datavorous/entrance-exam-dataset huggingface dataset:
Consider the following four conditions (i) - (iv) and select the correct pair of them as an adaptation to the environment in desert lizards.
The conditions :

(i) Burrowing in the soil to escape high temperature.
(ii) Losing heat rapidly from the body during high temperature.
(iii) Bask in the sun when the temperature is low.
(iv) Insulating body due to the thick fatty dermis.

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

The correct answer is:
(i), (iii)

Solution:
Adaptation is any attribute of the organism (morphological, physiological, behavioural) that enables the organism to survive and reproduce in its habitat. Many adaptations have evolved over a long evolutionary time and are genetically fixed.
Some organisms show behavioural responses to cope up with variations in their environment. Desert lizards lack the physiological ability that mammals have to deal with the high temperatures of their habitat, but manage to keep their body temperature fairly constant by behavioural means. They bask in the sun and absorb heat when their body temperature drops below the comfort zone, but move into shade when the ambient temperature starts increasing. Some species are capable of burrowing into the soil to hide and escape from the above-ground heat.

Topic: Biology, Organisms and Populations, NEET
Consider the following four conditions (I-IV) and select the correct pair of them as adaptation to environment in desert lizards.
The conditions :
I. Burrowing in soil to escape high temperature
II. Losing heat rapidly from the body during high temperature
III. Bask in sun when temperature is low
IV. Insulating body due to thick fatty dermis

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

The correct answer is:
(I) and (III)

Solution:
Some organims show behavioural responses to cope with variations in their environment. Desert lizzards bask in the sun and absorb heat when their body temperature drops below the comfort zone, but move into shade when the ambient temperature starts increasing. Some species are capable of burrowing into the soil to hide and escape from the above-ground heat.

Topic: Biology, Organisms and Populations, NEET, NEET 2011 (Screening)
Consider the following four conditions (i-iv) and select the correct pair of them as an adaptation to the environment in the desert lizards. The conditions
i. Burrowing in the soil to escape high temperatures.
ii. Losing heat rapidly from the body during high temperatures.
iii. Bask in the sun when the temperature is low.
iv. Insulating body due to the thick fatty dermis.

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

The correct answer is:
i and iii

Solution:
Desert lizards lack the physiological ability to cope up with the high temperatures. Hence, they have to manage by showing certain behavioral activities. Burrowing in the soil when the temperature is high and basking in the sun when the temperature is low are shown as adaptation.

Topic: Biology, Organisms and Populations, NEET
Consider the following four measures (A-D) that could be taken to successfully grow chick pea in an area where bacterial blight disease is common
(A) spray with Bordeaux mixture
(B) control of the insect vector of the disease pathogen
(C) use of only disease-free seeds
(D) use of varieties resistant to the disease
Which two of the above measures can control the disease?

Options:
A B and C
B A and B
C C and D
D A and D

The correct answer is:
C and D

Solution:
Bacterialblight of chick pea (Cicer arietinum) is caused by bacterium Xanthomonas campestris. The ejected tissues collapse and are digested by bacteria. Rapid spread of bacterial infection occurs through vascular strand. The stems and leaves of infected plant give blighted or burnt up appearance. Control measures include rogueing; 3-year crop rotation, disease free seeds, spray of copper fungicides and antibiotics besides showing disease resistant varieties.
Bordeaux mixture is most common fungicide. It was first discovered by Millardet in 1882 at university of Bordeaux France to control downy mildew of grapes. It is prepared by mixing copper sulphate, lime and water in the ratio of $4: 4: 50$.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2008 (Screening)
Consider the following four statements (1-4) regarding kidney transplant and select the two correct ones out of these.
(1) Even if a kidney transplant is proper the recipient may need to take immuno-suppresants for a long time
(2) The cell-mediated immune response is responsible for the graft rejection
(3) The B-lymphocytes are responsible for rejection of the graft
(4) The acceptance of rejection of a kidney transplant depends on specific interferons The two correct statements are

Options:
A (2) and (3)
B (3) and (4)
C (1) and (3)
D (1) and (2)

The correct answer is:
(1) and (2)

Solution:
To prevent graft (kidney transplant) rejection, the principle is to use chemicals, which inhibit the entire activity of the immune system. When this occurs graft rejection is delayed. It has been shown that immunosupppression may make the patients more preen to develop cancers. One way of overcoming the problems of radiation and immunosuppression is to suppress only the cells responsible for rejection, normally the killer T-cells. In this way, the rest of the patients immune system would continue to function normally. The most promising approach is to treat the patient with monoclonal antibodies that recognize and destroy the killer T-cells. A monoclonal antibody has been developed, which is a very effective at preventing rejection of transplanted kidneys.

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2010 (Screening)
Consider the following four statements (A-D) about certain desert animals such as kangaroo rat
A. they have dark colour and high rate of reproduction and excrete solid urine.
B. they do not drink water, breathe at a slow rate to conserve water and have their body covered with thick hairs.
C. they feed on dry seeds and do not require drinking water.
D. they excrete very concentrated urine and do not use water to regulate body temperature.
Which two of the above statements for such animals are true?

Options:
A C and D
B B and C
C C and A
D A and B

The correct answer is:
B and C

Solution:
- The kangaroo-rat is adapted to live in hot and dry conditions of the desert
- They can survive without drinking any water and get the needed moisture from their seed diet.
- They conserve water by lowering their metabolic activity which reduces loss of water through their skin and respiratory system.

Topic: Biology, Organisms and Populations, NEET, NEET 2008 (Screening)
Consider the following four statements and choose the correct option based on whether they are RIGHT or WRONG.

(I) The sporophyte in liverworts is more elaborate than in mosses.
(II) Salvinia is heterosporous.
(III) The life cycle in all seed-bearing plants is diplontic.
(IV) In Pinus, male and female cones are borne on different trees.

Options:
A I and III are correct
B I and IV are incorrect
C II and III are incorrect
D I and II are correct

The correct answer is:
I and IV are incorrect

Solution:
Salvinia is a genera of Pteridophyte which produces two types of spores - micro and macro. Such type of spores are called heterosporous. The microspore gives rise to male plant and the megaspore forms the female plant.
Liverworts, type of bryophytes, usually grow in moist shady habitats and its plant body is called as thalloid. The sporophyte of liverworts is differentiated into foot, setae and capsule. Mosses are another type of bryophyte whose sporophyte stage has similar three parts like liverworts but more elaborated.
Gymnosperms are heterosporous and the spores are made in sporangia that are borne on sporophylls called stobili or cones. Male cones are the ones bearing the male gametes and female cone bears female gamete. In Pinus, both the cones are found on the same tree unlike in Cycas.

Topic: Biology, Plant Kingdom, NEET
Consider the following four statements and select the option with both incorrect statements.

A. The sporophyte in liverworts is more elaborate than that in mosses.
B. Sellaginella is heterosporous.
C. The life cycle in all seed-bearing plants is diplontic.
D. In Pinus, male and female cones are borne on different trees.

Options:
A AÂ andÂ C
B A and D
C BÂ andÂ C
D AÂ andÂ D

The correct answer is:
A and D

Solution:
The sporophyte in liverworts is small and simple, and sporophyte in moss is more elaborate than liverworts. In Pinus, plant is monoecious, in which male and female cones are borne on the same plant.

Topic: Biology, Plant Kingdom, NEET
Consider the following four statements (IâIV) related to cell cycle, and select the correct option stating them as true [T] and false [F].

i. G1 phase is a metabolically active stage of the cell cycle.
ii. Interphase is the phase of actual cell division.
iii. The number of chromosomes doubles in S-phase.
iv. The cells that do not divide further exit the G1 - phase to enter a quiescent stage.

Options:
A

I	II	III	IV

T	F	F	F

B

I	II	III	IV

F	T	T	T

C

I	II	III	IV

F	F	T	T

D

I	II	III	IV

T	F	F	T

The correct answer is:

I	II	III	IV

T	F	F	T

Solution:
Interphase is the preparatory phase where the cell prepares itself for cell division by increasing the concentration of proteins. Replication of chromosomes takes place in the S phase where the number of chromatids double.

Topic: Biology, Cell Cycle and Cell Division, NEET
Consider the following four statements. Which of the following statements are incorrect?

(A) The sporophyte in liverworts is more elaborate than in mosses.
(B) Salvinia is heterosporous.
(C) The life cycle in all seed-bearing plants is diplontic.
(D) In Pinus, male and female cones are borne on different trees.

Options:
A Statements (A) and (C)
B Statements (A) and (B)
C Statements (A) and (D)
D Statements (B) and (C)

The correct answer is:
Statements (A) and (D)

Solution:
(A) The sporophyte is more developed in mosses rather than in liverwort.
(D) Pinus is monoecious in which male and female cones are borne on different branches.

Topic: Biology, Plant Kingdom, NEET
Consider the following graph:

X, Y and Z can be respectively

Options:
A Ne, Ar and Xe
B Ar, Xe and He
C Kr, Ar and Ne
D Ar, He and Ne

The correct answer is:
Kr, Ar and Ne

Solution:
Most probable speeds are inversely proportional to the atomic mass.

Topic: Chemistry, States of Matter, NEET
Consider the following haloalkanes
(1) CH3F
(2) CH3Cl
(3) CH3Br
(4) CH3I
The correct sequence of increasing order of dipole moments is

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

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

Solution:
CH3I1.64D<CH3Br1.79D<CH3F1.84D<CH3Cl1.94D
The charge separation in C-X bond decreases in the order
C-F>C-Cl>C-Br>C-I, due to decrease in the electronegativity as we more from F→Cl→Br→I .
CH3Cl has higher dipole moment than CH3F , because of larger bond length of C-Cl bond. The larger C-Cl bond length causes dipole moment (μ)=q×d to be larger for CH3Cl than CH3F . In case of other haloalkanes, the effect of larger C-X bond length cannot outweigh the effect of decreased charge separation.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Consider the following liquid - vapour equilibrium.
Liquid⇌Vapour
Which of the following relations is correct?

Options:
A dln⁡GdT2=∆HυRT2
B dln⁡PdT=-∆HυRT
C dln⁡PdT2=-∆HυT2
D dln⁡PdT=∆HυRT2

The correct answer is:
dln⁡PdT=∆HυRT2

Solution:
P=Ke-∆H/RT
ln P=ln⁡K-∆HRT
ddTln⁡P=∆HυRT2
∴dln⁡PdT=∆HυRT2

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2016 (Phase 1)
Consider the following molecule :

What are the number of sigma and pi bonds present in the above molecule respectively?

Options:
A 20, 6
B 18, 5
C 15, 5
D 14, 4

The correct answer is:
20, 6

Solution:
Total 20 sigma and 6 Ï bonds.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Consider the following plant tissues:
(A) Axillary buds
(B) Fascicular vascular cambium
(C) Interfascicular cambium
(D) Cork cambium
(E) Intercalary meristem
Identify the lateral meristems among the above.

Options:
A $(A),(B),(C)$ and $(E)$ only
B $(A),(B),(D)$ and $(E)$ only
C $(A),(C)$ and (D) only
D $(B), (C)$ and $(D)$ only

The correct answer is:
$(B), (C)$ and $(D)$ only

Solution:
Fascicular vascular cambium, interfascicular cambium and cork cambium are examples of lateral meristems.
The meristem which occurs between mature tissues is known as intercalary meristem.
Cells of shoot apical meristem form the axillary buds.
Both intercalary meristem and axillary buds are primary meristems

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2023 (Manipur)
Consider the following reaction:
$2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-483.64 \mathrm{~kJ}$. What is the enthalpy change for decomposition of one mole of water? (Choose the right option).

Options:
A $18 \mathrm{~kJ}$
B $100 \mathrm{~kJ}$
C $120.9 \mathrm{~kJ}$
D $241.82 \mathrm{~kJ}$

The correct answer is:
$241.82 \mathrm{~kJ}$

Solution:
$2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \longrightarrow 2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})$
$\Delta \mathrm{H}^{\circ}=483.64 \mathrm{~kJ}$
For decomposition of 2 moles of $\mathrm{H}_2 \mathrm{O}$, enthalpy change $=483.64 \mathrm{~kJ}$
For decomposition of 1 mole of $\mathrm{H}_2 \mathrm{O}$, enthalpy change $=\frac{483.64}{2} \mathrm{~kJ}$
$=241.82 \mathrm{~kJ}$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2023 (Manipur)
Consider the following reaction and select incorrect statement about gas (P)
Zn+HNO3(dilute)→Zn(NO3)2+P↑

Options:
A Gives neutral solution in water
B Contains more O2 than air
C Forms brown ring with FeSO4 solution
D None of these

The correct answer is:
Forms brown ring with FeSO4 solution

Solution:
Here statement in option C is incorrect while rest are correct.
4Zn+10HNO3(dilute)→4Zn(NO3)2+N2O+5H2O
N2O is neutral, contain 33.3% oxygen which is more than air.
NO gas forms brown ring complex with FeSO4 solution, whereas N2O gas does not.

