datavorous/entrance-exam-dataset huggingface dataset:
People living at sea level have around 5 million RBC per cubic millimetre of their blood, whereas those living at an altitude of 5400 metres have around 8 million. This is because at high altitude,

Options:
A Atmospheric O2 level is less and hence more RBCs are needed to absorb the required amount of O2 to survive
B There is more UV radiation which enhances RBC production
C People eat more nutritive food, therefore more RBCs are formed
D People get pollution - free air to breathe and more oxygen is available.

The correct answer is:
Atmospheric O2 level is less and hence more RBCs are needed to absorb the required amount of O2 to survive

Solution:
At the altitude of 5400 meters, the low atmospheric pressure of O2 will be too low so the solubility of oxygen in the blood will be very less hence the oxygen carried by each RBC will be too less. However, to fulfill the oxygen requirement of the body blood has to carry more oxygen to the body tissue, and this done by the increased number of RBCs.

Topic: Biology, Breathing and Exchange of Gases, NEET
People suffering from albinism cannot synthesize:

Options:
A Suberin
B Melanin
C Keratin
D Collagen

The correct answer is:
Melanin

Solution:
In albinism, genetic changes may cause a person to have light colored hair and skin. Albinism is an inherited disease characterized by a substantially lower rate of melanin production. Melanin is the pigment responsible for the color of the skin, and hair.

Topic: Biology, Principles of Inheritance and Variation, NEET
People who are known to be lactose intolerant do not produce to the lactose digesting enzyme, lactase. This happens because the lactase producing gene is shut off in their cells connected to the digestive system. Which of the following process is not occurring for the mentioned gene?

Options:
A Translation
B Mutation
C Replication
D Transcription

The correct answer is:
Transcription

Solution:
Gene regulation is the mechanism of controlling expression of genes in a cell's DNA. For a gene to work and produce its protein, it must get transcribed. If it is not transcribed then its protein will not be formed. Similarly, if lactase is not formed in a lactose intolerant individual that means the gene for making the enzyme is not getting transcribed in the intolerant person.

Topic: Biology, Molecular Basis of Inheritance, NEET
People who have migrated from the planes to an area adjoining Rohtang Pass about six months back

Options:
A have more RBCs and their haemoglobin has a lower binding affinity to $\mathrm{O}_2$
B are not physically fit to play games like football
C suffer from altitude sickness with symptoms like nausea, fatigue, etc.
D have the usual RBC count but their haemoglobin has very high binding affinuty to $\mathrm{O}_2$

The correct answer is:
have more RBCs and their haemoglobin has a lower binding affinity to $\mathrm{O}_2$

Solution:
As a person moves up a hill the $p \mathrm{O}_2$ and total atmospheric pressure decreases. Decrease in $p \mathrm{O}_2$ due to increasing altitude, stimulates the JG-cells of kidney to secrete erythropoietin hormone, increasing the number of RBCs (polycythemia) to compensate the supply of $\mathrm{O}_2$. At higher altitude, haemoglobin has lower binding affinity to $\mathrm{O}_2$ because the primary factor responsible for binding is $p_2$, which decreases at higher altitude.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2012 (Screening)
People with Kleinfelter's syndrome have 47 chromosomes, including three sex chromosomes (XXY). What is the term to describe the aberration that occurs during meiosis that results in abnormal chromosome number?

Options:
A Crossing over
B Non-disjunction
C Independent assortment
D Pairing of homologous chromosome.

The correct answer is:
Non-disjunction

Solution:
Non-disjunction is the failure of homologous chromosomes or sister chromatids to separate properly during cell division. There are three forms of non-disjunction: failure of a pair of homologous chromosomes to separate in meiosis I, failure of sister chromatids to separate during meiosis II, and failure of sister chromatids to separate during mitosis. Non-disjunction results in daughter cells with abnormal chromosome numbers (aneuploidy).

Topic: Biology, Principles of Inheritance and Variation, NEET
Peptide synthesis inside a cell takes place in

Options:
A mitochondria
B chromoplast
C ribosomes
D chloroplast

The correct answer is:
ribosomes

Solution:
The cellular factory responsible for synthesizing proteins (peptide synthesis) is the ribosome.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2011 (Screening)
Peptization is a process of

Options:
A precipitating colloidal particles
B purifying colloidal particles
C dispersing the precipitate into colloidal state
D None of the above

The correct answer is:
dispersing the precipitate into colloidal state

Solution:
The phenomenon of converting freshly precipitated mass into colloidal state by the action of solute or solvent is called peptization. For e.g., a colloid solution of ferric hydroxide is prepared by adding ferric chloride in it.

Topic: Chemistry, Surface Chemistry, NEET
Percent by mass of solute (molar mass = 25 g/mol) in its aqueous solution is 30. The mole fraction and molality of the solute in solution are

Options:
A Mole fraction=0.18 ; Molality=17.9
B Mole fraction=0.236 ; Molality=17.14
C Mole fraction=0. 236 ; Molality=15.96
D Mole fraction=0.38 ; Molality=17.14

The correct answer is:
Mole fraction=0.236 ; Molality=17.14

Solution:
Basis=100 gm solution
wt of solute=30 gm ; wt of water=70 gm
∴molality=3025× 10370=120070=1207
=17.14
moles of solute=3025=65
moles of H2O=7018=359
∴mole fraction=6/565+359=6/554+17545=6/522945
=65 × 45229=54229=0.236

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Percentage of free space in a body centred cubic unit cell is

Options:
A $30 \%$
B $32 \%$
C $34 \%$
D $28 \%$

The correct answer is:
$32 \%$

Solution:
Key Idea :
Packing fraction $=\frac{\text { Volume occupied by atoms in a unit cell }}{\text { Volume of the unit cell }}$
For body centred cube,
Packing fraction $=\frac{2 \times \frac{4}{3} \pi r^3}{\left(\frac{4 r}{\sqrt{3}}\right)^3}=\frac{\sqrt{3} \pi}{8}=0.68$
where, edge length $a=\frac{4 r}{\sqrt{3}}$
$\therefore$ Volume occupied $=68 \%$
and volume vacant $=32 \%$

Topic: Chemistry, Solid State, NEET, NEET 2008 (Screening)
Percentage of offsprings with genotype aaBbCc from a cross Aabbcc and AaBBCC would be:

Options:
A 50%
B 25%
C 12.5%
D 37.5%

The correct answer is:
25%

Solution:
AaBBCC×Aabbcc
aaBbCc=aa=14, Bb=44, Cc=44
i.e., 14×44×44=1664=14×100=25%

Topic: Biology, Principles of Inheritance and Variation, NEET
Pericardium is

Options:
A Double walled membranous bag
B Single walled membranous bag
C Double walled cartilaginous cavity
D Double walled coelomic cavity

The correct answer is:
Double walled membranous bag

Solution:
The pericardium is a tough double-layered fibro serous sac which covers the heart. The space between the two layers of the serous pericardium, the pericardial cavity, is filled with serous fluid which protects the heart from any kind of external jerk or shock.

Topic: Biology, Body Fluids and Circulation, NEET
Pericarp is formed of

Options:
A Endosperm
B Ovary wall
C Tapetum
D Epidermis

The correct answer is:
Ovary wall

Solution:
In fleshy fruits, pericarp forms the outer layer. This tissue develops from the ovary wall of the flower and surrounds the seed to protect it.
Endosperm is the food laden tissue, epidermis is the outer layer in the plants and tapetum is the layer that sorrounds the sporogenous tissues in anther.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Pericycle is present in

Options:
A monocot stem, but absent in dicot stem.
B dicot stem, but absent in monocot stem.
C monocot root, dicot root and dicot stem, but absent in monocot stem.
D monocot stem, but absent in monocot root, dicot root and dicot stem.

The correct answer is:
monocot root, dicot root and dicot stem, but absent in monocot stem.

Solution:
The pericycle is present in both dicot and monocot roots, especially in monocots it produces the lateral roots. The pericycle also gives rise to, cork cambium and the part of the vascular cambium. It is absent in monocot stem but present in the dicot stem.

Topic: Biology, Anatomy of Flowering Plants, NEET
Perigynous flower and diadelphous condition are found in the family

Options:
A Papilionaceae
B Caesalpinoidae
C Mimosoidae
D Solanaceae

The correct answer is:
Papilionaceae

Solution:
The tendency of cohesion of stamen results in diadelphous condition. In perigynous flowers, a hypanthium (a floral tube formed from the fusion of the stamens, petals, and sepals) is attached to the receptacle below the gynoecium and surrounds the ovary; the ovary is superior, and the free parts of the petals, sepals, and stamens are attached to the rim of the hypanthium. Perigynous flower and diadelphous condition are found in the family Papilionaceae.

Topic: Biology, Morphology of Flowering Plants, NEET
Perimetrium layer of the uterus is

Options:
A Thin and muscular
B Thick and glandular
C Thin and membranous
D Thick and fibrous

The correct answer is:
Thin and membranous

Solution:
3 Layers are found in the wall of uterus:
1. Perimetrium : External thin membranous layer.
2. Myometrium : Middle thick layer of smooth muscles which exhibits strong contraction duringpartiturition.
3. Endometrium: Inner glandular layer lining the uterine cavity that undergoes cyclical changes duringmenstrual cycle.

Topic: Biology, Human Reproduction, NEET
Periodic acid splits glucose and fructose into formaldehyde and formic acid, Ratio of moles of formic acid in glucose and fructose is

Options:
A 1:2
B 5:3
C 1:1
D 2:3

The correct answer is:
5:3

Solution:
CHO-CHOH-CHOHglucose-CHOH-CHOH-CH2OHâ5HIO45HCOOH+HCHO+5HIO3
CH 2 OH− C O || − CHOH fructose −CHOH−CHOH− CH 2 OH → 5 HIO 4 3HCOOH+2HCHO+5 HIO 3 + CO 2 + H 2 O
So molar ratio is 5:3

Topic: Chemistry, Biomolecules, NEET
Periodic classification of elements based on atomic volume curve was given by

Options:
A Newland
B Lother Mayer
C Dobereiner
D Mendeleev

The correct answer is:
Lother Mayer

Solution:
Lother Meyer gave periodic classification of elements based on atomic volume curve.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
Peripatus is a connecting link between

Options:
A Ctenophora and Platyhelminthes
B Mollusca and Echinodermata
C Annelida and Arthropoda
D Coelenterata and Porifera

The correct answer is:
Annelida and Arthropoda

Solution:
Peripatus is connecting link between Annelida and Arthropoda.

Topic: Biology, Animal Kingdom, NEET, NEET 2009 (Screening)
Periplaneta americana is a very common species of cockroach. It is

Options:
A Ureotelic
B Uricotelic
C Ammonotelic
D Guanotelic

The correct answer is:
Uricotelic

Solution:
Periplaneta americana is the scientific name of the cockroach. The excretory system is uricotelic as it releases nitrogen waste in the form of uric acid.

Topic: Biology, Excretory Products and their Elimination, NEET
Periplaneta americana is also known as American cockroach. Which of the following information about it is incorrect?

Options:
A Number of wings = 2 pairs.
B Number of legs = 3 pairs
C Type of circulatory system = Closed type
D Number of Ovaries = 2

The correct answer is:
Type of circulatory system = Closed type

Solution:
Circulatory system in cockroach: It is open type. Blood vessels are poorly developed, and they open into haemocoel or open spaces. Body fluid present in haemocoel is also known as haemolymph. Blood is colourless as there is no respiratory pigment.

Topic: Biology, Structural Organisation in Animals, NEET
Perisperm is remnant of:

Options:
A Integument
B Nucellus
C Embryo sac
D Testa

The correct answer is:
Nucellus

Solution:
In some seeds such as black pepper and beet, remnants of nucellus are also persistent. This residual, persistent nucellus is the perisperm.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Permanent hardness of water can be removed by which of the following process?

Options:
A Sodiumbicarbonate
B Sodium chloride
C Washing soda
D Sodalime

The correct answer is:
Washing soda

Solution:
Washing soda removes both the temporary and permanent hardness by converting soluble calcium and magnesium compounds into insoluble carbonates.
CaCl2+Na2CO3→CaCO3+2NaCl
CaSO4+Na2CO3→CaCO3+Na2SO4
Ca(HCO3)2+Na2CO3→CaCO3+2NaHCO3.

Topic: Chemistry, s Block Elements, NEET
Permanent method of birth control is

Options:
A oral contraceptive pills
B insertion of IUD
C use of condoms
D vasectomy

The correct answer is:
vasectomy

Solution:
Vasectomy is a permanent method of birth control. It is a method of sterilization done in males. It is a surgical procedure in which a small part of the vas deferens is either removed or tied up through an incision in the scrotum. This blocks the transportation of sperms. Thus, the semen which is produced contains no sperms.
Oral contraceptive pills, insertion of IUD and use of condoms are all temporary methods of contraception.

Topic: Biology, Reproductive Health, NEET
Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation for the reaction.
Which of the following represents the balanced ionic equation for the reaction?

Options:
A $8 \mathrm{MnO}_4^{-}+\mathrm{Br}^{-}+8 \mathrm{OH}^{-} \rightarrow 8 \mathrm{MnO}_2+\mathrm{BrO}_3^{-}+4 \mathrm{H}_2 \mathrm{O}$
B $2 \mathrm{MnO}_4^{-}+\mathrm{Br}^{-}+4 \mathrm{OH}^{-} \rightarrow 2 \mathrm{MnO}_2+\mathrm{BrO}_3^{-}+2 \mathrm{H}_2 \mathrm{O}$
C $4 \mathrm{MnO}_4^{-}+\mathrm{Br}^{-}+4 \mathrm{OH}^{-} \rightarrow 4 \mathrm{MnO}_2+\mathrm{BrO}_3^{-}+2 \mathrm{H}_2 \mathrm{O}$
D $8 \mathrm{MnO}_4^{-}+2 \mathrm{Br}^{-}+8 \mathrm{OH}^{-} \rightarrow 8 \mathrm{MnO}_2+2 \mathrm{BrO}_3^{-}+4 \mathrm{H}_2 \mathrm{O}$

The correct answer is:
$4 \mathrm{MnO}_4^{-}+\mathrm{Br}^{-}+4 \mathrm{OH}^{-} \rightarrow 4 \mathrm{MnO}_2+\mathrm{BrO}_3^{-}+2 \mathrm{H}_2 \mathrm{O}$

Solution:
Correct Option is : (C)
$4 \mathrm{MnO}_4^{-}+\mathrm{Br}^{-}+4 \mathrm{OH}^{-} \rightarrow 4 \mathrm{MnO}_2+\mathrm{BrO}_3^{-}+2 \mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, Redox Reactions, NEET
Permanganate(VII) ion, \(\mathrm{MnO}^{-}\) in basic solution oxidises iodide ion, \(\mathrm{I}^{-}\) to produce molecular iodine (I2) and manganese (IV) oxide (MnO2). Write a balanced ionic equation to represent this redox reaction.

Options:
A $\mathrm{MnO}_4^{-}+3 \mathrm{I}^{-}+4 \mathrm{OH}^{-} \rightarrow \mathrm{I}_2+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$
B $\mathrm{MnO}_4^{-}+3 \mathrm{I}^{-}+2 \mathrm{OH}^{-} \rightarrow \mathrm{I}_2+\mathrm{MnO}_2+\mathrm{H}_2 \mathrm{O}$
C $\mathrm{MnO}_4^{-}+2 \mathrm{I}^{-}+4 \mathrm{OH}^{-} \rightarrow \mathrm{I}_2+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$
D $\mathrm{MnO}_4^{-}+2 \mathrm{I}^{-}+2 \mathrm{OH}^{-} \rightarrow \mathrm{I}_2+\mathrm{MnO}_2+\mathrm{H}_2 \mathrm{O}$

The correct answer is:
$\mathrm{MnO}_4^{-}+3 \mathrm{I}^{-}+4 \mathrm{OH}^{-} \rightarrow \mathrm{I}_2+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$

Solution:
Correct Option is : (A)
$\mathrm{MnO}_4^{-}+3 \mathrm{I}^{-}+4 \mathrm{OH}^{-} \rightarrow \mathrm{I}_2+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, Redox Reactions, NEET
Pernicious anaemia is caused by the deficiency of:

Options:
A Vitamin B6
B Vitamin B12
C Vitamin B2
D Vitamin B1

The correct answer is:
Vitamin B12

Solution:
Correct Option is : 2
Vitamin B12

Topic: Chemistry, Biomolecules, NEET
Peroxide among the following compounds is:

Options:
A KO2
B BaO2
C MnO2
D NO2

The correct answer is:
BaO2

Solution:
The oxidation state of O in peroxide O 2 2â is â1.

The oxidation state of O in:

KO 2 = â1 2

BaO 2 = â1

MnO 2 = â2

NO 2 = â2

Therefore, BaO2 is a peroxide.

Topic: Chemistry, s Block Elements, NEET
Peroxisomes are associated with

Options:
A respiration
B photosynthesis
C photorespiration
D photophosphorylation

The correct answer is:
photorespiration

Solution:
Peroxisomes are microbodies, where the oxidation of organic compounds by oxygen in the presence of light in C3 plants takes place and is called photorespiration.
Photorespiration is the uptake of O2 and release of CO2 in light and results from the biosynthesis of glycolate in chloroplasts and subsequent metabolism of glycolate acid in the same leaf cell. The biochemical mechanism for photorespiration is also called glycolate metabolism.
Substrate of photorespiration is glycolate which is 2-C compound therefore it is also called the C2-cycle or glycolate cycle.
Loss of energy occurs during this process. The process of photorespiration involves the involvement of chloroplasts, peroxisomes and mitochondria. RuBP carboxylase also catalyzes another reaction which interferes with the successful functioning of the Calvin cycle.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Persistent nucellus in the seed is observed in

Options:
A Castor and Beet
B Black pepper and Beet
C Black pepper and Maize
D All of these

The correct answer is:
Black pepper and Beet

Solution:
Perisperm is the residues of nucellus are sometimes persistent in few seeds, such as in beet and black pepper. This type of nucellus is perisperm. This is a nutritive tissue that is present around the embryo of the seed.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2019 (All India)
Persistent nucellus in the seed is observed in

Options:
A Castor and Beet
B Black pepper and Beet
C Black pepper and Maize
D All of these

The correct answer is:
Black pepper and Beet

Solution:
Perisperm is the residues of nucellus are sometimes persistent in few seeds, such as in beet and black pepper. This type of nucellus is perisperm. This is a nutritive tissue that is present around the embryo of the seed.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Person A can clearly see objects between 25 cm and 200 cm. Choose the range of clear vision for a person B having muscles stronger than A but all other parameters of eye identical to that of A.

Options:
A 25Â cm to 200Â cm
B 18 cm to 200 cm
C 25Â cm to 300Â cm
D 18Â cm to 300Â cm

The correct answer is:
18 cm to 200 cm

Solution:
In the relaxed condition, person B will see the same as that of person A, i.e., 200 cm.
Under maximum strain, the least distance for distinct vision for A is 25 cm. As the muscles of person B are stronger than A, therefore, B can see objects closer than 25 cm.

Topic: Physics, Ray Optics, NEET
Person with blood group AB is considered as universal recipient because he has:

Options:
A no antigen on RBC and no antibody in the plasma
B both A and B antibodies in the plasma.
C both A and B antigens on RBC but no antibodies in the plasma.
D both A and B antigens in the plasma but no antibodies

The correct answer is:
both A and B antigens on RBC but no antibodies in the plasma.

Solution:
Person with blood group AB is considered as universal recipient because he has no antigen on RBC and no antibody in the plasma.

Blood Group	Antigens on RBCs	Antibodies in Plasma	Donor's Group

A	A	anti-B	A, O

B	B	anti-A	B, O

AB	A, B	nil	AB, A, B, O

O	nil	anti-A, B	O

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2014
Person with blood group AB is considered as universal recipient because he has:

Options:
A both A and B antigens on RBC but no antibodies in the plasma.
B both A and B antibodies in the plasma.
C no antigen on RBC and no antibody in the plasma
D both A and B antigens in the plasma but no antibodies

The correct answer is:
no antigen on RBC and no antibody in the plasma

Solution:
Person with blood group AB is considered as universal recipient because he has no antigen on RBC and no antibody in the plasma.

Blood Group	Antigens on RBCs	Antibodies in Plasma	Donorâs Group

A	A	anti-B	A, O

B	B	anti-A	B, O

AB	A, B	nil	AB, A, B, O

O	nil	anti-A, B	O

Topic: Biology, Body Fluids and Circulation, NEET
Persons with 'AB' blood group are called as "Universal recipients". This is due to:

Options:
A Absence of antigens A and B on the surface of RBCs
B Absence of antigens A and B in plasma
C Presence of antibodies, anti-A and anti-B, on RBCs
D Absence of antibodies, anti-A and anti-B, in plasma

The correct answer is:
Absence of antibodies, anti-A and anti-B, in plasma

Solution:
AB blood groups lack antibodies against antigen A & B, i.e. anti-A and anti-B, in plasma. Hence, they cannot target any particular blood group and have no chances of incompatibility reaction. Thus, called universal recipients.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2021
PGA as the first CO2 fixation product was discovered in photosynthesis of

Options:
A alga
B bryophytes
C gymnosperms
D angiosperms

The correct answer is:
alga

Solution:
PGA also is known as 3-phosphoglyceric acid was discovered in the photosynthesis of alga. Melvin Calvin was responsible for this discovery as he used radioactive C14 in algal (chlorella) photosynthetic studies which led to the first carbon-dioxide fixation product which is PGA. This acid also acts as an enzyme in the conversion of CO2 to organic acid.

Topic: Biology, Photosynthesis in Higher Plants, NEET
PGA as the first CO2 fixation product was discovered in photosynthesis of

Options:
A Bryophyte
B Gymnosperm
C Angiosperm
D Alga

The correct answer is:
Alga

Solution:
Dr. Melvin Calvin discovered the correct "path of carbon" in photosynthesis, i.e. sequence of biochemical reactions of carbon dioxide fixation into glucose. He carried out experiments on unicellular green algae like Chlorella and Scenedesmus, and used radioactive isotope of carbon, C14 as a tracer. 14CO2 was fed to photosynthesizing algal cells for definite time in seconds.The algal cells were then killed by dropping the suspension in hot methanol.The technique of paper chromatography was used to separate the intermediates formed. He discovered that the first stable product is a 3-carbon compound, Phosphoglyceric acid (PGA). Hence Dr. Calvin called it C3 cycle.

Topic: Biology, Photosynthesis in Higher Plants, NEET
PGA as the first $\mathrm{CO}_2$ fixation product was discovered in photosynthesis of

Options:
A bryophyte
B gymnosperm
C angiosperm
D algae

The correct answer is:
algae

Solution:
The use of radioactive ${ }^{14} \mathrm{C}$ by Melvin Calvin in algal (Chlorella) photosynthesis studies led to the discovery that the first $\mathrm{CO}_2$ fixation product was a 3-carbon organic acid. The first product identified was 3-phosphoglyceric acid (PGA).

Topic: Biology, Respiration in Plants, NEET, NEET 2010 (Screening)
PGA as the first CO2 fixation product was discovered in:

Options:
A Bryophyta
B Gymnosperm
C Angiospasm
D Algae

The correct answer is:
Algae

Solution:
Calvin and his colleagues used the unicellular green alga Chlorella and Scenedesmus to study the carbon-assimilation reactions of photosynthesis. They incubated C14 with illuminated suspensions of algae and followed the time course of appearance of reduction cycle compounds.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Ph− C || O − NH 2 → POCl 3 (A); Product (A) is

Options:
A Ph-NH2
B Ph-CH2-NH2
C Ph-C|HOH-NH2
D Ph-C=N

The correct answer is:
Ph-C=N

Solution:
Ph-C||O-NH2→POCl3Ph-CN

Topic: Chemistry, Amines, NEET