datavorous/entrance-exam-dataset huggingface dataset:
The number of N-CH2-N bonds in urotropine is

Options:

The correct answer is:
6

Solution:
Urotropine is hexamethylene tetramine CH26N4

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The number of natural numbers less than 7000 which can be formed by using the digits 0, 1, 3, 7, 9 (repetition of digits allowed) is equal to:

Options:
A 375
B 250
C 374
D 372

The correct answer is:
374

Solution:
∴ Total number formed less than

7000=3×5×5×5=375

∴ Total natural number formed less than

7000=375-1=374

(number zero has been excluded)

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The number of natural numbers n for which the equation x-8x=nn-10 has no real solution is equal to

Options:
A 2
B 3
C 4
D 5

The correct answer is:
5

Solution:
x2-8x-nn-10=0 has no real solution.
⇒64+4nn-10<0
⇒n2-10n+16<0
⇒n-8n-2<0
⇒2<n<8
⇒n=3,4,5,6,7
⇒ Number of values of n is 5

Topic: Mathematics, Quadratic Equation, JEE Main
The number of natural numbers $n$ in the interval $[1005,2010]$ for which the polynomial $1+x+x^{2}+x^{3}+\ldots . x^{n-1}$ divides the polynomial $1+x^{2}+x^{3}+x^{4}+\ldots .+x^{2010}$ is-

Options:
A 0
B 100
C 503
D 1006

The correct answer is:
503

Solution:
$$
\begin{array}{l}
1+x^{2}+x^{4}+\ldots \ldots x^{2010}=\frac{1\left(1-x^{2012}\right)}{1-x^{2}}=\frac{\left(1-x^{1006}\right)\left(1+x^{1006}\right)}{(1-x)(1+x)} \\
=\left(1+x^{1006}\right)\left(\frac{\left(1-x^{503}\right)}{(1-x)}\right)\left(\frac{\left(1+x^{503}\right)}{(1+x)}\right) \\
=\left(1+x^{1006}\right)\left(1+x+x^{2}+\ldots . x^{502}\right)\left(1-x+x^{2}-x^{3}+\ldots . x^{502}\right)
\end{array}
$$
this is divisible by $1+x+x^{2}+\ldots . . x^{n-1}$
if $n-1=502$
$$
n=503
$$

Topic: Mathematics, Binomial Theorem, JEE Main
The number of neutrons emitted when ${ }_{92}^{235} \mathrm{U}$ undergoes controlled nuclear fission to ${ }_{54}^{142} \mathrm{Xe}$ and ${ }_{38}^{90} \mathrm{Sr}$ is

Options:

The correct answer is:
3

Solution:
${ }_{92} \mathrm{U}^{235} \longrightarrow{ }_{54} \mathrm{Xe}^{142}+{ }_{38} \mathrm{Sr}^{90}+3{ }_0 n^1$
Nuclear chemistry
Basic application
II

Topic: Chemistry, Structure of Atom, JEE Main
The number of neutrons emitted when \( { }_{92}^{235} \mathrm{U} \) undergoes controlled nuclear fission to \( { }_{54}^{142} \mathrm{Xe} \) and \( { }_{38}^{91} \mathrm{Sr} \) is

Options:

The correct answer is:
3

Solution:
The nuclear reaction for the given case is,

9 2 U 2 3 5 + 0 n 1 → 9 2 U 2 3 6 → 5 4 Xe 1 4 2 + 3 8 Sr 9 1 + 3 0 n 1 Slow Unstable

It is clear from the reaction that 3 neutrons are emitted in the reaction.