Topic: Chemistry, d and f Block Elements, NEET
Consider the following reaction at 1000°C
(I) Zn(s) +  1 2 O 2 (g) → Δ ZnO(g); ΔG° =−360 kJ  mol −1
(II) C(s) +  1 2 O 2 (g)  → Δ CO(g); ΔG° = −460 kJ  mol −1
and choose the correct statement at 1000°C .

Options:
A ZnO is more stable than CO
B ZnO can be reduced to Zn by C
C ZnO and CO are formed at equal rate
D ZnO cannot be reduced to Zn by C

The correct answer is:
ZnO can be reduced to Zn by C

Solution:
Here the correct statement is ZnO can be reduced to Zn by C.
ZnO(g) + C(s) → Δ Zn(g) + CO(g) .

Topic: Chemistry, Thermodynamics (C), NEET
Consider the following reaction at equilibrium
2Fe2+(aq)+Cu2+→2Fe3+(aq)+Cu
When the reaction comes to equilibrium, what is the cell voltage?

Options:
A 0.43 V
B 1.11 V
C 0.78 V
D 0 V

The correct answer is:
0 V

Solution:
When the equilibrium is attained in cell reaction, the cell will no longer do any electrical work, i.e., ΔG=0
-ΔG=nEcellF
So 0=nEcellF
or Ecell=0 .

Topic: Chemistry, Electrochemistry, NEET
Consider the following reaction
CH3Br+Mg→etherA→HCHOB→HOHC. Compound C is

Options:
A Acetic acid
B Acetaldehyde
C Ethyl alcohol
D Formic acid

The correct answer is:
Ethyl alcohol

Solution:
CH 3 Br + Mg → Δ Ether CH 3 MgBr → 2)  H + /H 2 O 1) HCHO CH 3 CH 2 OH

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Consider the following reaction
CH3-CH=CH2→Br2/NaCl
Product of the reaction will be

Options:
A Only 1, 2-dibromopropane
B Only 1-bromo-2-chloropropane
C Only-2-bromo-1-chloropropane
D Mixture of 1,2-dibromopropane and 1-bromo-2-chloropropane

The correct answer is:
Mixture of 1,2-dibromopropane and 1-bromo-2-chloropropane

Solution:
In this reaction Mixture of 1,2-dibromopropane and 1-bromo-2-chloropropane is formed as follows

Topic: Chemistry, Hydrocarbons, NEET
Consider the following reaction

Compound 'A' will be

Options:
A
B
C
D All of these

The correct answer is:

Solution:
Here the compound A is and the reaction is as follows

Topic: Chemistry, Hydrocarbons, NEET
Consider the following reaction equilibrium

N2g+3H2(g)â2NH3(g)

Initially, 1 mole of N2 and 3 moles of H2 are taken in a 2 L flask. At equilibrium state if, the number of moles of N2 is 0.6, what is the total number of moles of all gases present in the flask?

Options:
A 0.8
B 1.6
C 3.2
D 6.4

The correct answer is:
3.2

Solution:
Number of moles of N2=0.6=1-x
∴ x=1-0.6=0.4
So, 3-3x=3-3×0.4=1.8,
and 2x=2×0.4=0.8
Therefore, the total number of moles at equilibrium.
=1-x+3-3x+2x
=0.6+1.8+0.8
=3.2 mol

Topic: Chemistry, Chemical Equilibrium, NEET
Consider the following reaction
Glucose → Reagent Mannose
The above reaction is known as

Options:
A Anomerisation
B Racemisation
C Epimerisation
D Conversion

The correct answer is:
Epimerisation

Solution:
Conversion of one epimer into other is known as epimerisation.

Topic: Chemistry, Biomolecules, NEET
Consider the following reaction in a sealed vessel at equilibrium with concentrations of $\mathrm{N}_2=3.0 \times 10^{-3} \mathrm{M}, \mathrm{O}_2=4.2 \times 10^{-3} \mathrm{M}$ and $\mathrm{NO}=2.8 \times 10^{-3} \mathrm{M}$.
$2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$
If 0.1 mol $\mathrm{L}^{-1}$ of $\mathrm{NO}_{(\mathrm{g})}$ is taken in a closed vessel, what will be degree of dissociation ( $\alpha$ ) of $\mathrm{NO}_{(\mathrm{g})}$ at equilibrium?

Options:
A 0.0889
B 0.8889
C 0.717
D 0.00889

The correct answer is:
0.717

Solution:
$\begin{aligned} & 2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \\ & \mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]}{[\mathrm{NO}]^2} \\ &=\frac{3 \times 10^{-3} \times 4.2 \times 10^{-3}}{2.8 \times 10^{-3} \times 2.8 \times 10^{-3}} \\ &=1.607 \\ & \mathrm{t}=0 \quad 2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \\ & 0.1-0.1 \alpha \quad 0.05 \alpha \quad 0.05 \alpha \\ & \mathrm{K}_{\mathrm{c}}=\frac{0.05 \alpha \times 0.05 \alpha}{(0.1-0.1 \alpha)^2} \\ & \mathrm{~K}_{\mathrm{c}}=\frac{0.05 \alpha \times 0.05 \alpha}{0.01(1-\alpha)^2} \\ & 1.607=\frac{(0.05)^2 \alpha^2}{0.01(1-\alpha)^2} \\ & \frac{\alpha^2}{(1-\alpha)^2}=\frac{1.607 \times(0.1)^2}{(0.05)^2}\end{aligned}$
$\begin{gathered}\frac{\alpha}{1-\alpha}=\frac{1.27 \times 0.1}{0.05} \\ \frac{\alpha}{1-\alpha}=2.54 \\ \alpha=2.54-2.54 \alpha \\ 3.54 \alpha=2.54 \\ \alpha=\frac{2.54}{3.54}=0.717\end{gathered}$

Topic: Chemistry, Chemical Equilibrium, NEET, NEET 2024
Consider the following reaction in aqueous solution.

5 Br − aq + BrO 3 − aq +6 H + aq →3 Br 2 aq +3 H 2 O l

If the rate of appearance of Br 2 at a particular time during the reaction is 0.025   Msec −1 , what is the rate of disappearance (in Msec −1) of Br − at that time?

Options:
A 0.025 Msec â1
B 0.042 Msec â1
C 0.075 Msec â1
D 0.125 Msec â1

The correct answer is:
0.042 Msec â1

Solution:
The two rates are related as â 1 5 d Br â dt = 1 3 d Br 2 dt
â d Br â dt = 5 3 d Br 2 dt = 5 3 Ã0.025=0.042 Msec â1

Topic: Chemistry, Chemical Kinetics, NEET
Consider the following reaction

In the above reaction ester formation takes place by

Options:
A Breaking of O-H bond of A and O-H bond of B
B Breaking of C-O bond of A and C-O bond of B
C Breaking of O-H bond of A and C-O bond of B
D Breaking of C-O bond of A and O-H bond of B

The correct answer is:
Breaking of O-H bond of A and C-O bond of B

Solution:
In the above reaction ester formation takes place by breaking of O-H bond of A and C-O bond of B.

Topic: Chemistry, Carboxylic Acid Derivatives, NEET
Consider the following reaction

In the above reaction reagent/catalyst will be

Options:
A CH3-CH2-Cl,Anyl.AlCl3
B CH2=CH2,conc.H2SO4
C CH3-CH2-OH,BF3
D All of these

The correct answer is:
All of these

Solution:

Topic: Chemistry, Hydrocarbons, NEET
Consider the following reaction
M+O2→MO2
(M = alkali metal) (stable superoxide)

Options:
A M can not be Li and Na
B M can not be Cs and Rb
C M can not be Li and Rb
D None of these

The correct answer is:
M can not be Li and Na

Solution:
The tendency to form superoxides among alkali metals increases down the group. Here Li and Na do not form stable superoxide.

Topic: Chemistry, s Block Elements, NEET
Consider the following reaction, Phenol $\stackrel{\mathrm{Zn} \text { dust }}{\longrightarrow} \mathrm{X} \underset{\text { Anhydrous } \mathrm{AlCl}_3}{\stackrel{\mathrm{CH}_3 \mathrm{Cl}}{\longrightarrow}} \mathrm{Y} \stackrel{\begin{array}{l}\text { Alkaline } \\ \mathrm{KMnO}\end{array}}{\longrightarrow} \mathrm{Z}$, the product, $Z$, is

Options:
A toluene
B benzaldehyde
C benzoic acid
D benzene

The correct answer is:
benzoic acid

Solution:
Key Idea (i) $Z n$ dust converts $-\mathrm{OH}$ group into- $\mathrm{H}$.
(ii) Reaction with $\mathrm{CH}_3 \mathrm{Cl}$ in presence of anthy. $\mathrm{AlCl}_3$ is called Friedel-Craft's alkylation.
(iii) Alkaline $\mathrm{KMnO}_4$ converts complete carbon chain which is directly attached with benzene ring, into - $\mathrm{COOH}$ group.

Topic: Chemistry, Carboxylic Acid Derivatives, NEET, NEET 2009 (Screening)
Consider the following reaction
Phenol â Zn dust X â CH 3 Cl Y â Alkaline KMnO 4 Z â H 3 O + the product Z is

Options:
A benzaldehyde
B benzoic acid
C benzene
D toluene

The correct answer is:
benzoic acid

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Consider the following reaction sequence of alkene

CH 3 CH= CHCH 3 → O 3 A → Zn H 2 O B

The product B is

Options:
A CH3CH2COCH3
B CH3COCH3
C CH3CHO
D CH3CH2CHO

The correct answer is:
CH3CHO

Solution:
â´ The product B is acetaldehyde

Topic: Chemistry, Hydrocarbons, NEET
Consider the following reaction,
the product $Z$, is

Options:
A $\mathrm{CH}_2=\mathrm{CH}_2$
B $\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{O}-\mathrm{CH}_2-\mathrm{CH}_3$
C $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{O}-\mathrm{SO}_3 \mathrm{H}$
D $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$

Solution:
Key Idea (i) $\mathrm{PBr}_3$ is a halogenating agent, ie, converts $-\mathrm{OH}$ group into $-\mathrm{Br}$.
(ii) Alc. $\mathrm{KOH}$ is a dehydrohalogenating agent.
(iii) $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{H}_2 \mathrm{O}$ converts an olefin into alcohol.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET, NEET 2009 (Screening)
Consider the following reaction :

x MnO 4 - + y C 2 O 4 2 - + z H + â¶ x Mn 2 + + 2 y CO 2 + z 2 H 2 O

The values of x, y and z in the reaction are, respectively :

Options:
A 2,5 and 16
B 5,2 and 8
C 5,2 and 16
D 2,5 and 8

The correct answer is:
2,5 and 16

Solution:
C 2 O 4 2 - ⟶ 2 CO 2 + 2 e - ] × 5 MnO 4 - + 8 H + + 5 e- ⟶ Mn 2 + + 4 H 2 O ] × 2
-----------------------------------------------------------------------------------
2 MnO 4 - + 1 6 H + + 5 C 2 O 4 2 - ⟶ 1 0 CO 2 + 2 Mn 2 + + 8 H 2 O
MnO 4 - ⇒ x = 2
C 2 O 4 2 - ⇒ y = 5
H + ⇒ z = 1 6

Topic: Chemistry, Redox Reactions, NEET
Consider the following reactions
A→(i)CH3MgBrB→Cu2-methyl-2-butene
The mass percentage of carbon in A is

Options:
A 33.33
B 6.67
C 16.66
D 66.67

The correct answer is:
66.67

Solution:
%carbon=Atomicmass×AtomicityMolarmass×100
=12×472×100=66.66%

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Consider the following reactions:

(i) C2O42-→CO2

(ii) SO42-→SO32-

(iii) MnO42-→MnO4-

(iv) Fe3+→Fe2+

Choose the correct answer: -

Options:
A iÂ &Â (ii) shows oxidation
B iiiÂ &Â iv shows reduction
C i & iii shows oxidation
D iiiÂ &Â iv shows oxidation

The correct answer is:
i & iii shows oxidation

Solution:
In C2O42-âCO2 oxidation number of carbon change from +3 to +4 .
In SO42-âSO32- oxidation number of sulphur change from +6 to+4 .
In MnO42-âMnO4- oxidation number of Mn change from +6 to+7 .
In Fe3+âFe2+ oxidation number of carbon change from +3 to +2.

Topic: Chemistry, Redox Reactions, NEET