datavorous/entrance-exam-dataset huggingface dataset:
The binding energy per nucleon of deuteron (H)12and helium nucleus (He)24is 1.1 MeV and 7 MeV, respectively. If two deuteron nuclei react to form a single helium nucleus , then the energy released is

Options:
A 13.9 MeV
B 26.9 MeV
C 23.6 MeV
D 19.2 MeV

The correct answer is:
23.6 MeV

Solution:
As given,
1H2 + 1H2â 2He4 +energy
The binding energy per nucleon of a deuteron (1H2) =1.1 MeV
â´ Total binding energy of one deuteron nucleus = 2 Ã 1.1=2.2 MeV
â´ The binding energy per nucleon of Helium (2He4) =7 MeV
â´ Total binding energy = 4 Ã 7 =28 MeV
Hence, energy released in the above process =28-2 Ã2.2
= 28 â 4.4 = 23.6 MeV

Topic: Physics, Nuclear Physics, NEET
The Binding energy per nucleon of Li37 and He24 nucleon are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction 37Li+ 11H→ 24He+ 24He+Q, the value of energy Q released is

Options:
A 19.6Â MeV
B -2.4Â MeV
C 8.4Â MeV
D 17.3 MeV

The correct answer is:
17.3 MeV

Solution:
Li74 +H1 1→He42 2
BE of products=(5.6 MeV×7)+0
=39.2 MeV
Ei=-39.2 MeV
BE of reactant=7.06×4×2
=56.48 MeV
Ef=-56.48 MeV

As nuclear energy decreases, some energy will be released.
Qrelease=Ei-Ef=-39.2--56.48=17.28 MeV

Topic: Physics, Nuclear Physics, NEET, NEET 2014
The Binding energy per nucleon of Li37 and He24 nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction 37Li+ 11Hâ2 24He+Q, the value of energy Q released is

Options:
A 19.6 MeV
B -2.4Â MeV
C 8.4Â MeV
D 17.3 MeV

The correct answer is:
17.3 MeV

Solution:
Binding energy of 2He4=4×7.06=28.24 MeV
Binding energy of 37Li=7×5.60=39.20 MeV
Li37+H11→He24+He24+Q
Therefore, Q=56.48−39.20=17.28 MeV

Topic: Physics, Nuclear Physics, NEET
The biodegradable polymer is:

Options:
A nylon-6, 6
B nylon 2-nylon 6
C nylon-6
D Buna-S

The correct answer is:
nylon 2-nylon 6

Solution:
Nylon-2-nylon-6 is biodegradable polymer. Another important biodegradable polymer is Poly β-hydroxybutyrate–co-β-hydroxy valerate (PHBV).

Topic: Chemistry, Polymers, NEET, NEET 2019 (All India)
The biological levels of organization of living things arranged from the simplest to most complex are,

Options:
A Biosphere, ecosystems, communities, populations, organisms, organ systems, organs, tissues, cells and organelles.
B Biosphere, communities, ecosystems, organisms, populations, organ systems, organs, tissues, cells and organelles.
C Organelle, cells, tissues, organs, organ systems, organisms, populations, communities, ecosystems, and biosphere.
D Cells, organelles, tissues, organs, organ systems, organisms, populations, communities, ecosystems, and biosphere.

The correct answer is:
Organelle, cells, tissues, organs, organ systems, organisms, populations, communities, ecosystems, and biosphere.

Solution:
The biological levels of organization of living things arranged from the simplest to most complex are organelle, cells, tissues, organs, organ systems, organisms, populations, communities, ecosystems, and biosphere.

Topic: Biology, Organisms and Populations, NEET
The biological scissors are:

i. Restriction enzymes are used to cut DNA into nucleotide monomers.
ii. Enzymes always cut the DNA at specific nucleotide sequences.
iii. Present variously recognizing and cutting different DNA sequences.

Select the option with the correct statement/s.

Options:
A i and ii
B ii and iii
C i and iii
D i, ii and iii

The correct answer is:
ii and iii

Solution:
Enzymes are majorly made of proteins which act as catalyst in increasing the reaction rate and producing products faster. There are huge varieties of enzyme specific for a particular substrate and they are categorized as per their activity. The enzyme group that break nucleic acids or long chain polynucleotides is called nucleases. Depending on where they make the break in the chain, there are two categories - exonucleases which break nucleotides at the terminal part of the chain and endonucleases which make the break within the chain. These enzymes make the break or 'cut' at a particular site or sequence (palindromic); they are also called restriction endonucleases. These biocatalysts are extensively used in the genetic engineering process for manipulation of genes. Because of their cutting ability like a scissors, they are commonly called as molecular scissors.

Topic: Biology, Biotechnology - Principles and Processes, NEET
The biome in which the mean annual temperature is approximately between 20Â°C and 25Â°C and the mean annual precipitation is approximately between 150 cm to 450 cm is

Options:
A Temperate forest
B Tropical forest
C Conifer forest
D Alpine tundra

The correct answer is:
Tropical forest

Solution:
Major biomes are formed due to annual variations and intensity of temperature and annual variations in precipitation. The approximate mean annual temperature (MT) in °C and mean annual precipitation (MP) in cm of various biomes are
Temperate forest: MT: 20–25 and MP: 75–225
Conifer forest biome: MT: 0–15 and MP: 50–250
Alpine tundra biome: MT: -10–0 and MP: 10–125
Tropical forest: MT: 20–25 and MP: 150–450.

Topic: Biology, Organisms and Populations, NEET
The biomolecules that are formed by the polymerization of a huge number of micromolecules possessing higher molecular weight is called ___

Options:
A Macromolecules
B Micromolecules
C Biomolecules
D None of the above

The correct answer is:
Macromolecules

Solution:
The biomolecules that are formed by the polymerization of a huge number of micromolecules possessing higher molecular weight is called macromolecules

Topic: Biology, Biomolecules (B), NEET
The biosynthesis of ribosomal RNA occurs in:

Options:
A Ribosomes
B Golgi apparatus
C Microbodies
D Nucleolus

The correct answer is:
Nucleolus

Solution:
The fluid inside the nucleus is called nucleoplasm or nuclear matrix. The nucleoplasm contains nucleolus and chromatin. The nucleolus is a spherical structure. The nucleolus is not a membrane-bound structure. Synthesis of ribosomal RNA (rRNA) takes place in the nucleolus. It also participate in the formation of signal recognition particles and play a role in the cell's response to stress. Nucleoli are made of proteins, DNA and RNA and form around specific chromosomal regions called nucleolar organizing regions.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2020 (Phase 2)
The biosynthetic phase of photosynthesis is

Options:
A directly dependent on the presence of light.
B directly dependent on the products of the light reaction.
C only dependent on CO2 and water.
D not dependent on ATP and NADPH.

The correct answer is:
directly dependent on the products of the light reaction.

Solution:
The products of the light reaction are ATP, NADPH and O2. Of these O2 diffuses out of the chloroplast while ATP and NADPH are used to drive the processes leading to the synthesis of food, more accurately, sugars. This is the biosynthetic phase of photosynthesis. This process does not directly depend on the presence of light but is dependent on the products of the light reaction, i.e., ATP and NADPH, besides CO2 and H2O. You may wonder how this could be verified; it is simple: immediately after light becomes unavailable, the biosynthetic process continues for some time, and then stops. If then, light is made available, the synthesis starts again.

Topic: Biology, Photosynthesis in Higher Plants, NEET
The birds of Galapagos island are

Options:
A seed eating finches
B insect eating finches
C cactus eating finches
D all of the above

The correct answer is:
all of the above

Solution:
Darwin's Finches (one of the best examples of Adaptive Radiation)
- Darwin visited the Galapagos Islands during his journey and observed an amazing diversity of creatures (especially birds).
- He realised that there were many varieties of small black birds (later on called Darwin's Finches) in the same island and also they have evolved on the island itself. The original birds were seed-eating from which many forms with altered beaks arose making them insectivorous and vegetarian finches.

Topic: Biology, Evolution, NEET
The Birkeland-Eyde process is a commercial method, the raw material used in this process is

Options:
A Air
B NH3
C NO2
D HNO3

The correct answer is:
Air

Solution:
The Birkeland-Eyde process is a commercial method, the raw material used in this process air. Dinitrogen is prepared commercially from air by liquefaction and fractional distillation. When liquid air is allwed to distil, dinitrogen having lower boiling point (77K) distils over first leaving behind liquid oxygen (boiling point is 90K.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
The birth control measure that is related to males is

Options:
A Oral pill
B Tubectomy
C Vasectomy
D None of these

The correct answer is:
Vasectomy

Solution:
Sterilization procedure in the male is called vasectomy and that in the female is called tubectomy.

Topic: Biology, Reproductive Health, NEET
The black pigment in the eye, which reduces the internal reflection, is located in

Options:
A Retina
B Iris
C Sclerotic
D Cornea

The correct answer is:
Retina

Solution:
The black pigment in the eye, which reduces the internal reflection is located in optic part of retina. It lies in contact with the choroid. It is thick and composed of four layers of cells. Beginning from the choroid side, it has layer of pigmented cells, a layer of receptor cells, a layer of bipolar nerve cells and a layer of ganglion cells. The pigment cells reinforce the light absorbing property of choroid in reducing the scattering of light in the eye.

Topic: Biology, Neural Control and Coordination, NEET
The blades of a windmill sweep out a circle of area A. Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m2 , v = 36 km/h and the density of air is 1.2 kg m–3. What is the electrical power produced ?

Options:
A 3000W
B 3500W
C 4000W
D 4500W

The correct answer is:
4500W

Solution:
Correct Option is : (D)
4500W

Topic: Chemistry, Work Power Energy, NEET
The bleaching action of bleaching powder is due to the formation of:

Options:
A CaCl2
B CaSO4
C HClO
D Ca(ClO3)2

The correct answer is:
HClO

Solution:
In presence of dil. Acids, bleaching powder loses oxygen.
2CaOCl2+H2SO4→CaCl2+CaSO4+2HClO
HClO→HCl+O
This oxygen is used for oxidation - bleaching.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
The block of mass 3M is attached to the pulley system as shown in the figure. At t=0, particle M falling vertically, strikes the block 3M with velocity 8 m s-1 and sticks to it. The speed of the combined mass just after the collision is

Options:
A 8m s-1
B 10m s-1
C 5m s-1
D 2 m s-1

The correct answer is:
2 m s-1

Solution:
As the tension in the string will be always zero, using momentum conservation just before and after collision

M×8=4M×V

V=2m s-1

Topic: Physics, Center of Mass Momentum and Collision, NEET
The blood group of the father is B and that of a daughter is AB. The genotype of the mother would be:

Options:
A IAIAÂ orÂ I0I0
B IAIB only
C IAIA or IAIB
D IAI0 only

The correct answer is:
IAIA or IAIB

Solution:
Blood group inheritance is a multiple allelic characters in which three alleles IA, IB and IO control the expression of four phenotype. In this inheritance the child receives one allele from mother and one allele from father.

Topic: Biology, Principles of Inheritance and Variation, NEET
The blood of cockroach contains no respiratory pigment. It means that:

Options:
A Cockroach does not respire
B Respiration is anaerobic
C O2 goes directly into tissues by diffusion
D O2 goes into tissue by intracellular capillary system

The correct answer is:
O2 goes directly into tissues by diffusion

Solution:
The blood of cockroach is present freely in haemocoel which is the body cavity of cockroach. It is a transparent fluid with no haemoglobin. O2 enters the haemocoel through the trachea and diffuses into the haemocoel directly.

Topic: Biology, Structural Organisation in Animals, NEET
The blood pressure values of four persons are given below

(i) Mr. A 120/80 mm Hg
(ii) Mr. B 80/120 mm Hg
(iii) Mrs. A 90/80 mm Hg
(iv) Mrs. B 150/80 mm Hg

Who among the following has normal blood pressure?

Options:
A Mr. A
B Mr. B
C Mrs. A
D Mrs. B

The correct answer is:
Mr. A

Solution:
Hypertension is the term for blood pressure that is higher than normal (120/80). In this measurement 120 mm Hg (millimetres of mercury pressure) is the systolic, or pumping, pressure and 80 mm Hg is the diastolic, or resting, pressure. If repeated checks of blood pressure of an individual is 140/90 (140 over 90) or higher, it shows hypertension. High blood pressure leads to heart diseases and also affects vital organs like brain and kidney.

Topic: Biology, Body Fluids and Circulation, NEET
The blood stem cells are related to which among the following options

Options:
A Liver
B Spleen
C Kidney
D Bone marrow

The correct answer is:
Bone marrow

Solution:
Bone marrow is a soft and spongy tissue present inside the bones. These tissues make blood-forming cells or blood stem cells. White blood cells to fight infections, red blood cells to carry oxygen throughout the body, platelets to control bleeding, macrophages for phagocytic functions are all produced in these tissues.

Topic: Biology, Structural Organisation in Animals, NEET
The 'blue baby' syndrome results from:

Options:
A excess of TDS (total dissolved solids)
B excess of chloride
C methaemoglobin
D excess of dissolved oxygen

The correct answer is:
methaemoglobin

Solution:
Infant methemoglobinemia is also called "blue baby syndrome." It is a condition where a baby's skin turns blue. This happens when there is not enough oxygen in the blood. Methemoglobinemia is a condition that some babies are born with (congenital) or some develop early in life (acquired).

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2006
The bob A of a pendulum released from horizontal to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. If the length of the pendulum is 1m, calculate the speed with which bob B starts moving. Neglect the size of the bobs and assume the collision to be elastic.

Options:
A 6.37m/s
B 4.42m/s
C 9.51m/s
D 9.81m/s

The correct answer is:
4.42m/s

Solution:
Correct Option is : (B)
4.42m/s

Topic: Chemistry, Work Power Energy, NEET
The bob of a simple pendulum executes a simple harmonic motion in water with a period t , while the period of oscillation of the bob is t 0 in air. Neglecting frictional force of water and given that the density of the bob is 4/3Ã1000 kg m-3. The correct relationship between t and t 0 is

Options:
A t = t 0
B t = t 0 / 2
C t = 2 t 0
D t = 4 t 0

The correct answer is:
t = 2 t 0

Solution:
t 0 = 2 π l / g ...............(i)
Due to upthrust of water on the top, its apparent weight decreases
upthrust = weight of liquid displaced
∴ Effective weight = m g - V σ g = V ρ g - V σ g

V ρ g ′ = V g ρ - σ , where σ is density of water
or g ′ = g ρ - σ ρ
∴ t = 2 π l / g ′ = 2 π l ρ g ρ - σ .............(ii)
∴ t t 0 = l ρ g ρ - σ × g l = ρ ρ - σ
= 4 × 1 0 0 0 / 3 4 0 0 0 3 - 1 0 0 0 = 2
or t = t 0 × 2 = 2 t 0.

Topic: Physics, Oscillations, NEET
The bob of a simple pendulum performs SHM with period T in air and with period T1 in water. Relation between T and T1 is (neglect friction due to water, density of the material of the bob is 98Ã103 k g /m3, density of water is1 g /cc)

Options:
A T1=3T
B T1=2T
C T1=T
D T1=T2

The correct answer is:
T1=3T

Solution:
T=2Ïlgeff
In air geff=g where as in water
geff=gÏ-ÏÏ=9.898Ã103-10398Ã103
=9.898-198=9.8Ã18Ã89=9.89
â´ T1=2Ïlgeff=2Ï9lg
⇒ T 1 = 9 T=3T

Topic: Physics, Oscillations, NEET
The body cavity is the cavity present between the body wall and the gut wall. In some animals, the body cavity is not lined by mesoderm. Such animals are called

Options:
A Acoelomates
B Pseudocoelomates
C Coelomates
D Haemocoelomates

The correct answer is:
Pseudocoelomates

Solution:
In Acoelomates, the body cavity is not lined by mesoderm.

Topic: Biology, Animal Kingdom, NEET
The body of the ovule fuses with funicle in the region called

Options:
A micropyle.
B integuments.
C hilum.
D chalaza.

The correct answer is:
hilum.

Solution:
The ovule is a small structure attached to the placenta by means of a stalk called funicle. The body of the ovule fuses with funicle in the region called hilum. Thus, hilum represents the junction between ovule and funicle. Each ovule has one or two protective envelopes called integuments. Integuments encircle the nucellus except at the tip where a small opening called the micropyle is organised. Opposite the micropylar end, is the chalaza, representing the basal part of the ovule.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The Bohr model for the H-atom relies on the Coulomb’s law of electrostatics. Coulomb’s law has not directly been verified for very short distances of the order of angstroms. Supposing Coulomb’s law between two opposite charge + q1, –q2 is modified to
What is the ground state energy of a hydrogen atom if the value of ε is 0.1 and R0is 1Å?

Options:
A –13.4 eV
B –11.4 eV
C –9.4 eV
D –7.4 eV

The correct answer is:
–11.4 eV

Solution:
Correct Option is : (B)
–11.4 eV

Topic: Chemistry, Atomic Physics, NEET
The Bohr model of atoms:

Options:
A assumes that the angular momentum of electrons is quantized
B uses Einstein's photo-electric equation
C predicts continuous emission spectra for atoms
D predicts the same emission spectra for all types of atoms

The correct answer is:
assumes that the angular momentum of electrons is quantized

Solution:
The both modes of atoms assumes that the angular momentum of electrons is quantred
$$
q= \pm h e
$$

Topic: Physics, Atomic Physics, NEET, NEET 2004
The boiling point of 0.2 mol kg-1 solution of X in water is greater than equimolal solution of Y in water. Which one of the following statements is true in this case?

Options:
A X is undergoing dissociation in water
B Molecular mass of X is greater than the molecular mass of Y.
C Molecular mass of X is less than the molecular mass of Y.
D Y is undergoing dissociation in water while X undergoes no change

The correct answer is:
X is undergoing dissociation in water

Solution:
∆Tb=i Kb.m
Kb & m are constant for solutions of X and Y. ∆Tb is greater for solution of X implies i (Van't Hoff factor for X > 1) i.e X undergoes dissociation.

Topic: Chemistry, Solutions, NEET, NEET 2015 (Phase 1)
The boiling point of 0.2 mol kg-1 solution of X in water is greater than equimolal solution of Y in water. Which one of the following statements is true in this case?

Options:
A Molecular mass of X is greater than the molecular mass of Y.
B Molecular mass of X is less than the molecular mass of Y.
C Y is undergoing dissociation in water while X undergoes no change
D X is undergoing dissociation is water.

The correct answer is:
X is undergoing dissociation is water.

Solution:
(ΔTb)x>(ΔTb)y
As same solvent is used so, Kb is same for both X and Y
As m is same (given)
So ix.kb.m>iy.kb.m
ix>iy
Hence x is undergoing dissociation in water.

Topic: Chemistry, Solutions, NEET
The boiling point of a glucose solution containing 12 g of glucose in 100 g of water is 100. 34°C . Boiling point of water is 100° C. The molal elevation constant of water is

Options:
A 0.51 ° C/Molal
B 51 ° C/Molal
C 5.1 ° C/Molal
D None of these

The correct answer is:
0.51 ° C/Molal

Solution:
ΔTb=100.34-100=0.34°C
∆Tb=kb×w2×1000M2×w1
kb=∆Tb×M2×w1w2×1000
kb=0.34×180×10012×1000=0.51

Topic: Chemistry, Solutions, NEET
The boiling point of an aqueous solution of a non-electrolyte is 100.52oC . Then freezing point of this solution will be
Given:   k f =1.86 K kg  mol −1 ,  k b =0.52 K kg  mol −1  for water

Options:
A 0 oC
B -1.86 oC
C 1.86 oC
D None of the above

The correct answer is:
-1.86 oC

Solution:
ΔTb=Kb×m i.e. 0.52=0.52×m
ΔTf=Kf×m=1.86×1=1.86 ;

Topic: Chemistry, Solutions, NEET
The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. What is the molar mass of the solute? (Kbfor benzene is 2.53 K kg mol–1)

Options:
A 88 g mol–1
B 40 g mol–1
C 72 g mol–1
D 58 g mol–1

The correct answer is:
58 g mol–1

Solution:
Correct Option is : (D)
58 g mol–1

Topic: Chemistry, Solutions, NEET
The bond angle formed by different hybrid orbitals are in the order

Options:
A sp2>sp3>sp
B sp3<sp2>sp
C sp3>sp2>sp
D sp>sp2>sp3

The correct answer is:
sp>sp2>sp3

Solution:
Hybridization Shape Bond angle
sp 180o
sp2 120o
sp3 109o28

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The bond angle H-X-H is the greatest in the compound :

Options:
A PH3
B CH4
C NH3
D H2O

The correct answer is:
CH4

Solution:
Molecule	Hybridization	Bond Angle
1.	PH3	sp3	98o
2.	CH4	sp3	109oÂ 28′
3.	NH3	sp3	107o
4.	H2O	sp3	104.5o

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The bond angle of H 2 Se is best described as

Options:
A Between 109 o and 120 o
B Greater than 120 o
C Less than that in H 2 S but not less than 90 o
D Less than 90 o

The correct answer is:
Less than that in H 2 S but not less than 90 o

Solution:
According to Drago's rule :
∠ HXH : H 2 O 1 0 4 o > H 2 S 9 2 o > H 2 Se 9 1 o > H 2 Te 8 9 · 5 o

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The bond angles of NH3, NH4+ and NH2- are in the order

Options:
A NH2->NH3>NH4+
B NH4+>NH3>NH2-
C NH3>NH2->NH4+
D NH>NH4+>NH2-

The correct answer is:
NH4+>NH3>NH2-

Solution:
As the number of lone pairs of electrons increases, bond angle decreases. Thus, the order of bond angle is
N⁡ H⁡ 4 + > N⁡ H⁡ 3 > N⁡ H⁡ 2 - no Ip 1 Ip 2 IP

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The bond dissociation energies of X2 Y2 and XY are in the ratio of 1 :0.5 :1. ΔH for the formation of XY is -200 kJ mol-1. What will be the bond dissociation energy of X2?

Options:
A 800 kJ mol-1
B 100 kJ mol-1
C 200 kJ mol-1
D 400 kJ mol-1

The correct answer is:
800 kJ mol-1

Solution:
Let B.E. of x2, y2 and xy are x kJ mol-1, 0.5x kJ mol-1 and x kJ mol-1 respectively
12X2+12Y2→XY; ΔH=-200 kJ mol-1
ΔH=-200=ΣB.EReactant-ΣB.EProduct
-200=12×x+12×0.5x-1×x⇒x=800kJ  mol
B.E of X2=800 kJ mol-1

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2018
The bond dissociation energies of X2,Y2 and XY are in the ratio of 1 : 0.5 : 1. ΔH for the formation of XY is –200 kJ mol-1. The bond dissociation energy of X2 will be

Options:
A 800 kJ mol-1
B 100 kJ mol-1
C 200 kJ mol-1
D 400 kJ mol-1

The correct answer is:
800 kJ mol-1

Solution:
he reaction for ÎfHÂ°(XY)
12X2(g)+12Y2(g)âXY(g)
Bond energies of X2,Y2 and XY are X,X2 and X respectively.
Hence ÎH=X2+X4-X=-200
On solving, we get -X2+X4=-200
X=800 kJ/mole

Topic: Chemistry, Thermodynamics (C), NEET
The bond having the highest bond energy is

Options:
A C=C
B C=S
C C=O
D P=N

The correct answer is:
C=O

Solution:
The bond having the highest bond energy is C=O due to more effective axial and sideways overlapping between atomic orbitals of carbon and those of oxygen atom. It is higher due to smaller size of oxygen atom. Oxygen is more electronegative than carbon atom, due to these factors CO has highest bond energy.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The bond length in O2+,O2,O2- and O22- follows the order

Options:
A O22->O2->O2>O2+
B O2+>O2>O2-<O22-
C O2>O2->O22->O2+
D O2->O22->O2+>O2

The correct answer is:
O22->O2->O2>O2+

Solution:
O2-2=Ï1s2<Ï*1s2<Ï2s2<Ï*2s2<Ï2px2<Ï2py2=Ï2pz2<Ï*2py2=Ï*2pz2
B.O.=nb-na2=6-42=1
O2-=Ï1s2<Ï*1s2<Ï2s2<Ï*2s2<Ï2px2<Ï2py2=Ï2pz2<Ï*2py2=Ï*2pz1
B.O.=6-32=1.5
O2=Ï1s2<Ï*1s2<Ï2s2<Ï*2s2<Ï2px2<Ï2pz2=Ï2py2<Ï*2py1=Ï*2pz1
B.O.=6-22=2
O2+=Ï1s2<Ï*1s2<Ï2s2<Ï*2s2<Ï2px2<Ï2pz2=Ï2py2<Ï*2py1=Ï*2pz0
B.O.=6-12=2.5
Bond lengthâ1Bond order
Bond length O2-2>O2- >O2>O2+

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET