Grass-G/jee-main-questions huggingface dataset:
Subject: Chemistry — Amines — Chemical Reactions of Amines
Difficulty: Tough

Question: Nitration of aniline in the strong acidic medium also gives m
-nitroaniline because
A In absence of substituents nitro group always goes to m- positions
B In spite of substituents nitro group always goes to only m -
positions
C In strong acidic medium aniline is present as anilinium ion
D In electrophilic substitution reactions amino group is meta
directive


Correct option: 1

Solution:
Nitration of aniline in strong acidic medium also gives
m-nitroaniline because in acidic (strong) medium aniline is present as
anilinium ion. Amino ${- NH}_{2}$ group is ortho-para directing but
${- NH}_{3}^{+}$ group is meta directing.
$C_{6}H_{5} - {NH}_{2} + H^{+}\overset{strong\ acid}{\longrightarrow}C_{6}H_{5} - {NH}_{3}^{+}\overset{{HNO}_{3}}{\longrightarrow}m - nitroaniline$
Subject: Chemistry — Stereoisomerism — R/S & D/L Naming.
Difficulty: Moderate

Question: Which of the following is the structure of (S)-Pentan-2-ol is?


Correct option: 3

Solution:
[IMAGE] the arrow is clockwise but least priority
group is on horizontal line of fischer projection.
Subject: Chemistry — Mole concept — Limiting reagent, % Excess, % Yield / Efficiency
Difficulty: Tough

Question: Equal weight of (At. wt. = 24) are
reacted to form the compound[IMAGE]. Then

1) X is the limiting reagent
2) Y is the limiting reagent
3) No reactant is left over and mass of
[IMAGE] formed is double the mass of 'X' taken
4) none of these

Correct option: 3

Solution:
[IMAGE] No. one is L.R.
Subject: Chemistry —  — 

Question: Suppose that a hypothetical H-like atom gives a red Green, blue, &
violet line spectrum.
Which jump according to figure would give off the red spectral line.

1) 3 - 1
2) 2 → 1
3) 4 ࢮ 1
4) 3 ࢮ 2


Solution:
In VIBGYOR, Red has maximum wavelength.
∴ Line having red color will have maximum energy
Hence
3 → 2
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Moderate

Question: N_2 and H_2 are taken in 1: 3 molar ratio in a closed vessel
to attained the following
equilibrium$N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2{NH}_{3}(g)$. Find
K_p for reaction at total pressure of 2P if [IMAGE] at
equilibrium is $\frac{P}{3}$.

1) $\frac{1}{3P^{2}}$
2) $\frac{4}{3P^{2}}$
3) $\frac{4P^{2}}{3}$
4) None

Correct option: 2

Solution:
$N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2{NH}_{3}(g)$.
At equilibrium $\frac{P}{3}$ P, [IMAGE] $K_{p} = \frac{\frac{2P}{3} \times \frac{2P}{3}}{\frac{P}{3} \times P^{3}} = \frac{1}{P^{2}} \cdot \frac{4}{3}$
$\Rightarrow$ $K_{p} = \frac{4}{3P^{2}}$
Subject: Chemistry — Physical Chemistry — 
Difficulty: Moderate

Question: Lowering of vapour pressure is highest for

1) Urea
2) 0.1M glucose
3) 0.1 M MgSO_4
4) 0.1 M BaCl_2

Correct option: 4

Solution:
of P^0 - P_s Ps is maximum for BaCl_2.
Subject: Chemistry — S-block — Oxides, Peroxides, Super Oxides, Hydroxides
Difficulty: Easy

Question: What is in the following reaction ?
MgCl2.6H2O + HCl + H2O

1) MgO
2) Mg
3) Mg(OH)2
4) Mg(OH)Cl.

Correct option: 1

Solution:
MgCl2.6H2O[IMAGE] MgO + 2 HCl + 5H2O
Subject: Chemistry — Physical Chemistry — Electrochemistry
Difficulty: Moderate

Question: 4 g of copper was dissolved in concentrated nitric acid. The
copper nitrate solution on strong heating gave 5 g of its oxide. The
equivalent weight of copper is

1) 23
2) 32
3) 12
4) 20

Correct option: 2

Solution:
present.
 --------------------------------------------
 Element Wt. At Wt. Wt./At Wt. Ratio
 ---------- ---- ----- ------------ ------------
 Cu 4 gm 63.5 4/63.5 = 0.625 [IMAGE] O 1 gm 16 1-16 = 0.0625 [IMAGE] --------------------------------------------
Empirical formula = [IMAGE] of oxide
In this oxide, oxidation no. of Cu = +2
[IMAGE] but Equivalent weight should be an integral no.
= 32
Subject: Chemistry — Inorganic Chemistry — P-block Halogen
Difficulty: Moderate

Question: The substance used in holmes singnals of the ship is a mixture of:
tgktkssa gksYe flXuy (holmes singnals)

1) CaC2 + Ca3P2
2) Ca3(PO4)2 + Pb3O4
3) H3PO4 + CaCl2
4) NH3 + HOCl

Correct option: 1

Solution:
The spontaneous combustion of phosphine is technically used in
Holme's signals. Containers containing calcium carbide and calcium
phosphide are pierced and thrown in the sea when the gases evolved burn
and serve as a signal.
QkWLQhu ds Lor ngu dks rduhdh:i ls gksYe ds lwpdksa (Holme's signals)
 
ik=kksa dks fNfnzr dj leqnz esa Qsad 
gksdj tyrh gSa rc
Subject: Chemistry — Physical Chemistry — Chemical Kinetics & Radioactivity
Difficulty: Moderate

Question: In a first order reaction, the concentration of the reactant,
decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the
concentration to change from 0.1 M to 0.025 M is

1) 30 minutes
2) 15 minutes
3) 7.5 minutes
4) 60 minutes

Correct option: 1

Solution:
The concentration fo the reactant decreases from 0.8 M to 0.4 M
in 15 minutes, i.e., $t_{1/2} = 15$ minute. Therefore, the concentration
of reactant will fall from 0.1 M to 0.025 in two half live. i.e.,
$2t_{1/2} = 2 \times 15 = 30$ minutes.
Subject: Chemistry — Physical Chemistry — Mole concept
Difficulty: Easy

Question: Excess of silver nitrate is added to a water sample to determine
the amount of chloride ion present in the sample. 1.4 g of silver
chloride is precipitated. The mass of chloride ion present in the sample
is:
Molar masses (g-mol-1): AgNO3 169.91, AgCl 143.25

1) 0.25 g
2) 0.35 g
3) 0.50 g
4) 0.75 g

Correct option: 2

Solution:
${AgNO}_{3} + {Cl}^{-} \longrightarrow AgCl \downarrow + {NO}_{3}^{-}$
1.4 g
$= \frac{1.4}{143.5}mol$
$= \frac{1.4}{143.25} \times 35.5gmCl = 0.35gm$
Subject: Chemistry — Inorganic Chemistry — p-block elements
Difficulty: Easy

Question: In Goldschmidt aluminothermic process, thermite contains

1) 3 parts of Al_2O_3 and 4 parts of Al
2) 3 parts of Fe_2O_3 and 2 parts of Al
3) 3 parts of Fe_2O_3 and 1 parts of Al
4) 1 parts of Fe_2O_3 and 1 parts of Al

Correct option: 3

Solution:
In Goldschmidt aluminothermic process, thermite contains 3
parts of Fe_2O_3 and 1 part
of Al.
Subject: Chemistry — Physical Chemistry — Redox Reaction
Difficulty: Moderate

Question: [IMAGE], the reaction is

1) Redox
2) Neutralization
3) Oxidation
4) Reduction

Correct option: 1

Solution:
[IMAGE] Oxidation and Reduction both occur so the reaction is redox.
Subject: Chemistry — Inorganic Chemistry — Periodic Table
Difficulty: Moderate

Question: The following statements concern elements in the periodic table.
Which of the following is true?

1) All the elements in Group 17 are gases.
2) The Group 13 elements are all metals.
3) Elements of Group 16 have lower ionization enthalpy values
compared to those of Group 15 in the correspondingperiods.
4) For Group 15 elements, the stability of +5 oxidation
stateincreases down the group.

Correct option: 3

Solution:
The group $g_{out\ } = \frac{Gm}{r^{2}}$ elements have
ionization energy that is much higher than that of the elements if
group $F = Y\alpha_{L}A\Delta t$ as $F = \left( 2 \times 10^{11}{Nm}^{- 2} \right)\left( 1.2 \times 10^{- 5}K^{- 1} \right)\left( 40 \times 10^{- 4}m^{2} \right)(10)$
the ground = 9.6 × 10^4 N
= 1 × 10^5 N
the group = 9.6 × 10^4 N
= 1 ×10^5 N
elements have half-filled configuration.
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Moderate

Question: Change in volume of the system does not alter the number of moles
in which of the following equilibriums:
fuEu esa fdl fy, vk;ru ifjorZu

1) N2(g) + O2(g)[IMAGE] 2NO(g)
2) PCl5(g)[IMAGE] PCl3(g) + Cl2(g)
3) N2(g) + 3H2(g)[IMAGE] 2NH3(g)
4) SO2Cl2(g)[IMAGE] SO2(g) + Cl2(g)

Correct option: 1

Solution:
In this reaction the ratio of number of moles of reactants to
products in same i.e., 2: 2, hence change in volume will not alter the
number of moles.
leku gksrk gS] vFkkZr~ 2: 2 gSA vr vk;ru ifjorZu
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Moderate

Question: The pH of a solution obtained by mixing 100 mL of 0.2 M
CH_3COOH with 100 mL of 0.2 M NaOH would be: (pK_a for CH_3COOH =
4.74)

1) 4.74
2) 8.87
3) 9.10
4) 8.57

Correct option: 2

Solution:
CH_3COOH + OH^- [IMAGE] CH3COO^- + H_2O
t = 0 20 20
t = eq - - 20
So, [CH_3 COO^-] $= \frac{20}{200}$ = 0.1M
pH = 7 +
$\frac{1}{2}{pK}_{a} + \frac{1}{2}logC = 7 + 2.37 + \frac{1}{2}$
log10^-1 = 7 + 2.37 - 0.5 = 8.87
Subject: Chemistry — Organic Chemistry — Hydrocarbons
Difficulty: Easy

Question: Which of the following graph of potential energy represent the
unimolecular elimination reaction ?


Correct option: 3

Solution:
It is two step reaction and first step is slow and rds.
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Tough

Question: The dissociation pressure of silver oxide at 445ºC is 207 atm.
Calculate $\Delta$Gº for the formation of 1 mole Ag_2O(s) from metal
and oxygen at this temperature. (log 207 = 2.315). Report your answer in
Kcal and round of the value to nearest integer.


Answer: 4

Solution:
T = 445ºC = 445 + 273 = 718K
P = 207atm
n = 1 mole
${Ag}_{2}O \rightleftharpoons 2Ag(s) + \frac{1}{2}O_{2}(g)$
$K_{p} = \sqrt{P_{O_{2}}}$ = (207)^1/2 = 14.39
$\Delta G^{\circ} = \Delta G^{0}$ + 2.303 Rt log K_eq
But at eq. $\Delta$G = 0
$\Delta$ G^0 = - 2.303 RT log K_eq.
But for formation
$2Ag(s) + \frac{1}{2}O_{2}(g) \rightleftharpoons {Ag}_{2}O(s)$
1 Mole
$\Delta G^{0}$ = - 2.303 × 8.312 × 718 log $\frac{1}{(207)^{1/2}}$
$\Delta G^{0} = \frac{2.303 \times 8.312 \times 718}{2}$ × log⁡207
$\Delta G^{0}$ = 6872.17 log 207
$\Delta G^{0}$ = 15915.75 J
$\Delta G^{0}$ = 3789.46 Cal
$\Delta G^{0}$ = 3.789 KCal
$\Delta G^{0}$ = 3.8 KCal
Subject: Chemistry — Organic Chemistry — Organic Chemistry
Difficulty: Moderate

Question: Least contributing resonating structure of nitroethene is


Correct option: 3

Solution:
In (C) two positive charge on adjacent position.
Subject: Chemistry — Inorganic Chemistry — Periodic Properties
Difficulty: Moderate

Question: How many of the following compounds are found to exist?
(i) BiF_5 (ii) T[IMAGE] I_3 (iii) PbO_2 (iv) SnCl_2



Solution:
6 (except (vi))
Subject: Chemistry — Inorganic Chemistry — Chemical Bonding
Difficulty: Tough

Question: In which one of the following species the central atom has the
type of hybridization which is not the same as that present in the other
three?

1) SF_4
2) $l_{3}^{-}$
3) ICl_5
4) PCl_5

Correct option: 3

Solution:
ICl_5 has sp^3d^2 hybridization.
$l_{3}^{-}$, SF_4, and PCl_5; all have sp3d hybridization
Subject: Chemistry — Organic Chemistry — Organic Chemistry
Difficulty: Moderate

Question: Which is the most stable carbocation?

1) iso-propyl
2) Triphenylmethyl cation
3) Ethyl cation
4) [IMAGE] -propyl cation

Correct option: 2

Solution:
Stability of carbonium ions. tertiary alkyl > secondary alkyl
> primary alkyl > methyl. More the number of alkyl groups, the greater
the dispersal of positive charge and therefore, more the stability of
carbonium ion is observed.
[IMAGE] Dispersal of positive charge increases with the increase in the number
of benzene ring.
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Moderate

Question: For a BH_2Cl_2 solution (chloride salt of a diacidic base B):
(Dissociation constants of base are [IMAGE] &


Correct option: 3

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

Question: Mottling of teeth caused by excessive intake of

1) Potassium
2) Iodine
3) Chlorine
4) Fluorine

Correct option: 4

Solution:
Explanation: Dental fluorosis is also known as mottling of
teeth. It is caused by excessive ingestion of Fluorine during enamel
formation.
Subject: Chemistry — Inorganic Chemistry — P-block
Difficulty: Moderate

Question: Identify the reaction which does not liberate hydrogen

1) Reaction of zinc with aqueous alkali
2) Electrolysis of acidified water using Pt electrodes.
3) Allowing a solution of sodium in liquid ammonia to stand.
4) Reaction of lithium hydride with B_2 H_6.

Correct option: 4

Solution:
Lithium hydride reacts
with$U_{total\ } = \frac{1}{2}nRT + \frac{5}{2}NRT - \frac{5}{2}NRT$
$= \frac{1}{2}nRT$
and the reaction is expressed as follows,
$a = a_{0}e^{\frac{- bc}{m}}$
The above reaction proves that the liberation of hydrogen gas is not
possible.
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Moderate

Question: The standard Gibbs energy change at 300 K for the reaction is
2A[IMAGE] B + C is 2494.2 J. At a given time, the
composition of the reaction mixture is = 2
and = 8.314

1) Forward direction because Q > KC
(2) Reverse direction because Q > KC
2) foijhr fn'kk
3) Forward direction because Q < KC
4) Reverse direction because Q < KC
300 K ij vfHkfØ;k 2A[IMAGE] B + C dh ekud fxCt 
gksrh

Correct option: 2

Solution:
DG = DGº + RT ln Q
= 2494.2 + 8.314 × 300 ln 4
= positive èkukREkd
$\Delta G = RT\mathcal{l}n\frac{Q}{K}$
Since, ΔG is positive so, Q > K, so reaction shifts in reverse
direction.
pwfda ΔG èkukRed gS, blfy, Q > K, vr
Subject: Chemistry — Organic Chemistry — Chemistry in everyday life
Difficulty: Easy

Question: Morphine is

1) Anesthetic
2) Analgesic
3) Antiseptic
4) Antibiotics

Correct option: 2

Solution:
Morphine is analgesic.
Subject: Chemistry — Inorganic Chemistry — D & F -Block Elements
Difficulty: Easy

Question: Which of the following statement is not true about Mohr's salt

1) It decolorizes [IMAGE]
2) It is a primary standard
3) It is a double salt
4) Oxidation state of iron is +2 in this salt

Correct option: 4

Solution:
Oxidation state of iron in Mohr's salt
Subject: Chemistry — Physical Chemistry — Redox Reaction
Difficulty: Moderate

Question: Of the four oxy-acids of chlorine the strongest oxidizing agent
in dilute aqueous solution is


Correct option: 4

Solution:
[IMAGE] is the strongest oxidizing agent. The
correct order of oxidizing power is
Subject: Chemistry — Surface Chemistry — Classification and Preparation of Colloid
Difficulty: Easy

Question: Colloidal solution of gold prepared by different methods of
different colors because of

1) different diameters of colloidal gold particles
2) variable valency of gold
3) different concentration of gold particles
4) impurities produced by different methods

Correct option: 1

Solution:
Different colloidal particle will provide different color to
the sol.
Subject: Chemistry — Organic Chemistry — Some Basic Principles and Techniques
Difficulty: Tough

Question: Total number of structural unsaturated 3º alcohols possible with
molecular formula C_6H_12O is

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

Correct option: 4

Solution:
(i) [IMAGE] (ii)
Subject: Chemistry —  — 

Question: Which is not true for CO molecule?

1) Bond order of CO is equal to N_2
2) Bond enthalpy of CO is equal
to N_2
3) CO is a strong ligand
4) CO is a good π acid


Solution:
Bond enthalpy of CO is greater than N_2 because CO has electro
negativity difference which provide strength to bond length.
Subject: Chemistry — Organic Chemistry — Organic Chemistry
Difficulty: Moderate

Question: Which gives mono substituted product

1) o -dinitrobenzene
2) m -dinitrobenzene
3) p -dinitrobenzene
4) Nitrobenzene

Correct option: 2

Solution:
[IMAGE] gives only mono-substitution product as
[IMAGE] group is meta directing and only one m -position
is possible in m-dinitrobenzene.
Subject: Chemistry — Organic Chemistry — Some Basic Principles and Techniques
Difficulty: Moderate

Question: In which species positive charge is delocalised ?


Correct option: 2

Solution:
Subject: Chemistry — Physical Chemistry — Some Basic Concepts of Chemistry
Difficulty: Easy

Question: The amount of arsenic pentasulphide (As_2S_5) that can be
obtained when 35.5arsenic acid (H_3AsO_4) is treated with excess H_2S
in the presence of conc. HCl (assuming 100 conversion) is

1) 0.333 mol
2) 0.125 mol
3) 0.25 mol
4) 0.50 mol

Correct option: 2

Solution:
H_3AsO_4 + H_2S + HCL [IMAGE] As_2S_5
Initial moles of H_3AsO_4 = $\frac{35.5}{142}\mathbf{=}\frac{1}{4}$
mol
Applying POAC on As atoms; moles of As_2S_5$= \frac{1}{8}$ mol
Subject: Chemistry — Physical Chemistry — Electrochemistry
Difficulty: Easy

Question: The conductivity of a saturated solution of ${BaSO}_{4}$is
$3.06 \times 10^{- 6}{ohm}^{- 1}{cm}^{- 1}$ and its
equivalentconductance is $1.53{ohm}^{- 1}{cm}^{- 1}{equivalent}^{- 1}$.
The $K_{sp}$ of the $BaSO_{4}$ is $X \times 10^{- 6}$ value of X is.


Answer: 4

Solution:
$\lambda m = \frac{1000K}{S} = \frac{1000 \times 3.06 \times 10^{- 6}}{S} = 1.53$
$S = 2 \times 10^{- 3}\frac{mol}{litre}$
$K_{sp({BaSO}_{4})} = S^{2} = {(2 \times 10^{- 3})}^{2} = 4 \times 10^{- 6}$
Subject: Chemistry — Physical Chemistry — Thermochemistry
Difficulty: Moderate

Question: If the bond dissociation energies of XY, X_2 and Y_2 (all
diatomic molecules) are in the ratio of 1: 1: 0.5 and $\Delta_{f}H$ for
the formation of XY is $- 200kJ{mole}^{- 1}$. The bond dissociation
energy of X_2 will be

1) $100\ k\ J\ {mole}^{- 1}$
2) $800\ k\ J\ {mole}^{- 1}$
3) $300\ k\ J\ {mole}^{- 1}$
4) $400\ k\ J\ {mole}^{- 1}$

Correct option: 2

Solution:
$XY \longrightarrow X_{(g)} + Y_{(g)};\Delta H = + akJ/mole$....(i)
$X_{2} \longrightarrow 2X;\Delta H = + akJ/mole$...(ii)
$Y_{2} \longrightarrow 2Y;\Delta H = + 0.5akJ/mole$...(iii)
$\frac{1}{2} \times (ii) + \frac{1}{2} \times (iii) - (i),$ gives
$\frac{1}{2}X_{2} + \frac{1}{2}Y_{2} \longrightarrow XY$
$\Delta H = ( + \frac{a}{2} + \frac{0.5}{2}a - a)kJ/mole$
$+ \frac{a}{2} + \frac{0.5a}{2} - a = - 200$
$a = 800$
Subject: Chemistry — Organic Chemistry — Hydrocarbon
Difficulty: Moderate

Question: Which of the following compounds on hydrolysis gives propyne ?

1) CaC2
2) Mg2C3
3) Al4C3
4) Cu2Cl2

Correct option: 2

Solution:
$2{Mg}^{+ 2}\left( \overset{¯}{C} \equiv C - C^{3 -} \right)\overset{H_{3}O^{+}}{\rightarrow}Mg(OH)_{2} + CH \equiv C - {CH}_{3}(\ Propyne)$
Subject: Chemistry — Inorganic Chemistry — P-Block Element
Difficulty: Moderate

Question: Mark the oxide which is amphoteric in character

1) CO_2
2) SiO_2
3) SnO_2
4) CaO

Correct option: 3

Solution:
SnO_2 + 2NaOH [IMAGE] NA_2 SnO_3 +
H_2O
SnO_2 + 4HC[IMAGE] SnCl_4 + 2H_2O
Subject: Chemistry — Physical Chemistry — Redox Reaction
Difficulty: Moderate

Question: In the following reaction, [IMAGE]

1) Bromine is oxidized and carbonate is reduced
2) Bromine is reduced and water is oxidized
3) Bromine is neither reduced nor oxidized
4) Bromine is both reduced and oxidized

Correct option: 1

Solution:
[IMAGE] In this reaction bromine is oxidized as
well as reduced.
Subject: Chemistry —  — 

Question: How many of the following compounds give benzene on reaction with
PhMgBr?
(a) [IMAGE] (b)



Solution:
Subject: Chemistry — Coordination Compounds — Valence Bond Theory and Crystal Field Theory (Part II)
Difficulty: Tough

Question: Amongst $Ni(CO)_{4},\left\lbrack Ni(CN)_{4} \right\rbrack^{2 -}$
and ${NiCl}_{4}^{2 -}$

1) $Ni(CO)4$and$NiCI42–$ are diamagnetic and
$\lbrack Ni(CN)4\rbrack 2–$ is paramagnetic.
2) $NiCI42–$and$\lbrack Ni(CN)4\rbrack 2–$ are diamagnetic and
$Ni(CO)4$ is paramagnetic.
3) $Ni(CO)4$and$\lbrack Ni(CN)4\rbrack 2–$ are diamagnetic and
$NiCI42$- is paramagnetic.
4) $Ni(CO)4$is diamagnetic and $NiCI42–$ and
$\lbrack Ni(CN)4\rbrack 2–$ are paramagnetic.

Correct option: 3

Solution:
In complex, $\lbrack Ni(CO)4\rbrack$ nickel is in zero
oxidation state. The $CO$ is strong field ligand and, therefore compels
for the pairing of electrons. The hybridisation scheme is as shown in
figure.
$Niº(\lbrack Ar\rbrack\ 3d8\ 4s2)$
[IMAGE] Four pairs of electrons from four CO.
$\mathbf{sp}\mathbf{3}$hybrid orbitals
It is tetrahedral and as all electrons are paired so diamagnetic.
In the diamagnetic and square planar complex$\lbrack Ni(CN)4\rbrack 2–$,
the nickel is in $+ 2$ oxidation state and the ion has the electronic
configuration $3d8$. The hybridisation scheme is as shown in figure.
$Ni2 +,\ \lbrack Ar\rbrack 3d8$
[IMAGE] $d\mathbf{sp}2$hybrid orbitals
In the paramagnetic and tetrahedral complex
$\mathbf{\lbrack}\mathbf{NiCl}\mathbf{4}\mathbf{\rbrack}\mathbf{2}\mathbf{–}$,
the nickel is in $\mathbf{+ 2}$ oxidation state and the ion has the
electronic configuration $\mathbf{3}\mathbf{d}\mathbf{8}$. The
hybridisation scheme is as shown in figure.
$Ni2 +,\ \lbrack Ar\rbrack 3d8$
[IMAGE] $\mathbf{sp}\mathbf{3}$hybrid orbitals
gy ladqy $\lbrack Ni(CO)4\rbrack$ esa fudy vkWDlhdj.k voLFkk
gSaA
$Niº(\lbrack Ar\rbrack\ 3d8\ 4s2)$
[IMAGE] pkj $\mathbf{CO}$ ls pkj bysDVªkWu;qXe
curs gSaA
$\mathbf{\ }\mathbf{sp}\mathbf{3}$ adfjr prq"Qyd gS rFkk lHkh bysDVªkWu blfy, ç aA
$\mathbf{\lbrack}\mathbf{Ni}\mathbf{(}\mathbf{CN}\mathbf{)}\mathbf{4}\mathbf{\rbrack}\mathbf{2}\mathbf{–}$
çfrpqEcdh; rFkk oxZ leryh
$\mathbf{+}\mathbf{2}\mathbf{\ }$vkWDlhdj.k voLFkk rFkk vk;u
$\mathbf{3}\mathbf{d}\mathbf{8}$ bysDVªkWfud foU;kl j
x;s fp=kkuqlkj ladj.k O;oLFkk fn
$Ni2 +,\ \lbrack Ar\rbrack 3d8$
$\lbrack NiCl4\rbrack 2–,$ vuqpqEcdh; rFkk prq"Qydh
$+ 2$ vkWDlhdj.k voLFkk rFkk bysDVªkWfud foU;kl $3d8$ gSA
fp=kuqlkj ladj.k xbZ gSaA
$Ni2 +,\ \lbrack Ar\rbrack 3d8$
Subject: Chemistry — Physical Chemistry — Ionic Equilibrium
Difficulty: Easy

Question: 8 × 10-6 M AgNO3 solution is gradually added in 1 L of 10-4 M
KCl solution. Upto what volume of AgNO3 solution being added (in L),
precipitation of AgCl will not take place? (Ksp of AgCl = 2 × 10-10)
ikru uk gks ik;s (AgCl


Answer: 1

Solution:
$2 \times 10^{- 10} = \left( \frac{8 \times 10^{- 8} \times V}{V + 1} \right)\left( \frac{10^{- 4}}{V + 1} \right)$
(V + 1)^2 = 4V
(V -1)^2 =
Subject: Chemistry — Physical Chemistry — Redox Reaction
Difficulty: Moderate

Question: In the reaction [IMAGE] the
[IMAGE] acts as

1) Reducing agent
2) Oxidizing agent
3) Bleaching agent
4) None of the above

Correct option: 2

Solution:
Subject: Chemistry — Physical Chemistry — States of matter
Difficulty: Moderate

Question: Initially, the root mean square (rms) velocity of N_2 molecules
atcertain temperature is u. If this temperature is doubled and allthe
nitrogen molecules dissociate into nitrogen atoms, then the new rms
velocity will be

2) 2u
3) 4u
4) 14u


Solution:
The root mean square velocity for nitrogen initially is
calculated as, (d+200)
The dissociation of all the nitrogen molecules into nitrogen atoms
results in the new root mean square velocity, that
is,$t = \sqrt{\frac{2d}{2}}$
Subject: Chemistry — Organic Chemistry — Organic compounds containing Nitrogen
Difficulty: Tough

Question: Starting from propanoic acid, the following reactions were
carried out
[IMAGE] What is the compound Z


Correct option: 2

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

Question: For the chemical equilibrium, CaCO_3 (s) $\rightleftharpoons$
CaO(s) +CO_2 (g), [IMAGE] can be determined for which
one of the following plots


Correct option: 1

Solution:
For the reaction,
CaCO_3 (g) = CaO(s) + CO_2 (g)
$K_{p} = P_{\infty},$ and $K_{c} = \left\lbrack {CO}_{2} \right\rbrack$
( $\because$ [CaCO_3 ] = 1 and   for solids$\ \rbrack$
According to Arrhenius equation we have
$K = Ae^{- \Delta H^{*},/RT}$
 Taking logarithm, we have
log⁡K_p = log A $- \frac{\Delta H_{r}^{0}}{RT(2.303)}$
This is an equation of straight line. When log K_p is plotted against
1/T. we get a straight line.
[IMAGE] The intercept of this line = log A, slope = - 
2.303R
Knowing the value of slope from the plot and universal gas constant R,
[IMAGE] can be calculated.
(Equation of straight line: Y = mx + C. Here,
Subject: Chemistry — Physical Chemistry — Chemical Equilibrium
Difficulty: Moderate

Question: The equilibrium
${SO}_{2}(g) + \frac{1}{2}O_{2}(g) \rightleftharpoons {SO}_{3}(g)$ is
established in a container of 4L at a particular temperature. If the
number of moles of SO_2, O_2 and SO_3 at equilibrium are 2, 1 and 4
respectively then find the value of equilibrium constant.


Answer: 4

Solution:
${SO}_{2}(g) + \ \ \ \ \ \ \frac{1}{2}O_{2}(g)\ \ \ \rightleftharpoons \ \ \ \ {SO}_{3}(g)$
No. of mole 2 1 4
Conc. $\frac{2}{4}$ $\frac{1}{4}$ $\frac{4}{4}$
$K_{C} = \frac{\left\lbrack {SO}_{3} \right\rbrack}{\left\lbrack {SO}_{2} \right\rbrack\left\lbrack O_{2} \right\rbrack^{1/2}} = \frac{1}{(1/2)(1/4)^{1/2}} = \frac{1}{(1/2) \times (1/2)} = 4$