datavorous/entrance-exam-dataset huggingface dataset:
Which of the following options describes a method used for the identification of primary, secondary and tertiary amines correctly?

Options:
A In the Hinsberg Test, the amines act as nucleophiles and attack the electrophile (sulphonyl chloride). This leads to the displacement of the chloride and the generation of the sulphonamides.Primary amine sulphonamides are soluble in alkali whereas secondary amine sulphonamides are not soluble in alkali.No such reaction takes place for tertiary amines.
B The Hinsberg test is a quantitative test used to determine the concentration of amines in a solution.
C The Hinsberg test is a method for the synthesis of amines from aromatic compounds.
D The Hinsberg test is a titration method used to measure the acidity of amines.

The correct answer is:
In the Hinsberg Test, the amines act as nucleophiles and attack the electrophile (sulphonyl chloride). This leads to the displacement of the chloride and the generation of the sulphonamides.Primary amine sulphonamides are soluble in alkali whereas secondary amine sulphonamides are not soluble in alkali.No such reaction takes place for tertiary amines.

Solution:
Correct Option is : (A)
In the Hinsberg Test, the amines act as nucleophiles and attack the electrophile (sulphonyl chloride). This leads to the displacement of the chloride and the generation of the sulphonamides.Primary amine sulphonamides are soluble in alkali whereas secondary amine sulphonamides are not soluble in alkali.No such reaction takes place for tertiary amines.

Topic: Chemistry, Amines, NEET
Which of the following options is incorrect about bond length?

Options:
A F2
B HCl2
C HBr
2
D Hl2

The correct answer is:
F2

Solution:
Correct Option is : 1
F2

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Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Which of the following options is the correct description of Carbylamine reaction?

Options:
A The Carbylamine reaction involves the identification of a primary amine by treating it with chloroform and a strong base, resulting in the release of a foul-smelling isocyanide.
B The Carbylamine reaction is a method for converting primary amines into secondary amines through a series of chemical reactions involving nitrous acid.
C The Carbylamine reaction is a test used to distinguish between primary, secondary, and tertiary amines based on their reaction with bromine water.
D The Carbylamine reaction is a synthesis method used to prepare aromatic amines from corresponding nitro compounds.

The correct answer is:
The Carbylamine reaction involves the identification of a primary amine by treating it with chloroform and a strong base, resulting in the release of a foul-smelling isocyanide.

Solution:
Correct Option is : (A)
The Carbylamine reaction involves the identification of a primary amine by treating it with chloroform and a strong base, resulting in the release of a foul-smelling isocyanide.

Topic: Chemistry, Amines, NEET
Which of the following options is the correct description of Diazotization?

Options:
A Diazotization is a chemical process used to convert primary amines into secondary amines by treating them with nitrous acid.
B Aromatic primary amines react with nitrous acid (prepared in situ from NaNO2and a mineral acid such as HCl) at low temperatures (273-278 K) to form diazonium salts. This conversion of aromatic primary amines into diazonium salts is known as diazotization.
C Diazotization is a method used to selectively oxidize secondary amines to their corresponding ketones.
D Diazotization is a process in which tertiary amines are converted into quaternary ammonium salts by reacting them with alkyl halides.

The correct answer is:
Aromatic primary amines react with nitrous acid (prepared in situ from NaNO2and a mineral acid such as HCl) at low temperatures (273-278 K) to form diazonium salts. This conversion of aromatic primary amines into diazonium salts is known as diazotization.

Solution:
Correct Option is : (B)
Aromatic primary amines react with nitrous acid (prepared in situ from NaNO2and a mineral acid such as HCl) at low temperatures (273-278 K) to form diazonium salts. This conversion of aromatic primary amines into diazonium salts is known as diazotization.

Topic: Chemistry, Amines, NEET
Which of the following options is the correct description of Hofmann’s bromamide reaction?

Options:
A Hofmann's bromamide reaction is a method used to convert primary amines into secondary amines by treating them with bromine in the presence of a strong base.
B When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, a primary amine with one carbon atom less than the original amide is produced. This degradation reaction is known as Hoffmann bromamide reaction.
C Hofmann's bromamide reaction is a method for converting secondary amines into tertiary amines by reacting them with bromine and a strong acid.
D Hofmann's bromamide reaction is a chemical process used to selectively oxidize primary alcohols to carboxylic acids using bromine and heat.

The correct answer is:
When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, a primary amine with one carbon atom less than the original amide is produced. This degradation reaction is known as Hoffmann bromamide reaction.

Solution:
Correct Option is : (B)
When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, a primary amine with one carbon atom less than the original amide is produced. This degradation reaction is known as Hoffmann bromamide reaction.

Topic: Chemistry, Amines, NEET
Which of the following organisms are indicator of SO2 pollution?

Options:
A Mosses
B Puffballs
C Lichens
D Mushrooms

The correct answer is:
Lichens

Solution:
Lichens are widely used as environmental indicators or bio-indicators. If air is very badly polluted with sulphur dioxide there may be no lichens present, just green algae may be found. If the air is clean, shrubby, hairy and leafy lichens become abundant.

Topic: Biology, Biological Classification, NEET
Which of the following pairs are correctly matched?
\(\begin{array}{|l|l|l|}
\hline & \text { Animals } & \begin{array}{c}
\text { Morphological } \\
\text { feature }
\end{array} \\
\hline \text { (i) } & \text { Crocodile } & \begin{array}{l}
\text { 4-chambered } \\
\text { heart }
\end{array} \\
\hline \text { (ii) } & \text { Sea urchin } & \text { Parapodia } \\
\hline \text { (iii) } & \text { Obelia } & \text { Metagenesis } \\
\hline \text { (iv) } & \text { Lemur } & \text { Thecodont } \\
\hline
\end{array}\)

Options:
A (ii), (iii) and (iv)
B only (i) and (iv)
C only (i) and (ii)
D (i), (iii) and (iv).

The correct answer is:
(i), (iii) and (iv).

Solution:
Crocodile belongs to Reptilia class which has four chambered heart. Lemur teeth are embedded in the sockets of two which is known as thecodont. In Obelia, polyps reproduce medusae asexually and medusae form the polyps sexually. Such alternation of sexual phases in the life cycle is called metagenesis. Parapodia are present in Annelida, while sea urchin is a Echinodermata.

Topic: Biology, Animal Kingdom, NEET, NEET 2007
Which of the following pairs of an animal and a plant represents endangered organisms in India?

Options:
A banyan and black duck
B Bentinckia nicobarica and red panda
C tamarind and rhesus monkey
D Cinchona and leopard

The correct answer is:
Cinchona and leopard

Solution:
Endangered species is a plant or animal species defined by IUCN as being in immediate danger of extinction because its members have reached a critical level or their habitats have been drastically reduced. A plant, Bentinckia condapanna/nicobarica (member of family Arecaceae) and the animal, Red panda, both are declared as endangered in India. Tamarind, Banyan etc. are not endangered in India.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2006
Which of the following pairs of organisms are correctly matched with their description?

1. Pteridophytes which are heterosporous: Selaginella, Salvinia

2. Gymnosperms with branched stem: Pinus, Cedrus

3. Algae with haplo-diplontic life cycle: Ectocarpus, Polysiphonia

4. Bryophytes with protonema stage: Polytrichum, Sphagnum

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

The correct answer is:
1, 2, 3, 4

Solution:
In majority of the pteridophytes all the spores are of similar kinds; such plants are called homosporous. Genera like Selaginella and Salvinia which produce two kinds of spores, macro (large) and micro (small) spores, are known as heterosporous. The stems of gymnosperms are unbranched (Cycas) or branched (Pinus, Cedrus). The predominant stage of the life cycle of a moss is the gametophyte which consists of two stages. The first stage is the protonema stage, which develops directly from a spore. It is a creeping, green, branched and frequently filamentous stage. The second stage is the leafy stage, which develops from the secondary protonema as a lateral bud. Common examples of mosses are Funaria, Polytrichum and Sphagnum. While most algal genera are haplontic, some of them such as Ectocarpus, Polysiphonia, kelps are haplo-diplontic. Fucus, an alga is diplontic.

Topic: Biology, Plant Kingdom, NEET
Which of the following parts of the plant is used as commercial wood?

Options:
A Primary xylem
B Primary phloem
C Secondary xylem
D Secondary phloem

The correct answer is:
Secondary xylem

Solution:
Wood is a porous and fibrous structural tissue found in the stems and roots of trees and other woody plants. It is an organic material - a natural composite of cellulose fibers that are strong in tension and embedded in a matrix of lignin that resists compression, this is the product of old secondary xylem

Topic: Biology, Anatomy of Flowering Plants, NEET
Which of the following phase include meiosis during spermatogenesis?

Options:
A Phase of multiplication
B Phase of maturation division
C Phase of growth
D Spermiation

The correct answer is:
Phase of maturation division

Solution:
Some spermatogonia (termed as primary spermatocytes) periodically undergo meiosis.
A primary spermatocyte completes meiosis I (reduction division) and gives rise to 2 equal-sized cells called secondary spermatocytes (n = 23). Both secondary spermatocytes undergo meiosis II (equational division) to produce 2 equal spermatids (n) each.

Topic: Biology, Human Reproduction, NEET
Which of the following phenols is isolated from defensive secretion of grasshopper species?

Options:
A 4 - Methylphenol
B Benzene-1, 3- diol
C 2, 5 - dichlorophenol
D Phenol

The correct answer is:
2, 5 - dichlorophenol

Solution:
The defensive froth emitted by the grasshopper Romalea microptera contains several odorous compounds (phenols, terpenes, benzoquinone), including a chlorinated aromatic compound, 2,5-dichlorophenol. This compound, which is repellent to ants and therefore defensively useful to the grasshopper, probably stems from herbicide or herbicide derivative ingested by the insect with its diet. Although there is precedent for the defensive employ by one species of chemical agents produced by another, no instance was known involving secondary utilization of a pesticide dispensed by man.

Topic: Chemistry, Alcohols Phenols and Ethers, MHT CET, MHT CET 2020 (13 Oct Shift 2)
Which of the following primary and secondary valencies are not correctly marked against the compound?
[ p = Primary valency; s = Secondary valency]

Options:
A [Cr(NH3)6]Cl3,p=3,s=6
B K2[PtCl4],p=2,s=4
C [Pt(NH3)2Cl2],p=2,s=4
D [Cu(NH3)4]SO4,p=4,s=4

The correct answer is:
[Cu(NH3)4]SO4,p=4,s=4

Solution:
In [Cu(NH3)4]SO4 primary valency is 2 and secondary valency is 4. In rest other complexes the values of primary and secondary valencies are correctly given.

Topic: Chemistry, Coordination Compounds, NEET
Which of the following properties are not seen in Funaria?

I. They have unicellular rhizoids.
II. The gametophytic phase of life cycle includes protonema and leafy stage
III. They are higher members of Bryophyta.
IV. Sporophyte consists of root, stem, and leaves.

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

The correct answer is:
I and IV

Solution:
Funaria belongs to division bryophyte. It is a type of moss. Bryophytes include liverworts and mosses. Out of these, mosses are more evolved members of bryophytes. Their plant body is not differentiated in root, stem and leaves. The gametophyte has two stages: The first stage is the protonema stage, which develops directly from a spore. It is a creeping, green, branched and frequently filamentous stage. The second stage is the leafy stage, which develops from the secondary protonema as a lateral bud. They consist of upright, slender axes bearing spirally arranged leaves. They are attached to the soil through multicellular and branched rhizoids. This stage bears sex organs. The sporophyte is made up of capsule, seta and foot.

Topic: Biology, Plant Kingdom, NEET
Which of the following rare gas is the abundant in air?

Options:
A He
B Ne
C Ar
D Kr

The correct answer is:
Ar

Solution:
The $\%$ by volume of rare gases are
\begin{array}{ccc}
\hline & \% by volume & \% by weight \\
\hline He & 0.0005 & 0.000037 \\
\hline Ne & 0.0015 & 0.001 \\
\hline Ar & 0.932 & 1.285 \\
\hline Kr & 0.0001 & 0.00028 \\
\hline Xe & 0.0001 & 0.00004 \\
\hline
\end{array}
So, Ar is most abundant in air.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JIPMER, JIPMER 2019
Which of the following reaction is not feasible?

Options:
A

B

C

D

The correct answer is:

Solution:
Reaction in option (b) is not feasible as it has a primary alcohol and the dehydration of a $1^{\circ}$ alcohol is preferable at temperatures of about $443 \mathrm{~K}$.
Secondary and tertiary alcohols are able to undergo dehydration at mild conditions too.

Topic: Chemistry, Alcohols Phenols and Ethers, AP EAMCET, AP EAMCET 2023 (15 May Shift 2)
Which of the following reactions is not feasible based on the standard electrode potentials?

Options:
A FeBr2(s)
B No Reaction
C Fe2Br3(s)
D FeBr3(s)

The correct answer is:
No Reaction

Solution:
Correct Option is : (B)
No Reaction

Topic: Chemistry, Electrochemistry, NEET
Which of the following reactions is not feasible based on the standard electrode potentials?

Options:
A Fe2+(aq) + Ag+(aq)
B No Reaction
C Fe(s) + Ag+(aq)
D AgFe(s)

The correct answer is:
Fe(s) + Ag+(aq)

Solution:
Correct Option is : (C)
Fe(s) + Ag+(aq)

Topic: Chemistry, Electrochemistry, NEET
Which of the following reactions is not feasible based on the standard electrode potentials? Br2(aq) + Fe2+(aq) → ?

Options:
A FeBr2(s)
B No Reaction
C FeBr3(s)
D Fe(s) + Br2(aq)

The correct answer is:
No Reaction

Solution:
Correct Option is : (B)
No Reaction

Topic: Chemistry, Electrochemistry, NEET
Which of the following reactions is not feasible based on the standard electrode potentials?

Options:
A Cu2+(aq) + Ag(s)
B No Reaction
C Cu+(aq) + Ag(s)
D Ag2Cu(s)

The correct answer is:
Cu+(aq) + Ag(s)

Solution:
Correct Option is : (C)
Cu+(aq) + Ag(s)

Topic: Chemistry, Electrochemistry, NEET
Which of the following reactions proceeds via a secondary free radical?

Options:
A

B $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2 \underset{\text { UV light }}{\stackrel{\mathrm{HBr}}{\longrightarrow}}$ $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Br}$
C $\mathrm{C}_6 \mathrm{H}_6 \stackrel{\mathrm{Br}_2 / \mathrm{FeBr}_3}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_5 \mathrm{Br}$
D $\mathrm{C}_6 \mathrm{H}_6 \underset{\text { UV light }}{\stackrel{\mathrm{Br}_2}{\longrightarrow}} \mathrm{CH}_3-\mathrm{CH}_2 \mathrm{Br}$

The correct answer is:
$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2 \underset{\text { UV light }}{\stackrel{\mathrm{HBr}}{\longrightarrow}}$ $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Br}$

Solution:
$\mathrm{H}-\mathrm{Br} \stackrel{\mathrm{UV}}{\longrightarrow} \dot{\mathrm{H}}+\mathrm{Br}$.
$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2+\mathrm{B} \dot{\mathrm{r}} \longrightarrow \mathrm{CH}_3-\mathrm{CH}-\mathrm{CH}_2 \mathrm{Br}$
$\mathrm{CH}_3 \mathrm{CH}-\mathrm{CH}_2 \mathrm{Br} \stackrel{\dot{\mathrm{H}}}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}$

Topic: Chemistry, Hydrocarbons, AP EAMCET, AP EAMCET 2003
Which of the following reactions proceeds via a secondary free radical?

Options:
A

B $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2 \underset{\text { UV light }}{\stackrel{\mathrm{HBr}}{\longrightarrow}}$ $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Br}$
C $\mathrm{C}_6 \mathrm{H}_6 \stackrel{\mathrm{Br}_2 / \mathrm{FeBr}_3}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_5 \mathrm{Br}$
D $\mathrm{C}_6 \mathrm{H}_6 \underset{\text { UV light }}{\stackrel{\mathrm{Br}_2}{\longrightarrow}} \mathrm{CH}_3-\mathrm{CH}_2 \mathrm{Br}$

The correct answer is:
$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2 \underset{\text { UV light }}{\stackrel{\mathrm{HBr}}{\longrightarrow}}$ $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Br}$

Solution:
$\mathrm{H}-\mathrm{Br} \stackrel{\mathrm{UV}}{\longrightarrow} \dot{\mathrm{H}}+\mathrm{Br}$.
$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2+\mathrm{B} \dot{\mathrm{r}} \longrightarrow \mathrm{CH}_3-\mathrm{CH}-\mathrm{CH}_2 \mathrm{Br}$
$\mathrm{CH}_3 \mathrm{CH}-\mathrm{CH}_2 \mathrm{Br} \stackrel{\dot{\mathrm{H}}}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}$

Topic: Chemistry, Hydrocarbons, TS EAMCET, TS EAMCET 2003
Which of the following reactions will NOT give primary amine as the product?

Options:
A CH3CONH2→(ii) H3O+(i) LiAlH4Product
B CH3CONH2→Br2/KOHProduct
C CH3CN→(ii) H3O+(i) LiAlH4Product
D CH3NC→(ii) H3O+(i) LiAlH4Product

The correct answer is:
CH3NC→(ii) H3O+(i) LiAlH4Product

Solution:
Lithium aluminium hydride is a strong reducing agent. It converts amides, cyanides and isocyanides to amines.

CH3CONH2→(ii) H3O+(i) LiAlH4CH3CH2NH2primary amine

CH3CN→(ii) H3O+(i) LiAlH4CH3CH2NH2Primary amine

CH3NC→(ii) H3O+(i) LiAlH4CH3NHCH3Secondary amine

In Hoffmann bromamide reaction, primary amides converts to primary amines with loss of one carbon.

CH3CONH2→Br2/KOHCH3NH2

Topic: Chemistry, Amines, NEET, NEET 2023 (All India)
Which of the following represents fundamental forces of nature?

Options:
A Gravitational force; Coulomb's force; Strong surface tension force; Weak Vander Waal's force
B Gravitational force; Electromagnetic force; Strong viscous force; Weak nuclear force
C Gravitational force; Magneto static force; Strong nuclear force; Weak frictional force
D Gravitational force; Electromagnetic force; Strong nuclear force; Weak nuclear force

The correct answer is:
Gravitational force; Electromagnetic force; Strong nuclear force; Weak nuclear force

Solution:
Four fundamental forces are,

1 The strong nuclear force which is short-range force acts between nucleon.

2 The weak nuclear force which is also short-range force responsible for radioactive decay law.

3 The electromagnetic force having infinite range.

4 Gravitational force which is long-range force and weakest force in nature.

Topic: Physics, Units and Dimensions, TS EAMCET, TS EAMCET 2019 (03 May Shift 2)
Which of the following represents the correct combination without any exception?

Options:
A

Characteristics	Class
Mouth ventral; 5 -7 pairs of gills; skin with placoid scales; persistent notochord	Chondrichthyes

B

Characteristics	Class
Sucking and circular mouth; jaws present, integument without scales; paired appendages	Cyclostomata

C

Characteristics	Class
Body covered with feathers; skin moist and glandular; hindlimbs form wings; lungs with air sacs	Aves

D

Characteristics	Class

Mammary gland; hair on body; pinnae; two pairs of limbs	Mammalia

The correct answer is:

Characteristics	Class

Mammary gland; hair on body; pinnae; two pairs of limbs	Mammalia

Solution:
Class Pisces is divided into two groups: Chondrichthyes and Osteichthyes. Chondrichthyes are cartilaginous fishes. These have a ventral mouth, placoid scales while air bladder and operculum are absent. Thus they differ from Osteichthyes. These do not show 5-7 pairs of gill slits. Cyclostomatas are chordates with circular mouth but jaws are absent and hence are known as Agnathas. Class Aves belongs to tetrapods in which fore limbs are modified into wings. All mammals have mammary glands, hair on body and two pairs of limbs. All mammals have pinnae except aquatic mammals.

Topic: Biology, Animal Kingdom, NEET
Which of the following secondary metabolites are wrongly matched with their categories?

A. Toxin - Ricin
B. Drug - Curcumin
C. Alkaloid - Codeine
D. Lectin - Abrin

Options:
A Only A
B A and B both
C A and C both
D Only D

The correct answer is:
Only D

Solution:
List of secondary metabolites:
Toxins - abrin, ricin
Alkaloids - morphine, codeine
Drugs - vinblastine, curcumin
Lectin - concanavalin A
Abrin is a toxin, not a lectin. Lectins are a class of proteins mostly of plant origin which bind to specific sugars and cause agglutination. One such example is concanavalin A.

Topic: Chemistry, Biomolecules, NEET
Which of the following set of data has least standard deviation?

Options:
A 10,20,30,40
B 2,4,6,8
C 3,6,9,12
D 1,2,3,4

The correct answer is:
1,2,3,4

Solution:
We know that, Standard deviation measures the spread of a data distribution. The more spread out a data distribution is, the greater its standard deviation.

Here, from the options we see that the set 1, 2, 3, 4 has the least spread as data values are least far from their mean (Mean=2.5)

Hence, it has least Standard Deviation.

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2021 (19 Aug Shift 1)
Which of the following statement, about the advantage of roasting of sulphide ore before reduction is not true?

Options:
A The $\Delta \mathrm{G}_f^{\text {o }}$ of the sulphide is greater than those for $\mathrm{CS}_2$ and $\mathrm{H}_2 \mathrm{~S}$.
B The $\Delta \mathrm{G}_f^0$ is negative for roasting of sulphide ore to oxide
C Roasting of the sulphide to the oxide is thermodynamically feasible
D Carbon and hydrogen are suitable reducing agents for metal sulphides

The correct answer is:
The $\Delta \mathrm{G}_f^{\text {o }}$ of the sulphide is greater than those for $\mathrm{CS}_2$ and $\mathrm{H}_2 \mathrm{~S}$.

Solution:
The sulphide ore is roasted to oxide before reduction because the $\Delta \mathrm{G}$ of most of the sulphides are greater than those of $\mathrm{CS}_2$ and $\mathrm{H}_2 \mathrm{~S}$, therefore neither $\mathrm{C}$ nor $\mathrm{H}$ can reduce metal sulphide to metal. Further, the standard free energies of formation of oxide are much less than those of $\mathrm{SO}_2$. Hence oxidation of metal sulphides to metal oxide is thermodynamically favourable.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET, NEET 2007
Which of the following statement is correct?

Options:
A $\Delta G$ is equal to $\Delta G^{\circ}$ when the system is at the standard state.
B $\Delta G^{\circ}$ is zero when the system is at equilibrium.
C $\Delta G$ measures activation energy of a reaction.
D When $\Delta G$ is positive, the reaction should proceed forward to form more product.

The correct answer is:
$\Delta G$ is equal to $\Delta G^{\circ}$ when the system is at the standard state.

Solution:
(a) $\Delta G=\Delta G^{\circ}$ for a standard system of STP condition $\left(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)$.
The corrected statements of options-(a), (c) and (d) will be like,
b) At equilibrium state, $\Delta G=0$.
(c) $\Delta G$ measures spontaneity of a reaction.
d) When $\Delta G>0$, the reaction will proceed in backward direction.

Topic: Chemistry, Thermodynamics (C), TS EAMCET, TS EAMCET 2020 (14 Sep Shift 2)