datavorous/entrance-exam-dataset huggingface dataset:
Acetylene can be obtained by the reaction

Options:
A $\mathrm{HCOOK} \xrightarrow{\text { electrolysis }}$
B $\mathrm{CHI}_3+6 \mathrm{Ag}+\mathrm{CHI}_3 \xrightarrow{\Delta}$
C $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \xrightarrow[443^{\circ} \mathrm{C}]{\text { Conc. } \mathrm{H}_2 \mathrm{SO}_4}$
D $\mathrm{Be}_2 \mathrm{C}+\mathrm{H}_2 \mathrm{O} \rightarrow$

The correct answer is:
$\mathrm{CHI}_3+6 \mathrm{Ag}+\mathrm{CHI}_3 \xrightarrow{\Delta}$

Solution:
$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH} \xrightarrow{\mathrm{PCl}_3}$

Topic: Chemistry, Hydrocarbons, JEE Main
Acetylene can be obtained by the reaction?

Options:
A HCOOKâElectrolysisÂ Â Â Â
B CHI3+Agâ â
C CH3CH2OHâConc.H2SO4Â Â 443Â K
D Be2C+H2Oâ

The correct answer is:
CHI3+Agâ â

Solution:
Iodoform and Silver (Ag) undergo dehalogenation reaction to produce acetylene.
(A) HCOOK+H2OââH2+CO2+KOH
(B) 2CHI3Iodoform+6AgSilverââCHâ¡CHAcetylene+6AgI
(C) CH3CH2OHâConc.H2SO4 443KCH2=CH2ethylene+H2O
(D) Be2C+H2OâCH4Methane

Topic: Chemistry, Hydrocarbons, JEE Main
Acetylene gives

Options:
A White precipitate with $\mathrm{AgNO}_3$ and red precipitate with $\mathrm{Cu}_2 \mathrm{Cl}_2$
B White precipitate with $\mathrm{Cu}_2 \mathrm{Cl}_2$ and red precipitate with $\mathrm{AgNO}_3$
C White precipitate with both the reagents
D Red precipitate with both the reagents

The correct answer is:
White precipitate with $\mathrm{AgNO}_3$ and red precipitate with $\mathrm{Cu}_2 \mathrm{Cl}_2$

Solution:
$\begin{gathered}\mathrm{CH}_3 \equiv \mathrm{CH}+\underset{\text { (Ammonica) }}{\mathrm{AgNO}_3} \rightarrow \mathrm{Ag}-\underset{\text { White ppt. }}{\mathrm{C}} \equiv \mathrm{C}-\mathrm{Ag} \\ \mathrm{CH}_3 \equiv \mathrm{CH}+\underset{\text { (Ammonica) }}{\mathrm{CuCl}_2} \rightarrow \mathrm{Cu}-\underset{\text { Red ppt. }}{\mathrm{C}} \equiv \mathrm{Cu}-\mathrm{Cu}\end{gathered}$

Topic: Chemistry, Hydrocarbons, JEE Main
Acid catalyzed hydration of alkenes except ethene leads to the formation of

Options:
A Primary alcohol
B Secondary or tertiary alcohol
C Mixture of primary and secondary alcohols
D Mixture of secondary and tertiary alcohol

The correct answer is:
Secondary or tertiary alcohol

Solution:
Acid catalysed hydration of alkenes except ethene leads to the formation of secondary or tertiary alcohol. Acid catalysed hydration of ethene leads to the formation of primary alcohol. A molecule of water is added to $\mathrm{C}=\mathrm{C}$ double bond where $-\mathrm{OH}$ group is added to more substituted $\mathrm{C}$ atom.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Acid catalyzed hydration of alkenes except ethene leads to the formation of

Options:
A
primary alcohol

B
secondary or tertiary alcohol

C
mixture of primary and secondary alcohols

D
mixture of secondary and tertiary alcohols

The correct answer is:
mixture of secondary and tertiary alcohols

Solution:
No Solution Available

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2005
Acid rain is mainly caused by the emissions of which of the following gases?
I. Sulphur dioxide
II. Carbon dioxide
III. Nitrogen dioxide
IV. Methane

Options:
A I and II
B I, II and III
C I and III
D I, II, III and IV

The correct answer is:
I and III

Solution:
Out of the gases,
I. $\mathrm{SO}_2$, II. $\mathrm{CO}_2$, III. $\mathrm{NO}_2$ and IV. $\mathrm{CH}_4$
Only $\mathrm{SO}_2$ and $\mathrm{NO}_2$ are mainly responsible for acid rain.
$\left.\begin{array}{l}2 \mathrm{SO}_2(g)+\mathrm{O}_2(g)+\underset{\text { Rain water }}{2 \mathrm{H}_2 \mathrm{O} \longrightarrow} 2 \mathrm{H}_2 \mathrm{SO}_4(a q) \\ 4 \mathrm{NO}_2(g)+\mathrm{O}_2(g)+\underset{\text { Rain water }}{2 \mathrm{H}_2 \mathrm{O} \longrightarrow} 4 \mathrm{HNO}_3(a q)\end{array}\right] \begin{gathered}\text { Acid rain } \\ (\mathrm{pH} < 5.6)\end{gathered}$

Topic: Chemistry, Environmental Chemistry, TS EAMCET, TS EAMCET 2020 (14 Sep Shift 2)
Acid rains are produced by

Options:
A Excess \( \mathrm{NO}_{2} \) and \( \mathrm{SO}_{2} \) from burning fossil fuels
B Excess production of \( \mathrm{NH}_{3} \) by industry and coal gas
C Excess release of carbon monoxide by incomplete combustion
D Excess formation of \( \mathrm{CO}_{2} \) by combustion and animal respiration

The correct answer is:
Excess \( \mathrm{NO}_{2} \) and \( \mathrm{SO}_{2} \) from burning fossil fuels

Solution:
The nitrogen and oxygen present in the atmosphere combine to give oxides of nitrogen, which dissolve in rain water and give nitric acid.

Topic: Chemistry, Environmental Chemistry, JEE Main
Acid strength given below the correct decreasing order will be :

Options:
A NO2CH2COOH>FCH2COOH>CNCH2COOH>ClCH2COOH
B NO2CH2COOH>NCCH2COOH>FCH2COOH>ClCH2COOH
C CNCH2COOH>O2NCH2COOH>FCH2COOH>ClCH2COOH
D FCH2COOH>NCCH2COOH>NO2CH2COOH>ClCH2COOH

The correct answer is:
NO2CH2COOH>NCCH2COOH>FCH2COOH>ClCH2COOH

Solution:
Acidic character α stability of conjugate base (anion)
Stability of anion α–I
Hence greater the –I effect greater will be the acidity
Order of -I-effect -NO2>CN>-F>-Cl Hence, order of acidity will be
O2NCH2COOH>NC CH2COOH>FCH2 COOH>Cl CH2 COOH

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Acid strength given below the correct decreasing order will be :

Options:
A NO2CH2COOH>FCH2COOH>CNCH2COOH>ClCH2COOH
B NO2CH2COOH>NCCH2COOH>FCH2COOH>ClCH2COOH
C CNCH2COOH>O2NCH2COOH>FCH2COOH>ClCH2COOH
D FCH2COOH>NCCH2COOH>NO2CH2COOH>ClCH2COOH

The correct answer is:
NO2CH2COOH>NCCH2COOH>FCH2COOH>ClCH2COOH

Solution:
Acidic character α stability of conjugate base (anion)
Stability of anion α–I
Hence greater the –I effect greater will be the acidity
Order of -I-effect -NO2>CN>-F>-Cl Hence, order of acidity will be
O2NCH2COOH>NC CH2COOH>FCH2 COOH>Cl CH2 COOH

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Acid strength of oxy acids of chlorine follows the order

Options:
A $\mathrm{HClO} \lt \mathrm{HClO}_2 \lt \mathrm{HClO}_3 \lt \mathrm{HClO}_4$
B $\mathrm{HClO}_4 \lt \mathrm{HClO}_3 \lt \mathrm{HClO}_2 \lt \mathrm{HClO}$
C $\mathrm{HClO}_4 \lt \mathrm{HClO}_3 \lt \mathrm{HClO} \lt \mathrm{HClO}_2$
D None of these

The correct answer is:
$\mathrm{HClO} \lt \mathrm{HClO}_2 \lt \mathrm{HClO}_3 \lt \mathrm{HClO}_4$

Solution:
$\stackrel{+7}{\mathrm{HClO}_4} \gt \stackrel{+5}{\mathrm{HClO}}_3 \gt \stackrel{+3}{\mathrm{HClO}}_2 \gt \stackrel{+1}{\mathrm{HCl}}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Acidic ferric chloride solution on treatment with excess of potassium ferrocyanide gives a Prussian blue coloured colloidal species. It is:

Options:
A HFeFe(CN)6
B K5FeFe(CN)62
C Fe4Fe(CN)63
D KFeFe(CN)6

The correct answer is:
KFeFe(CN)6

Solution:
If we add ferric chloride over potassium ferrocyanide, it forms a white precipitate that will become blue. The reaction is

FeCl3+K4Fe(CN)6⟶KFeFe(CN)6+3KCl

Being small quantity of ferric chloride, not all potassium is displaced.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (27 Aug Shift 1)
Acidic strength of Boron trihalide are in order of

Options:
A $\quad \mathrm{BF}_3 \lt \mathrm{BCl}_3 \lt \mathrm{BBr}_3 \lt \mathrm{Bl}_3$
B $\quad \mathrm{Bl}_3 \lt \mathrm{BBr}_3 \lt \mathrm{BCl}_3 \lt \mathrm{BF}_3$
C $\mathrm{BBr}_3 \lt \mathrm{BCl}_3 \lt \mathrm{BF}_3 \lt \mathrm{Bl}_3$
D $\quad \mathrm{BF}_3 \lt \mathrm{Bl}_3 \lt \mathrm{BCl}_3 \lt \mathrm{BBr}_3$

The correct answer is:
$\quad \mathrm{BF}_3 \lt \mathrm{BCl}_3 \lt \mathrm{BBr}_3 \lt \mathrm{Bl}_3$

Solution:
Concentration of Lewis acid of boron tri halides is increased in following order.
$B F_3 \lt \mathrm{BCl}_3 \lt \mathrm{BBr}_3 \lt B I_3$

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Acidified potassium dichromate is treated with hydrogen sulphide. In the reaction, the oxidation number of chromium

Options:
A Increases from $+3$ to $+6$
B Decreases from $+6$ to $+3$
C Remains unchanged
D Decreases from $+6$ to $+2$

The correct answer is:
Decreases from $+6$ to $+3$

Solution:
Decreases from $+6$ to $+3$.

Topic: Chemistry, Redox Reactions, JEE Main
Acidified potassium dichromate is treated with hydrogen sulphide. In the reaction, the oxidation number of chromium

Options:
A Increases from $+3$ to $+6$
B Decreases from $+6$ to $+3$
C Remains unchanged
D Decreases from $+6$ to $+2$

The correct answer is:
Decreases from $+6$ to $+3$

Solution:
Decreases from $+6$ to $+3$.

Topic: Chemistry, Redox Reactions, VITEEE, VITEEE 2019
Acidified potassium permanganate solution is decolourised by

Options:
A Bleaching powder
B Microcosmic salt
C Mohr salt
D White vitriol

The correct answer is:
Mohr salt

Solution:
Mohr salt is $\left(\mathrm{FeSO}_4\right)\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 .6 \mathrm{H}_2 \mathrm{O}$.

Topic: Chemistry, s Block Elements, JEE Main
Acrilan fibre used for cloth, carpets and blankets, it the polymer of

Options:
A acrylonitrile
B ethylacrylate
C styrene
D monochlorotrifluoro ethane

The correct answer is:
acrylonitrile

Solution:
Acrilan fibre used for cloth, carpets and blankets, it the polymer of acrylonitrile
CH2=CH| CNAcrylonitrile→-CH2-CH| CN-nPolyacrylonitrile

Topic: Chemistry, Polymers, JEE Main
Across a metallic conductor of non-uniform cross section a constant potential difference is applied. The quantity which remains constant along the conductor is:

Options:
A Current
B Drift velocity
C Electric field
D Current density

The correct answer is:
Current

Solution:
Current flow across the metallic conductor remains constant because the area of cross-section of conductor is non-uniform. So, Current density will be different but the number of flow of electrons will be same.

Topic: Physics, Current Electricity, NEET, NEET 2015 (Phase 1)
Action and reaction forces act on

Options:
A The same body
B The different bodies
C The horizontal surface
D Nothing can be said

The correct answer is:
The different bodies

Solution:
The correct options are
B on different bodies
According to Newton's third law of motion, for every action there is an equal and opposite reaction. These action reaction forces act on two different bodies at the same time.

Topic: Physics, Laws of Motion, JEE Main
Action of soap is due to emulsification and micelle formation. Comment.

Options:

None

Solution:
Soap is sodium or potassium salt of a higher fatty acid and may be represented as \(\mathrm{RCOO}^{-} \mathrm{Na}^{+}\)(e.g., sodium stearate \(\mathrm{CH}_3\left(\mathrm{CH}_2\right)_{16} \mathrm{COO}^{-} \mathrm{Na}^{+}\), which is a major component of many bar soaps). When dissolved in water, it dissociates into \(R \mathrm{COO}^{-}\)and \(\mathrm{Na}^{+}\)ions. The \(\mathrm{RCOO}^{-}\)ions, however, consist of two parts - a long hydrocarbon chain \(R\) (also called non-polar 'tail') which is hydrophobic (water repelling), and a polar group \(\mathrm{COO}^{-}\)(also called polar- ionic 'head'), which is hydrophilic (water loving).
The \(\mathrm{RCOO}^{-}\)ions are, therefore, present on the surface with their \(\mathrm{COO}^{-}\)groups in water and the hydrocarbon chains \(R\) staying away from it and remain at the surface. But at critical micelle concentration, the anions are pulled into the bulk of the solution and aggregate to form a spherical shape with their hydrocarbon chains pointing towards the centre of the sphere with \(\mathrm{COO}^{-}\)part remaining outward on the surface of the sphere. An aggregate thus formed is known as 'ionic micelle'.
The cleansing action of soap is due to the fact that soap molecules form micelle around the oil droplet in such a way that hydrophobic part of the stearate ions is in the oil droplet and hydrophilic part projects out of the grease droplet like the bristles. Since the polar groups can interact with water, the oil droplet surrounded by stearate ions is now pulled in water and removed from the dirty surface. Thus soap helps in emulsification and washing away of oils and fats. The negatively charged sheath around the globules prevents them from coming together and forming aggregates.
(a) Grease on cloth
(b) Stearate ions (from soap) arranging around the grease droplets
(c) Micelle formed

Topic: Chemistry, Surface Chemistry
Action of water in the presence of sulphuric acid with the following alkenes produces:

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Activated sludge should have the ability to settle quickly so that it can

Options:
A
be rapidly pumped back from sedimentation tank to aeration tank

B
absorb pathogenic bacteria present in waste water while sinking to the bottom of the settling tank

C
be discarded and anaerobically digested

D
absorb colloidal organic matter

The correct answer is:
be rapidly pumped back from sedimentation tank to aeration tank

Solution:
$\ldots$ During secondary treatment of waste water, when $\mathrm{BOD}$ of sewage has reduced, the effluent is passed into setting tank. Here, the bacterial flocs settle and the sediment is called activated sludge.
In the aeration tank a small part of the sludge is used as an inoculum and the remaining part is passed into large tanks called anaerobic sludge digesters. That's why activated sludge should have the ability to settle quickly.

Topic: Biology, Microbes in Human Welfare
Activation energy $\left(E_a\right)$ and rate constants $\left(k_1\right.$ and $\left.k_2\right)$ of a chemical reaction at two different temperatures $\left(T_1\right.$ and $\left.T_2\right)$ are related by

Options:
A $\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$
B $\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$
C $\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}+\frac{1}{T_1}\right)$
D None of the above

The correct answer is:
$\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

Solution:
According to Arrhenius equation, activation energy $\left(E_a\right)$ and rate constants $\left(k_1\right.$ and $\left.k_2\right)$ of a chemical reaction at two different temperatures ( $T_1$ and $\left.T_2\right)$ are related as
$$
\begin{aligned}
\ln \frac{k_2}{k_1} & =\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \\
& =-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2012 (Mains)
Activation energy is given by the formula

Options:
A $\log \frac{K_2}{K_1}=\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_1 T_2}\right]$
B $\log \frac{K_1}{K_2}=-\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_1 T_2}\right]$
C $\log \frac{K_1}{K_2}=-\frac{E_a}{2.303 R}\left[\frac{T_1-T_2}{T_1 T_2}\right]$
D None of these

The correct answer is:
$\log \frac{K_1}{K_2}=-\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_1 T_2}\right]$

Solution:
\(\log _e K_2=\log _e A-\frac{E_A}{R T_2} \ldots \ldots \ldots \ldots . .(1)\)
\(\log _e K_1=\log _e A-\frac{E_A}{R T_1} \ldots \ldots \ldots \ldots . . . . .(2)\)
Eqn. (1) - (2) we get,
\(\log \frac{K_2}{K_1}=\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_1 T_2}\right]\)

Topic: Chemistry, Chemical Kinetics, JEE Main
Activities of three radioactive substances A, B and C are represented by the curves A, B and C, in the figure. Then their half-lives T12A:T12B:T12C are in the ratio:

Options:
A 2: 1: 1
B 3: 2: 1
C 2: 1: 3
D 4: 3: 1

The correct answer is:
2: 1: 3

Solution:
R=R0e-λt

lnR=-λ·nt+lnR0

Slope=ln2t1/2=λ

T1A:T1B:T1C=2:1:3

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2020 (05 Sep Shift 1)
Activity of a radioactive sample decreases to $\left(\frac{1}{3}\right)^{r d}$ of its original value in 3 days. Then in 9 days its activity reduces to

Options:
A $\left(\frac{1}{18}\right)$ of the original value
B $\left(\frac{1}{9}\right)$ of the original value
C $\left(\frac{1}{27}\right)$ of the original value
D $\left(\frac{1}{3}\right)$ of the original value

The correct answer is:
$\left(\frac{1}{27}\right)$ of the original value

Solution:
Activity after time $t$ given that half-life is $T$ :
$A=\frac{A_0}{2^{t / T}}$
$\frac{A_0}{3}=\frac{A_0}{2^{3 / T}}$
Activity in 9 days,
$A=\frac{A_0}{2^{t / T}}=\frac{A_0}{2^{9 / T}}=\frac{A_0}{\left(2^{3 / T}\right)^3}=\frac{A_0}{27}$

Topic: Physics, Nuclear Physics, JEE Main
Adaptation in organisms can be

Options:
A Morphological
B Physiological
C Behavioural
D All of these

The correct answer is:
All of these

Solution:
Adaptation is an attribute of the organism (morphological, physiological, behavioural) that enables the organism to survive and reproduce in its habitat. Many adaptations have evolved over a long evolutionary time and are genetically fixed. However, these may remain epigenetic (not fixed genetically).

Topic: Biology, Organisms and Populations, NEET
Adaptations to plants against dry environmental conditions is/are

Options:
A Stomata arranged in deep pits
B Stomata remain open during day time
C CAM pathway
D More than one option is correct

The correct answer is:
More than one option is correct

Solution:
Many desert plants have a thick cuticle on their leaf surfaces and have their stomata arranged in deep pits (sunken) to minimize water loss through transpiration. They also have a special photosynthetic pathway (CAM) that enables their stomata to remain closed during day time. Some desert plants like Opuntia, have no leaves and the photosynthetic function is taken over by the flattened stems.

Topic: Biology, Organisms and Populations, NEET
Adding of $\mathrm{Cl}_2$ to benzene in the presence of $\mathrm{AlCl}_3$ is an example of

Options:
A Addition
B Halogenation
C Substitution
D Elimination

The correct answer is:
Substitution

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Adding surfactants in non polar solvent, the micelles structure will look like

(a)

(b)

(c)

(d)

Options:
A

b

B

c

C

a

D d

The correct answer is:

a

Solution:
Micelles are formed by self-assembly of amphiphilic molecules. The structures contain hydrophilic/polar region (head) and hydrophobic/nonpolar region (tail). Micelles are formed in aqueous solution whereby the polar region faces the outside surface of the micelle and the nonpolar region forms the core.

Non-polar part will interact with non-polar solvent.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Addition of arsenic in small amounts to pure germanium will result in the formation of

Options:
A n-type semiconductor
B p-type semiconductor
C germanium arsenide
D a superconducting alloy

The correct answer is:
n-type semiconductor

Solution:
Arsenic in a group 15 element having '5' valence electrons. When arsenic is added to germanium, only four of five electrons involved in bonding. The remaining fifth electron remains loosely held with arsenic atom. Thus, each arsenic atom has one loosely held electron, which acts as charge carrier to form n-type semiconductor.

Topic: Chemistry, Solid State, NEET
Addition of bromine to 1, 3-butadiene gives

Options:
A 1,2 addition product only
B 1, 4 addition product only
C Both 1,2 and 1, 4 addition products
D No reaction

The correct answer is:
Both 1,2 and 1, 4 addition products

Solution:
1,3-butadiene accompanies the bromine molecule to form two products.
They are 1,2-add products and 1,4-surplus products.
At low temperatures, the 1,2-pair is predominant but at higher temperatures, the 1,4-pair is predominant. Products A and B will be on this basis.
\(\underset{\substack{\text{3,4-Dibromobut-1-ene} \\ \text{(1,2-addition product)} \\ \text{(A)}}}{\mathrm{H}_2 \mathrm{C}=\underset{\Large\mathrm{H}}{\mathrm{C}}-\underset{\substack{| \\ \Large \mathrm{Br}}}{\stackrel{\Large\mathrm{H}}{\mathrm{C}}}-\underset{\substack{| \\ \Large \mathrm{Br}}}{\mathrm{C}}\mathrm{H}_2}\)>
\(\underset{\substack{\text{1,4-Dibromobut-2-ene} \\ \text{(1,4-addition product)} \\ \text{(B)}}}{\mathrm{H}_2 \underset{\substack{| \\ \Large \mathrm{Br}}}{\mathrm{C}}-\underset{\Large\mathrm{H}}{\mathrm{C}}=\underset{\Large\mathrm{H}}{\mathrm{C}}-\underset{\substack{| \\ \Large \mathrm{Br}}}{\mathrm{C}}\mathrm{H}_2}\)

Topic: Chemistry, Hydrocarbons, JEE Main
Addition of CdCl2 to AgCl yields solid solutions, where the divalent cations Cd2+ occupy the Ag+ sites. Which one of the following statement in true?

Options:
A The number of cationic vacancies is equal in number to that of divalent ions added
B The number of cationic vacancies is one-half of the number of that of divalent ions added
C The number of anionic vacancies is equal in number to that of divalent ions added
D No cationic or anionic vacancies are produced

The correct answer is:
The number of cationic vacancies is equal in number to that of divalent ions added

Solution:
To maintain the electrical neutrality, the number of cationic vacancies is equal to the number of divalent ions added.

Topic: Chemistry, Solid State, NEET
Addition of excess aqueous ammonia to a pink coloured aqueous solution of MCl2.6H2OX and NH4Cl gives an octahedral complex Y in the presence of air. In aqueous solution, complex Y behaves as 1:3 electrolyte. The reaction of X with excess HCl at room temperature results in the formation of a blue coloured complex Z. The calculated spin only magnetic moment of X and Z is 3.87 B.M., whereas it is zero for complex Y.

Among the following options, which statement is incorrect?

Options:
A The hybridization of the central metal ion in Y is d2sp3
B Addition of solver nitrate to Y gives only two equivalents of silver chloride
C When X and Z are in equilibrium at 0°C, the colour of the solution is pink
D Z is a tetrahedral complex

The correct answer is:
Addition of solver nitrate to Y gives only two equivalents of silver chloride

Solution:
X and Z are in equilibrium at 0oC , equilibrium is shifted towards X, making colour of solution pink.

Topic: Chemistry, Coordination Compounds, JEE Main
Addition of inert gas to system

N2(g)+3H2(g)⇌2NH3(g)

at equilibrium at constant volume. Then

Options:
A N2 and H2 are formed in abundance
B N2 , H2 and NH3 will have the same molar concentration
C The production of ammonia increases
D No change in the equilibrium

The correct answer is:
No change in the equilibrium

Solution:
At constant volume on adding inert gas partial pressure of all gases remain unchanged, hence no change in equilibrium constant.

Topic: Chemistry, Chemical Equilibrium, NEET
Addition of phosphate fertilizers to water bodies causes:

Options:
A Deposition of calcium phosphate
B Enhanced growth of algae
C Increase in fish population
D Increase in amount of dissolved oxygen in water

The correct answer is:
Enhanced growth of algae

Solution:
Addition of phosphate fertilizers or other fertilizers to water bodies causes enrichment in nutrients, this phenomenon is called Eutrophication. Increase in nutrients enhanced growth of algae (vegetation) which cause consume of dissolved oxygen. Hence decrease in quantity of dissolved oxygen, effects the life of aqua-animals or other vegetation. So Eutrophication is a types of water pollution.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2015 (11 Apr Online)
Addition of sodium hydroxide solution to a weak acid (HA) results in a buffer of pH 6. If ionization constant of HA is 10-5, the ratio of salt to acid concentration in the buffer solution will be:

Options:
A 10 : 1
B 4 : 5
C 5 : 4
D 1 : 10

The correct answer is:
10 : 1

Solution:
pH=pka+log  Salt Acid (∵ [Salt]=[Anion])
⇒6=5+log⁡ SaltAcid
⇒1=log ⁡SaltAcid
⇒log10=log ⁡SaltAcid
Saltacid=101

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2017 (08 Apr Online)
Adenine is one of the two purine bases involved in the formation of nucleotides of the nucleic acids. The molecular formula of adenine is CxHxNx the value of 'x' is

Options:

The correct answer is:
5

Solution:
The molecular formula of adenine is C5H5N5 . So, value of 'x' is 5 .

Topic: Chemistry, Biomolecules, JEE Main