by anion. In coordination sphere
the ligands are named alphabetically followed by name of metal ion and
then its oxidation state in Roman numeral.
In [Co(NH3)4Cl(ONO)]+Cl-, cobalt is in +3 oxidation state and the
complex is cationic. ONO ligands attach to metal ion through O atom;
so --O-- is used while naming as nitrito before the name of metal
ion. So IUPAC name given in option (D) is correct.
Subject: Chemistry — Organic Chemistry — Reduction Oxidation
Difficulty: Tough

Question: How many of the following may produce salicylic acid on
hydrolysis under appropriate conditions ?
(i) [IMAGE] (ii)


Answer: 4

Solution:
[IMAGE] on acid hydrolysis)