Grass-G/jee-main-questions huggingface dataset:
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: Write the major product of following reactions.
Ph - Br + CH_3 - Br [IMAGE] (report mixture)
Ph - Br + CH_3 - Br

1) Ph - Ph - CH_3 - CH_3 + Ph - CH_3
2) Ph - CH_3 + CH_3 - CH_3 + Ph - CH_3
3) Ph - Ph + CH_3 - CH_3 + Ph - OH
4) None of these

Correct option: 1

Solution:
Ph - Br + CH_3 - Br
[IMAGE] Ph - Ph + CH_3 - CH_3 + Ph - CH_3
These are wurtz reaction oqVZ~t
Subject: Chemistry — Physical Chemistry — Thermodynamics
Difficulty: Easy

Question: The enthalpies of neutralization of a weak base AOH and a strong
base BOH by HCI are -12250 mol respectively.
When one mole of HCI is added to a solution containing 1 mole of AOH and
1 mole of BOH, the enthalpy change was -12500 cal/ mol. In what ratio
the acid is distributed between AOH and BOH?

1) 2: 1
2) 2: 3
3) 1: 2
4) None of these

Correct option: 1

Solution:
$- 12250x - 13000(1 - x) = - 12500$
$750x = 500 \Rightarrow x = 2/3$
So, required ratio is = $\frac{2}{1}$
Subject: Chemistry —  — 

Question: are INTEGER ANSWER TYPE Questions.
(The answer of each of the questions is 2 digits integer)]


Answer: 04

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

Question: The pH of which salt solution is independent of its
concentration ?
A. (CH_3COO)C_5H_5NH
B. NaH_2PO_4
C. Na_2HPO_4
D. NH_4CN

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

Correct option: 1

Solution:
pH of amphiprotic salt solutions and weak acid-weak base salt
solutions is independent of their concentration.
Subject: Chemistry — Stereoisomerism — Definition and Properties of Enantiomers, Diastereomers, Meso
Difficulty: Moderate

Question: Which of the following pairs of compounds are enantiomers


Correct option: 1

Solution:
[IMAGE] non-super impossible mirror image stereoisomers.
Subject: Chemistry — Equivalent Concept — Eudiometric/Eudiometric Titration, Calculation of Available
Difficulty: Tough

Question: A 0.2 g sample containing copper (II) was analyzed
eudiometrically, where copper (II)is reduced tocopper (I) by iodide
ions. $2{Cu}^{2 +} + 4I^{-} \longrightarrow 2CuI + I_{2}$
If 20 mL of 0.1 M Na2S2O3 solution is required for titration of the
liberated iodine, then the percentage of copper in the sample will be


Answer: 63.5

Solution:
2 moles of Cu2+ = 1 mole of I2
= 2 moles of hypo.
so moles of hypo used = 20 × 10-3 × 0.1 = 2 m moles = moles of copper
hence
$\% of\ copper = \frac{2 \times 10^{- 3} \times 63.5}{0.2} \times 10\% = 63.5\%$
Subject: Chemistry —  — 

Question: CrO_3 dissolves in aqueous NaOH to give

1) Cr_2O_7 ^2
2) CrO_4 ^2
3) Cr(OH)_3
4) Cr(OH)_2


Solution:
[IMAGE] gives the chromate ions in the basic medium
Subject: Chemistry — Physical Chemistry — 
Difficulty: Moderate

Question: The vapour pressure lowering caused by the addition of 100 g of
sucrose (molecular mass = 342 to 1000 g of water if the vapour
pressure of pure water at 25ºC is 23.8 mm Hg.

1) 1.25 mm Hg
2) 0.125 mm Hg
3) 1.15 mm Hg
4) 00.12 mm Hg

Correct option: 2

Solution:
Given molecular mass of sucrose = 342
Moles of sucrose [IMAGE] = 0.292 mole
Moles of water N [IMAGE] = 55.5 moles and vapour pressure
of pure water P^0 = 23.8 mm Hg
According to Raoult's law
[IMAGE] = 0.125 mm Hg.
Subject: Chemistry — Physical Chemistry — Mole Concept
Difficulty: Moderate

Question: The hydrated salt Na_2CO_3.xH_2O undergoes 63 loss in mass on
heating and becomes anhydrous.Calculate the value of x.
 Na_2CO_3.xH_2O 63 x


Answer: 10

Solution:
[IMAGE] Let 100 g of Na2CO3. xH2O be present
[IMAGE] mole of H2O formed 6678 + 1134 x = 1800 x
666 x = 6678
[IMAGE] gy- [IMAGE] Na2CO3. xH2O [IMAGE] 100 g 
[IMAGE] H2O 6678 + 1134 x = 1800 x
666 x = 6678
Subject: Chemistry — Atomic Structure — Spectrum
Difficulty: Moderate

Question: The difference between the wave number of 1st line of Balmer
series and last line of paschen series for Li2+ ion is:
Li2+ fy, ckWej Js.kh dh 1st Js.kh dh vfUre
js[kk

1) $\frac{R}{36}$
2) $\frac{5R}{36}$
3) 4R
4) $\frac{R}{4}$

Correct option: 4

Solution:
For 1st line of Balmer series
ckej Js.kh dh 1^st fy,
${\overline{v}}_{1} = R_{H}(3)^{2}\left\lbrack \frac{1}{(2)^{2}} - \frac{1}{(3)^{2}} \right\rbrack = 9R\left( \frac{5}{36} \right) = \frac{5}{4}R$
For last line of Pachen series
ik'pu Js.kh dh vfUre fy,
${\overline{v}}_{2} = R_{H}(3)^{2}\left\lbrack \frac{1}{(3)^{2}} - \frac{1}{(\infty)^{2}} \right\rbrack = R$
so,${\overline{v}}_{1} - {\overline{v}}_{2} = \frac{5}{4}R - R = \frac{R}{4}$
Subject: Chemistry — Physical Chemistry — 
Difficulty: Moderate

Question: MX2 dissociates into M2+ and X- ions in an aqueous
solution, with a degree of dissociation $(\alpha)$ of 0.5 The ratio of
the observed depres sion of freezing point of the aqueous solution to
the value of the de pression of freezing point in the absence of ionic
diss ociation is
MX2,d 0.5 dh,d fo;kstu ek=k (degree of
dissociation) α ds lkFk M2+ rFkk X- gksrk gSA
 fgekad voueu (depression of freezing point)
rFkk (depression)


Answer: 2

Solution:
${MX}_{2}\ \ \ \ \ \ \ \ \rightleftharpoons \ \ \ \ M^{2 +} + 2X^{-}$
$m_{0}\left( \begin{matrix}
1 - \alpha)\ \ \ \ m_{0}\alpha\ \ \ \ 2m_{0}\alpha
\end{matrix} \right.\ $ $;m = m_{0}(1 + 2\alpha)$
$\therefore m = m_{0}(1 + 2 \times 0.5) = 2m_{0}$ (as given) k
$\frac{\left( - \Delta T_{f} \right)_{observed\ }}{\left( - \Delta T_{f} \right)_{undissociated\ }} = i = \frac{m}{m_{0}} = 2$
Subject: Chemistry —  — HCP & CCP structures
Difficulty: Tough

Question: Metallic magnesium has a hexagonal close-packed structure and a
density of 1.74 g/cm3. Assume magnesium atoms to be spheres of radius r.
74.1 of the space are occupied by atoms. Calculate the volume of each
atom and the atomic radius r. (Mg = 24.31)


Answer: 1.60

Solution:
Packing efficiency of hcp
$= \frac{6\ \times \ vol.\ \ of\ 1\ atom}{Vol.\ \ of\ unit\ cell}$
Volume of unit cell $= \frac{Z \times M}{N_{A} \times d}$
$= \frac{6 \times 24.31}{6.023 \times 10^{23} \times 1.74}$
$= 13.9179 \times 10^{- 23}{cm}^{3}$
So, volume of 1 atom
$= \frac{Packing\ fraction \times Vol.\ of\ unit\ cell}{6}$
$= \frac{0.741 \times 13.9179 \times 10^{- 23}}{6}$
$= 1.718 \times 10^{- 22}{cm}^{3}$
Volume of 1 atom $= \frac{4}{3}\pi r^{3}$
Radius of 1 atom
$= (r) = {(\frac{Volume\ \ of\ 1\ atom \times 3}{4\pi})}^{1/3}$
$= {(0.41055 \times 10^{- 23})}^{1/3} = 1.60Å.$
Subject: Chemistry — Inorganic Chemistry — P-Block Element
Difficulty: Moderate

Question: The metal which does not form ammonium nitrate by reaction with
dilute nitric acid is

1) Al
2) Fe
3) Pb
4) Mg

Correct option: 3

Solution:
Lead form nitric oxide with dil. HNO_3.
3Pb + 8 HNO_3 [IMAGE] 3Pb(NO_3)_2 + 2 NO +
4H_2O
Subject: Chemistry — Organic Chemistry — Organic Chemistry
Difficulty: Moderate

Question: Which one of the following is least reactive in a nucleophilic
substitution reaction?


