datavorous/entrance-exam-dataset huggingface dataset:
The standard reduction potentials at \( 298 \mathrm{~K} \) for the following half reactions are given against each
\[
\begin{array}{ll}
\mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e} \rightleftharpoons \mathrm{Zn}(\mathrm{s}) & -0.762 \\
\mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{e} \rightleftharpoons \mathrm{Cr}(\mathrm{s}) & -0.740 \\
2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e} \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g}) & 0.000 \\
\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e} \rightleftharpoons \mathrm{Fe}^{2+}(\mathrm{aq}) & 0.770
\end{array}
\]
Which is the strongest reducing agent?

Options:
A \( \mathrm{Fe}^{2+} \) (aq)
B \( \mathrm{H}_{2}(\mathrm{~g}) \)
C \( \operatorname{Cr}(\mathrm{s}) \)
D \( \operatorname{Zn}(\mathrm{s}) \)

The correct answer is:
\( \operatorname{Zn}(\mathrm{s}) \)

Solution:
For an element or an Ion to act as a reducing agent, it should itself get oxidised. Hence, the species which has the highest negative value of reduction potential will be most easily oxidised. As Zn2+ has the highest negative value for reduction, it will have the highest positive value for oxidation and thus will be the best reducing agent among the given options.

The correct order of reducing character is:Zn2+(E°=-0.762)>Cr3+(E°=−0.740)>H+(E°=0)>Fe3+(E°=0.770)

Topic: Chemistry, Electrochemistry, JEE Main
The standard reduction potentials for $\mathrm{Zn}^{2+} / \mathrm{Zn}, \mathrm{Ni}^{2+} / \mathrm{Ni}$, and $\mathrm{Fe}^{2+} / \mathrm{Fe}$ are $-0.76,-0.23$ and $-0.44 \mathrm{~V}$ respectively. The reaction $\mathrm{X}+\mathrm{Y}^{2+} \rightarrow \mathrm{X}^{2+}+\mathrm{Y}$ will be spontaneous when:

Options:
A
$X=N i, Y=F e$

B
$X=N i, Y=Z n$

C
$X=F e, Y=Z n$

D
$X=Z n, Y=N i$

The correct answer is:
$X=Z n, Y=N i$

Solution:
$\begin{aligned}
& \mathrm{Zn}+\mathrm{Fe}^{+2} \rightarrow \mathrm{Zn}^{+2}+\mathrm{Fe} \\
& \mathrm{Fe}+\mathrm{Ni}^{+2} \rightarrow \mathrm{Fe}^{2+}+\mathrm{Ni} \\
& \mathrm{Zn}+\mathrm{Ni}^{2+} \rightarrow \mathrm{Zn}^{+2}+\mathrm{Ni}
\end{aligned}$
All these are spontaneous

Topic: Chemistry, Electrochemistry, JEE Main
The standard reduction potentials for $\mathrm{Zn}^{2+} / \mathrm{Zn}, \mathrm{Ni}^{2+} / \mathrm{Ni}$, and $\mathrm{Fe}^{2+} / \mathrm{Fe}$ are $-0.76,-0.23$ and $-0.44 \mathrm{~V}$ respectively. The reaction $\mathrm{X}+\mathrm{Y}^{2+} \rightarrow \mathrm{X}^{2+}+\mathrm{Y}$ will be spontaneous when:

Options:
A
$X=N i, Y=F e$

B
$X=N i, Y=Z n$

C
$X=F e, Y=Z n$

D
$X=Z n, Y=N i$

The correct answer is:
$X=Z n, Y=N i$

Solution:
$\begin{aligned}
& \mathrm{Zn}+\mathrm{Fe}^{+2} \rightarrow \mathrm{Zn}^{+2}+\mathrm{Fe} \\
& \mathrm{Fe}+\mathrm{Ni}^{+2} \rightarrow \mathrm{Fe}^{2+}+\mathrm{Ni} \\
& \mathrm{Zn}+\mathrm{Ni}^{2+} \rightarrow \mathrm{Zn}^{+2}+\mathrm{Ni}
\end{aligned}$
All these are spontaneous

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2012 (Offline)
The standard reduction potentials of $\mathrm{Zn}^{2+}\left|\mathrm{Zn}, \mathrm{Cu}^{2+}\right| \mathrm{Cu}$ and $\mathrm{Ag}^{+} \mid \mathrm{Cu}$ and $\mathrm{Ag}^{+} \mid \mathrm{Ag}$ are respectively $-0.76,0.34$ and $0.8 \mathrm{~V}$. The following cells were constructed :
(1) $\mathrm{Zn}\left|\mathrm{Zn}^{2+}\right|\left|\mathrm{Cu}^{2+}\right| \mathrm{Cu}$
(2) $\mathrm{Zn}\left|\mathrm{Zn}^{2+}\right|\left|\mathrm{Ag}^{+}\right| \mathrm{Ag}$
(3) $\mathrm{Cu}\left|\mathrm{Cu}^{2+} \| \mathrm{Ag}^{+}\right| \mathrm{Ag}$
What is the correct order of $E_{\text {cell }}^{\circ}$ of these cells?

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

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

Solution:
Given
$\mathrm{Zn}^{2+} \longrightarrow \mathrm{Zn} \quad E^{\circ}=-0.78 \mathrm{~V}$
$\mathrm{Cu}^{2+} \longrightarrow \mathrm{Cu} \quad E^{\circ}=0.34 \mathrm{~V}$
$\mathrm{Ag}^{+} \longrightarrow \mathrm{Ag} \quad E^{\circ}=0.8 \mathrm{~V}$
Cell reaction of (1) is
$\mathrm{Zn}+\mathrm{Cu}^{2+} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Cu}$
$E_{\text {cell }}^{\circ}=E_{\text {oxidation }}+E_{\text {reduction }}$
$=+0.78+0.34$
$=+1.12 \mathrm{~V}$
Cell reaction of (2) is
$\mathrm{Zn}+\mathrm{Ag}^{+} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Ag}$
$E_{\text {cell }}^{\circ}=0.78+0.8$
$=+1.58 \mathrm{~V}$
Cell reaction of (3) is :
$\mathrm{Cu}+\mathrm{Ag}^{+} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{Ag}$
$E_{\text {cell }}^{\circ}=-0.34+0.8$
$=+0.46$
So, the correct order of $E_{\text {cell }}^{\circ}$ of these cell is -
$\mathrm{b}>\mathrm{a}>\mathrm{c} .$

Topic: Chemistry, Electrochemistry, JEE Main
The starting material for convenient preparation of deuterated hydrogen peroxide D2O2 in laboratory is:

Options:
A K2 S2O8
B 2-ethylanthraquinol
C BaO2
D BaO

The correct answer is:
K2 S2O8

Solution:
Peroxodisulphate, obtained by electrolytic oxidation of acidified sulphate solutions at high current density, on hydrolysis yields hydrogen peroxide.

2HSO4- (aq) →Electrolysis HO3SOOSO3H (aq) →Electrolysis 2HSO4- (aq) + 2H+ (aq) + H2O2 (aq)

This method is now used for the laboratory preparation of D2O2.

K2 S2O8 (s)+2D2O (l)→2KDSO4 (aq.)+D2O2 (l)

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2023 (01 Feb Shift 2)
The state of heart when it is not pumping blood effectively enough to meet the needs of the body is called:

Options:
A Heart attack
B Heart block
C Heart failure
D Cardiac arrest

The correct answer is:
Heart failure

Solution:
Heart attack means when the heart muscle is suddenly damaged by an adequate blood supply. Heart block means when the action potential of heart beat is not passed to the ventricles properly. Cardiac arrest means when the heart stop beating at all.
Heart failure means the state of heart when it is not pumping blood effectively enough to meet the needs of the body. It is sometimes called congestive heart failure because congestion of the lungs is one of the main symptoms of this disease. Heart failure is not the same as cardiac arrest (when the heart stops beating) or a heart attack (when the heart muscle is suddenly damaged by an inadequate blood supply).

Topic: Biology, Body Fluids and Circulation, NEET
The state of hybridization of $\mathrm{C}_2, \mathrm{C}_3, \mathrm{C}_5$ and $\mathrm{C}_6$ of the hydrocarbon -
is in the following sequence :

Options:
A $\mathrm{sp}, \mathrm{sp}^2, \mathrm{sp}^3$ and $\mathrm{sp}^2$
B $\mathrm{sp}, \mathrm{sp}^3, \mathrm{sp}^2$ and $\mathrm{sp}^3$
C $\mathrm{sp}^3, \mathrm{sp}^2, \mathrm{sp}^2$ and $\mathrm{sp}$
D $\mathrm{sp}, \mathrm{sp}^2, \mathrm{sp}^2$ and $\mathrm{sp}^3$

The correct answer is:
$\mathrm{sp}, \mathrm{sp}^3, \mathrm{sp}^2$ and $\mathrm{sp}^3$

Solution:
$$
\begin{aligned}
& \mathrm{C}_2 \text { having } \rightarrow 2-\sigma \text { bond } \rightarrow \mathrm{sp} \\
& \mathrm{C}_3 \text { having } \rightarrow 4-\sigma \text { bond } \rightarrow \mathrm{sp}^3 \\
& \mathrm{C}_5 \text { having } \rightarrow 3-\sigma \text { bond } \rightarrow \mathrm{sp}^2 \\
& \mathrm{C}_6 \text { having } \rightarrow 4-\sigma \text { bond } \rightarrow \mathrm{sp}^3
\end{aligned}
$$

Topic: Chemistry, Biomolecules, NEET, NEET 2009 (Mains)
The statement A→B→A is equivalent to :

Options:
A A→A∧B
B A→A∨B
C A→A↔B
D A→A→B

The correct answer is:
A→A∨B

Solution:
A→B→A

≡A→~B∨A

≡~A∨~B∨A

≡~A∨A∨~B

≡T∨~B≡T

∴ T∨B=T

≡~A∨A∨B

≡~A∨A∨B

≡A→A∨B

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2021 (25 Feb Shift 1)
The statement about cerium which is not true:

Options:
A The common oxidation states of cerium are \( +3 \) and \( +4 \).
B The \( +3 \) oxidation state of cerium is more stable than \( +4 \) oxidation state.
C Cerium (III) acts as an oxidising agent.
D The \( +4 \) oxidation state of cerium is known in solution.

The correct answer is:
Cerium (III) acts as an oxidising agent.

Solution:
+4 oxidation state of cerium is also known in solutions.
4 f⁡ 1 5 d 0 6 S 2
4 f⁡ 2 5 d 0 6 S 2
The high charge on Ce4+ leads to high hydration, hydrated Ce4+ is hydrolysed giving polymeric species & H+ . Ce(4+) is stable in strongly acidic solutions & used as oxidising agent in volumetric analysis.

Topic: Chemistry, d and f Block Elements, JEE Main
The statement among the following that is a tautology is:

Options:
A A∨A∧B
B A∧A∨B
C B→A∧A→B
D A∧A→B→B

The correct answer is:
A∧A→B→B

Solution:
A∧A→B→B

=A∧~A∨B→B

=A∧~A∨A∧B→B

=A∧B→B

=~A∧B∨B

=~A∨~B∨B

=T

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2021 (24 Feb Shift 1)
The statement B⇒~A∨B is not equivalent to :

Options:
A B⇒A⇒B
B A⇒A⇔B
C A⇒~A⇒B
D B⇒~A⇒B

The correct answer is:
A⇒A⇔B

Solution:
Given statement is B⇒~A∨B

Truth table for given statement:

A	B	~A	~A∨B	B⇒~A∨B

T	T

F

T	T

T	F	F	F	T

F	T	T	T	T

F	F	T	T	T

It is a tautology.

Truth table of given options:

A⇒B	~A⇒B	B⇒A⇒B	A⇒~A⇒B	B⇒~A⇒B

T	T	T	T	T

F	T	T	T	T

T	T	T	T	T

T	F	T	T	T

Option A: B⇒A⇒B is a tautology.

Option C: A⇒~A⇒B is a tautology.

Option D: B⇒~A⇒B is a tautology.

Hence, it is very clear that given statement is not equivalent to option B.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2023 (29 Jan Shift 2)
The statement ~p∨~p∧q is equivalent to

Options:
A ~p∧q
B p∧q∧~p
C ~p∧q∧q
D ~p∨q

The correct answer is:
p∧q∧~p

Solution:
Given,

~p∨~p∧q

Now we know that,

By demorgan law, ~A∨B≡~A∧~B

Now using above formula we get,

~p∨~p∧q

≡~p∧p∧q as ~~A≡A

≡p∧q∧~p

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2023 (10 Apr Shift 2)
The statement ~p⇔q is not equivalent to

Options:
A ~pâq
B pâ§~qâ¨qâ§~p
C p∨q∧~p∨~q
D pâ¨qâpâ§q

The correct answer is:
p∨q⇒p∧q

Solution:
Option A is theorem hence A is equivalent to ~p⇔q
Option B ~p⇔q means only p or only q
Which is same as p∧~q∨q∧~p
Option C p∨q∧~pv~q
≡p∨q∧~p∧q
Means p∨q happens but p∧q does not happen. It means only p or only q happens
Option D ~p⇔q≡p∨q∨~p∧q
≡~p∨q⇒p∧q
but not equivalent to p∨q⇒p∧q

Topic: Mathematics, Mathematical Reasoning, JEE Main
The statement p⇔q is not equivalent to

Options:
A pâ¨q⇒pâ§q
B pâ§qâpâ¨q
C pâ¨qâpâ§q
D ~p∨q∨p∧q

The correct answer is:
p∧q⇒p∨q

Solution:
A p∨q⇒p∧q
which is same as p⇔q
B p∧q⇒p∨q is tautology because p∧q is subset of p∨q
C p∨q⇔p∧q
which is same as p⇔q
D ~p∨q∨p∧q
which is same as p⇔q

Topic: Mathematics, Mathematical Reasoning, JEE Main
The statement ∼p↔~q is

Options:
A A tautology
B A fallacy
C Equivalent to p↔q
D Equivalent to ~p↔q

The correct answer is:
Equivalent to p↔q

Solution:
p	q	~q	p↔~q	~p↔~q	p↔q

T	T	F	F	T	T

T	F	T	T	F	F

F	T	F	T	F	F

F	F	T	F	T	T

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2014 (06 Apr)
The statement ~p→q→p is equivalent to

Options:
A p→p→q
B p→p∨q
C q→p
D q→p→q

The correct answer is:
q→p

Solution:
∵s→r≡~s∨r
∴~p→q→p≡p∨q→p
≡p∨~q∨p≡~q∨p=q→p

Topic: Mathematics, Mathematical Reasoning, JEE Main
The statement $p \rightarrow(q \rightarrow p)$ is equivalent to

Options:
A
$p \rightarrow(p \rightarrow q)$

B
$p \rightarrow(p \vee q)$

C
$p \rightarrow(p \wedge q)$

D
$p \rightarrow(p \leftrightarrow q)$

The correct answer is:
$p \rightarrow(p \vee q)$

Solution:
$$
\begin{aligned}
& p \rightarrow(q \rightarrow p)=\sim p \vee(q \rightarrow p) \\
& =\sim p \vee(\sim q \vee p) \quad \text { since } p \vee \sim p \text { is always true } \\
& =\sim p \vee p \vee q=p \rightarrow(p \vee q) .
\end{aligned}
$$

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2008
The statement $p \rightarrow(q \rightarrow p)$ is equivalent to :

Options:
A
$p \rightarrow q$

B
$p \rightarrow(p \vee q)$

C
$p \rightarrow(p \rightarrow q)$

D
$p \rightarrow(p \wedge q)$

The correct answer is:
$p \rightarrow(p \vee q)$

Solution:
Since truth value of $p \rightarrow(q \rightarrow p)$ and $p \rightarrow(p \vee q)$ are same, hence $p \rightarrow(q \rightarrow p)$ is equivalent to $p \rightarrow(p \vee q)$.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2013 (22 Apr Online)
The statement ~p⇔q∧p is equivalent to

Options:
A p∧q
B q⇔p
C p∧~q
D ~p∧q

The correct answer is:
p∧~q

Solution:
p	q	p⇔q	~p⇔q	~p⇔q∧p	p∧~q	~p∧q

T	T	T	F	F	F	F

T	F	F	T	T	T	F

F	T	F	T	F	F	T

F	F	T	F	F	F	F

Topic: Mathematics, Mathematical Reasoning, JEE Main
The statement p⇒q∧p is negation of the statement

Options:
A p⇒q
B pâ§q
C ~p⇒q
D ~pâ§q

The correct answer is:
~p⇒q

Solution:
∵~s⇒t≡s∧~t
∵~p⇒q∧p≡p∧~q∧p
≡p∧~q∨~p
≡p∧~q∨p∧~p
≡p∧~p∨f
≡p∧~q≡~p⇒q

Topic: Mathematics, Mathematical Reasoning, JEE Main
The statement p→q→p→p→p∨q is :

Options:
A equivalent to p∧q∨~q
B a contradiction
C equivalent to p∨q∧~p
D a tautology

The correct answer is:
a tautology

Solution:
p	q	q→p	p∨q	r:p→q→p	s:p→(p∨q)	r→s

T	T	T	T	T	T	T

T	F	T	T	T	T	T

F	T	F	T	T	T	T

F	F	T	F	T	T	T

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2020 (05 Sep Shift 2)
The statement p∧~q⇒p⇒~q is

Options:
A equivalent to ~p∨~q
B a tautology
C equivalent to p∨q
D a contradiction

The correct answer is:
a tautology

Solution:
Using logical equivalence condition p→q≡~p∨q, given statement can be expressed as,

p∧~q→p→~q

≡ ~p∧~q∨~p∨~q.......(1)

We know De Morgan's law of logical equivalence is,

~p∧q≡~p∨~q

Applying De Morgan's law in equation(1) we get,

≡ ~p∨q∨~p∨~q ∵~~p≡p

≡ ~p∨~p∨q∨~q

≡~p∨t≡t (tautology)

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2023 (25 Jan Shift 1)
The statement (p∧(~q))∨((~p) ∧ q) ∨((~p) ∧ (~q)) is equivalent to _____

Options:
A ~p∨q
B ~p∨~q
C p∨~q
D p∨q

The correct answer is:
~p∨~q

Solution:
Given,

(p∧(~q))∨((~p) ∧ q) ∨((~p) ∧ (~q))

=((~p) ∧ q) ∨((~p) ∧ (~q))∨(p∧(~q))

=~p∧q∨~q∨p∧~q

We know that ~A∨A=T, where T is tautology,

=~p∧T∨p∧~q

=~p∨p∧~q

=~p∨p∧~p∨~q

=T∧~p∨~q

=~p∨~q

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2023 (13 Apr Shift 2)
The statement p→q→~p→q→q is

Options:
A A tautology
B Equivalent to ~p→q
C Equivalent to p→~q
D A fallacy

The correct answer is:
A tautology

Solution:
p	q	p→q	~p	~p→q	~p→q→q	p→q→~p→q→q

T	T	T	F	T	T	T

T	F	F	F	T	F	T

F	T	T	T	T	T	T

F	F	T	T	F	T	T

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2017 (02 Apr)
The statement p∧q⇒p∧r is equivalent to

Options:
A q⇒p∧r
B p⇒p∧r
C p∧r⇒p∧q
D p∧q⇒r

The correct answer is:
p∧q⇒r

Solution:
Given,

p∧q⇒p∧r

{Using A⇒B≡~A∨B}

=~p∧q∨p∧r

{Using Demorgan's law ~A∧B≡~A∨~B}

=~p∨~q∨p∧r

=~p∨p∧r∨~q

=~p∨p∧~p∨r∨~q

=~p∨r∨~q

=~p∨~q∨r

=~p∧q∨r

=p∧q⇒r

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2022 (29 Jul Shift 1)
The statement p⇒q∨p⇒r is NOT equivalent to:

Options:
A p∧~r⇒q
B ~q⇒~r∨p
C p⇒q∨r
D p∧~q⇒r

The correct answer is:
~q⇒~r∨p

Solution:
Given,

p⇒q∨p⇒r

Now using the formula A⇒B≡~A∨B

We get, ~p∨q∨~p∨r

≡~p∨q∨r

≡p⇒q∨r, so option C is true.

Now solving option 1 we get,

p∧~r⇒q

≡~p∧~r∨q

≡~p∨r∨q

≡~p∨r∨q

≡~p∨r∨~p∨q

≡p⇒q∨p⇒r, so option 1 is also true,

Now solving option 4 we get,

p∧~q⇒r

≡~p∧~q∨r

≡~p∨q∨r

≡~p∨q∨r

≡~p∨q∨~p∨r

≡p⇒q∨p⇒r, so option 4 is also correct.

Hence option B is only wrong.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2022 (29 Jul Shift 2)
The statement ~p⇔~q∧q is:

Options:
A a tautology
B a contradiction
C equivalent to p⇒q∧q
D equivalent to p⇒q∧p

The correct answer is:
equivalent to p⇒q∧p

Solution:
~p⇔~q∧q =p⇔q∧q

Proof: Solving the given ~p⇔~q

=~p→~q∧~q→p

=~~p∨~q∧q∨p

=~~p∧q∧p∨q

=p∧q∨~p∨q ...1

Now, p⇔q≡p→q∧q→p

=~p∨q∧~q∨p

=~p∨q∧~q∨~p∨q∧p

=~p∧~q∨q∧~q∨~p∧p∨q∧p

=~p∧~q∨c∨c∨q∧p where c is contradiction

=~p∧~q∨p∧q

=~p∨q∨p∧q ...2

So from equation 1 & 2 we get

~p⇔~q∧q≡p⇔q∧q

p	q	p↔q	p↔q∧q	p→q	p→q∧q	p→1∧p

T	T	T	T	T	T	T

T	F	F	F	F	F	F

F	T	F	F	T	T	F

F	F	T	F	T	F	F

∴ ~p⇔~q∧q is equivalent to p⇒q∧p.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2022 (26 Jul Shift 1)
The statement that is INCORRECT about Ellingham diagram is

Options:
A provides idea about the reaction rate.
B provides idea about free energy change.
C provides idea about changes in the phases during the reaction.
D provides idea about reduction of metal oxide.

The correct answer is:
provides idea about the reaction rate.

Solution:
Ellingham diagram is a plot between ΔG° and T and does not give any information regarding rate of reaction

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2021 (27 Jul Shift 1)
The statement that is INCORRECT about the interstitial compounds is:

Options:
A They have high melting points.
B They are very hard.
C They are chemically reactive.
D They have metallic conductivity.

The correct answer is:
They are chemically reactive.

Solution:
The principal physical and chemical characteristics of interstitial compounds are as follows:

(i) They have high melting points, higher than those of pure metals.

(ii) They are very hard, some borides approach diamond in hardness.

(iii) They retain metallic conductivity.

(iv) They are chemically inert.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The statement that is NOT correct is

Options:
A Energies of stationary states in hydrogen like atoms is inversely proportional to the square
of the principal quantum number.
B The radius of the first orbit of \( \mathrm{He}^{+} \)is half that of the first orbit of hydrogen atom.
C Angular quantum number signifies the shape of the orbital.
D Total number of nodes for \( 3 \mathrm{~s} \) orbital is three

The correct answer is:
Total number of nodes for \( 3 \mathrm{~s} \) orbital is three

Solution:
The total number of nodes of an orbital is the sum of angular and radial nodes and is given in terms of the $\mathrm{n}$ and $\mathrm{I}$
quantum number by the following equation. Number of nodes $=n-l-1$
for 3 s-orbital, $\mathrm{n}=3$
$\mathrm{I}=0$
Nodes $=3-0-1=2$
Because for s-orbitals, $\mathrm{l}=0$, therefore, there is no angular nodes only radial nodes are present, that is, 2 radial nodes.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The statement that is not true about ozone is :

Options:
A in the stratosphere, CFCs release chlorine free radicals CI which reacts with O3 to give chlorine
B in the atmosphere, it is depleted by CFCs.
C in the stratosphere, it forms a protective shield against UV radiation.
D It is a toxic gas and its reaction with NO gives NO2

The correct answer is:
in the stratosphere, CFCs release chlorine free radicals CI which reacts with O3 to give chlorine

Solution:
In presence of sunlight CFC’s molecule divides & release chlorine free radical, which react with ozone give chlorine monoxide radical (ClO) and oxygen

CF2Cl2g⟶UVC˙lg+C˙F2Clg

Cl*(g)+O3(g)⟶ClO*(g)+O2(g)

ClO∙(g)+O(g)⟶Cl'(g)+O2(g)

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2020 (02 Sep Shift 1)
The Statement that is TRUE among the following is

Options:
A
The contrapositive of $3 x+2=8 \Rightarrow x=2$ is $x \neq 2$ $\Rightarrow 3 x+2 \neq 8$.

B
The converse of $\tan \mathrm{x}=0 \Rightarrow x=0$ is $x \neq 0 \Rightarrow$ $\tan x=0$.

C
$p \Rightarrow q$ is equivalent to $p \vee \sim q$.

D
$p \vee q$ and $p \wedge q$ have the same truth table.

The correct answer is:
The contrapositive of $3 x+2=8 \Rightarrow x=2$ is $x \neq 2$ $\Rightarrow 3 x+2 \neq 8$.

Solution:
Only statement given in option
(a) is true.
(b) The converse of $\tan \mathrm{x}=0 \Rightarrow x=0$ is $x=0 \Rightarrow \tan x=0$
$\therefore$ Statement (b) is false
(c) $\sim(p \Rightarrow q$ is dquivalent to $p \wedge \sim q$
$\therefore$ Statement given in option (c) is false.
(d) No, $p \vee q$ and $p \wedge q$ does not have the same truth value.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2012 (07 May Online)
The statement that is true for the long form of the periodic table is

Options:
A It reflects the sequence of filling the electrons in the order of sub-energy levels s, p, d and f
B It helps to predict the stable valency states of the elements
C It reflects trends in physical and chemical properties of the elements
D It helps to predict the relative ionicity of the bonds between any two elements

The correct answers are:
It reflects the sequence of filling the electrons in the order of sub-energy levels s, p, d and f, It helps to predict the stable valency states of the elements, It reflects trends in physical and chemical properties of the elements, It helps to predict the relative ionicity of the bonds between any two elements

Solution:
It reflects trends in physical and chemical properties of the elements.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The statements
(i) In filling a group of orbitals of equal energy, it is energetically preferable to assign electrons to empty orbitals rather than pair them into a particular orbital.
(ii) When two electrons are placed in two different orbitals, energy is lower if the spins are parallel. are valid for

Options:
A Aufbau principle
B Hund's rule
C Pauli's exclusion principle
D Uncertainty principle

The correct answer is:
Hund's rule

Solution:
Follow Hund's multiplicity rules.

Topic: Chemistry, Structure of Atom, JEE Main
The statement(s) that are correct about the species \(\mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Na}^{+}\)and \(\mathrm{Mg}^{2+}\).
(A) All are isoelectronic
(B) All have the same nuclear charge
(C) \(\mathrm{O}^{2-}\) has the largest ionic radii
(D) \(\mathrm{Mg}^{2+}\) has the smallest ionic radii
Choose the most appropriate answer from the options given below :

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

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

Solution:

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2024 (05 Apr Shift 1)
The statement/s which are true about antagonists from the following is/are:
A. They bind to the receptor site.
B. Get transferred inside the cell for their action.
C. Inhibit the natural communication of the body.
D. Mimic the natural messenger.
Choose the correct answer from the options given below:

Options:
A B only
B A, C and D
C A and C
D A and B

The correct answer is:
A and C

Solution:
Drugs that bind to the receptor site and inhibit its natural function are called antagonists. Antagonists are molecules that bind to receptors but do not activate them. Instead, they block the action of other molecules, including natural ligands and agonists, that normally bind to the receptor and activate it. Antagonists inhibit the natural communication of the body by preventing the binding of the natural ligands, which can result in physiological effects such as decreased heart rate, blood pressure, and inflammation.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The states of hybridization of boron and oxygen atoms in boric acid $\left(\mathrm{H}_3 \mathrm{BO}_3\right)$ are respectively

Options:
A
$\mathrm{sp}^2$ and $\mathrm{sp}^2$

B
$\mathrm{sp}^3$ and $\mathrm{sp}^3$

C
$\mathrm{sp}^3$ and $\mathrm{sp}^2$

D
$\mathrm{sp}^2$ and $\mathrm{sp}^3$

The correct answer is:
$\mathrm{sp}^2$ and $\mathrm{sp}^3$

Solution:
No Solution Available

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2004
The steady-state current through L1 after the switch has been closed for a very long time is

Options:
A V0R
B V0L1RL1+L2
C V0L2RL1+L2
D None of these

The correct answer is:
V0L2RL1+L2

Solution:
After a very long time t→∞,

I0=VR ....(i)

Since the inductors are in parallel,

L1dI1dt=L2dI2dt

After a long time,

L1I1=L2I2 .... (ii)

The total current I0 through the battery is divided into the inductors, hence

I0=I1+I2 ..... (iii)
From equations (ii) & (iii)
I1=(L2L1+L2)I0&I2=(L1L1+L2)I0
The current through L1 is

I1=(L2L1+L2)I0
=(L2L1+L2)V0R

Topic: Physics, Alternating Current, JEE Main
The step that proved to be the main challenging obstacle in the production of human insulin by genetic engineering is:

Options:
A Removal of C-peptide from active insulin
B Getting insulin assembled into a mature form
C Addition of C-peptide to proinsulin
D Splitting A and B polypeptide chains

The correct answer is:
Getting insulin assembled into a mature form

Solution:
Getting insulin assembled into a mature form proved to be the main challenging obstacle in the production of human insulin by genetic engineering.

Topic: Biology, Biotechnology and Its Applications, NEET