datavorous/entrance-exam-dataset huggingface dataset:
The collision of an electron with kinetic energy $5.5 \mathrm{eV}$ and a hydrogen atom in its ground state can be described as

Options:
A completely inelastic
B may be completely inelastic
C may be partially elastic
D elastic

The correct answer is:
elastic

Solution:
Given, kinetic energy of electron $=5.5 \mathrm{eV}$.
The energy required to excite a hydrogen atom from ground state $(n=1)$ to first excited state $(n=2)$ is
$\begin{aligned}
\Delta E & =E_2-E_2 \\
& =-3.4-(-13.6)=10.2 \mathrm{eV}
\end{aligned}$
When an electron collides with hydrogen atom in the ground state with energy of $5.5 \mathrm{eV}$ which is less than $10.2 \mathrm{eV}$, then it will not be able to excite hydrogen atom in to first excited state. Therefore, electron will not give any energy to the hydrogen atom. Hence, the total kinetic energy of electron remains conserved. Hence, its collision with hydrogen atom is elastic.

Topic: Physics, Atomic Physics, JEE Main
The collision of an electron with kinetic energy $5.5 \mathrm{eV}$ and a hydrogen atom in its ground state can be described as

Options:
A completely inelastic
B may be completely inelastic
C may be partially elastic
D elastic

The correct answer is:
elastic

Solution:
Given, kinetic energy of electron $=5.5 \mathrm{eV}$.
The energy required to excite a hydrogen atom from ground state $(n=1)$ to first excited state $(n=2)$ is
$\begin{aligned}
\Delta E & =E_2-E_2 \\
& =-3.4-(-13.6)=10.2 \mathrm{eV}
\end{aligned}$
When an electron collides with hydrogen atom in the ground state with energy of $5.5 \mathrm{eV}$ which is less than $10.2 \mathrm{eV}$, then it will not be able to excite hydrogen atom in to first excited state. Therefore, electron will not give any energy to the hydrogen atom. Hence, the total kinetic energy of electron remains conserved. Hence, its collision with hydrogen atom is elastic.

Topic: Physics, Atomic Physics, TS EAMCET, TS EAMCET 2019 (03 May Shift 1)
The colloidal solution of mercury in water can be easily obtained by

Options:
A Mechanical precipitation
B Bredig's arc method
C Repeated washing
D Ultrasonic dispersion

The correct answer is:
Ultrasonic dispersion

Solution:
Correct option is 4. Ultrasonic dispersion
A colloidal solution of certain elements such as mercury and sulphur is obtained by passing their vapors through cold water containing a stabilizer ( an ammonium salt or citrate).

Topic: Chemistry, Surface Chemistry, JEE Main
The color of $\mathrm{CoCl}_{3} .5 \mathrm{NH}_{3} \cdot \mathrm{H}_{2} \mathrm{O}$ is

Options:
A red
B orange
C orange - yellow
D pink

The correct answer is:
pink

Solution:
$\mathrm{CoCl}_{3} .5 \mathrm{NH}_{3} \mathrm{H}_{2} \mathrm{O}$ is pink in colour

Topic: Chemistry, Coordination Compounds, JEE Main
The colour change of an indicator HIn in acid base titrations is given below

Which of the following statement is correct?

Options:
A In a strong alkaline solution colour Y will be observed
B In a strong acidic solution colour Y will be observed
C Concentration of In- is higher than that of HIn at the equivalence point
D In a strong alkaline solution colour X is observed

The correct answer is:
In a strong alkaline solution colour Y will be observed

Solution:
It is an acidic indicator therefore will remain in ionized form in strong alkaline solution (opposite ion effect) so it will show the colour due to Inaq- which is colour Y.

Topic: Chemistry, Redox Reactions, JEE Main
The colour coding on a carbon resistor is shown in the given figure. The resistance value of the given resistor is:

Options:
A (7500±375)Ω
B (5700±285)Ω
C (7500±755)Ω
D (5700±375)Ω

The correct answer is:
(7500±375)Ω

Solution:
Colour	Letter as an aid to memory	Number	Multiplier	Colour	Tolerance

Black	B	0	100	Gold	5%

Brown	B	1	101	Silver	10%

Red	R	2	102

No fourth band

20%

Orange	O	3	103

Yellow	Y	4	104

Green	G	5	105

Blue	B	6	106

Violet	V	7	107

Grey	G	8	108

White	W	9	109

A set of coloured co-axial rings or bands is printed on the resistor which reveals the following facts :
1. The first band indicates the first significant figure.
2. The second band indicates the second significant figure.
3. The third band indicates the power of ten with which the above two significant figures must be multiplied to get the resistance value in ohms.
4. The fourth band indicates the tolerance or possible variation in percent of the indicated value. If the fourth band is absent, it implies a tolerance of ±20%

R=(7500±375)Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (27 Aug Shift 2)
The colour of KMnO4 is due to: [where M→ metal, L→ ligand]

Options:
A σ− σ * transition
B MâL charge transfer transition
C d-d transition
D LâM charge transfer transition

The correct answer is:
L→M charge transfer transition

Solution:
Charge transfer from O 2 - to empty d -orbitals of metal ion (Mn 7+ ).
The colour in KMnO 4 arises from an electronic transition. In the permanganate ion (MnO 4 − ), manganese is in the +7 oxidation state, so it has no d electrons. Photons promote an electron from the highest energy molecular orbital to an empty d -orbital on the manganese. When a photon of light is absorbed, this charge transfer takes place from L to M.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2015 (04 Apr)
The colour of KMnO4 is due to: [where M→ metal, L→ ligand]

Options:
A σ− σ * transition
B MâL charge transfer transition
C d-d transition
D LâM charge transfer transition

The correct answer is:
L→M charge transfer transition

Solution:
Charge transfer from O 2 - to empty d -orbitals of metal ion (Mn 7+ ).
The colour in KMnO 4 arises from an electronic transition. In the permanganate ion (MnO 4 â ), manganese is in the +7 oxidation state, so it has no d electrons. Photons promote an electron from the highest energy molecular orbital to an empty d -orbital on the manganese. When a photon of light is absorbed, this charge transfer takes place from L to M.

Topic: Chemistry, Coordination Compounds, JEE Main
The colour of the transition metal ions is due to

Options:
A d-d transition
B Change in geometry
C Variable oxidation states
D None of these

The correct answer is:
d-d transition

Solution:
The colour of transition metal ions is due to d-d transition. When a ligand approaches metal ion, its d -orbital does not remains degenerate.
They split into two sets, one consisting of lower energy and the other consisting higher energy level.
Thus electrons can jump from lower energy d -orbitals to higher energy d -orbitals. Such transition are known as d-d transition.
The required amount of energy during d-d transition is obtained by absorption of light of a particular wavelength in the region of visible light.
Note: Sometimes the colour of transition metal ions is also due to charge transfer spectra.

Topic: Chemistry, d and f Block Elements, NEET
The coloured part of the eye which controls the amount of light entering the eye is a modified part of the

Options:
A sclerotic and choroid
B retina
C choroid
D choroid and retina

The correct answer is:
choroid

Solution:
Human eye is made up of three layers. Outer layer is fibrous layer that consists of Sclera and Cornea. The middle layer consists of Choroid which continues as ciliary apparatus. The innermost layer is known as Retina. Choroid is the main vascular layer of the eye. The main function of this is to prevent scattering of light that enters the eye so that the light can focus on the posterior portion to form clear images. It is also the coloured portion of the eye as it is bluish in colour due to presence of pigment cells.

Topic: Biology, Neural Control and Coordination, NEET
The combination of gates shown below yields

Options:
A
OR gate

B
NOT gate

C
XOR gate

D
NAND gate

The correct answer is:
OR gate

Solution:
Truth table for given combination is
This comes out to be truth table of OR gate

Topic: Physics, Semiconductors, JEE Main, JEE Main 2010
The combination of NAND gates shown here in the figure give output C and C'. C and C' are equivalent to

Options:
A OR gate and AND gate respectively
B AND gate and NOT gate respectively
C AND gate and OR gate respectively
D OR gate and NOT gate respectively

The correct answer is:
OR gate and AND gate respectively

Solution:
C=A¯⋅B¯¯
=A+B
Thus its OR gate
C'=A⋅B¯¯=A⋅B
Thus its AND gate

Topic: Physics, Semiconductors, JEE Main
The combination of 'NANR' gates shown here under (figure) are equivalent to

Options:
A An OR gate and an AND gate respectively
B An AND gate and a NOT gate respectively
C An AND gate and an OR gate respectively
D An OR gate and a NOT gate respectively.

The correct answer is:
An OR gate and an AND gate respectively

Solution:
$C=\overline{\bar{A}} \overline{\bar{B}}=\overline{\bar{A}+\bar{B}}=A+B$ (De morgan's theorem)
Hence output $C$ is equivalent to $O R$ gate.
$C=\overline{\overline{A B} \cdot \overline{A B}}=\overline{\overline{A B}+\overline{A B}}=A B+A B=A B$
In this case output $C$ is equivalent to AND gate.

Topic: Physics, Semiconductors, JEE Main
The combination of plots which does not represent isothermal expansion of an ideal gas is
(A)
(B)
C
(D)

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

The correct answer is:
B and D

Solution:
PV= constant (isothermal expansion)
P∝1V. The graph will be a straight line, y=mx+c. So, the above graph is correct.
PV= constant (isothermal expansion)
On increasing or decreasing P, PV becomes constant. So, the above graph is correct.
A, C graphs are correct.
PV= constant (isothermal expansion)
P∝1V. The graph will be a straight line, y=mx+c. So, the above graph is incorrect.
ΔU=nCvdT
and for isothermal process, ΔU=0. So, the above graph is incorrect.
Options B, D are incorrect.

Topic: Chemistry, Thermodynamics (C), JEE Main
The combination of plots which does not represent isothermal expansion of an ideal gas is
(A)
(B)
C
(D)

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

The correct answer is:
B and D

Solution:
PV= constant (isothermal expansion)
P∝1V. The graph will be a straight line, y=mx+c. So, the above graph is correct.
PV= constant (isothermal expansion)
On increasing or decreasing P, PV becomes constant. So, the above graph is correct.
A, C graphs are correct.
PV= constant (isothermal expansion)
P∝1V. The graph will be a straight line, y=mx+c. So, the above graph is incorrect.
ΔU=nCvdT
and for isothermal process, ΔU=0. So, the above graph is incorrect.
Options B, D are incorrect.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2019 (12 Jan Shift 2)
The combined equation of the asymptotes of the hyperbola $2 x^2+5 x y+2 y^2+4 x+5 y=0$

Options:
A $2 x^2+5 x y+2 y^2=0$
B $2 x^2+5 x y+2 y^2-4 x+5 y+2=0$
C $2 x^2+5 x y+2 y^2+4 x+5 y-2=0$
D $2 x^2+5 x y+2 y^2+4 x+5 y+2=0$

The correct answer is:
$2 x^2+5 x y+2 y^2+4 x+5 y+2=0$

Solution:
\(2 x^2+5 x y+2 y^2+4 x+5 y=0 \rightarrow \text { equation of hyperbola }\)
Equation of Asymptoms & Hyperbola differ by a constant
So, lets take equation of asymptotes as
\(2 x^2+5 x y+2 y^2+4 x+5 y+c=0\) (pair of lines)
\(\begin{aligned}
& \Rightarrow \Delta=o=a b c+2 f g h-a f^2-\mathrm{bg}^2-\mathrm{ch}^2=0 \\
& \text { Here, } a=2, h=\frac{5}{2}, \mathrm{c}=\mathrm{c}, \mathrm{b}=\mathrm{z}, \mathrm{g}=2, \mathrm{f}=\frac{5}{2} \\
& \mathrm{abc}+2 \mathrm{fgh}-\mathrm{af}^2-\mathrm{bg}^2-\mathrm{ch}^2=0 \\
& 4 \mathrm{c}+25-\frac{25}{2}-8-\mathrm{c}\left(\frac{25}{4}\right)=0 \\
& 4 \mathrm{c}+\frac{25}{2}-8-\frac{25 c}{4}=0 \\
& \frac{9}{2}+\frac{(16 x-25 c)}{4}=0 \\
& \frac{9}{2}=\frac{9 c}{4} \\
& \Rightarrow \mathrm{c}=2 ., \text { So, } 2 x^2+5 x y+2 y^2+4 x+5 y+2=0 .
\end{aligned}\)

Topic: Mathematics, Hyperbola, JEE Main
The combustion of benzene l gives CO2g and H2Ol . Given that heat of combustion of benzene at constant volume is -3263.9 kJmol-1 heat of combustion (in kJ mol-1) of benzene at constant pressure will be

R=8.314JK-1mol-1

Options:
A -3267.6
B 4152.6
C -452.46
D 3260

The correct answer is:
-3267.6

Solution:
C6H6l+152O2gâ6CO2g+3H2Ol
Îng=6-152=-32
ÎH=ÎU+ÎngRT
ÎH=-3263900-32Ã8.314Ã298J
=-3267616 J
=-3267.616 kJ

Topic: Chemistry, Thermodynamics (C), JEE Main
The combustion of benzene l gives CO2g and H2Ol. Given that heat of combustion of benzene at constant volume is -3263.9 kJ mol-1 at 25°C; the heat of combustion in kJ mol-1 of benzene at constant pressure will be
R=8.314 JK-1 mol-1

Options:
A -3267.6
B 4152.6
C -452.46
D 3260

The correct answer is:
-3267.6

Solution:
C6H6l+152O2g→6CO2g+3H2O(I) ∴ ∆ng= -32
Use, ΔH=ΔU+ΔngRT
=-3263.5×103-32×8.314×298×10-3
=-3267.6 kJ mol-1

Topic: Chemistry, Thermodynamics (C), JEE Main
The combustion of benzene l gives CO2g and H2Ol. Given that heat of combustion of benzene at constant volume is -3263.9 kJ mol-1 at 25°C; the heat of combustion in kJ mol-1 of benzene at constant pressure will be
R=8.314 JK-1 mol-1

Options:
A -3267.6
B 4152.6
C -452.46
D 3260

The correct answer is:
-3267.6

Solution:
C6H6l+152O2g→6CO2g+3H2O(I) ∴ ∆ng= -32
Use, ΔH=ΔU+ΔngRT
=-3263.5×103-32×8.314×298×10-3
=-3267.6 kJ mol-1

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2018 (08 Apr)
The combustion of sodium in excess air yields a higher oxide. What is the oxidation state of the oxygen in the product? Neglect the negative sign.

Options:

The correct answer is:
1

Solution:
Na+O2âNa2O,ââNa2O+12O2âNa2O2

It exists as 2Na+ and O22â

So oxidation state of O is â1

Topic: Chemistry, Redox Reactions, JEE Main
The combustion of sodium in excess air yields a higher oxide. What is the oxidation state of the oxygen in the product?

Options:

The correct answer is:
-1

Solution:
Na+O2âNa2O,ââNa2O+12O2âNa2O2

It exists as 2Na+ and O22â

So oxidation state of O is â1

Topic: Chemistry, s Block Elements, JEE Main
The common chord of the circle $x^2+y^2+4 x+1=0$ and $x^2+y^2+6 x+2 y+3=0$ is

Options:
A $x+y+1=0$
B $5 x+y+2=0$
C $2 x+2 y+5=0$
D $3 x+y+3=0$

The correct answer is:
$x+y+1=0$

Solution:
$S_1=x^2+y^2+4 x+1=0$
$\begin{aligned} & S_2=x^2+y^2+6 x+2 y+3=0 \\ & \text { Common chord } \equiv S_1-S_2=0 \Rightarrow 2 x+2 y+2=0 \\ & \Rightarrow x+y+1=0 .\end{aligned}$

Topic: Mathematics, Circle, JEE Main
The common difference of the A.P. b1, b2,....,bm is 2 more than common difference of A.P. a1, a2, ....., an. If a40=-159, a100=-399 and b100=a70, then b1 is equal to :

Options:
A 81
B -127
C -81
D 127

The correct answer is:
-81

Solution:
Let a1,a1+d1,a1+2d.................first A.P.

a40=a1+39d=-159.....................(1)

a100=a1+99d=-399....................(2)

d=-4,a1=-3

now

b100=a70

⇒b1+99D=a1+69d

b1+99(-2)=-3+69(-4)

(∵Accoding to question D=d+2)

⇒b1=-81

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The common feature that is shared by Funaria, Dryopteris and Ginkgo is

Options:
A Presence of archegonia
B Well developed vascular tissues
C Independent gametophyte
D Independent sporophyte

The correct answer is:
Presence of archegonia

Solution:
Funaria is a bryophyte in which female sex organ is archegonia, vascular tissue is absent and main lant body is independent gametophyte. Dryopteris is a pteridophyta in which female sex organ is archegonia, vascular tissue are developed and main plant body is independent sporophyte. Ginkgo is a gymnosperm in which female sex organ is archegonia, well developed vascular tissue and independent sporophyte occurs.

Topic: Biology, Plant Kingdom, NEET
The common name of $\mathrm{K}\left[\mathrm{PtCl}_{3}\left(\eta^{2} \cdot \mathrm{C}_{2} \mathrm{H}_{4}\right)\right]$ is

Options:
A potassium salt
B Zeise's salt
C complex salt
D None of these

The correct answer is:
Zeise's salt

Solution:
Zeise's salt is common name of $\mathrm{K}\left[\mathrm{P} \mathrm{C}_{3}\left(\mathrm{\eta}^{2}=\mathrm{C}_{2} \mathrm{H}_{4}\right)\right]$

Topic: Chemistry, Coordination Compounds, JEE Main
The common positive oxidation states for an element with atomic number 24, are:

Options:
A +2 to +6
B +1 and +3 to +6
C +1 and +3
D +1 to +6

The correct answer is:
+2 to +6

Solution:
Cr(Z=24)

[Ar]4 s13 d5Cr shows common oxidation states starting from +2 to +6.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2021 (17 Mar Shift 2)
The common roots of the equations $x^{12}-1=0, x^4+x^2+1=0$ are

Options:
A $\pm \omega$
B $\pm \omega^2$
C $\pm \omega, \pm \omega^2$
D None of these

The correct answer is:
$\pm \omega, \pm \omega^2$

Solution:
$x^{12}-1=\left(x^6+1\right)\left(x^6-1\right)=\left(x^6+1\right)\left(x^2-1\right)\left(x^4+x^2+1\right)$
Common roots are given by $x^4+x^2+1=0$
$\therefore \quad x^2=\frac{-1 \pm i \sqrt{3}}{2}=\omega, \omega^2 \text { or } \omega^4, \omega^2 \quad\left(\omega^3=1\right)$
or $x= \pm \omega^2, \pm \omega$

Topic: Mathematics, Complex Number, JEE Main
The common tangent to the circles x2+y2=4 and x2+y2+6x+8y-24=0 also passes through the point:

Options:
A 4,-2
B -4,6
C 6,-2
D -6,4

The correct answer is:
6,-2

Solution:
The centre and radius of the circle x2+y2+2gx+2fy+c=0 are respectively -g, -f and g2+f2-c.

Let, for the circle S1:x2+y2=4 centre C1:0, 0, r1=2 units.

And, for the circle S2:x2+y2+6x+8y-24=0 centre

C2-3, -4, r2=32+42+24=7 units.

The distance between the points x1, y1 and x2, y2 is x1-x22+y1-y22

Hence, the distance between the centres of the circles is C1C2=-3-02+-4-02=5 units.

Also, r1-r2=5 units.

Since, C1C2=r1-r2, hence both the circles touches each other internally.

Therefore, the common tangent is same as the radical axis of the two circles i.e. S1-S2=0

Thus, the common tangent is x2+y2-4-x2+y2+6x+8y-24=0

⇒ -6x-8y+20=0

⇒ 3x+4y-10=0.

Clearly, the point 6,-2 lies on the common tangent.

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The commonest disaccharide has the molecular formula

Options:
A $\mathrm{C}_{10} \mathrm{H}_{18} \mathrm{O}_9$
B $\mathrm{C}_{10} \mathrm{H}_{20} \mathrm{O}_{10}$
C $\mathrm{C}_{18} \mathrm{H}_{22} \mathrm{O}_{11}$
D $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$

The correct answer is:
$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$

Solution:
Disaccharides are the carbohydrates which on hydrolysis gives two same or different carbohydrates. Their general formula is $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$.

Topic: Chemistry, Biomolecules, JEE Main
The complete combustion of 0.492 g of an organic compound containing 'C', 'H' and 'O' gives 0.793 g of CO2 and 0.442 g of H2O. The percentage of oxygen composition in the organic compound is____.(nearest integer)

Options:

The correct answer is:
46

Solution:
Organic compound +O2⟶ CO2g + H2O 0.793 gram 0.442 gram

Weight of the organic compound=0.492 gram

Number of moles of CO2=Number of moles of C=0.79344Weight of carbon=0.7934412=0.216g

Weight of hydrogen=0.442182=0.0491g

Weight of oxygen=0.492-0.216-0.0491=0.2269g

% of O=0.22490.492×100=46.11≈46

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (28 Jun Shift 2)
The complete set of values of α for which the lines x-12=y-23=z-34 and x-32=y-5α=z-7α+2 are coplanar is

Options:
A 2,3
B 0,3
C -2,3
D R

The correct answer is:
R

Solution:
Any point on the first line is 1+2λ,2+3λ,3+4λ,
Any point on the second line is 3+2μ,5+αμ,7+2μ+αμ
Now, as they are intersecting 1+2λ=3+2μ and 2+3λ=5+αμ
So, λ=μ+1, hence 2+3μ+1=5+αμ
⇒α-3μ=0
Hence for λ=1,μ=0, the point 3,5,7 lies on both the lines irrespective of α.

Hence, ∀α∈R lines are concurrent.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The complete solution set of the inequality cos-1â¡(cos4)>3x2-4x is

Options:
A 0,2+6π-83
B 2-6π-83,0
C (-2, 2)
D 2-6π-83,2+6π-83

The correct answer is:
2-6π-83,2+6π-83

Solution:
As, cos-1â¡(cosâ¡4)=cos-1â¡cosâ¡2Ï-4=2Ï-4
â 2Ï-4>3x2-4x
â 3x2-4x-2Ï-4<0
â 2-6Ï-83<x<2+6Ï-83

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The complex $\left[\operatorname{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]$ has the IUPAC name

Options:
A diamminesilver (I) dicyanosilver (I)
B diamminesilver (I) dicyanoargentate (I)
C dicyanosilver (I) diammineargentate (I)
D none of these

The correct answer is:
diamminesilver (I) dicyanoargentate (I)

Solution:
Both options are same. The IUPAC name of complex $\left[\operatorname{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\left[\operatorname{Ag}(\mathrm{CN})_{2}\right]^{-}$is diamminesilver $(\mathrm{I})$ dicyanoargentate (I).

Topic: Chemistry, Coordination Compounds, JEE Main
The complex having square planar geometry is

Options:
A [Ni CO) 4
B MnCl42-
C CuCl42-
D CuNH342+

The correct answer is:
CuNH342+

Solution:
Complex	O.S	HBZ	Geometry	Unpaired ‘e’	Magnetic property
NiCO4	0	sp3	Tetrahadral	0	Diamagnetic
MnCl42-	+2	sp3	Tetrahadral	5	Paramagnetic
CuCl42-	+2	sp3	Tetrahadral	1	Paramagnetic
CuNH342+ Shifting of e 3d→4p	+2	sp2	Square planar	1	Paraagnetic

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The complex having the highest spin only magnetic moment is

Options:
A FeCN63-
B FeH2O62+
C MnF64-
D NiCl42-

The correct answer is:
MnF64-

Solution:
The spin only magnetic moment for a complex can be calculated using formula

μ=nn+2

(a) FeCN63-

The oxidation number Fe in FeCN63- is +3 .The electronic configuration of Fe3+ is -Ar3d54s0

As CN- is a strong ligand pairing will occur.

FeCN63-

∴n=1⇒μ=11+2=3

(b) FeH2O62+

The oxidation number of Fe in

FeH2O62+ is +2 .The electronic configuration of Fe2+ is -Ar3d64s0

As H2O is weak ligand, so pairing will not occur.

FeH2O62-

n=4

μ=44+2=24

(c) MnF64-

The oxidation number of Mn is +2. It's
configuration is -Ar3d54s0

As F- is a weak field ligand, so pairing will not occur.

MnF64-

n=5

μ=55+2=35

(d) NiCl42-

In NiCl42-, the oxidation number of Ni will be +2. It's configuration is -Ar3d84s0

As Cl- is a weak field ligand, so pairing will not occur.

NiCl42-

n=2

μ=22+2=8

Thus, MnF64- will have the highest spin only magnetic moment.

Topic: Chemistry, Coordination Compounds, JEE Main
The complex ion $\left[\operatorname{Pt}\left(\mathrm{NO}_2\right)(\mathrm{Py})\left(\mathrm{NH}_3\right)\left(\mathrm{NH}_2 \mathrm{OH}\right)\right]^{+}$ will give

Options:
A
2 isomers (Geometrical)

B
3 isomers (Geometrical)

C
6 isomers (Geometrical)

D
4 isomers (Geometrical)

The correct answer is:
3 isomers (Geometrical)

Solution:
Complexes of the type $\mathrm{M}_{A B C D}$ may exist in three isomeric form. Similarily $\left.\left(\mathrm{Pt}\left(\mathrm{NO}_2\right)(\mathrm{Py})\left(\mathrm{NH}_3\right)\left(\mathrm{NH}_2 \mathrm{OH}\right)\right]^{+}\right)$may exist in three isomeric form.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2012 (26 May Online)
The complex ion that will lose its crystal field stabilization energy upon oxidation of metal to +3 state is:

Options:
A Niphen32+
B Fephen32+
C Cophen32+
D Znphen32+

The correct answer is:
Fephen32+

Solution:
Fe+2Ar3d6→Fe+3Ar3d5
CFSE=6×-0.4Δ0=-2.4Δ0
CFSE=-2.0 Δ0

In all other cases, the electron is removed from eg orbital and is hence, this will lead to decrease in pairing energy.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The complex number with argument $\frac{5 \pi^c}{6}$ at a distance of 2 units from the origin is

Options:
A $\sqrt{3}-\mathrm{i}$
B $\sqrt{3}+\mathrm{i}$
C $-\sqrt{3}-\mathrm{i}$
D $-\sqrt{3}+\mathrm{i}$

The correct answer is:
$-\sqrt{3}+\mathrm{i}$

Solution:
Let $\mathrm{z}=\mathrm{a}+\mathrm{ib}$ and we have
$\frac{\mathrm{b}}{\mathrm{a}}=\tan \left(\frac{5 \pi}{6}\right)$ and $\sqrt{\mathrm{a}^2+\mathrm{b}^2}=2$
$\therefore \frac{\mathrm{b}}{\mathrm{a}}=\tan \left(\pi-\frac{\pi}{6}\right)=-\tan \frac{\pi}{6}=\frac{-1}{\sqrt{3}}$ and $\mathrm{a}^2+\mathrm{b}^2=4$
$\therefore \mathrm{b}=\frac{-\mathrm{a}}{\sqrt{3}}$ and $\mathrm{a}^2+\mathrm{b}^2+4$
$\therefore \mathrm{a}^2+\frac{\mathrm{a}^2}{3}=4 \Rightarrow 4 \mathrm{a}^2=12 \Rightarrow \mathrm{a}^2=3 \Rightarrow \mathrm{a}= \pm \sqrt{3}$
Also $\mathrm{b}=\frac{-\mathrm{a}}{\sqrt{3}}=\frac{ \pm \sqrt{3}}{\sqrt{3}}= \pm 1$
Since complex number lies in $2^{\text {nd }}$ quadrant, $\mathrm{a}=-\sqrt{3}$ and $\mathrm{b}=1 \Rightarrow \mathrm{z}=-\sqrt{3}+\mathrm{i}$

Topic: Mathematics, Complex Number, JEE Main