datavorous/entrance-exam-dataset huggingface dataset:
Which of the following is part of bioprocess engineering?

Options:
A Techniques for formation of hybrid DNA in genetic engineering process
B Techniques for alteration of sequence of genetic material
C Maintenance of sterile ambiance in chemical engineering process to produce large quantities of biotechnological products
D Methods of isolation of only desirable genes from host cell required for recombinant DNA technology

The correct answer is:
Maintenance of sterile ambiance in chemical engineering process to produce large quantities of biotechnological products

Solution:
The two core techniques that enabled the birth of modern biotechnology are
(i) Genetic engineering: It involves the techniques to alter the chemistry of genetic material (DNA and RNA) to introduce these into host organisms and thus change the phenotype of the host organism.
(ii) Bioprocess engineering: It is the process of maintaining sterile ambience in chemical engineering processes to enable the growth of only the desired microbe/eukaryotic cell in large quantities for the manufacture of biotechnological products like antibiotics, vaccines, enzymes, etc.

Topic: Biology, Biotechnology - Principles and Processes, NEET
Which of the following is present in maximum percentage in living matter?

Options:
A Carbon
B Nitrogen
C Oxygen
D Potassium

The correct answer is:
Oxygen

Solution:
Oxygen is the most abundant element present in the living tissue. In human body, it forms 65.5%. It is the main component of all biomolecules such as carbohydrate, proteins fats and nucleic acid.

Topic: Chemistry, Biomolecules, NEET
which of the following is properties of co normal of ellipse

Options:
A The points on the curve at which the normals pass through a common point are called co-normal points
B The algebraic sum of the slopes of three concurrent normals is zero
C The sum of the ordinates of the co-normal points is zero
D All of the above

The correct answer is:
The points on the curve at which the normals pass through a common point are called co-normal points

Solution:
In an ellipse, the concept of co-normal points refers to points at which the normal lines to the ellipse intersect at a common point. These points have specific properties related to their eccentric angles.
Eccentric Angle ( $\theta$ ): For a point on an ellipse given by the parametric form $(a \cos \theta, b \sin \theta)$ ,$\theta$ is called the eccentric angle.
When considering co-normal points, a known result in the geometry of ellipses is that the sum of the eccentric angles of these points is equal to an odd mulliple of $\pi$. This is derived from the geometric properties and symmetry of the ellipse.
Therefore, the given statement is correct, and the solution provided is indeed "(A) true".

Topic: Mathematics, Ellipse, JEE Main
Which of the following is put into Anaerobic sludge digester for further sewage treatment?

Options:
A Floating debris
B Effluents of primary treatment
C Activated sludge
D Primary sludge

The correct answer is:
Activated sludge

Solution:
A small part of the activated sludge is pumped back into the aeration tanks to serve as inoculums. The remaining major part of the sludge is pumped into large tanks called anaerobic sludge digesters.

Anaerobic sludge digesters: Anaerobic methanogenic bacteria, digest the waste present in sludge. During this digestion, bacteria produce a mixture of gases such as methane, hydrogen sulphide, and carbon dioxide. These gases form biogas and can be used as a source of energy as it is inflammable.

Topic: Biology, Microbes in Human Welfare, NEET, NEET 2020 (Phase 1)
Which of the following is reducing sugar?

Options:
A

B

C

D

The correct answer is:

Solution:
Carbohydrate molecules have hemiacetal or hemiketal groups can reduces Tollen's reagent and Fehling's solution are reducing. All monosaccharides are reducing sugars. If any sugar is having free –OH group at anomeric carbon then it will be a reducing sugar.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2022 (26 Jul Shift 1)
Which of the following is referred to as "Evil Quartet' reference to the loss of biodiversity?

Options:
A Species richness, extinctions, deforestation, era
B Habitat loss and fragmentation, over-exploitation, alien species invasion, co-extinction
C Overexploitation, grazing, decomposition, extinction
D Habitat destruction, co-extinction, deforestation species richness

The correct answer is:
Habitat loss and fragmentation, over-exploitation, alien species invasion, co-extinction

Solution:
Biodiversity includes all the flora and fauna on Earth. All the varieties of organisms come under biodiversity. Ecologist say that without any doubt there is decline in the biological wealth which is more rapid due to the activities of us humans. The causes for such high rate of decline is referred to as 'Evil Quartet'. The four main reasons are also known as 'the four horsemen of the ecological apocalypse'. The reasons include overexploitation, loss of habitat, alien species invasion and co- extinction of species.

Topic: Biology, Biodiversity and its Conservation, NEET
Which of the following is responsible for Earth's magnetic field?

Options:
A The bar magnet buried deep inside the Earth
B Revolution of Earth
C Convection currents in Earth's core
D The presence of magnetite rocks formed from volcanoes

The correct answer is:
Convection currents in Earth's core

Solution:
The Earth has its magnetic field obtained from the flow of hot metal core i.e. convective currents in the Earth's core.

Topic: Physics, Magnetic Properties of Matter, JEE Main
Which of the following is responsible for the depletion of the ozone layer in the upper strata of the atmosphere?

Options:
A Chlorine
B Ferrocenes
C Fullerenes
D Freons

The correct answer is:
Freons

Solution:
Freons or chlorofluoro carbons are responsible for depletion of the ozone layer in the upper strata of the atmosphere. They are used as propellants, aerosol spray caps, refrigerants, fire fighting reagents etc. They are stable and chemically inert compounds. They absorb UV-radiation and break down liberating free atomic chlorine which causes decomposition of ozone through free radical reaction. This results in the depletion of the ozone layer.
Freons are mainly freon-1
(CFCL3) and freon − 12 (CF2Cl2). They form free radical of chlorine in the presence of UV- radiation. Such free radical decomposes O3 as follows

Topic: Chemistry, Environmental Chemistry, NEET
Which of the following is satisfied by an ideal solution?

Options:
A Formation of an azeotropic mixture
B \( \Delta \mathrm{S}_{\mathrm{mix}}=0 \)
C Raoult's law is obeyed under particular set of conditions only
D \( \Delta \mathrm{H}_{\operatorname{mix}}=0 \)

The correct answer is:
\( \Delta \mathrm{H}_{\operatorname{mix}}=0 \)

Solution:
The solutions which obey Raoult's law over the entire range of concentration are called Ideal solutions. I other words when the forces of attraction between A-A, B-B is similar to the A-B, then A and B will be ideal solution.

In case of ideal solutions, there is no enthalpy change when the components are mixed.

For ideal solution, Δ H mix = 0 .

Topic: Chemistry, Solutions, JEE Main
Which of the following is similar to graphite

Options:
A $\underline{B}$
B $\mathrm{BN}$
C $\mathrm{B}_{2} \mathrm{H}_{6}$
D $\mathrm{B}_{4} \mathrm{C}$

The correct answer is:
$\mathrm{BN}$

Solution:
Graphite and boron nitride have similar structure.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Which of the following is strongest Bronsted base?

Options:
A
B
C
D

The correct answer is:

Solution:
Bronsted base are those which can donate its lone pair to H+ ion. The more easily the compound can donate its lone pair to H+ strong will be a basic character.

Option D, is the strongest Bronsted base due to the availability of a lone pair which does not involve in any resonance.

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Which of the following is strongly acidic?

Options:
A Phenol
B o-cresol
C p-nitrophenol
D $p$-cresol

The correct answer is:
p-nitrophenol

Solution:
The acidity of phenols is due to the greater resonance stabilization of the phenoxide ion relative to phenol. Therefore, electron withdrawing groups (EWG) like - $\mathrm{NO}_{2}$ which stabilize the phenoxide ion more by dispersing the negative charge will tend to increase the acidity of phenols.
On the other hand, electron donating groups (EDG) like-alkyl group destabilize the phenoxide ion by intensifying the negative charge relative to phenol tend to decrease the acidic strength of phenols.
As methyl group has $+1$ effect and it is stronger at o-position than at p-position, (+ / effect decreases with distance) o-cresol is a weaker acid than $p$-cresol.
Thus, the order of acidic strength.
$p$ - nitrophenol $>$ phenol $>p$ - cresol $>0$-cresol

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Which of the following is structure of a separating funnel?

Options:
A
B
C
D

The correct answer is:

Solution:
When an organic compound is present in an aqueous medium, it is separated by shaking it with an organic solvent in which it is more soluble than in water. The organic solvent and the aqueous solution should be immiscible with each other so that they form two distinct layers which can be separated by a separatory funnel.

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2022 (27 Jun Shift 1)
Which of the following is the buffer solution of strong acidic nature

Options:
A $\mathrm{HCOOH}+\mathrm{HCOO}^-$
B $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COO}^{-}$
C $\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+\mathrm{C}_2 \mathrm{O}_4^{2-}$
D $\mathrm{H}_3 \mathrm{BO}_3+\mathrm{BO}_3^{3-}$

The correct answer is:
$\mathrm{HCOOH}+\mathrm{HCOO}^-$

Solution:
It is a buffer solution of strong acid and its weak conjugate base.

Topic: Chemistry, Ionic Equilibrium, JEE Main
Which of the following is the correct combination of the class of vertebrates and properties exhibited by it?

Options:
A Internal fertilization: Aves and Reptilia
B Oviparity: Chondrichthyes and Osteichthyes
C Poikilothermy: Amphibia and Mammalia
D Three chambered heart: Amphibia and Aves

The correct answer is:
Internal fertilization: Aves and Reptilia

Solution:
Most of the members of class Chondrichthyes (cartilaginous fishes) are viviparous, while those belonging to class Osteichthyes (bony fishes) are oviparous. Amphibians cannot maintain a constant internal temperature (poikilothermic), while mammals can maintain a constant internal temperature (homeothermic). Amphibians have a three-chambered heart, while birds have a four-chambered heart. Both birds and reptiles have internal fertilization.

Topic: Biology, Animal Kingdom, NEET
Which of the following is the correct description of follicular phase of menstrual cycle?

Options:
A Primary follicles in the ovary grow to become a fully mature Graafian follicle and simultaneously the endometrium of uterus regenerates through proliferation.
B The corpus luteum secretes large amounts of progesterone which is essential for maintenance of the endometrium.
C Both LH and FSH attain a peak level about 10th day leading to rupture of Graafian follicle.
D These changes in the ovary and the uterus are induced by changes in the levels of adrenal gland.

The correct answer is:
Primary follicles in the ovary grow to become a fully mature Graafian follicle and simultaneously the endometrium of uterus regenerates through proliferation.

Solution:
Phases of the menstrual cycle:
I. Menstrual phase (3–5 days): Breakdown of the endometrial lining of the uterus and its blood vessels occurs that comes out of the vagina as menstrual fluid.
II. Follicular phase: Primary follicles grow to become Graafian follicles. Regeneration of lost endometrium starts with hormonal changes, and a gradual increase in the release of gonadotropins like LH and FSH takes place, which stimulates the follicular development and estrogen release by the growing follicles.
III. Ovulatory phase (14th day): LH surge occurs with a peak concentration of LH during the mid-cycle, that ruptures the Graafian follicle and releases an ovum (ovulation).
IV. Luteal phase (next 14 days): The remnants of the Graafian follicle are transformed into the corpus luteum that secretes large amounts of progesterone to maintain the endometrium (necessary for implantation of the fertilized ovum).

Topic: Biology, Human Reproduction, NEET
Which of the following is the correct electron dot structure of \( \mathrm{N}_{2} \mathrm{O} \) molecule ?

Options:
A \( : \mathrm{N}=\mathrm{N}=\ddot{\mathrm{O}} \)
B \( : \mathrm{N}=N^{+}-\ddot{\mathrm{O}}: \)
C \( : \ddot{N}=N=N=O-O-N=\ddot{O} \)
D \( : \ddot{N}-\mathrm{N}=\ddot{O} \)

The correct answer is:
\( : \mathrm{N}=N^{+}-\ddot{\mathrm{O}}: \)

Solution:
Electron dot structure of $\mathrm{N}_{2} \mathrm{O}$ molecule
First draw the main skeleton structure of $\mathrm{N}_{2} \mathrm{O}$ having $\mathrm{N}-\mathrm{N}-\mathrm{O}$ connectivity with single bonds without showing any formal charge, that is,
Total number of valence electrons $=(5 \times 2)+(6 \times 1)=16 e^{-} s$
Four electrons are used to form two bonds in the skeleton structure
Remaining valence electrons $=16-4=12$
The central nitrogen atom needs two more lone pairs, terminal nitrogen atom needs three more lone pairs and oxygen
atom needs three more lone pairs to complete their octet.
Therefore, total 16 electrons are needed. But we have 12 electrons.
Hence, 2 additional bonds are formed, which can be placed in three different ways,
The most appropriate structure of $N_{2} O$ is
The resonating structures of $\mathrm{N}_{2} \mathrm{O}$ are as follows:

Topic: Chemistry, Aldehydes and Ketones, KCET, KCET 2017
Which of the following is the correct expression and value for the heat current (in watts) through the compound bar formed by soldering an iron bar (L1 = 0.1 m, A1 = 0.02 ${{m}^{2}}$, K1 = 79 W m⁻¹ K⁻¹) and a brass bar (L2 = 0.1 m, A2 = 0.02${{m}^{2}}$, K2 = 109 W m⁻¹ K⁻¹) together, with the free ends of the iron bar and brass bar maintained at 373 K and 273 K, respectively?

Options:
A 542.0 W
B 678.5 W
C 789.3 W
D 916.1 W

The correct answer is:
916.1 W

Solution:
Correct Option is : (D)
916.1 W

Topic: Chemistry, Thermal Properties of Matter, NEET
Which of the following is the correct matching of the events occurring during menstrual cycle ?

Options:
A Menstruation - Breakdown of myometrium and ovum not fertilized
B Ovulation - LH and FSH attain peak level and sharp fall in the secretion of progesterone
C Proliferative phase - Rapid regeneration of myometrium and maturation of Grafian follicle
D Development of corpus luteum - Secretory phase and increased secretion of progesterone.

The correct answer is:
Development of corpus luteum - Secretory phase and increased secretion of progesterone.

Solution:
No Solution Available

Topic: Biology, Human Reproduction, NEET, NEET 2009 (Mains)
Which of the following is the correct matching pair of a sanctuary and its main protected wild animal ?

Options:
A Gir-Lion
B Sariska-Tiger
C Sunderban-Rhino
D Kaziranga-Musk deer

The correct answer is:
Gir-Lion

Solution:
Gir forest is important for protecting lion. The ecosystem of Gir, with its diverse flora and fauna, is result of the efforts of the Government, forest department wildlife activists and NGOs. The forest area of Gir and its lions were declared as "protected" in the early 1900's by the then Nwab of the princely state of Junagadh. This initiative assisted in the conservation of the lions whose population had plummeted to only 15 through slaughter for trophy hunting.

Topic: Biology, Biodiversity and its Conservation, NEET
Which of the following is the correct order of decreasing $\mathrm{SN}^2$ reactivity?

Options:
A
$\mathrm{RCH}_2 \mathrm{X}>\mathrm{R}_3 \mathrm{CX}>\mathrm{R}_2 \mathrm{CHX}$

B
$\mathrm{RCH}_2 \mathrm{X}>\mathrm{R}_2 \mathrm{CHX}>\mathrm{R}_3 \mathrm{CX}$

C
$\mathrm{R}_3 \mathrm{CX}>\mathrm{R}_2 \mathrm{CHX}>\mathrm{RCH}_2 X$

D
$\mathrm{R}_2 \mathrm{CHX}>\mathrm{R}_3 \mathrm{CX}>\mathrm{RCH}_2 \mathrm{X}$
$(\mathrm{X}=$ a halogen $)$

The correct answer is:
$\mathrm{RCH}_2 \mathrm{X}>\mathrm{R}_2 \mathrm{CHX}>\mathrm{R}_3 \mathrm{CX}$

Solution:
More is the steric hindrance at the carbon bearing the halogen, lesser is the $\mathrm{S}_{\mathrm{N}} 2$ reactivity.
Hence, (B) is correct.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Which of the following is the correct order of decreasing $\mathrm{SN}^2$ reactivity?

Options:
A
$\mathrm{RCH}_2 \mathrm{X}>\mathrm{R}_3 \mathrm{CX}>\mathrm{R}_2 \mathrm{CHX}$

B
$\mathrm{RCH}_2 \mathrm{X}>\mathrm{R}_2 \mathrm{CHX}>\mathrm{R}_3 \mathrm{CX}$

C
$\mathrm{R}_3 \mathrm{CX}>\mathrm{R}_2 \mathrm{CHX}>\mathrm{RCH}_2 X$

D
$\mathrm{R}_2 \mathrm{CHX}>\mathrm{R}_3 \mathrm{CX}>\mathrm{RCH}_2 \mathrm{X}$
$(\mathrm{X}=$ a halogen $)$

The correct answer is:
$\mathrm{RCH}_2 \mathrm{X}>\mathrm{R}_2 \mathrm{CHX}>\mathrm{R}_3 \mathrm{CX}$

Solution:
More is the steric hindrance at the carbon bearing the halogen, lesser is the $\mathrm{S}_{\mathrm{N}} 2$ reactivity.
Hence, (B) is correct.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2007
Which of the following is the correct order of increasing number of lone pair of electrons on the central atom?

Options:
A $\mathrm{IF}_5 < \mathrm{XeF}_2 < \mathrm{IF}_7 < \mathrm{ClF}_3$
B $\mathrm{IF}_7 < \mathrm{ClF}_3 < \mathrm{XeF}_2 < \mathrm{IF}_5$
C $\mathrm{IF}_7 < \mathrm{XeF}_2 < \mathrm{ClF}_2 < \mathrm{IF}_5$
D $\mathrm{IF}_7 < \mathrm{IF}_5 < \mathrm{ClF}_3 < \mathrm{XeF}_2$

The correct answer is:
$\mathrm{IF}_7 < \mathrm{IF}_5 < \mathrm{ClF}_3 < \mathrm{XeF}_2$

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Which of the following is the correct order of melting points of group \( 16 \) elements?

Options:
A \( \mathrm{O} < \mathrm{S} < \mathrm{Se} < \mathrm{Te} < \mathrm{Po} \)
B \( \mathrm{O} < \mathrm{Se} < \mathrm{S} < \mathrm{Te} < \mathrm{Po} \)
C \( \mathrm{O} < \mathrm{S} < \mathrm{Se} < \mathrm{Te}>\mathrm{Po} \)
D \( \mathrm{O} < \mathrm{S} < \mathrm{Se}>\mathrm{Te} < \mathrm{Po} \)

The correct answer is:
\( \mathrm{O} < \mathrm{S} < \mathrm{Se} < \mathrm{Te}>\mathrm{Po} \)

Solution:
Melting point is the temperature where a solid is converted to liquid, both solids and liquids are in equilibrium.

Parameters which affect melting point are:

Ionic bonds.

Intermolecular force.

Shape of molecules.

Size of molecules.

One of the weak intermolecular force is Vanderwaal's force.

Down the group, the size of the molecule increases Thus, Vanderwaal's force increases and the melting point also increases.

But Polonium (Po)shows less melting point than Tellurium (Te) due to decrease in Vanderwaal's radius.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Which of the following is the correct order of radius?

Options:
A \( \mathrm{H}^{-}>\mathrm{H}>\mathrm{H}^{+} \)
B \( \mathrm{Na}^{+}>\mathrm{F}^{-}>\mathrm{O}^{2-} \)
C \( \mathrm{F}^{-}>\mathrm{O}^{2-}>\mathrm{Na}^{+} \)
D \( \mathrm{Al}^{3+}>\mathrm{Mg}^{2+}>\mathrm{N}^{3-} \)

The correct answer is:
\( \mathrm{H}^{-}>\mathrm{H}>\mathrm{H}^{+} \)

Solution:
The size of anion is always higher than that of the parentatom due to the decrease in effective nuclear charge on anion
\( \left(Z_{\text {eff }}\right) \) while, the size of cation is always smaller than parent atom due to lesser \( \left(Z_{\text {eff }}\right) \) than parent atom. Hence, the
correct given order is \( H^{-}>H>H^{+} \)

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Which of the following is the correct order of reactivity towards the oxidation of primary alcohols to aldehydes?

Options:
A $\mathrm{PCC}\gt\mathrm{KMnO}_4\gt\mathrm{CrO}_3$
B $\mathrm{CrO}_3\gt\mathrm{KMnO}_4\gt\mathrm{PCC}$
C $\mathrm{KMnO}_4\gt\mathrm{PCC}\gt\mathrm{CrO}_3$
D $\mathrm{PCC}\gt\mathrm{CrO}_3\gt\mathrm{KMnO}_4$

The correct answer is:
$\mathrm{PCC}\gt\mathrm{CrO}_3\gt\mathrm{KMnO}_4$

Solution:
$\mathrm{PCC}\gt\mathrm{CrO}_3\gt\mathrm{KMnO}_4$
Explanation: PCC is a mild oxidizing agent that selectively oxidizes primary alcohols to aldehydes without further oxidation. $\mathrm{CrO}_3$ in aqueous acidic conditions is a stronger oxidizer and can oxidize alcohols to carboxylic acids. $\mathrm{KMnO}_4$ is the strongest and typically oxidizes alcohols all the way to carboxylic acids.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Which of the following is the correct order of stability of allotropes of carbon?

Options:
A Diamond \( < \) graphite \( < \) fullerene
B Fullerene \( < \) graphite \( < \) diamond
C Graphite \( < \) fullerene \( < \) diamond
D Fullerene \( < \) diamond \( < \) graphite

The correct answer is:
Fullerene \( < \) diamond \( < \) graphite

Solution:
This problem is based on the stability of allotropes of carbon. Use the physical properties of allotropes of carbon. Use enthalpy of formation to solve the problem..
We know diamond is the hardest known material that's why many students get confused that diamond is the most stable material but this is wrong. The most stable allotrope of carbon is graphite.
The fullerene is least stable among these three is due to strained molecule. Graphite has maximum stability due to high negative value of enthalpy of formation.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Which of the following is the correct order of stability of allotropes of carbon ?

Options:
A Diamond < graphite < fullerene
B Fullerene < graphite < diamond
C Graphite < fullerene < diamond
D Fullerene < diamond < graphite

The correct answer is:
Fullerene < diamond < graphite

Solution:
This problem is based on the stability of allotropes of carbon. Use the physical properties of allotropes of carbon. Use enthalpy of formation to solve the problem..
We know diamond is the hardest known material that's why many students get confused that diamond is the most stable material but this is wrong. The most stable allotrope of carbon is graphite.
The fullerene is least stable among these three is due to strained molecule. Graphite has maximum stability due to high negative value of enthalpy of formation.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Which of the following is the correct plot for the probability density ψ2r as a function of distance 'r' of the electron form the nucleus for 2s orbital?

Options:
A
B
C
D

The correct answer is:

Solution:
Ψ2 vs r plot for 2s orbital of single-electron species will be

Here Ψ2 represent probability density of finding an electron wave at a position around nucleus. The 2s orbital will also have a nodal plane.

n-l-1=2-0-1=1

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Which of the following is the correct prediction about observed B−F bond length in BF3 molecule?

Options:
A BâF bond length in BF3 is found to be less than theoretical value because the electronegativity values of B2.04 and F4.0 suggest the bond length to be ionic and hence, the attraction between oppositely charged ions must decrease the bond length.
B BF3 and BF4â have equal BâF bond length.
C The decrease in the BâF bond length in BF3 is due to delocalised p Ï - p Ï bonding between vacant 2p orbital of B and filled 2p orbital of F
D The correct B−X bond length order is B−F>B−Cl>B−Br>B−I

The correct answer is:
The decrease in the BâF bond length in BF3 is due to delocalised p Ï - p Ï bonding between vacant 2p orbital of B and filled 2p orbital of F

Solution:
It has partial double character in BâF bond due to p Ï - p Ï back bonding.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main