Correct option: 4

Solution:
The non reactivity of the chlorine atom in vinyl chloride may
be explained from the molecular orbital point of view as follows. If the
chlorine atom has sp^2 hybridization the C -Cl bond will be a σ bond
and the two lone pairs of electron would occupy the other two
[IMAGE] orbitals. This would leave a p -orbital
containing a lone pair and this orbital could not conjugate with the π
bond of the ethylenic link. Thus two M.O.S. will be required to
accommodate these four π electrons. Furthermore since chlorine is more
electronegative than carbon, the electron will tend to be found in the
chlorine atom has now lost full control of the one pair and so, is less
negative than it would have been had there been no conjugation. Since
two carbon atoms have acquired a share in the lone pair, each atom
acquires a small negative charge. Hence, owing to delocalization of
bonds (through conjugation), the vinyl chloride molecule has an
increased stability. Before the chlorine atom can be displaced by some
other group the lone pair must be localized again on the chlorine atom.
This requires energy and so the chlorine is more firmly bound than has
no conjugation occurred.
Subject: Chemistry — Physical Chemistry — 
Difficulty: Moderate

Question: If two substances A and B have [IMAGE] = 1: 2
and have mole fraction in solution 1: 2 then mole fraction of A in
vapours.

1) 0.33
2) 0.25
3) 0.52
4) 0.2

Correct option: 2

Solution:
Relationship between mole fraction of a component in the vapour
phase and total vapour pressure of an ideal solution.
Subject: Chemistry — Salt Analysis — Dry test
Difficulty: Moderate

Question: Which of the following cannot evolve more than one gas (vapour)
if heated in dry test tube.
b.k ufydk

1) ${NaNO}_{3}(s)$
2) ${MgCO}_{3}(s)$
3) ${FeSO}_{4}(s)$
4) $\left( {NH}_{4} \right)_{2}{Cr}_{2}O_{7}(s)$

Correct option: 2

Solution:
$2{NaNO}_{3}(s)\frac{800^{\circ}C}{\Delta} \longrightarrow {Na}_{2}O(s) + N_{2}(g) + \frac{5}{2}O_{2}(g)$
${MgCO}_{3}(s)\overset{\Delta}{\longrightarrow}MgO(s) + {CO}_{2}(g)$
$2{FeSO}_{4}(s)\overset{\Delta}{\longrightarrow}{Fe}_{2}O_{3}(s) + {SO}_{2}(g) + {SO}_{3}(g)$
$\left( {NH}_{4} \right)_{2}{Cr}_{2}O_{7}(s)\overset{\Delta}{\longrightarrow}N_{2}(g) + 2H_{2}O(g) + {Cr}_{2}O_{3}(s)$
Subject: Chemistry — Physical Chemistry — Some Basic Concepts of Chemistry
Difficulty: Moderate

Question: A 1.5 m solution of acetic acid (I) is mixed with 3 m solution of
the acetic acid (II) to prepare 2m solution (m is molality of solution).
Select the correct statement(s)

1) Mass ratio of solvents mixed$\left( \frac{I}{II} \right)$ is
$\frac{1}{2}$
2) Mass ratio of solvents mixed$\left( \frac{I}{II} \right)$ is
$\frac{2}{1}$
3) Mass ratio of solvents mixed$\left( \frac{I}{II} \right)$ is
$\frac{109}{59}$
4) Mass ratio of solvents mixed$\left( \frac{I}{II} \right)$ is
$\frac{59}{109}$

Correct option: 3

Solution:
Let mass of solvent in 1.5 m solution = m_1 Kg and let mass of
solvent in 3 m solution = m_2 Kg.
Therefore molality of resulting solution
$= \frac{1.5m_{1} + 3m_{2}}{m_{1} + m_{2}} = 2$
$\therefore\frac{m_{1}}{m_{2}} = \frac{2}{1}$
Therefore required ratio $= \frac{109}{59}$
Subject: Chemistry — Physical Chemistry — Gaseous state
Difficulty: Moderate

Question: At P is drawn select
correct graph.


Correct option: 3

Solution:
Boyle temperature gas behaves as ideal gas (Z = 1) in a range of
low pressure but at high pressure Z>
Subject: Chemistry — Physical Chemistry — Electrochemistry
Difficulty: Easy

Question: In a metal oxide, there is 20 oxygen by weight. It equivalent
weight is


Answer: 32

Solution:
Subject: Chemistry — Physical Chemistry — Gaseous State
Difficulty: Tough

Question: 15 mL of a gaseous hydrocarbon was exploded with 72 mL of oxygen.
The volume of gases on cooling was found to be 57 mL, 30 mL of which was
absorbed by aq. KOH and the rest was absorbed in a solution of alkalline
pyrogallol. Then the formula of hydrocarbon is:
 15 mL 72 mL 
 57 mL 30 mL KOH

1) C_2H_2 (2) C_2H_4
2) 45: 8
3) C_2H_6
4) C_3H_6

Answer: 2

Solution:
Subject: Chemistry — Inorganic Chemistry — p block
Difficulty: Easy

Question: Acedic strength of Boron trihalide is in order of


Correct option: 1

Solution:
Concentration of Lewis acid of boron tri halides is increased
in following order.
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: Which acid will decarboxylate with greatest difficulty ?


Correct option: 1

Solution:
Bridge has carbon sp^2.
lsrq 'kh"kZ okys dkcZu sp^2 gksrs gSaA
Subject: Chemistry —  — 

Question: How many number of mole of R-MgX consume in given structure?



Solution:
08
Subject: Chemistry — Physical Chemistry — 
Difficulty: Easy

Question: Equimolar solutions in the same solvent have

1) same boiling point but different freezing point
2) same freezing point but different boiling point
3) same boiling and same freezing points
4) differnet boiling and freezing points
eku foyk;d

Correct option: 3

Solution:
According to Raoult's law equimolal solutions of all the
substances in the same solvent will show equal elevation in boiling
points as well as equal depression in freezing point.
gy- vuqlkj lHkh inkFkZ
Subject: Chemistry — Physical Chemistry — Thermodynamics
Difficulty: Easy

Question: The molar heat capacity of water at constant pressure is
$75\ {J\ K}^{- 1}{mol}^{- 1}$. When 1.0 kJ of heat is supplied to 100 g
of water which is free to expand the increase in temperature of water is

1) 6.6 K
2) 1.2 K
3) 2.4 K
4) 4.8 K

Correct option: 3

Solution:
Heat capacity of water per gram$= \frac{75}{18} = 4.17J$
$Q = mst;1000 = 100 \times 4.17 \times T$
$T = \frac{1000}{100 \times 4.17} = 2.4K$
Subject: Chemistry — Surface Chemistry — Coagulation, Protection And application of colloid
Difficulty: Moderate

Question: The volume of colloidal particle VC as compared to the volume of
a solute particle in a true solution VS could be. Find 10^Vc/ Vs^


Answer: 0.001

Solution:
For true solution the diameter range is 1 to 10 and for
colloidal solution diameter range is 10 to 10,000 
$\frac{V_{0}}{V_{8}} = \frac{(4/3)\pi r_{c}^{3}}{(4/3)\pi r_{s}^{3}} = {(\frac{r_{c}}{r_{s}})}^{3}$
Ratio of diameters = (10/1)3 = 103
Vc/ Vs = 103 =0.001
Subject: Chemistry — Alcohol, Phenol and Ethers — Uses of alcohol, Phenol and Ethers
Difficulty: Moderate

Question: The Bouveault-Balance reduction involves
ckWosYV&csysal

1) $C_{2}H_{5}OH/Na$
2) ${\ LiAlH}_{4}$
3) $C_{2}H_{5}{MgX}^{-}$
4) $Zn/HCl$

Correct option: 1

Solution:
$\begin{matrix}
C_{3}H_{7}{COOC}_{2}H_{5} \\
\ Ethylbuty\ rate\ 
\end{matrix}\overset{NaC_{2}H_{5}OH}{\rightarrow}\begin{matrix}
C_{3}H_{7}{CH}_{2}OH \\
\ Butylalcohol\ 
\end{matrix}$
Subject: Chemistry —  — 

Question: How many molecules have two lone pairs on the central atom?
H_2O, SF_4, I_3¯, XeF_5¯, XeO_3, XeOF_4, PCl_3,
NCl_3, ClF_3, XeF_2, NO_2¯, CO_3^2



Solution:
(i) H_2O
[IMAGE] (ii) ClF_3
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Moderate

Question: For the reaction equilibrium, N2O4 (g)[IMAGE] 2NO2
(g) the concentrations of N2O4 and NO2 at equilibrium are 4.8 × 10-2
and 1.2 × 10-2 mol L-1 respectively. The reaction for Kc =___ ×
10-3


Answer: 3

Solution:
= 4.8 × 10-2 mol L-1, = 1.2 × 10-2 mol L-1
$K_{c} = \frac{\left. \ {\lbrack NO}_{2} \right\rbrack^{2}}{\left\lbrack N_{2}O_{4} \right\rbrack} = \frac{1.2 \times 10^{- 2} \times 1.2 \times 10^{- 3}}{4.8 \times 10^{- 2}}$
= 0.3 × 10-2 = 3 × 10-3 mol L-1
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Moderate

Question: In a saturated solution of the sparingly soluble strong
electrolyte AgIO3 (Molecular mass = 283), the equilibrium which sets in
is:
AgIO3(s) [IMAGE] Ag+(aq) + IO3- (aq)
If the solubility product constant Ksp of AgIO3 at a given temperature
is 1.0 × 10-8, what is the mass of AgIO3 contained in 100 mL of its
saturated solution?;fn fn, x, rki ij AgIO3 xq.kuQy fLFkjkad (Ksp) 1.0 × 10-8


Answer: 2.83

Solution:
AgIO3(s)[IMAGE] Ag+(aq) + IO-3(aq) [s =
Ksp = s2
or = 1.0 × 10-4 × 283 g/L = 2.83 × 10-3
Subject: Chemistry — Physical Chemistry — 
Difficulty: Easy

Question: In the Bragg's equation for diffraction of X-rays, n
represents for

1) Quantum number
2) An integer
4) Moles

Correct option: 2

Solution:
Bragg's equation is $n\lambda = 2\theta\ $
Where n is an integer i.e. 1,2,3,4 etc.
Subject: Chemistry — Physical Chemistry — Electrochemistry
Difficulty: Easy

Question: Identify the correct statement

1) Iron corrodes in oxygen-free water.
2) Iron corrodes more rapidly in salt water because its
electrochemical potential is higher.
3) Corrosion of iron can be minimized by forming a contact with
another metal with a higher reduction potential.
4) Corrosion of iron can be minimized by forming an impermeable
barrier at its surface.

Correct option: 4

Solution:
The statement "corrosion of iron can be minimized by forming
an impermeable barrier at its surface." is correct.
Subject: Chemistry — Physical Chemistry — Electrochemistry
Difficulty: Moderate

Question: If the [IMAGE] electrode is diluted to 100 times
then the change in e.m.f. is increase of _____________ mV

1) In crease of 59 mV
2) Decrease of 59 mV
3) Increase of 29.5 mV
4) Decrease of 29.5 mV

Answer: 59

Solution:
Subject: Chemistry — IUPAC and Structural Isomerism — Nomenculture
Difficulty: Moderate

Question: The IUPAC name of

1) 5, 5-Diethyl-4, 4-dimethylpentane
2) 3-Ethyl-4,4-dimethylheptane
3) 1, 1-Diethyl-2, 2-dimethylpentane
4) 4, 4-Dimethyl-5, 5-diethylpentane

Correct option: 2

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

Question: The solubility of CaF2 (Ksp = 5.3 × 10-9) in 0.1 M solution of
NaF would be: Assume no reaction of cation/anion. n × 10-7. Find the
value of n.
NaF ds 0.1 M CaF2 (Ksp = 5.3 × 10-9).......
gksxh u ekusaA n × 10-7. n


Answer: 5.3

Solution:
(a) NaF[IMAGE] Na+ + F-
0.1 0.1 0.1
CaF2 [IMAGE] Ca2+ + 2F-
x (2x + 0.1) 0.1
Ksp = x (0.1)2 = 5.3 × 10-9
So vr, x = 5.3 × 10-7 M.
Subject: Chemistry — Inorganic Chemistry — Chemical Bonding
Difficulty: Easy

Question: Select which type of overlapping is responsible for
$\pi$-character in Si-N bond of H_3SiNCO?


Correct option: 4

Solution:
$\pi$-character in Si-N bond is due to
$3d\ \pi\ \leftarrow \ 2P\ \pi$ back bonding.
Subject: Chemistry — Physical Chemistry — Electrochemistry
Difficulty: Moderate

Question: When a lead storage battery is discharged

1) [IMAGE] is evolved
2) Lead sulphate is consumed
3) Lead is formed
4) Sulphuric acid is consumed

Correct option: 4

Solution:
[IMAGE] Sulphuric acid is consumed on discharging.
Subject: Chemistry — Organic Chemistry — Chemistry in everyday life
Difficulty: Moderate

Question: Asthma patient use a mixture of..... for respiration

1) $O_{2}$ and $N_{2}O$
2) $O_{2}$ and $He$
3) $O_{2}$ and $NH_{3}$
4) $O_{2}$ and $CO$

Correct option: 2

Solution:
A mixture of $O_{2}$ and He is used for respiration as helium is
inert and light gas and diffuse rapidly.
Subject: Chemistry — Organic Chemistry — Chemistry in everyday life
Difficulty: Tough

Question: The structure given below is known as

1) Penicillin F
2) Penicillin G
3) Penicillin K
4) Ampicillin

Correct option: 2

Solution:
It is the known structure of penicillin G.
Subject: Chemistry — Chemical Bonding — Hybridization
Difficulty: Moderate

Question: Among the triatomic molecules/ions, BeCl2, N3-, N2O, NO2+, O3,
SCl2, ICl2-, I3- and XeF2, the total number of linear
molecules(s)/ion(s) where the hybridization of the central atom does not
have contribution from the d-orbital(s) is


Answer: 4

Solution:
Subject: Chemistry — Chemical Bonding — Resonance and Bond order Calculation
Difficulty: Easy

Question: The average charge on each O atom and average bond order of I-O
bond in IO65- is

1) -1 and 1.67
2) - 5/6 and 1.67
3) -5/6 and 1.33
4) -5/6 and 1.167

Correct option: 4

Solution:
Subject: Chemistry — Physical Chemistry — Atomic Structure
Difficulty: Moderate

Question: has four choices

1) ,
2) ,
3) ,
4) out of which ONLY ONE is correct]
Atomic masses: [H = 1, D = 2, Li = 7, C = 12, N = 14, O = 16, F =
19, Na = 23, Mg = 24, Al = 27, Si = 28, P = 31, S = 32, Cl = 35.5, K =
39, Ca = 40, Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65, As = 75, Br
= 80, Ag = 108, I = 127, Ba = 137, Hg = 200, Pb = 207]
ijek. = 1, D = 2, Li = 7, C = 12, N = 14, O = 16, F =
19, Na = 23, Mg = 24, Al = 27, Si = 28, P = 31, S = 32, Cl = 35.5, K =
39, Ca = 40, Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65, As = 75, Br
= 80, Ag = 108, I = 127, Ba = 137, Hg = 200, Pb = 207]
1. For the electronic configuration of Mn (Z =25), which of the
following statement is
correct ?

Correct option: 4

Solution:
(2)
[IMAGE] A 100
(1) [IMAGE] for 3p & 4s electrons. Hence, total = 6+2 =8
(2) m = 0 for all s-electron and two p - electron and one d- electron.
Hence total = 13
(3) No. of unpaired electrons =
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Moderate

Question: In above question, concentration of Triethyl ammonium ion
([C_6NH_16^+]) in resulting solution will be

1) 100 K_b
2) 200 K_b
3) 10 K_b
4) K_b

Correct option: 4

Solution:
$K_{b} = \frac{\left\lbrack {OH}^{-} \right\rbrack\left\lbrack C_{6}{NH}_{16}^{+} \right\rbrack}{\left\lbrack C_{6}{NH}_{15} \right\rbrack} \Rightarrow \frac{\frac{10^{- 2}}{2}\left\lbrack C_{\theta}{NH}_{16}^{+} \right\rbrack}{\frac{10^{- 2}}{2}} \Rightarrow \left\lbrack C_{6}{NH}_{16}^{+} \right\rbrack = K_{b}$
Subject: Chemistry —  — HCP & CCP structures
Difficulty: Moderate

Question: The shortest distance between Ist and Vth layer of HCP
arrangement is

1) $8\sqrt{\frac{2}{3}}r$
2) $4\sqrt{\frac{3}{2}}r$
3) $16\frac{\sqrt{2}}{3}r$
4) $8\sqrt{\frac{3}{2}}r$

Correct option: 1

Solution:
The shortest distance between Ist and Vth layer of HCP
arrangement is
$= 2C = 2 \times 4\sqrt{\frac{2}{3}}r = 8\sqrt{\frac{2}{3}}r$
Subject: Chemistry — Physical Chemistry — Periodic Table
Difficulty: Easy

Question: The chemical formula of Phosphorous acid is

1) H3PO4 (2) H3PO3
3) H3PO2
4) H2PO3

Correct option: 2

Solution:
Subject: Chemistry —  — 

Question: How many monochloro isomeric products (Including stereoisomers) is
obtained in given reaction?



Solution:
[IMAGE] Total monochloro products are 5.
Subject: Chemistry — Coordination Compounds — Valence Bond Theory + Crystal Field Theory (Part-I)
Difficulty: Tough

Question: All the metal ions contain t2g6 eg0 configurations. Which of the
following complex will be paramagnetic?


Correct option: 1

Solution:
(1) [FeCl(CN)4(O2)]4-; O2 is O22-; CN- is strong field
ligand, so compels for the pairing of electrons.
Fe3+→ 3d5 (2) K4[Fe(CN)6]; CN- is strong field ligand so compels for the
 pairing of electrons.
Fe2+→ 3d6 (3) [Co(NH3)6]Cl3; NH3 is strong field ligand and 3d6 configuration
 has higher CFSE compelling for the pairing of electrons.
Co3+→ 3d6 (4) [Fe(CN)5(O2)]5-; O2 is O22-; CN- is strong field ligand
so compels for the pairing of electrons
Subject: Chemistry —  — 
Difficulty: Moderate

Question: Which of the following curves represents the Henry's law?


Correct option: 1

Solution:
where m = mass of gas absorbed by
given volume of the solvent.
P = pressure of gas
Subject: Chemistry — Physical Chemistry — 
Difficulty: Easy

Question: Two solutions A and B are separated by semi permeable membrane.
If liquid flows from A to B then

1) A is less concentrated than B
2) A is more concentrated than B
3) Both have same concentration
4) None of these

Correct option: 1

Solution:
Osmosis occurs from dilute solution to concentrate solution.
Therefore solution A is less concentrated than B.
Subject: Chemistry — Chemical Kinetics — Method to monitor the progress of Reactions (Titration Method
Difficulty: Moderate

Question: The decomposition of N2O5 in chloroform was followed by
measuring the volume of O2 gas evolved: 2N2O5(CCI4) → 2N2O4(CCI4) +
O2(g). The maximum volume of O2 gas obtained was 100 cm3. In 500
minutes, 90 cm3 of O2 were evolved. The first order rate constant (in
min-1) for the disappearance of N2O5 is

1) $\frac{2303}{500}$
2) $\frac{2303}{500}\log\frac{100}{90}$
3) $\frac{2303}{500}\log\frac{90}{100}$
4) $\frac{100}{10 \times 500}$

Correct option: 1

Solution:
$kt = ln(\frac{C_{0}}{C_{1}})$
$2N_{2}O_{5} \longrightarrow 2N_{2}O_{4} + O_{2}$
$\begin{matrix}
t = 0 & 200{cm}^{3}\ \ \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ \ \ \ 0
\end{matrix}$
$\begin{matrix}
t = t & 20{cm}^{3} & 180{cm}^{3} & 90{cm}^{3}
\end{matrix}$
$\begin{matrix}
t = \infty & 0 & \ \ \ \ \ \ \ \ 200{cm}^{3} & 100{cm}^{3}
\end{matrix}$
Initial volume of $N_{2}O_{5} = 200{cm}^{3}$
Because Max. volume of$= O_{2} = 100{cm}^{3}$
$\therefore K \times 500 = ln(\frac{200}{20}) \Rightarrow k = \frac{ln10}{500} = \frac{2.303}{500}$
Subject: Chemistry — P-block — Oxides, Hydroxides, Oxyacids, Borax
Difficulty: Easy

Question: $B(OH)_{3} + NaOH \rightleftharpoons Na\lbrack B(OH)_{4}\rbrack$
How can this reaction be made to proceed in forward direction?

1) Addition of cis - 1, 2-diol
2) Addition of boron
3) Addition of trans-1, 2-diol
4) Addition of Na2HPO4

Correct option: 1

Solution: