datavorous/entrance-exam-dataset huggingface dataset:
Consider an electron in front of metallic surface at a distance d (treated as an infinite plane surface). Assume the force of attraction by the plate is given as
What is the work done in taking a charge to an infinite distance from a plate, given that the distance (d) is 0.1 nm? Calculate the work done in electron volts.

Options:
A 3.2 eV
B 2.6 eV
C 3.6 eV
D 4.1 eV

The correct answer is:
3.6 eV

Solution:
Correct Option is : (C)
3.6 eV

Topic: Chemistry, Dual Nature of Matter, NEET
Consider an ideal gas with following distribution of speeds.
Calculate T ( 3.0× 10-26kg)

Options:
A 273K
B 296K
C 331K
D 308K

The correct answer is:
296K

Solution:
Correct Option is : (B)
296K

Topic: Chemistry, Kinetic Theory of Gases, NEET
Consider an interference pattern between two coherent sources. If I1 and I2 be intensities at points where the phase difference are Ï 3 and 2 Ï 3 respectively, then the intensity at maxima is

Options:
A I2-3I12
B I1-3I22
C 3I2+I12
D 3I1-I22

The correct answer is:
3I1-I22

Solution:
I1=IA+IB+2IAIBcosπ3...(1)
I2=IA+IB+2IAIBcos2π3 ...(2)
⇒ I1+I2=2IA+IB...(3)
From (1)
⇒ I1=I1+I22+IAIB
I1-I1+I22=IAIB
I1-I22=IAIB
Imax=IA+IB2
=IA+IB+2IAIB ...(4)
Imax=I1+I22+I1-I2
=3I1-I22

Topic: Physics, Wave Optics, NEET
Consider an interference pattern between two coherent sources. If I1 and I2 be intensities at points where the phase difference are Ï 3 and 2 Ï 3 respectively, then the intensity at maxima is

Options:
A I2-3I12
B I1-3I22
C 3I2+I12
D 3I1-I22

The correct answer is:
3I1-I22

Solution:
I 1 = I A + I B + 2 I A I B cos Ï 3 ...(1)
I 2 = I A + I B + 2 I A I B cos 2 Ï 3 ...(2)
â I 1 + I 2 = 2 I A + I B ...(3)
From (1)
â I 1 = I 1 + I 2 2 + I A I B
I 1 - I 1 + I 2 2 = I A I B
I 1 - I 2 2 = I A I B
I max = I A + I B 2
= I A + I B + 2 I A I B ...(4)
I max = I 1 + I 2 2 + I 1 - I 2
= 3 I 1 - I 2 2

Topic: Physics, Wave Optics, NEET
Consider an ion with the following structure: [H2C=C=NH2]+. For this ion, we can define two planes: one plane containing H-C-H group, the other plane contaning H-N-H group. What is the relationship between these planes ?

Options:
A They are at 1200
B They are perpendicular to each other
C They are in the same plane
D More information is required

The correct answer is:
They are perpendicular to each other

Solution:
Like allene, the two pi- bonds are perpendicular to one another.

Topic: Chemistry, General Organic Chemistry, NEET
Consider Fraunhofer diffraction pattern obtained with a single slit illuminated at normal incidence. At the angular position of the first diffraction minimum, the phase difference (in radians) between the wavelets from the opposite edges of the slit is

Options:
A π
B 2π
C π4
D π2

The correct answer is:
2π

Solution:
For Ist minimum θ=λa .
and ϕ=2πλ.Δx=2πλasin⁡θ
ϕ=2πλ.aθ=2πλaλa
∴ ϕ=2π

Topic: Physics, Wave Optics, NEET
Consider that the Earth is revolving around the Sun in a circular orbit with a period T. The area of the circular orbit is directly proportional to

Options:
A T23

B T13

C T43

D T12

The correct answer is:
T43

Solution:
From Kepler's law T2 ∝ R3
∵ Area ∝ R2
â∴ Area ∝ T43

Topic: Physics, Gravitation, NEET
Consider the above reaction and identify the missing reagent/chemical.

Options:
A B2H6
B Red Phosphorus
C CaO
D DIBAL-H

The correct answer is:
CaO

Solution:
Given reaction is soda lime decarboxylation which takes place with NaOH and CaO. In this reaction carbanion is formed as intermediate.

Topic: Chemistry, Carboxylic Acid Derivatives, NEET, NEET 2021
Consider the change in oxidation state of Bromine corresponding to different emf values as
shown in the diagram below

Then the species undergoing disproportionation is

Options:
A Br2
B BrO4-
C BrO3-
D HBrO

The correct answer is:
HBrO

Solution:
HBrO+1→Br20,EHBrO/Br20=1.595V
HBrO+1→Br+5O3-,EBrO3-/HBrO0=1.5V
Ecell0 for the disproportionation of HBrO,
Ecell0=EHBrO/Br20-EBrO3-/HBrO0
=1.595-1.5=0.095V=+ve
Hence, option (C) is correct answer as Ecell0 is positive in it.

Topic: Chemistry, Electrochemistry, NEET
Consider the change in oxidation state of Bromine corresponding to different emf values, as shown in the diagram below:
Then, what is the species undergoing disproportionation?

Options:
A Br2
B HBrO
C BrO4-
D BrO3-

The correct answer is:
HBrO

Solution:
Calculate Ecello corresponding to each compound undergoing disproportionation reaction. The reaction for which Ecello comes out +ve is spontaneous.
HBrO→Br2 Eo=1.595, SRP Cathode
2HBrO→Br2+BrO3-
Ecello= SRP (cathode) - SRP (Anode)
=1.595-1.5
=0.095 V
Ecello>0⇒ ΔGo<0 [spontaneous]

Topic: Chemistry, Electrochemistry, NEET, NEET 2018
Consider the circuit in Fig 3.10. How much energy is absorbed by electrons from the initial state of no current (ignoring thermal motion) to the state of drift velocity?

Options:
A 5× 10-17J
B 1× 10-17J
C 2× 10-17J
D 3× 10-17J

The correct answer is:
2× 10-17J

Solution:
Correct Option is : (C)
2× 10-17J

Topic: Chemistry, Current Electricity, NEET
Consider the compounds, BCl₃ and CCl₄. How will they behave with water? Justify.

Options:
A BCl₃ will react with water to form HCl and a boron-containing compound. CCl₄ will not react with water.
B CCl₄ will react with water to form HCl and a carbon-containing compound. BCl₃ will not react with water.
C Both BCl₃ and CCl₄ will react with water to form HCl and boron- or carbon-containing compounds, respectively.
D Neither BCl₃ nor CCl₄ will react with water.

The correct answer is:
BCl₃ will react with water to form HCl and a boron-containing compound. CCl₄ will not react with water.

Solution:
Correct Option is : (A)
BCl₃ will react with water to form HCl and a boron-containing compound. CCl₄ will not react with water.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
Consider the decomposition of N2O5 as
N2O5→2NO2+12O2
The rate of reaction is given by
-d[N2O5]dt=12d[NO2]dt=2d[O2]dt=k1[N2O5]
Therefore, -d[N2O5]dt=k1[N2O5]
+d[NO2]dt=2k1[N2O5]=k1'[N2O5]
d[O2]dt=12k1[N2O5]=k1"[N2O5]
Choose the correct option.

Options:
A 4k1=2k1'=k1"
B 4k1=k1'=2k1"
C k1=2k1'=k1"
D 2k1=k1'=4k1"

The correct answer is:
2k1=k1'=4k1"

Solution:
-d[N2O5]dt=12d[NO2]dt=2d[O2]dt
k1[N2O5]=12k1'[N2O5]=2×k1"[N2O5]
k1=12k1'=2k1"
=2k1=k1'=4k1"

Topic: Chemistry, Chemical Kinetics, NEET
Consider the diagram below and choose the correct option regarding the structures (a-d).