Topic: Physics, Nuclear Physics, JEE Main
The number of neutrons present in the more abundant isotope of boron is ' \(x\) '. Amorphous boron upon heating with air forms a product, in which the oxidation state of boron is ' \(y\) '. The value of \(x+y\) is _______

Options:
A 3
B 9
C 4
D 6

The correct answer is:
9

Solution:
More abundant isotope $=\mathrm{B}^{11}$
[Number of neutrons $=6$ ]
$\begin{aligned}
& \mathrm{x}=6 \\
& \mathrm{~B}+\mathrm{O}_2 \rightarrow \mathrm{B}_2 \mathrm{O}_3
\end{aligned}$
Oxidation state of $\mathrm{B}$ in $\mathrm{B}_2 \mathrm{O}_3=+3$
So, $y=3$
Hence $x+y=9$

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2024 (05 Apr Shift 1)
The number of nitrogen atoms in a semicarbazone molecule of acetone is_____

Options:

The correct answer is:
3

Solution:
Semicarbazone molecule of acetone

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The number of non-ionisable hydrogen atoms present in the final product obtained from the hydrolysis of PCl5 is :

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2021 (26 Aug Shift 2)
The number of non-ionisable protons present in the product B obtained from the following reaction is___C2H5OH+PCl3→C2H5Cl+A

A+PCl3→B

Options:

The correct answer is:
2

Solution:
C2H5OH+PCl3⟶C2H5Cl+H3PO3

H3PO3+PCl3⟶H4P2O5+HCl

A is H3PO3(Phosphorous acid)

The basicity of A is two

B is H4P2O5(Pyrophosphorous acid)

Structure of H4P2O5 is as given below

Hydrogen which are directly attached to the phosphorous in the above structure are non-ionisable.

The Number of non-ionisable hydrogen atoms are two in the above structure.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2022 (26 Jul Shift 2)
The number of non-negative integral solutions of the equation x+y+3z=33 is

Options:
A 120
B 135
C 210
D 520

The correct answer is:
210

Solution:
x+y=33-3z
Here, z will vary from 0 to 11
The number of required solutions
=∑z=011C33-3z+2-12-1
=∑z=011C34-3z1=∑z=01134-3z
=34×12−3⋅ 11×12 2
=34×12-33×6=210.

Topic: Mathematics, Permutation Combination, JEE Main
The number of non-negative integral solutions of the equation x+y+z+5t=15 is

Options:
A 196
B 224
C 312
D 364

The correct answer is:
224

Solution:
x+y+z+5t=15, t∈0,1,2,3
⇒x+y+z=15-5t
Number of non-negative integral solutions of
x+y+z=15-5t is equal to 17-5tC2=17-5t16-5t2
Hence, the desired number is equal to
∑t=0317-5t16-5t2
=17×162+12×112+7×62+2×12
=136+66+21+1
=224

Topic: Mathematics, Permutation Combination, JEE Main
The number of non-polar molecules from the following is
HF,H2O,SO2,H2,CO2,CH4,NH3,HCl,CHCl3,BF3

Options:

The correct answer is:
4

Solution:
The molecules which have perfect geometry and no lone pairs are zero dipole moment molecules. Those molecules termed as non-polar molecules. Amongst the given compounds, CH4, CO2,H2,BF3 are non polar molecules.

H2O has two lone pairs and two bond pairs. SO2 has two sigma bond pairs and one lone pair. CHCl3 has net dipole moment because the bond moments do not cancel with each other. NH3 have one lone pair and three bond pairs. In HF bond moment is the dipole moment. Hence, all these molecules have some net dipole moment.

The non-polar molecules are CO2, H2, CH4 and BF3

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The number of normals drawn to the parabola $y^{2}=4 x$ from the point $(1,0)$ is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
1

Solution:
Given curve is $y^{2}=4 x$.
Also, point $(1,0)$ is the focus of the parabola.
It is clear from the graph that only one normal is possible.

Topic: Mathematics, Parabola, JEE Main
The number of numbers between 2,000 and 5,000 that can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is not allowed) and are multiple of 3 is

Options:
A 36
B 30
C 24
D 48

The correct answer is:
36

Solution:
When a number is divisible by 3, then sum of digits of a number is also divisible by 3.

Number can be formed from (0,1,2,3) or (0,2,3,4).
Number of 4 digits number

=3×3×2×1+3×3×2×1

=36.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2018 (16 Apr Online)
The number of numbers, lying between 99 and 1000 that can be made from the digits 2, 3, 7, 0, 8 and 6 when the digits occur only once in each number, is

Options:
A 100
B 90
C 120
D 80

The correct answer is:
100

Solution:
Required number of numbers =6P3−5​P2=120−20=100.

{Since, between 99 & 1000, the numbers are of 3 digits, but those numbers should be omitted in which 0 comes at hundreds place}.

Topic: Mathematics, Permutation Combination, JEE Main
The number of numbers, strictly between 5000 and 10000 can be formed using the digits 1,3,5,7,9 without repetition, is

Options:
A 6
B 12
C 120
D 72

The correct answer is:
72

Solution:
Given,

Numbers 1,3,5,7,9

Now to form number between 5000-10000

_⏟5,7,9_⏟4 ways_⏟3 ways_⏟2 ways

Here at 1000's place should be taken by 5, 7 & 9

So for 1000's place there will be 3 ways,

Now for hundredth place there will be 4 ways,

For ten's place there will be 3 ways,

And for unit place there will be 2 ways,

So total ways will be 3×4×3×2=72 ways.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (25 Jan Shift 2)
The number of octahedral sites per sphere in a fcc structure is

Options:
A $8$
B $4$
C $2$
D $1$

The correct answer is:
$1$

Solution:
Number of octahedral sites $=$ Number of sphere in the packing.
$\therefore$ Number of octahedral sites per sphere $=1$.

Topic: Chemistry, Solid State, JEE Main
The number of octahedral voids per lattice site in a lattice is_. (Rounded off to the nearest integer)

Options:

The correct answer is:
1

Solution:
If number of lattice points are N.

then effective octahedral voids =N

So, octahedral voids / lattice site =1

Topic: Chemistry, Solid State, JEE Main, JEE Main 2021 (26 Feb Shift 2)
The number of onto functions from the set $\{1,2, \ldots, 11\}$ to the set $\{1,2, \ldots, 10\}$ is

Options:
A $5 \times 11 !$
B $10 !$
C $\frac{11 !}{2}$
D $10 \times 11 !$

The correct answer is:
$10 \times 11 !$

Solution:
Let $\quad A=\{1,2, \ldots, 11\}$
$\begin{array}{ll}\therefore & n(A)=11 \text { and } B=\{1,2, \ldots, 10\} \\ \therefore & n(B)=10\end{array}$
$\therefore \text{Hence number of onto function}$ $=^{n(A)} C_{n(B)} \times n(B) ! \times n(B)$
$={ }^{11} C_{10} \times 10 ! \times 10$
$=(11 \times 10 !) \times 10=11 ! \times 10$

Topic: Mathematics, Functions, JEE Main
The number of optical isomers in following compound is: ________

Options:

The correct answer is:
32

Solution:
Total chiral centre $=5$
No. of optical isomers $=2^5=32$.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (08 Apr Shift 1)
The number of optically active products obtained from the complete ozonolysis of the given compound is

Options:
A 4
B 2
C 0
D 1

The correct answer is:
0

Solution:
Ozonolysis : CH3CHO (inactive)

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The number of optically active structural isomers of C4H9Br is __________

Options:

The correct answer is:
1.00

Solution:
C 4 H 9 Br:
* represents chiral carbon

Topic: Chemistry, General Organic Chemistry, JEE Main
The number of orbitals associated with quantum numbers n=5,ms=+12 is:

Options:
A 11
B 25
C 50
D 15

The correct answer is:
25

Solution:
For n=5 5s, 5p, 5d, 5f  and 5g and subshell is available
In 5s→1 orbital
5p→3 orbital
5d→5 orbital
5f→7 orbital
5g→9 orbital
In each subshell 1 electron of every orbital will belong to m 5 =+ 1 2 So, Total number of orbitals that will be associate with m 5 =+ 1 2 will be 1+3+5+7+9=25

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2020 (07 Jan Shift 1)
The number of ordered pairs a,x satisfying the equation sec2⁡a+2x+a2-1=0, -π<x<π is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Given equation is sec2⁡a+2x+a2-1=0
⇒tan2⁡a+2x+a2=0
⇒tan⁡a+2x=0 and a=0
⇒tan⁡2x=0
⇒x=0,π2,-π2.
So, 0,0,0,π2,0,-π2 are ordered pairs a,x satisfying the equation.

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of ordered pairs $(\mathrm{m}, \mathrm{n})$, where $\mathrm{m}, \mathrm{n} \in\{1,2,3, \ldots \ldots, 50\}$, such that $6^{\mathrm{m}}+9^{\mathrm{n}}$ is a multiple of 5 is $-$

Options:
A 1250
B 2500
C 625
D 500

The correct answer is:
1250

Solution:
$\mathrm{m}$ can be any value and $\mathrm{n}$ will be odd number then sum is multiple of 5 $50 \times 25=1250$

Topic: Mathematics, Sets and Relations, JEE Main
The number of ordered pairs of positive integers a,b, such that their Least Common Multiple is the given positive integer 72×113×194, is equal to

Options:

The correct answer is:
315

Solution:
Required number of ordered pairs
=2Ã2+12Ã3+12Ã4+1
=5Ã7Ã9=315

Topic: Mathematics, Permutation Combination, JEE Main
The number of ordered pairs r,k for which 6.Cr35=k2-3.Cr+136, where k is an integer is

Options:
A 3
B 2
C 6
D 4

The correct answer is:
4

Solution:
Given, 6.Cr35=k2-3.Cr+136

⇒36r+1×35Crk2-3=6×35Cr, using properties of binomial coefficient.
⇒k2-3=r+16

⇒k2=3+r+16

r can be 5,35

For r=5,k=±2

r=35,k=±3

Hence, number of solution =4

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2020 (07 Jan Shift 2)
The number of ordered pairs $(\mathrm{x}, \mathrm{y})$ for which $\mathrm{A}=$ $\left(\begin{array}{lll}1 & 2 & 1 \\ 2 & 2 & x \\ y & 1 & 2\end{array}\right)$ is a singular and symmetric matrix is

Options:
A $1$
B $0$
C $2$
D $3$

The correct answer is:
$0$

Solution:
Given $A=\left[\begin{array}{lll}1 & 2 & 1 \\ 2 & 2 & x \\ y & 1 & 2\end{array}\right]$
Since $A$ is singular i.e. $|A|=0$
$$
\begin{aligned}
\Rightarrow & \left|\begin{array}{lll}
1 & 2 & 1 \\
2 & 2 & x \\
y & 1 & 2
\end{array}\right|=0 \\
\Rightarrow & 1(4-x)-2(4-x y)+1(2-2 y)=0 \\
\Rightarrow & 4-x-8+2 x y+2-2 y=0
\end{aligned}
$$
$\Rightarrow \quad-2-x-2 y+2 x y=0$ ...(i)
$$
\Rightarrow \quad-2-x-2 y+2 x y=0
$$
Since $A$ is symmetric
$$
\begin{aligned}
& \Rightarrow A^T=A \\
& \Rightarrow\left[\begin{array}{lll}
1 & 2 & \mathrm{y} \\
2 & 2 & 1 \\
1 & \mathrm{x} & 2
\end{array}\right]=\left[\begin{array}{lll}
1 & 2 & 1 \\
2 & 2 & \mathrm{x} \\
\mathrm{y} & 1 & 2
\end{array}\right]
\end{aligned}
$$
Comparing above, we get
$y=1, x=1$ ...(i)

Topic: Mathematics, Matrices, JEE Main
The number of ordered pairs x,y of real numbers satisfying the system of equations sin⁡x=sin⁡2y and cos⁡x=sin⁡y, where 0≤x,y≤π, is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
sin⁡x=sin⁡2y and cos⁡x=sin⁡y

Squaring and adding, we get,
1=2sin⁡ycos⁡y2+sin2⁡y
1-sin2⁡y=4sin2⁡ycos2⁡y
cos2⁡y1-4sin2⁡y=0
cos⁡y=0 or sin⁡y=±12
y=π2 or π6 or 5π6
y=π2⇒x=0
y=π6⇒x=π3
⇒ two ordered pairs 0,π2 & π3,π6

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of ordered pairs (x,y) satisfying the equations y=2sinâ¡x and y=5x2+2x+3 is/are

Options:
A 0
B 1
C 2
D Infinite

The correct answer is:
0

Solution:
2sinâ¡x=5x2+2x+3
â 2sinâ¡x=4x2+x+12+2
But, 2sinâ¡xâ¤2
and 4x2+x+12+2>2, so it has no solution

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of oxygen atoms bonded to one phosphorous atom in $\mathrm{P}_4 \mathrm{O}_6$ is

Options:
A $4$
B $3$
C $6$
D $5$

The correct answer is:
$3$

Solution:
$\mathrm{P}_4 \mathrm{O}_6$ has following structure.
Thus, every P-atom is linked to 3 oxygen atoms.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The number of oxygen present in a nucleotide formed from a base, that is present only in RNA is

Options:

The correct answer is:
9

Solution:
Nucleotide with uracil base present in RNA.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2022 (26 Jun Shift 1)
The number of $\mathrm{P}-\mathrm{O}$ bonds in $\mathrm{P}_4 \mathrm{O}_6$ is:

Options:
A
9

B
6

C
12

D
18

The correct answer is:
12

Solution:
$$
\text { The number of } \mathrm{P}-\mathrm{O} \text { bonds in } \mathrm{P}_4 \mathrm{O}_6=12
$$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The number of P-O-P bonds in H4P2O7, HPO33, and P4O10 are respectively

Options:
A 0, 3, 6
B 0, 3, 4
C 1, 2, 4
D 1, 3, 6

The correct answer is:
1, 3, 6

Solution:
In H4P2O7, each phosphorus undergo sp3 hybridisation. It has one P-O-P linkage in it.

HPO33

In HPO33, each phosphorus undergo sp3 hybridisation. It has three P-O-P linkages in it.

P4O10

In P4O10, each phosphorus undergo sp3 hybridisation. It has six P-O-P linkages in it.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2023 (15 Apr Shift 1)
The number of P-OH bonds and the oxidation state of phosphorus atom in pyrophosphoric acid H4P2O7, respectively are:

Options:
A Five and four.
B Four and five.
C Five and five.
D Four and four.

The correct answer is:
Four and five.

Solution:
As we can clearly see in the structure,

the number of P−OH bonds is 4.

The oxygen is an electronegative element and will always carry a negative charge. Now, it depends on its surrounding atoms about what charge it carries. In case of a hydroxyl group, the hydrogen carries a +1 charge. So, the opposite charge generated on the oxygen atom along with magnitude will be -1.
In the case of only oxygen atoms, we know that the oxidation number of oxygen is equal to -2. So, the charge generated on the adjacent atom of oxygen will be +2.
Thus, the oxidation state of phosphorus atom will be generated because of:
Oxidation state on phosphorous = [2×(P−OH)+(1×=O)+(1×−O)] oxidation states
Thus, the oxidation state of both the phosphorus atoms will be:
PO.S=2×(+1)+1×(+2)+1×(+1)=+5

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2017 (09 Apr Online)
The number of pairs a,b of real numbers, such that whenever α is a root of the equation x2+ax+b=0, α2-2 is also a root of this equation, is :

Options:
A 6
B 8
C 4
D 2

The correct answer is:
6

Solution:
Consider the equation x2+ax+b=0

It has two roots (not necessarily real α and β)

Either α=β or α≠β

Case (1) If α=β, then it is repeated root. Given than α2-2 is also root

So, α=α2-2

⇒ α+1α-2=0

⇒ α=-1 or α=2

When α=-1 then a,b=2,1

α=2 then a,b=-4,4

Case (2) If α≠β, then four possibilities are there

(I) α=α2-2 and β=β2-2

Here α,β=2,-1 or -1,2

Hence a,b=-α+β,αβ

=-1,-2

(II) α=β2-2 and β=α2-2

Then α-β=β2-α2=β-αβ+α

Since α≠β we get α+β=β2+α2-4

α+β=α+β2-2αβ-4

Thus -1=1-2αβ-4 which implies

αβ=-1. Therefore a,b=-α+β,αβ

=1,-1

(III) α=α2-2=β2-2 and α≠β

⇒ α=-β

Thus α=2,β=-2

α=-1,β=1

Therefore a,b=0,-4 and 0,-1

(IV) β=α2-2=β2-2 and α≠β is same as (III)

Therefore we get 6 pairs of a,b

Which are 2,1,-4,4,-1,-2,1,-1,0,-4,0,-1

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2021 (01 Sep Shift 2)
The number of pairs of the solution having the same value of the osmotic pressure from the following is
(Assume 100% ionization)

A. 0.500 M C2H5OH(aq) and 0.25 M KBr(aq)
B. 0.100 M K4Fe(CN)6(aq) and 0.100 M FeSO4NH42SO4(aq)

C. 0.05 M K4Fe(CN)6 (aq) and 0.25 M NaCl (aq)
D. 0.15 M NaCl(aq) and 0.1 M BaCl2(aq)
E. 0.02 M KCl.MgCl2.6H2O(aq) and 0.05 M KCl(aq)

Options:

The correct answer is:
4

Solution:
We know,
π=iCRT

π∝iC

A. 0.500 M C2H5OH(aq)iC=1×0.5=0.5 and 0.25 M KBr(aq)iC=2×0.25=0.5

B. 0.100 M K4Fe(CN)6(aq)iC=5×0.1=0.5 and 0.100 M FeSO4NH42SO4(aq)iC=5×0.1=0.5

C. 0.05 M K4Fe(CN)6iC=5×0.05=0.25 (aq) and 0.25 M NaCliC=2×0.25=0.5 (aq)

D. 0.15 M NaCl(aq)iC=2×0.15=0.3 and 0.1 M BaCl2(aq)iC=3×0.1=0.3

E. 0.02 M KCl.MgCl2.6H2O(aq)iC=5×0.02=0.1 and 0.05 M KCl(aq)iC=2×0.05=0.1

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (25 Jan Shift 2)
The number of paramagnetic species among the following is B2,Li2,C2,C2-,O22-,O2+ and He2+

Options:

The correct answer is:
4

Solution:
The species that contain unpaired electrons are paramagnetic.

Electronic configuration of Li2 molecule is σ1 s2σ*1 s2σ2 s2.

The electronic configuration of B2 molecule is

σ1s2σ⋆1s2σ2s2σ⋆2s2π2px1=π2py1

The presence of unpaired electrons in the two π bonding molecular orbitals explains its paramagnetic nature.

The electronic configuration of C2 KKσ2s2σ*2s2π2px2π2py2

O2: KKσ2 s2σ2 s2σ2pz2π2px2=π2py2π*2px1=π*2py1

Hence, O2 molecule is paramagnetic.

The electronic configuration of C2- is σ1s2σ⋆1s2σ2s2σ⋆2s2π2px2=π2py2σ2pz1

Hence, it is paramagnetic substance.

O22-: KKσ2 s2σ2 s2σ2pz2π2px2=π2py2π*2px2=π*2py2

O2+: KKσ2 s2σ2 s2σ2pz2π2px2=π2py2π*2px1=π*2py

Hence, the above species is paramagnetic.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2022 (28 Jul Shift 1)
The number of paramagnetic species among the following is
O2,CO,N2,C2,CsO2,BaO2,CO2

Options:

The correct answer is:
2

Solution:
O2 and CsO2 are paramagnetic.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The number of paramagnetic species from the following is

Ni(CN)42-,Ni(CO)4,NiCl42-

Fe(CN)64-,CuNH342+

Fe(CN)63- and FeH2O62+

Options:

The correct answer is:
4

Solution:
Ni(CN)42-:Ni+2=

It is diamagnetic as CN- is a strong field ligand and forces electron pairing.

Ni(CO)4:Ni=

It is diamagnetic as carbonyl group is a strong field ligand and forces electron pairing.

NiCl42-:Ni2+=

It is paramagnetic as chloride group is a weak field ligand.

Fe(CN)64-:Fe2+ :-

It is diamagnetic as –CN is a strong field ligand which causes pairing of electrons.

CuNH34+2:Cu+2⇒ Here Copper ion has one unpaired electron, thus it is paramagnetic as it is 3d9 configuration.

Fe(CN)63- : Fe+3:

It is paramagnetic, here CN- is a strong field ligand

FeH2O62+:Fe2+:

It is paramagnetic, here H2O , here a weak field ligand

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (25 Jan Shift 1)
The number of pentagons in C60 and trigons (triangles) in white phosphorus, are;

Options:
A 12 and 3
B 20 and 4
C 20 and 3
D 12 and 4

The correct answer is:
12 and 4

Solution:
C60 is a molecule that consists of 60-C atoms, arranged as 12 pentagons and 20 hexagons.
It has 4 trigons.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2019 (10 Apr Shift 2)
The number of peptide bond(s) in the following molecule is/are

Options:

The correct answer is:
2

Solution:
Peptide bond is the amide linkage between two α - amino acids.

Topic: Chemistry, Biomolecules, JEE Main
The number of permutations by taking all letters and keeping the vowels of the word COMBINE in the odd places is

Options:
A 96
B 144
C 512
D 576

The correct answer is:
576

Solution:
Hints: Vowels : O, I, E
No. of Odd place : 4
No of ways $={ }^4 \mathrm{P}_3 \times 4 !=576$

Topic: Mathematics, Permutation Combination, JEE Main
The number of permutations of alphabets of the word “ENSHRINE” in which no two alike alphabets are together is equal to

Options:

The correct answer is:
5760

Solution:
EENNSHRI
Let the event E: E's are together and event N: N's are together
Required permutations=Total permutations -nE∪N
=8!2!2!-7!2!+7!2!-6!
=10080-2520-2520+720=5760

Topic: Mathematics, Permutation Combination, JEE Main
The number of permutations of the alphabets of the word “GOOGLE” in which O’s are together but G’s are separated, is

Options:
A 24
B 48
C 72
D 36

The correct answer is:
36

Solution:
Keep both O’s together, then apart from G’s other alphabets can be arranged in 3!=6 ways. In the gaps, two G’s can be placed in C42 ways.

OOLE

Hence, the total number of ways=3!⋅C24=36

Topic: Mathematics, Permutation Combination, JEE Main
The number of permutations, of the digits 1, 2, 3, …, 7 without repetition, which neither contain the string 153 nor the string 2467, is _______ .

Options:

The correct answer is:
4898

Solution:
Given,

The number 1,2,3,....,7

Now total numbers of 7 digit number will be 7! without repetition,

Now total numbers which contain string 154 will be 5!

Total numbers which contain string 2367 will be 4!

And number which contain both string 2367 & 154 will be 2

So, total number which does not contain string 154 or 2367 will be 7!-5!+4!-2

=5040-120+24-2

=5040-142=4898

Hence this is the required answer.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (10 Apr Shift 1)
The number of photoelectrons emitted for light of a frequency $\mathrm{v}$ (higher than the threshold frequency $\mathrm{v}_0$ ) is proportional to :

Options:
A Frequency of light (v)
B $v-v_0$
C Threshold frequency $\left(\mathrm{v}_0\right)$
D Intensity of light

The correct answer is:
Intensity of light

Solution:
No. of photoelectrons emitted is independent of frequency but depends on intensity.

Topic: Physics, Dual Nature of Matter, NEET, NEET 2009 (Mains)
The number of photons emitted per second by a 10 W sodium vapour lamp assuming that 90% of the consumed energy is converted into light, is [Given, the wavelength of sodium light is 590 nm, Planck's constant, h=6.63Ã10-34J s]

Options:
A 0.267×1018
B 0.267×1019
C 0.267×1020
D 0.267×1017

The correct answer is:
0.267×1020

Solution:
Energy of photon
E=hcÎ»=6.63Ã10-34Ã3Ã108590Ã10-9=6.63Ã359Ã10-18
Light energy produced per second =90100Ã10=9 W
â´ Number of photons emitted per second
=9Ã596.63Ã3Ã10-18=2.67Ã1019

Topic: Physics, Dual Nature of Matter, JEE Main
The number of photons falling per second on a completely darkened plate to produce a force of 6.62Ã10-5 N is nÃ1022. If the wavelength of the light falling is 5Ã10-7 m, then what is the value of n? h=6.62Ã10-34 J s

Options:

The correct answer is:
5

Solution:
The energy carried by n photons per second
P=n.hcλ
As photons get absorbed by dark plate, the force experienced by the plate
F=1c. nhcλ
Or F×c=n.hcλ as power P=force×speed
Or n=Fλh=6.62×10-5×5×10-76.62×10-34=5×1022

Topic: Physics, Dual Nature of Matter, JEE Main