datavorous/entrance-exam-dataset huggingface dataset:
Out of the following, which type of interaction is responsible for the stabilisation of α-helix structure of proteins?

Options:
A Covalent bonding
B Ionic bonding
C Hydrogen bonding
D vander Waals forces

The correct answer is:
Hydrogen bonding

Solution:
Hydrogen bonding is responsible for the stacking of α-helix structure of protein.

The α-helix structure of proteins is stabilized by intramolecular H-bonding between C−O of one amino acid residue and the N−H of the fourth amino acid residue in the chain.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2021 (24 Feb Shift 1)
Out of the given ten pairs of combination total pair(s), which results in zero overlapping is: (Assuming \( Z \) is overlapping axis). \( \left(\mathrm{p}_{\mathrm{y}}+\mathrm{s}\right),\left(\mathrm{d}_{\mathrm{xy}}+\mathrm{s}\right),\left(\mathrm{d}_{\mathrm{xy}}+\mathrm{p}_{\mathrm{y}}\right) \) \( \left(\mathrm{p}_{\mathrm{z}}+\mathrm{d}_{\mathrm{xy}}\right),\left(\mathrm{p}_{\mathrm{z}}+\mathrm{s}\right),\left(\mathrm{p}_{\mathrm{y}}+\mathrm{P}_{\mathrm{z}}\right) \) \( (\mathrm{s}+\mathrm{s}),\left(\mathrm{s}+\mathrm{p}_{\mathrm{y}}\right),\left(\mathrm{p}_{\mathrm{y}}+\mathrm{p}_{\mathrm{y}}\right) \)

Options:

The correct answer is:
4

Solution:
No overlapping is possible in the opposite symmetry orbitals (along Z axis) i.e., S+Py, Py+Pz, dxy+S, dxy+py.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Out of the given two compounds, the vapour pressure of $B$ at a particular temperature is

Options:
A higher than that of $A$
B lower than that of $A$
C higher or lower than $A$ depending on the size of the vessel
D same as that of $A$

The correct answer is:
higher than that of $A$

Solution:
Vapour pressure of o-nitrophenol is more than vapour pressure of $\$ p \$-n i t r o p h e n o l$ due to the intramolecular hydrogen bonding. As a result, it is more volatile.

Topic: Chemistry, Solutions, KCET, KCET 2011
Out of $\mathrm{TiF}_6^{2-}$ and $\mathrm{CoF}_6^{3-}, \mathrm{Cu}_2 \mathrm{Cl}_2$ and $\mathrm{NiCl}_4^{2-}$ ( $\mathrm{Z}$ of $\mathrm{Ti}=22, \mathrm{Co}=27, \mathrm{Cu}=29, \mathrm{Ni}=28)$ the colourless species are :

Options:
A $\mathrm{CoF}_6^{3-}$ and $\mathrm{NiCl}_4^{2-}$
B $\mathrm{TiF}_6^{2-}$ and $\mathrm{CoF}_6^{3-}$
C $\mathrm{Cu}_2 \mathrm{Cl}_2$ and $\mathrm{NiCl}_4^{2-}$
D $\mathrm{TiF}_6^{2-}$ and $\mathrm{Cu}_2 \mathrm{Cl}_2$

The correct answer is:
$\mathrm{TiF}_6^{2-}$ and $\mathrm{Cu}_2 \mathrm{Cl}_2$

Solution:
In both $\mathrm{TiF}_6^{2-}$ and $\mathrm{Cu}_2 \mathrm{Cl}_2$, these no $\mathrm{d}-$ electrons or no unpaired electrons so, these are colourless.

Topic: Chemistry, Coordination Compounds, NEET, NEET 2009 (Mains)
Out of two copper spheres of the same size, x is hollow while y is solid. If they are charged at the same potential, what can be said about the charges on them?

Options:
A Sphere x will have more charge
B Sphere y will have more charge
C Charge on both the spheres are equal
D Charges on both the sphere are zero

The correct answer is:
Charge on both the spheres are equal

Solution:
Charges on a conductor accumulate only on the surface. Therefore, a hollow sphere or a solid one, both will have exact same charge distribution and the exact same charge for given potential.

â´ Charge on both spheres is equal.

Topic: Physics, Current Electricity, JEE Main
Outermost electronic configurations of four elements A,B,C,D are given below:

(A) 3 s2

(B) 3s23p1

(C) 3 s23p3

(D) 3 s23p4

The correct order of first ionization enthalpy for them is

Options:
A A<B<C<D
B B<A<D<C
C B<D<A<C
D B<A<C<D

The correct answer is:
B<A<D<C

Solution:
(A) 3 s2→Mg

(B) 3 s23p1→Al

(C) 3 s23p3→P

(D) 3 s23p4→S

Al has one unpaired electron in its highest energy orbital 3p, and Mg's highest energy orbital 3s has the paired electrons.

It is energetically favourable for all the electrons in an orbital to be paired, which means that breaking up this pair would require more energy.

Compared to 16th group, 15 group have high ionisation energy. It is because of the stable half filled configuration of he 15th group elements.

P-3s23p3

S-3s23p4

Therefore, P has more ionisation energy than S.

Hence, the correct order of first ionization enthalpy is B<A<D<C

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2022 (27 Jul Shift 2)
Ovary to fruit formation in absence of fertilization refers to:

Options:
A Parthenocarpy
B Apogamy
C Apospory
D Parthenogenesis

The correct answer is:
Parthenocarpy

Solution:
Fruit is a characteristic feature found in flowering plants which develops from ovary after fertilization. If fruit develops without fertilization then it is called as parthenocarpic fruit. Fruit basically is part of the ovary which is not formed by fusion of gametes or cells. The ploidy level of fruit will remain the same as that of the female reproductive organ. In other words, ploidy of plant and fruit will be same.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Overlapping leads to the formation of \( \sigma \) bonding molecular orbital in:

Options:
A
B
C
D

The correct answer is:

Solution:
1. S + py orbital gives non-bonding molecular orbitals.

2. S + pz orbital give bonding molecular orbitals.

3. Lateral overlaps give Pi molecular orbitals.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
\( \overrightarrow{\mathrm{a}} \) and \( \overrightarrow{\mathrm{b}} \) are two unit vectors and \( \theta \) is the angle between them. Then \( \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}} \) is a unit vector if :

Options:
A \( \theta=\pi / 3 \)
B \( \theta=\pi / 4 \)
C \( \theta=\pi / 2 \)
D \( \theta=2 \pi / 3 \)

The correct answer is:
\( \theta=2 \pi / 3 \)

Solution:
Given a→=1, b→=1

And a→+b→=1

We know that

a→+b→2=a→2+b→2+2a→b→cosθ

⇒a→+b→2=1+1+21.1 cos θ=1 ​
∴ cos θ = - 1 / 2
∴ θ = 2 π 3

Topic: Mathematics, Vector Algebra, JEE Main
Oxalic acid, $\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4$, reacts with paramagnet ion according to the balanced equation $5 \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4(\mathrm{aq})+$ $2 \mathrm{MnO}_{4^{-}}(\mathrm{aq}) \rightleftharpoons 2 \mathrm{Mn}^{2+}(\mathrm{aq})+10 \mathrm{CO}_2(\mathrm{~g})+8 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$. The volume in $\mathrm{mL}$ of $0.0162 \mathrm{M} \mathrm{KMnO}_4$ solution required to react with $25.0 \mathrm{~mL}$ of $0.022 \mathrm{M} \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4$ solution is :

Options:
A $13.6$
B $18.5$
C $33.8$
D $84.4$

The correct answer is:
$13.6$

Solution:
In the reaction
\(\begin{aligned}
& \mathrm{MnO}_4^{-} \rightarrow \mathrm{Mn}^{+2} \\
& \text {thus nf }=5 \\
& \mathrm{C}_2^{+3} \mathrm{O}_4^{2-} \rightarrow \mathrm{CO}_2^{+4} \\
& 2 \cdot \mathrm{C}^{+3} \rightarrow \mathrm{C}^{+4} \\
& \mathrm{nf}=2 \times 1=2
\end{aligned}\)
thus by equation
\(\begin{gathered}
\mathrm{KMnO}_4 \mathrm{M}_1 \mathrm{~V}_1 \mathrm{nf}_1=\mathrm{M}_2 \mathrm{~V}_2 \mathrm{nf}_2 \quad \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \\
0.0162 \mathrm{M} \times \mathrm{V}_1 \times 5=0.022 \mathrm{M} \times 25.0 \mathrm{~mL} \times 2 \\
\mathrm{~V}_1=13.58 \mathrm{~mL} \simeq 13.6 \mathrm{~mL}
\end{gathered}\)

Topic: Chemistry, Redox Reactions, JEE Main
Oxalic acid reacts with concentrated $\mathrm{H}_2 \mathrm{SO}_4$ to give a mixture of two gases. When this mixture is passed through caustic potash, one of the gases is absorbed. What is the product formed by the absorbed gas with caustic potash?

Options:
A $\mathrm{K}_2 \mathrm{SO}_0$
B $\mathrm{K}_2 \mathrm{HCO}_3$
C $\mathrm{K}_2 \mathrm{CO}_3$
D $\mathrm{KOH}$

The correct answer is:
$\mathrm{K}_2 \mathrm{CO}_3$

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Oxalic acid when heated with conc. $\mathrm{H}_2 \mathrm{SO}_4$, gives out

Options:
A $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{CO}_2$
B $\mathrm{CO}$ and $\mathrm{CO}_2$
C Oxalic sulphate
D $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{~S}$

The correct answer is:
$\mathrm{CO}$ and $\mathrm{CO}_2$

Solution:
\(\begin{array}{ll} \mathrm{COOH} \\ ~| \\ \mathrm{COOH} \end{array} \xrightarrow{\text { conc. } \mathrm{H}_2 \mathrm{SO}_4} \mathrm{H}_2 \mathrm{O}+\mathrm{CO}+\mathrm{CO}_2\)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Oxidation number of iodine varies from

Options:
A -1 to +1
B -1 to +7
C +3 to +5
D -1 to +5

The correct answer is:
-1 to +7

Solution:
lodine shows -1 to +7 oxidation state.

Topic: Chemistry, Redox Reactions, JEE Main
Oxidation number of $\mathrm{Mn}$ in $\mathrm{K}_2 \mathrm{MnO}_4$ and $\mathrm{MnSO}_4$ are respectively

Options:
A $+7,+2$
B $+6,+2$
C $+5,+2$
D $+2,+6$

The correct answer is:
$+6,+2$

Solution:
$\begin{array}{ll}\mathrm{K}_2 \mathrm{MnO}_4 & \mathrm{MnSO}_4 \\ 2+x-8=0 & x+6-8=0 \\ x=+6 & x=+2 .\end{array}$

Topic: Chemistry, Redox Reactions, JEE Main
Oxidation number of $\mathrm{P}$ in $\mathrm{PO}_4^{3-}$, of $\mathrm{S}$ in $\mathrm{SO}_4^{2-}$ and that of $\mathrm{Cr}$ in $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ are respectively :

Options:
A $+3,+6$ and +6
B $+5,+6$ and +6
C $+3,+6$ and +5
D $+5,+3$ and +6

The correct answer is:
$+5,+6$ and +6

Solution:
$\mathrm{PO}_4^{3-}$, oxidation no. of $\mathrm{P} \Rightarrow+5$ $\mathrm{SO}_4^{2-}$, oxidation no. of $\mathrm{S} \Rightarrow+6$ $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$, oxidation no. of $\mathrm{Cr} \Rightarrow+6$

Topic: Chemistry, Redox Reactions, NEET, NEET 2009 (Mains)
Oxidation number of potassium in K2O,K2O2 and KO2, respectively, is:

Options:
A +2,+1 and +12
B +1,+1 and +1
C +1,+4 and +2
D +1,+2 and +4

The correct answer is:
+1,+1 and +1

Solution:
K 2 O K +1 â O â2 â K +1 oxygen is more E.N. than potassium
K 2O 2 K +1 â O â1 â O â1 â K +1
K O 2

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2020 (07 Jan Shift 1)
Oxidation of cyclohexene in presence of acidic potassium permaganate leads to

Options:
A glutaric acid
B adipic acid
C pimelic acid
D succinic acid

The correct answer is:
adipic acid

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Oxidation of one molecule of glucose will yield 36 ATP in proportion of

Options:
A All 36Â ATP in mitochondria
B 6Â ATP outside the mitochondria and 30 inside mitochondria.
C 2 ATP in glycolysis and 34 ATP in Kreb's cycle
D 2 ATP outside the mitochondria and 34 inside mitochondria.

The correct answer is:
2 ATP outside the mitochondria and 34 inside mitochondria.

Solution:
2 ATP are synthesized in glycolysis through phosphorylation and the remaining 34 ATP are synthesized in mitochondria through Kreb's cycle and ETS.

Topic: Biology, Respiration in Plants, NEET
Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, what is the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K)?

Options:
A 8.96 L
B 4.48 L
C 6.72 L
D 2.24 L

The correct answer is:
8.96 L

Solution:
Ethene, CO2, H2 are gases.

So total moles of gas =1+2+1=4
By 2 Farady=4 mole of gas produced
So, by 0.2 Farady =0.4 mole gas produced
ν=nRTP=0.4×0.0821×2731=8.96 L

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2016 (10 Apr Online)
Oxidation of toluene with $\mathrm{CrO}_3$ in the presence of $\left(\mathrm{CH}_3 \mathrm{CO}\right) \mathrm{O}_2$ gives a product ' $A$ ' which on treatment with aqueous $\mathrm{NaOH}$ produces

Options:
A $\mathrm{C}_6 \mathrm{H}_3 \mathrm{CHO}$
B $\left(\mathrm{C}_6 \mathrm{H}_3 \mathrm{CO}\right)_2 \mathrm{O}$
C $\mathrm{C}_6 \mathrm{H}_3 \mathrm{COONa}$
D 2, 4-diacetyl toluene

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_3 \mathrm{COONa}$

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Oxidation state of +1 for phosphorus is found in

Options:
A $\mathrm{H}_3 \mathrm{PO}_3$
B $\mathrm{H}_3 \mathrm{PO}_4$
C $\mathrm{H}_3 \mathrm{PO}_2$
D $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7$

The correct answer is:
$\mathrm{H}_3 \mathrm{PO}_2$

Solution:
$\begin{aligned} & \mathrm{H}_3 \mathrm{PO}_2 \\ & 3+x+(-2 \times 2)=0 \\ & x=+1\end{aligned}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Oxidation state of Fe(Iron) in complex formed in Brown ring test.

Options:

The correct answer is:
1

Solution:
The chemical test called a brown ring test is conducted to find out the presence of Nitrate ions in any solution, in which the solutions form a brown ring in the test tube.

The above test is also called the brown ring test, in which the brown ring complex compound is formulated as [Fe(H2O)5NO]SO4.

Here the oxidation state of Iron in the complex, can be found as:

We can assume that the oxidation state of Iron metal is x.

Here the oxidation state of the Sulphate ion is −2, the oxidation state of water is zero and the oxidation state of Nitroso NO is +1.Hence, the oxidation state of Iron can be calculated as:
⇒x+(+1)+5(0)=+2⇒x=+1

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2024 (29 Jan Shift 2)
Oxidation state of iron and chromium in chromite ore is:

Options:
A 2, 3
B 3, 2
C 2, 2
D 3,3

The correct answer is:
2, 3

Solution:
Chromite ore is FeCr2O4 containing Fe+2 and Cr+3
FeCr 2 O 4 =FeO⋅ Cr 2 O 3

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Oxidation state of Mn in the compound formed by alkaline oxidative fusion of MnO2 by KOH is:-

Options:
A 4
B 5
C 6
D 7

The correct answer is:
6

Solution:
Mn+4O2 alkaline oxidative MnO 4 2â Fusion
By the alkaline oxidative fusion of MnO2 by KOH the product formed is MnO42-
-2 = xâ+â4 -2
xâ=â6

Topic: Chemistry, d and f Block Elements, JEE Main
Oxidation state of oxygen in hydrogen peroxide is

Options:
A $-1$
B $+1$
C $0$
D $-2$

The correct answer is:
$-1$

Solution:
In all peroxide oxygen shows - 1 oxidation state.

Topic: Chemistry, Redox Reactions, JEE Main
Oxidation state of S in H2S2O8 is

Options:
A 8
B 6
C 4
D 7

The correct answer is:
6

Solution:
In the structure of H2S2O8 , two oxygen atoms are involved in the peroxide linkage,hence oxidation number will be -1(each). Remaining oxygen atoms will have -2 charge. This we can calculate the oxidation state of sulphur as follows :-

2(+1) + 2(x)+ 2(-1)+ 6(-2) =0

x= +6

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), TS EAMCET, TS EAMCET 2020 (09 Sep Shift 1)
Oxidation state of sulphur in anions $\mathrm{SO}_3^{2-}, \mathrm{S}_2 \mathrm{O}_4^{2-}$ and $\mathrm{S}_2 \mathrm{O}_6^{2-}$ increases in the orders :

Options:
A
$\mathrm{S}_2 \mathrm{O}_6^{2-} < \mathrm{S}_2 \mathrm{O}_4^{2-} < \mathrm{SO}_3^{2-}$
B
$\mathrm{SO}_6^{2-} < \mathrm{S}_2 \mathrm{O}_4^{2-} < \mathrm{S}_2 \mathrm{O}_6^{2-}$
C
$\mathrm{S}_2 \mathrm{O}_4^{2-} < \mathrm{SO}_3^{2-} < \mathrm{S}_2 \mathrm{O}_6^{2-}$
D
$\mathrm{S}_2 \mathrm{O}_4^{2-} < \mathrm{S}_2 \mathrm{O}_6^{2-} < \mathrm{SO}_3^{2-}$

The correct answer is:
$\mathrm{S}_2 \mathrm{O}_4^{2-} < \mathrm{SO}_3^{2-} < \mathrm{S}_2 \mathrm{O}_6^{2-}$

Solution:
$\mathrm{In} \mathrm{SO}_3^{--}$
$$
\begin{aligned}
& x+3(-2)=-2 ; x=+4 \\
& \text { In } \mathrm{S}_2 \mathrm{O}_4^{--} \\
& 2 x+4(-2)=-2 \\
& 2 x-8=-2 \\
& 2 x=6 ; \quad x=+3 \\
& \text { In } \mathrm{S}_2 \mathrm{O}_6^{2-} \\
& 2 x+6(-2)=-2 \\
& 2 x=10 ; \quad x=+5
\end{aligned}
$$
hence the correct order is
$$
\mathrm{S}_2 \mathrm{O}_4{ }^{--} < \mathrm{SO}_3^{--} < \mathrm{S}_2 \mathrm{O}_6{ }^{--}
$$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2013 (22 Apr Online)
Oxidation states of the metal in the minerals haematite and magnetite, respectively, are

Options:
A II, III in haematite and III in magnetite
B II, III in haematite and II in magnetite
C II in haematite and II, III in magnetite
D III in haematite and II, III in magnetite

The correct answer is:
III in haematite and II, III in magnetite

Solution:
Hematite is a common iron oxide with a formula of Fe2O3 so Fe in it has +3 oxidation state.
Magnetite is a rock mineral and one of the main iron ores, with the chemical formula Fe3O4 or Fe2O3.FeO .so it is a mixed oxide with oxidation states of Fe as +3 and +2.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below:
$$
\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \stackrel{\frac{1}{2} \Delta_{\mathrm{ass}} \mathrm{H}^{\ominus}}{\longrightarrow} \mathrm{Cl}(\mathrm{g}) \stackrel{\Delta_{\mathrm{eg} H^{\ominus}}}{\longrightarrow} \mathrm{Cl}^{-}(\mathrm{g}) \stackrel{\Delta_{\mathrm{m} \mathrm{on}^{H^{\ominus}}}^{\longrightarrow}}{\longrightarrow} \mathrm{Cl}^{-}(\mathrm{aq}) .
$$
The energy involved in the conversion of $\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g})$ to $\mathrm{Cl}^{-}(\mathrm{g})$ (using the data, $\left.\Delta_{\text {diss }} H_{\mathrm{Cl}_2}^{\ominus}=240 \mathrm{kJmol}^{-1}, \Delta_{\mathrm{eg}} H_{\mathrm{cl}}^{\ominus}=-349 \mathrm{kJmol}^{-1}, \Delta_{\mathrm{hyd}} \mathrm{H}_{\mathrm{Cl}}^{\ominus}=-381 \mathrm{kJmol}^{-1}\right)$ will be

Options:
A
$+152 \mathrm{kJmol}^{-1}$

B
$-610 \mathrm{kJmol}^{-1}$

C
$-850 \mathrm{kJmol}^{-1}$

D
$+120 \mathrm{kJmol}^{-1}$

The correct answer is:
$-610 \mathrm{kJmol}^{-1}$

Solution:
$$
\begin{aligned}
& \text { For the process } \frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \longrightarrow \mathrm{Cl}_{\mathrm{aq}}^{-} \\
& \Delta \mathrm{H}=\frac{1}{2} \Delta \mathrm{H}_{\text {diss }} \text { of } \mathrm{Cl}_2+\Delta_{\mathrm{eg}} \mathrm{Cl}+\Delta_{\mathrm{hyd}} \mathrm{Cl}^{-} \\
& =+\frac{240}{2}-349-381 \\
& =-610 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2008
Oxiran is

Options:
A Ethylene oxide
B Diethyl ether
C Ethyl glycolate
D Glycolic ester

The correct answer is:
Ethylene oxide

Solution:
Oxiran is ethylene oxide, $\mathrm{CH}_2-\mathrm{CH}_2$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Oxygen is not produced during photosynthesis by

Options:
A Cycas
B Nostac
C Green sulphur bacteria
D Chara

The correct answer is:
Green sulphur bacteria

Solution:
Green sulfur bacteria are nonmotile and capable of anoxygenic photosynthesis. In contrast to plants, green sulfur bacteria mainly use sulfide ions as electron donors.

Cycas belongs to Gymnospermaea, which show oxygenic photosynthesis. Chara is a Chlorophyceae member to show oxygenic photosynthesis. Cyanobacteria like Nostoc, Anabaena have evoleved oxygenic photosynthetic mechanism.

Topic: Biology, Biological Classification, NEET, NEET 2018
Oxygenic photosynthesis occurs in :

Options:
A Chlorobium
B Chromatium
C Oscillatoria
D Rhodospirillum

The correct answer is:
Oscillatoria

Solution:
No Solution Available

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2009 (Mains)
Ozone depletion in the stratosphere is mainly caused by

Options:
A SO2
B NO2
C NO
D Chlorofluorocarbons

The correct answer is:
Chlorofluorocarbons

Solution:
Chlorofluorocarbons destroy ozone. When CFCs reach the stratosphere, they split and produce reactive free redicals.

Topic: Chemistry, Environmental Chemistry, JEE Main
Ozone is obtained from oxygen

Options:
A By oxidation at high temperature
B By oxidation using a catalyst
C By silent electric discharge
D By conversion at high pressure

The correct answer is:
By silent electric discharge

Solution:
$3 \mathrm{O}_2 \underset{\text { electric discharge }}{\stackrel{\text { silent }}{\rightleftharpoons}} 2 \mathrm{O}_3$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Ozone with $K I$ solution produces

Options:
A $\mathrm{Cl}_2$
B $I_2$
C $\mathrm{HI}$
D $1 O_3$

The correct answer is:
$I_2$

Solution:
$\begin{aligned} & \mathrm{O}_3 \rightarrow \mathrm{O}_2+[\mathrm{O}] \\ & 2 \mathrm{KI}+\mathrm{H}_2 \mathrm{O}+[\mathrm{O}] \rightarrow 2 \mathrm{KOH}+\mathrm{I}_2 \\ & \overline{2 \mathrm{KI}+\mathrm{H}_2 \mathrm{O}+\mathrm{O}_3 \rightarrow 2 \mathrm{KOH}+\mathrm{I}_2}+\mathrm{O}_2\end{aligned}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of :

Options:
A
a vinyl group

B
an isopropyl group

C
an acetylenic triple bond

D
two ethylenic double bonds

The correct answer is:
a vinyl group

Solution:
Vinyl group
$$
\left(\mathrm{CH}_2=\mathrm{CH}-\right)
$$
on ozonolysis give formaldehyde

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2011
P1 and P2 are corresponding points on the ellipse x216+y29=1 and its auxiliary circle respectively. If the normal at P1 to the ellipse meets OP2 in Q (where O is the origin), then the length of OQ is equal to

Options:
A 3 units
B 9 units
C 4 units
D 7 units

The correct answer is:
7 units

Solution:
Let, P1=4cosâ¡Î¸,3sinâ¡Î¸
and P2=4cosâ¡Î¸,4sinâ¡Î¸
Now, equation of P1Q is 4xcosâ¡Î¸-3ysinâ¡Î¸=7
Equation of P2Q is xcosâ¡Î¸=ysinâ¡Î¸
Q=7cosâ¡Î¸,7sinâ¡Î¸
âOQ=7

Topic: Mathematics, Ellipse, JEE Main
$\mathrm{P}_2 \mathrm{O}_5$ is heated with water to give

Options:
A Hypophosphorus acid
B Orthophosphorus acid
C Hypophosphoric acid
D Orthophosphoric acid

The correct answer is:
Orthophosphoric acid

Solution:
$\mathrm{P}_2 \mathrm{O}_5+3 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}_3 \mathrm{PO}_4$ orthophosphoric acid

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
P 4 O 6 reacts with water according to equation P 4 O 6  +  6H 2 O→ 4H 3 PO 3 . Calculate the volume of 0.1 M NaOH (in mL ) solution required to neutralise the acid formed by dissolving 1.1 g of P 4 O 6 in H 2 O?

Options:

The correct answer is:
400.00

Solution:
P 4 O 6 + 6H 2 Oâ 4H 3 PO 3 ......(i)
Neutralisation :
H 3 PO 3 â+â2NaOHâ Na 2 HPO 3 â+â 2H 2 O âÃâ4 .......(ii)
Adding Eqs. (i) and (ii)
P 4 O 6 + 8 NaOH â¶ 4 Na 2 HPO 3 + 2 H 2 O 1 mol 8 mol .....(iii)
Number of moles of P 4 O 6 ,
n = m M = 1 . 1 2 2 0 = 1 2 0 0 mol
(Molar mass P 4 O 6 =(4Ã31)+(6Ã16)=220
âµ Product formed by 1 mole of P 4 O 6 is neutralised by 8 moles NaOH
â´ Product formed by 1 2 0 0 moles of P 4 O 6 will be neutralised by NaOH
= 8 Ã 1 2 0 0 = 8 2 0 0 moles NaOH
Given, Molarity of NaOHâ=â0.1âMâ=â0.1âmol/L
Molarity = Number of moles Volume in litres
Volume = Number of moles Molarity
= 8 2 0 0 Ã 1 0 . 1 = 0 . 4 L or 4 0 0 mL
â´ 400âmLâNaOH is required.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
$\mathrm{P}_4 \mathrm{O}_{10}$ is not used with to dry ${ }{\mathrm{NH}_3}$ gas because

Options:
A $\mathrm{P}_4 \mathrm{O}_{10}$ is basic and $\mathrm{NH}_3$ is acidic
B $\mathrm{P}_4 \mathrm{O}_{10}$ is acidic and $\mathrm{NH}_3$ is basic
C $P_4 O_{10}$ is not a drying agent
D $\mathrm{P}_4 \mathrm{O}_{10}$ reacts with moisture in $\mathrm{NH}_3$

The correct answer is:
$\mathrm{P}_4 \mathrm{O}_{10}$ is acidic and $\mathrm{NH}_3$ is basic

Solution:
$\mathrm{P}_4 \mathrm{O}_{10}$ is not used with to dry ${ }{\mathrm{NH}_3}$ gas because ${ }{\mathrm{P}_4 \mathrm{O}_{10}}$ is acidic and ${ }{\mathrm{H}_3}$ is basic

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
$P(A)=\frac{1}{4}, P(B)=\frac{5}{8}$ and $P(A \cup B)=\frac{3}{4}$, then $P(A \cap B)=$

Options:
A $\frac{1}{8}$
B $0$
C $\frac{3}{4}$
D $1$

The correct answer is:
$\frac{1}{8}$

Solution:
$P(A \cap B)=\frac{2}{8}+\frac{5}{8}-\frac{6}{8}=\frac{1}{8}$.

Topic: Mathematics, Probability, JEE Main
$P(A)=\frac{3}{8}, \quad P(B)=\frac{5}{8}$ and $P(A \cup B)=\frac{3}{4}$, then $P\left(\frac{A}{B}\right)=$
If $A$ and $B$ are two events such that

Options:
A $\frac{2}{5}$
B $\frac{2}{3}$
C $\frac{3}{5}$
D None of these

The correct answer is:
$\frac{2}{5}$

Solution:
$P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{(3 / 8)+(5 / 8)-(3 / 4)}{(5 / 8)}=\frac{2}{5}$

Topic: Mathematics, Probability, JEE Main
P and Q are two distinct points on the parabola, y2=4x, with parameters t and t1, respectively. If the normal at P passes through Q, then the minimum value of t12 , is

Options:
A 8
B 4
C 6
D 2

The correct answer is:
8

Solution:
Given Pt2, 2t and Qt12, 2t1, are parametric points on parabola
y2=4x
Slope of normal at P=-t

Since normal passes through Q,

⇒ slope of normal at P = slope of PQ

⇒ -t=2t1-tt12-t2

⇒ -t=2t1+t

⇒t1=-t-2t

⇒t12=t2+4t2+4 ...1

t2+4t2≥2t2.4t2≥4,

Using AM≥GM

∴ t12≥4+4 from 1

⇒ t12≥8

∴ Minimum value=8

Topic: Mathematics, Parabola, JEE Main, JEE Main 2016 (10 Apr Online)