Options:
A

a	b	c	d

Colleterial gland	Vas deferens	Ejaculatory duct	Anal cercus

B

a	b	c	d

Phallic gland	Vas deferens	Ejaculatory duct	Anal cercus

C

a	b	c	d

Phallic gland	Ejaculatory deferens	Vas duct	Caudal style

D

a	b	c	d

Phallic gland	Vas deferens	Ejaculatory duct	Titilator

The correct answer is:

a	b	c	d

Phallic gland	Vas deferens	Ejaculatory duct	Anal cercus

Solution:
The diagram given above is of the male reproductive system of cockroach. The labelled structures are phallic gland, vas deferens, ejaculatory duct, and anal cercus.

Topic: Biology, Structural Organisation in Animals, NEET
Consider the diagram given below and choose the correct option.

Options:
A Structure(a) = Urethral meatus,
Function (a) = Storage of urine
B Structure(b) = Seminal vesicle,
Function (b) = Attraction between opposite sex
C Structure(c) = Prostate gland,
Function(c) = Fuel of sperm
D Structure(d) = Bulbourethral gland,
Function(d) = Lubrication of the penis

The correct answer is:
Structure(d) = Bulbourethral gland,
Function(d) = Lubrication of the penis

Solution:
Bulbourethral gland A pea-sized gland in the male located behind and to the side of the urethra that discharges a component of seminal fluid into the urethra.

Topic: Biology, Human Reproduction, NEET
Consider the earth as a uniform sphere of mass M and radius R. Imagine a straight smooth tunnel made through the earth which connects any two points on its surface. Determine the time that a particle would take to go from one end to the other through the tunnel.

Options:
A 2πR3GM
B ÏR3GM
C π2R3GM
D None of these

The correct answer is:
πR3GM

Solution:
Suppose at some instant the particle is at radial distance r from centre of earth O. Since, the particle is constrained to move along the tunnel, we define its position as distance x from C. Hence, equation of motion of the particle is,
max = Fx
The gravitational force on mass m at distance r is,
F = GMmr R 3 (towards O)
Therefore, F x = - F sin Î¸ = - GMmr R 3 x r
= - GMmr R 3 Â· x
Since, Fx â - x, motion is simple harmonic in nature. Further,
m a x = - GMm R 3 Â· x or a x =- GM R 3 Â· x
â´ Time period of oscillation is,
T = 2 Ï | x a x | = 2 Ï R 3 GM
The time taken by particle to go from one end to the other is T 2.
t = T 2 = Ï R 3 GM

Topic: Physics, Oscillations, NEET
Consider the elements: Cs, Ne, I, and F. Identify the element that exhibits only negative oxidation state.

Options:
A Cs
B Ne
C I
D F

The correct answer is:
Ne

Solution:
Correct Option is : (B)
Ne

Topic: Chemistry, Redox Reactions, NEET
Consider the equations

P=LimΔS→0FΔS and P1-P2=ρgz

In an elevator accelerating upward,

Options:
A both the equations are valid
B first is valid but not the second
C second is valid but not the first
D both are invalid

The correct answer is:
first is valid but not the second

Solution:
Equation 1 is universal equation of pressure but the second equation is valid only when it is in state of rest, while moving upward the pressure difference is ΔP=zρg+a

Topic: Physics, Mechanical Properties of Fluids, NEET
Consider the equilibrium
CO2(g)⇌CO(g)+12O2(g)
The equilibrium constant K is given by (when α <<< 1 )

Options:
A K=α3/22
B K=α32
C K=α3/22
D K=α3/23

The correct answer is:
K=α3/22

Solution:
K=[Î±]Î±21/2(1-Î±)
=Î±3/22
(α < < 1)

Topic: Chemistry, Chemical Equilibrium, NEET
Consider the figure given:
Cell ‘A’ has ECell= 2 V and Cell ‘B’ has ECell= 1.1 V which of the two cells ‘A’ or ‘B’ will act as an electrolytic cell. Which electrode reactions will occur in this cell?

Options:
A Cell ‘A’ will act as electrolytic cell as it has lower emf
B Cell ‘B’ will act as electrolytic cell as it has lower emf
C Both Cell ‘A’ and ‘B’ will act as electrolytic cell
D Neither Cell ‘A’ nor Cell ‘B’ will act as electrolytic cell

The correct answer is:
Cell ‘B’ will act as electrolytic cell as it has lower emf

Solution:
Correct Option is : (B)
Cell ‘B’ will act as electrolytic cell as it has lower emf

Topic: Chemistry, Electrochemistry, NEET
Consider the figure given:
How will concentration of Zn2+ions and Ag+ions be affected when the cell functions?

Options:
A Concentration of Zn2+ions will increase and concentration of Ag+ions will increase
B Concentration of Zn2+ions will decrease and concentration of Ag+ions will increase
C Concentration of Zn2+ions will increase and concentration of Ag+ions will decrease
D Concentration of Zn2+ions will decrease and concentration of Ag+ions will decrease

The correct answer is:
Concentration of Zn2+ions will increase and concentration of Ag+ions will decrease

Solution:
Correct Option is : (C)
Concentration of Zn2+ions will increase and concentration of Ag+ions will decrease

Topic: Chemistry, Electrochemistry, NEET
Consider the figure given:
How will the concentration of Zn2+ions and Ag+ions be affected after the cell becomes ‘dead’?

Options:
A When Ecell= 0 equilibrium is reached and concentration of Zn2+ions and Ag+ions will increase.
B When Ecell= 0 equilibrium is reached and concentration of Zn2+ions and Ag+ions will not change.
C When Ecell= 0 equilibrium is reached and concentration of Zn2+ions and Ag+ions will decrease.
D When Ecell= 0 equilibrium is reached and concentration of Zn2+ions increases and Ag+ions decreases.

The correct answer is:
When Ecell= 0 equilibrium is reached and concentration of Zn2+ions and Ag+ions will not change.

Solution:
Correct Option is : (B)
When Ecell= 0 equilibrium is reached and concentration of Zn2+ions and Ag+ions will not change.

Topic: Chemistry, Electrochemistry, NEET
Consider the figure given:
If cell ‘A’ has ECell= 0.5 V and cell ‘B’ has ECell= 1.1 V then what will be the reactions at anode and cathode?

Options:
A At cathode : Zn → Zn2++ 2e–and Cu2++ 2e–→ Cu
B At cathode : Zn → Zn2++ 2e–and at anode : Cu2++ 2e–→ Cu
C At anode : Zn → Zn2++ 2e–and Cu2++ 2e–→ Cu
D At anode : Zn → Zn2++ 2e–and at cathode : Cu2++ 2e–→ Cu

The correct answer is:
At anode : Zn → Zn2++ 2e–and at cathode : Cu2++ 2e–→ Cu

Solution:
Correct Option is : (D)
At anode : Zn → Zn2++ 2e–and at cathode : Cu2++ 2e–→ Cu

Topic: Chemistry, Electrochemistry, NEET
Consider the figure given: Is silver plate the anode or cathode?

Options:
A Ag is both the cathode and anode
B Ag is neither the cathode nor anode
C Ag is the cathode
D Ag is the anode

The correct answer is:
Ag is the cathode

Solution:
Correct Option is : (C)
Ag is the cathode

Topic: Chemistry, Electrochemistry, NEET
Consider the figure given:
What will happen if salt bridge is removed?

Options:
A It does not have any effect on the cell
B Cell will stop functioning
C The cell voltage will increase.
D The cell voltage will decrease.

The correct answer is:
Cell will stop functioning

Solution:
Correct Option is : (B)
Cell will stop functioning

Topic: Chemistry, Electrochemistry, NEET
Consider the figure given:
When will the cell stop functioning?

Options:
A when ECell= 0
B when ECellis not equal to 0
C when ECellis equal to 1
D None of the above

The correct answer is:
when ECell= 0

Solution:
Correct Option is : (A)
when ECell= 0

Topic: Chemistry, Electrochemistry, NEET
Consider the first reaction of TCA cycle

What is TRUE about compound A?

Options:
A First product of TCA cycle
B Tricarboxylic acid and six carbon compound
C It undergoes reorganisation in the presence of enzyme aconitase to form cis-aconitate
D All of these

The correct answer is:
All of these

Solution:
The TCA cycle is also known as the citric acid cycle or Krebs cycle and was discovered by Hans Kreb in 1937. The presence of citrate synthase enzyme results in the formation of citrate, i.e. 6C compound which is the first product of the Krebs cycle. In the presence of Aconitase enzymes, result in the production of isocitric acid by removing and adding water from citrate. This enzyme helps in the isomerization reaction. So the correct answer is D.

Topic: Biology, Respiration in Plants, NEET
Consider the following alkyl halides
1. (CH3)3CCH2Br
2. ClCH2CH=CH2
3. ClCH2CH2CH3
4. BrCH2CH2CH3
Arrange these alkyl halides in decreasing order of reactivity in Williamson reaction.

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

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

Solution:
The Williamson reaction involves nucleophilic substitution of alkoxide ion for halide ion. It is strictly analogous to the preparation of alcohol by treatment of alkyl halide with aqueous hydroxide. Allyl halides undergo nucleophilic substitution at faster rate than aliphatic halides and bromides are more reactive than the corresponding chloride.
Neopentyl-type halides are inert to nucleophilic substitution (SN2) because they show maximum steric hindrance to the approaching nucleophilic.
Here the reactivity order is 2 > 4 > 3 > 1.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Consider the following carbocations
1. C6H5C+H2
2. C6H5C+HCH3
3. (C6H5)2C+H
4. (C6H5)3C+
The correct sequence of increasing order of their stabilities is

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

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

Solution:
Resonance effect (+ M effect) are always more stabilizing than the inductive effects (+I effect), because +M groups neutralize the +ve charge on the carbon atom more effectively than the +I groups.
Triphenyl carbocation in the most stable in this series, because its +ve charge is dispersed by resonance (+M effect of -C6H5 group).

The correct order of stability is
1 < 2 < 3 < 4

Topic: Chemistry, General Organic Chemistry, NEET
Consider the following complex ions P, Q and R.
P=[FeF6]-3,Q=[V(H2O)6]2+ and R=[Fe(H2O)6]2+
The correct order of the complex ions, according to their spin only magnetic moment
values (in B.M.) is

Options:
A R<Q<P
B Q<R<P
C R<P<Q
D Q<P<R

The correct answer is:
Q<R<P

Solution:
P=[FeF6]3- i.e. Fe+3=[Ar]3d5 so, n = 5
Magnetic moment =n(n+2)BM=5(5+2)=35BM
Q=[V(H2O)6]+2,V+2=[Ar]3d3 , n = 3
Magnetic moment =n(n+2)BM=3(3+2)=15BM
R=[Fe(H2O)6]+2 i.e. Fe+2=[Ar]3d6 , n = 4
Magnetic moment =4(4+2)=24BM
Hence the order is P > R > Q

Topic: Chemistry, Coordination Compounds, NEET
Consider the following compounds
1. Phenol
2. o-chlorophenol
3. m-chlorophenol
4. p-chlorophenol
Place these compounds in the decreasing order of acidity

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

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

Solution:
Although Cl is electron donating by resonance, yet its electron withdrawing acid-strengthening inductive effect (that decreases with increasing separation of Cl and OH) predominates over electron donating, acid weakening resonance effect. This makes all chlorophenols more acidic than phenol. As –I effect decreases with distance so the decreasing order of acidity is 2 > 3 > 4 > 1.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Consider the following compounds

Hyper conjugation occurs in

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

The correct answer is:
III only

Solution:
Only (III) has α -H in conjugation with free radical so hyper conjugation is possible.

Topic: Chemistry, General Organic Chemistry, NEET
Consider the following compounds

Hyperconjugation occurs in:

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

The correct answer is:
III only

Solution:
Hyper conjugation possible in only due to presence of α-H .

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2015 (Phase 1)
Consider the following compounds:
$\text { (i) } \mathrm{C}_6 \mathrm{H}_5 \mathrm{COCl}$
The correct decreasing order of their reactivity towards hydrolysis is:

Options:
A (i) $>$ (ii) $>$ (iii) $>$ (iv)
B (iv) $>$ (ii) $>$ (i) $>$ (iii)
C (ii) $>$ (iv) $>$ (i) $>$ (iii)
D (ii) $>$ (iv) $>$ (iii) $>$ (iv)

The correct answer is:
(ii) $>$ (iv) $>$ (i) $>$ (iii)

Solution:
Electron withdrawing group increases the positive charge and electron releasing group decreases the negative change. The degree of hydrolysis increases as the magnitude of positive charge on carbonyl group increases. Among these $\mathrm{NO}_2$ \& $\mathrm{CHO}$ are electron withdrawing group from which $\mathrm{NO}_2$ has more $-\mathrm{I}$ effect than-CHO. On the other hand $\mathrm{CH}_3$ is a electron releasing group.
Thus, the order of reactivity towards hydrolysis is
Related Theory
The mesomeric effect is negative $(-M)$ when the substituent is an electron-withdrawing group and the effect is positive $(+M)$ when the substituent is an electron donating group.

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2007
Consider the following compounds

Order of basicity of these compounds in decreasing …… is

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

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

Solution:
Basicity decreases as in the following order 1 > 3 > 2 > 4.
Here (1) is more basic than (3) as in (3) oxygen atom is electron withdrawing. (2) is more basic than (4) as in (4) lone pair is delocalized. (3) and (4) are less basic than (1) and (2) due to more s%.

Topic: Chemistry, General Organic Chemistry, NEET
Consider the following compounds

The correct decreasing order of their reactivity towards hydrolysis is

Options:
A (i) > (ii) > (iii) > (iv)
B (iv) > (ii) > (i) > (iii)
C (ii) > (iv) > (i) > (iii)
D (ii) > (iv) > (iii) > (i)

The correct answer is:
(ii) > (iv) > (i) > (iii)

Solution:
The ease of hydrolysis depends upon the magnitude of the +ve charge on the carbonyl group. Electron-withdrawing groups increase the magnitude of positive charge and electron donating groups decrease the magnitude of positive charge. Hence, the decreasing order of reactivity towards hydrolysis is

Topic: Chemistry, Carboxylic Acid Derivatives, NEET
Consider the following diagram in which an electrochemical cell is coupled to an electrolytic cell. What will be the polarity of electrodes ‘A’ and ‘B’ in the electrolytic cell?

Options:
A ‘A’ will have positive polarity and ‘B’ will have negative polarity
B ‘A’ will have positive polarity and ‘B’ will have positive polarity
C ‘A’ will have negative polarity and ‘B’ will have positive polarity
D ‘A’ will have negative polarity and ‘B’ will have negative polarity

The correct answer is:
‘A’ will have negative polarity and ‘B’ will have positive polarity

Solution:
Correct Option is : (C)
‘A’ will have negative polarity and ‘B’ will have positive polarity

Topic: Chemistry, Electrochemistry, NEET
Consider the following diazonium ions :

The order of reactivity towards diazo-coupling with phenol in the presence of dil. NaOH is -

Options:
A I < IV < II < III
B I < III < IV < II
C III < I < II < iV
D III < I < IV < II

The correct answer is:
I < III < IV < II

Solution:
Diazonium ion acts as an electrophile, electron-withdrawing groups on its phenyl ring increases its electrophilicity.

Topic: Chemistry, Amines, NEET
Consider the following dioxide of group 14
1. CO2
2. SiO2
3. GeO2
4. SnO2
5. PbO2
The basicity of the dioxide alters in the order

Options:
A CO2>SiO2>GeO2>SnO2>PbO2
B CO2<SiO2<GeO2<SnO2<PbO2
C CO2<SiO2<GeO2<PbO2<SnO2
D CO2<SiO2<SnO2<PbO2<GeO2

The correct answer is:
CO2<SiO2<GeO2<SnO2<PbO2

Solution:
The basicity of the dioxide alters in the order
CO2<SiO2<GeO2<SnO2<PbO2
The basicity increases on moving down the group. Thus, CO2 and SiO2 are purely acidic. GeO2 is not as strongly acidic as SiO2 . It is weakly acidic and weakly basic. SnO2 and PbO2 are definitely amphoteric.

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
Consider the following E° values:
E°Li+|Li=-3.05 V, E°Cu2+|Cu=+0.34 V
Under similar conditions, the potential for the reaction Cu+2Li+→Cu2++2Li, is

Options:
A -3.39 V
B +3.39 V
C -2.69 V
D +2.69 V

The correct answer is:
-3.39 V

Solution:
E Cell o = E red o Cathode − E red o Anode
Given
EÂ°Li+|Li=-3.05 V, EÂ°Cu2+|Cu=+0.34 V
For the reaction, Cu+2Li+âCu2++2Li,
EÂ°=-3.05+-0.34=-3.39 V

Topic: Chemistry, Electrochemistry, NEET
Consider the following electrochemical cell at 298 K
Ag(s)|AgI(s)|I-(aq)||Cl-(aq)|Hg2Cl2|Hg(l)Pt(s)
[Given ECl-|Hg2Cl2|Hgo=0.26V.EAg+/Ago=0.8V,Ksp(AgI)=10-16 and 2.303RTF=0.06 ]
The overall reaction occurring in the above cell is

Options:
A 12Hg2Cl2(s)+Ag(s)+I-(aq)→Hg(l)+Cl-(aq)+AgI(s)
B Ag+(aq)+I-(aq)→AgI(s)
C Hg22+(aq)+I2Cl-(aq)→Hg2Cl2(s)
D Cell reaction is not possible

The correct answer is:
12Hg2Cl2(s)+Ag(s)+I-(aq)→Hg(l)+Cl-(aq)+AgI(s)

Solution:
Anode Ag+I-→AgI+e-
Cathode Hg2Cl2+2e→2Hg+2Cl-
Hence, The overall reaction occurring in the above cell is
12Hg2Cl2(s)+Ag(s)+I-(aq)→Hg(l)+Cl-(aq)+AgI(s)

Topic: Chemistry, Electrochemistry, NEET
Consider the following features seen in a plant:

I: Male and female reproductive organs are generally found in separate flowers.
II: The male flowers having a number of long filaments terminating in exposed stamens.
III: The female flowers having long feather-like stigmas.

The flowers of this plant would most likely be pollinated by:

Options:
A Wind
B Water
C Bees
D Birds

The correct answer is:
Wind

Solution:
Pollination is the process of transfer of pollen grain from anther to stigma. If it is between different flower of different plant, then it is called cross pollination. From the given information, since male and female flowers are separate, it will follow cross pollination. The flower is not attractive but has long filament exposing the stamens and stigma is feathery. This suggests the agent of pollination to be wind rather than animal. As there is no mention of mucilage secretion on pollen, it is not water pollinated.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET