datavorous/entrance-exam-dataset huggingface dataset:
The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The molar solubility of oxygen in water is ______ ×10-5 mol dm-3.

(Round off to the Nearest Integer).

[Given : Henry's law constant =KH=8.0×104 kPa for O2. Density of water with dissolved oxygen=1.0 kg dm-3]

Options:

The correct answer is:
1389

Solution:
P=KH·x

or, 20×103=8×104×103×nO2nO2+nwater

or, 14000=nO2nO2+nwater =nO2nwater

Means 1 mole water =18 gm=18 ml dissolves

14000 moles O2. Hence, molar solubility

=1400018×1000=172 mol dm-3

=1388.89×10-5 mol dm-3≈1389 ×10-5 mol dm-3

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The ozone hole over Antarctica is mainly due to

Options:
A CH4
B CFC
C SO2
D Hydrogen

The correct answer is:
CFC

Solution:
Chlorofluorocarbons (CFCs) are nontoxic, nonflammable chemicals containing atoms of carbon, chlorine, and fluorine. They are used in the manufacture of aerosol sprays, blowing agents for foams and packing materials, as solvents, and as refrigerants. Chlorofluorocarbons (CFCs), hydrochlorofluorocarbons (HCFCs), and halogens destroy the earth's protective ozone layer, which shields the earth from harmful ultraviolet (UV-B) rays generated from the sun.

Topic: Biology, Environmental Issues, NEET
The p. d. f. of a random variable x is given by fx=14a,0<x<4a,a>0 =0, otherwise and Px<3a2=kPx>5a2 then K = …….

Options:
A 1
B 12
C 18
D 32

The correct answer is:
1

Solution:
We have p. d. f. of a random variable x is given by
fx=14a,0<x<4a.a>0
and also, px<3a2=kpx>5a2
⇒â«03a2fxdx=kâ«5a24afxdx
⇒â«03a214adx=kâ«5a24a14adx
⇒14aâ«03a2dx=k4aâ«5a24adx
⇒x03a2=kx5a24a
⇒3a2-0=k4a-5a2
⇒3a2=k3a2
⇒k=1

Topic: Mathematics, Probability, JEE Main
The pH at the equivalent point for the titration of 0.10 M KH2BO3 with 0.1 M HCl is
K a âofâ H 3 BO 3 =12 .8Ã10 â10
Report your answer by rounding it up to nearest whole number.

Options:

The correct answer is:
5

Solution:
First there will be an acid-base reaction between KH2BO3 and HCl .
conc. will become half after reaction of an equal volume of KH2BO3 and HCl
So 0.15=0.05
Now,
H+=KaC=12.8×10-10×0.05=8×10-6
pH=-log⁡8×10-6≈5.0

Topic: Chemistry, Ionic Equilibrium, JEE Main
The $p H$ of $0.1 \mathrm{M}$ solution of the following salts increases in the order

Options:
A $\mathrm{NaCl} \lt \mathrm{NH}_4 \mathrm{Cl} \lt \mathrm{NaCN} \lt \mathrm{HCl}$
B $\mathrm{HCl} \lt \mathrm{NH}_4 \mathrm{Cl} \lt \mathrm{NaCl} \lt \mathrm{NaCN}$
C $\mathrm{NaCN} \lt \mathrm{NH}_4 \mathrm{Cl} \lt \mathrm{NaCl} \lt \mathrm{HCl}$
D $\mathrm{HCl} \lt \mathrm{NaCl} \lt \mathrm{NaCN} \lt \mathrm{NH}_4 \mathrm{Cl}$

The correct answer is:
$\mathrm{HCl} \lt \mathrm{NaCl} \lt \mathrm{NaCN} \lt \mathrm{NH}_4 \mathrm{Cl}$

Solution:
$\mathrm{NH}_4 \mathrm{Cl}$ undergoes cationic hydrolysis hence $\mathrm{pH}$ is $\gt7$ because the solution due to cationic hydrolysis
$\mathrm{NaCN}$ undergoes anionic hydrolysis hence $\mathrm{pH}$ is $\gt7$.
$\mathrm{HCl}$ is strong acid and $\mathrm{NaCl}$ is neutral solution.
Hence the $p H$ of given solutions will increases.
$\mathrm{HCl} \lt \mathrm{NaCl} \lt \mathrm{NaCN} \lt \mathrm{NH}_4 \mathrm{Cl}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The pH of a 0.02 M NH4Cl solution will be [Given: KbNH4OH=10-5 and log⁡2=0.301]

Options:
A 4.65
B 2.65
C 4.35
D 5.35

The correct answer is:
5.35

Solution:
NH4Cl is a salt of a strong acid and a weak base.
pH=12pKw-pKb-logC
=1214-5-log⁡2×10-2
=129-log⁡2-log⁡10-2
=129-log⁡2+2
=10.72=5.35

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The pKa of acetic acid and pKb of NH4OH are 4.75 and 4.75 respectively. Find the pH of ammonium acetate solution?

Options:

The correct answer is:
7

Solution:
Ammonium acetate is a salt of weak acid and weak base.
So pH is given as
pH=12pKw+pKa-pKb
=1214+4.75-4.75
=7

Topic: Chemistry, Ionic Equilibrium, JEE Main
The pπ-pπ bond is present in

Options:
A XeO3
B SO42–
C SO2
D All of these

The correct answer is:
SO2

Solution:
SO2 has pÏ-pÏ and pÏ-dÏ bond while XeO3,SO42â have only pÏ-dÏ bond

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The $p-V$ diagram of a Carnot's engine is shown in the graph below. The engine uses 1 mole of an ideal gas as working substance. From the graph, the area
enclosed by the $p$ - $V$ diagram is [The heat supplied to the gas is $8000 \mathrm{~J}$ ]

Options:
A $1200 \mathrm{~J}$
B $2000 \mathrm{~J}$
C $3000 \mathrm{~J}$
D $1000 \mathrm{~J}$

The correct answer is:
$1200 \mathrm{~J}$

Solution:
[Hint Area enclosed by $p-V$ diagram is equal to
work done] Total work done by gas in one carnot cycle is given by.
$W=\mu R T_1 \ln \frac{V_2}{V_1}-\mu R T_2 \ln \frac{V_3}{V_4}$

Topic: Physics, Thermodynamics, JEE Main
The P-V diagram of a diatomic ideal gas system going under cyclic process as shown in figure. The work done during an adiabatic process CD is (use γ=1.4) :

Options:
A –500 J
B –400 J
C 400 J
D 200 J

The correct answer is:
–500 J

Solution:
Adiabatic process is from C to D
WD=P2 V2-P1 V11-γ
=PDVD-PCVC1-γ
=200(3)-(100)(4)1-1.4
=-500 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The pair from the following pairs having both compounds with net non-zero dipole moment is

Options:
A 1, 4-Dichlorobenzene, 1, 3-Dichlorobenzene
B cis-butene, trans-butene
C CH2Cl2, CHCl3
D Benzene, anisidine

The correct answer is:
CH2Cl2, CHCl3

Solution:
The dipole moment is the vector sum of all the bond moments. The CH2Cl2 and CHCl3 both have net non-zero dipole moment because the bond moments are not cancelled each other due to their tetrahedral structure. The 1, 4-dichlorobenzene, trans-butene, benzene have zero dipole moment due to cancellation of bond moments. The 1, 3-dichlorobenzene, cis-butene, anisidine has non-zero dipole moment.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (10 Apr Shift 1)
The pair having the same magnetic moment is:
At. No. :Cr=24, Mn=25, Fe=26, Co=27

Options:
A MnÂ H2O62+Â andÂ CrH2O62+
B CoCl42-Â andÂ FeH2O62+
C CrH2O62+Â andÂ CoCl42-
D CrH2O62+ and FeH2O62+

The correct answer is:
CrH2O62+ and FeH2O62+

Solution:
The compounds having the same number of unpaired electrons will have same magnetic moment.

Magnetic moment μ = nn+2

where n is the number of unpaired electrons

[ Cr ( H 2 O ) 6 ] +2
C r 2+ .(3 d 4 )  with weak field ligand so unpaired electrons are 4

So, μ= 44+2 = 24
F e +2 ( 3 d 6 ) Weak field ligand so unpaired electrons are 4

μ = 44+2 = 24

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2016 (03 Apr)
The pair having the same magnetic moment is

[At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27]

Options:
A CoCl42- and FeH2O6]2+
B CrH2O62+ and CoCl42-
C CrH2O62+ and FeH2O62+
D MnH2O62+ and CrH2O62+

The correct answer is:
CrH2O62+ and FeH2O62+

Solution:
In option A: CoCl42-,Co2+3d7 with W.F.L.,
and [Fe(H2O)6]2+,Fe2+(3d6) with W.F.L.,
In option B: [Cr(H2O)6]2+,Cr2+(3d4) with W.F.L.,
and [CoCl4]2-,Co2+(3d7) with W.F.L.,
In option C: [Cr(H2O)6]2+,Cr2+(3d4) with W.F.L.,
and [Fe(H2O)6]2+,Fe2+(3d6) with W.F.L.,
Here both complexes have same unpaired electrons i.e. = 4
In option D: [Mn(H2O)6]2+,Mn2+(3d5) with W.F.L.,
and [Cr(H2O)6]2+,Cr2+(3d4) with W.F.L.,

Topic: Chemistry, Coordination Compounds, JEE Main
The pair in which both the species have the same magnetic moment (spin only) is:

Options:
A CrH2O62+ and FeH2O62+
B CoOH42– and FeNH362+
C MnH2O62+ and CrH2O62+
D CrH2O62+ and CoCl42–

The correct answer is:
CrH2O62+ and FeH2O62+

Solution:
No. unpaired e-

CrH2O62+⇒Cr2+⇒3d4⇒t2g1,1,1, eg1,0	4

FeH2O62+⇒Fe2+⇒3d6⇒t2g1,1,1, eg1,1	4

FeNH362+⇒Fe2+⇒3d6⇒t2g2,1,1, eg1,1	4

CoOH42-⇒Co2+⇒3d7⇒ eg2,2, t21,1,1	3

CoCI42-⇒Co2+⇒3d7⇒ e2,2, t21,1,1	3

MnH2O62+⇒MN2+⇒3d5⇒ t2g1,1,1, eg1,1	5

So CrH2O62+ and FeH2O62+	have same magnetic moment (spin only)

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (04 Sep Shift 1)
The pair, in which ions are isoelectronic with Al3+ is

Options:
A Br- and Be2+
B Cl- and Li+
C S2- and K+
D O2- and Mg2+

The correct answer is:
O2- and Mg2+

Solution:
Isoelectronic species are known as atoms or ions that have the same number of electrons.

Species	Al3+	Mg2+	Be2+	O2-	Br-	Cl-	Li+	S2-	K+

No. of e-	10	10	2	10	36	18	2	18	18

So Mg2+ & O2- are isoelectronic with Al3+

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2022 (25 Jun Shift 1)
The pair in which phosphorous atom have a formal oxidation state of +3 is

Options:
A Pyrophosphorous and pyrophosphoric acids
B Orthophosphorous and pyrophosphorous acids
C Pyrophosphorous and hypophosphoric acids
D Orthophosphorous and hypophosphoric acids

The correct answer is:
Orthophosphorous and pyrophosphorous acids

Solution:
Acid Formula Formal oxidation state of phosphorous
Pyrophosphorous acids H4P2O5 +3
Pyrophosphoric acid H4P2O7 +5
Orthophosphorous acid H3PO3 +3
Hypophosphoric acid H4P2O6 +4
Both pyrophosphorous and orhtophosphorous acids have +3 formal oxidation state.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The pair in which phosphorous atoms have a formal oxidation state of + 3 is:

Options:
A Orthophosphorous and hypophospheric acids
B Pyrophosphorous and pyrophosphoric acids
C Orthophosphorous and pyrophosphorous acids
D Pyrophosphorous and hypophosphoric acids

The correct answer is:
Orthophosphorous and pyrophosphorous acids

Solution:
Acid	Formula	Formal oxidation state of phosphorous
Pyrophosphorous acid	H4P2O5	+3
Pyrophosphoric acid	H4P2O7	+5
Orthophosphorous	H3PO3	+3
Hypophosphoric acid	H4P2O6	+4

Both pyrophosphorous and orthophosphorous acid have +3 formal oxidation state

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2016 (03 Apr)
The pair of compounds having metals in their highest oxidation state is

Options:
A Mn2O7 and CrO2Cl2.
B FeCN63- and CuCN42-.
C NiCl42- and CoCl42-.
D FeCl4- and Co2O3.

The correct answer is:
Mn2O7 and CrO2Cl2.

Solution:
Mn2O7
2x+7(−2)=0   x=7
CrO2Cl2
x+2(−2)+2(−1)    x=6

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2017 (08 Apr Online)
The pair of compounds which have different hybridisation but same shape

Options:
A SO3,ClF3
B BF3,PCl3
C XeF2,CO2
D XeF4,SF4

The correct answer is:
XeF2,CO2

Solution:
XeF2→sp3d , Linear
CO2→sp , Linear

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The pair of lanthanides in which both elements have high third-ionization energy is:

Options:
A Dy,Gd
B Lu,Yb
C Eu,Yb
D Eu,Gd

The correct answer is:
Eu,Yb

Solution:
Europium has an atomic number of 63 and a stable electronic configuration of [Xe]4f76s2.

Ytterbium, on the other hand, has an atomic number of 70 and a stable electronic configuration of [Xe]4f146s2.

Both have high 3rd ionisation energy due to half filled & fully filled configuration respectively.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2023 (13 Apr Shift 1)
The pair of metal ions that can give a spin only magnetic moment of 3.9 BM for the complex [M(H2O)6]Cl2 , is:

Options:
A Co2+ and Fe2+
B V2+ and Fe2+
C Cr2+ and Mn2+
D V2+ and Co2+

The correct answer is:
V2+ and Co2+

Solution:
If the complex is MH2O6Cl2, M will have +2 oxidation state and hence, it has already lost 2 electrons from s- orbital. Since it has μ=3.9 = n( n+2 ) ⇒n=3
3 unpaired so, the configuration can either be:

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (12 Jan Shift 1)
The pair of oxides, which can be leached out, when powdered bauxite containing certain impurities is digested with concentrated solution of $\mathrm{NaOH}$ at $473-523 \mathrm{~K}$ temperature and $35-36$ bar pressure are

Options:
A $\mathrm{TiO}_2, \mathrm{SiO}_2$
B $\mathrm{SiO}_2, \mathrm{Al}_2 \mathrm{O}_3$
C $\mathrm{SiO}_2, \mathrm{Fe}_2 \mathrm{O}_3$
D $\mathrm{Al}_2 \mathrm{O}_3, \mathrm{Fe}_2 \mathrm{O}_3$

The correct answer is:
$\mathrm{SiO}_2, \mathrm{Al}_2 \mathrm{O}_3$

Solution:
Bauxite is the most important ore of aluminium contains only $30-60 \% \mathrm{Al}_2 \mathrm{O}_3$, the rest being a mixture of $\mathrm{SiO}_2, \mathrm{Fe}_2 \mathrm{O}_3$ and $\mathrm{TiO}_2$. The powdered bauxite is treated with concentrated solution of $\mathrm{NaOH}$ resulting in the dissolution of $\mathrm{Al}_2 \mathrm{O}_3$ and $\mathrm{SiO}_2$ while the other oxides $\left(\mathrm{Fe}_2 \mathrm{O}_3\right.$ and $\mathrm{TiO}_2$ ) being basic remain insoluble in $\mathrm{NaOH}$. Hence, only $\mathrm{SiO}_2$ and $\mathrm{Al}_2 \mathrm{O}_3$ leached out.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, AP EAMCET, AP EAMCET 2018 (24 Apr Shift 1)
The pair of quantities having same dimensions is:

Options:
A Impulse and Surface Tension
B Angular momentum and Work
C Work and Torque
D Young's modulus and Energy

The correct answer is:
Work and Torque

Solution:
(1)
$$
\begin{aligned}
\text { Torque }=\text { F.S. } & =\left[\mathrm{MLT}^{-2}\right][\mathrm{L}] \\
& =\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]
\end{aligned}
$$
correct option is (3)
(2) Angular moment $=$ M.I. $\times$ angular velocity
$$
\begin{aligned}
=\left[\mathrm{ML}^2\right]\left[\mathrm{T}^{-1}\right] & =\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right] \\
\text { Work }=F . S . & =\left[\mathrm{MLT}^{-2}\right][\mathrm{L}] \\
& =\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]
\end{aligned}
$$
(3) Work $=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$ from
Related Theory
(1)
$$
\begin{aligned}
& \text { Impluse }=\text { force } \times \text { time } \\
&=\left[M L T^{-2}\right][T]=\left[M L T^{-1}\right] \\
& \text { Surface tension }=\frac{\text { Force }}{\text { length }}= \\
& \frac{\left[M L T^{-2}\right]}{[L]}=\left[M T^{-1}\right]
\end{aligned}
$$

Topic: Physics, Units and Dimensions, JEE Main
The pair of species having identical shapes for molecules of both species is

Options:
A
$\mathrm{XeF}_2, \mathrm{CO}_2$

B
$\mathrm{BF}_3, \mathrm{PCl}_3$

C
$\mathrm{PF}_5, \mathrm{IF}_5$

D
$\mathrm{CF}_4, \mathrm{SF}_4$

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

Solution:
Both $\mathrm{XeF}_2$ and and $\mathrm{CO}_2$ have a linear structure.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2003
The pair of xenon compounds, which have a same number of lone pairs of electrons on the central atom is

Options:
A XeO3,XeF6
B XeF2, XeF4
C XeF4, XeO3
D XeF4, XeOF4

The correct answer is:
XeO3,XeF6

Solution:
1 XeO3: Xenon contains 8 electrons in its valence shell, 6 electrons are used in bonding with three oxygen atoms and 2 electrons remain unbonded, i.e., 1 lone pair of electrons.

2 XeF6: Xenon contains 8 electrons in its valence shell, 6 electrons are used in bonding with six fluorine atoms and 2 electrons remain unbonded, i.e., 1 lone pair of electrons.

3 XeF2: Xenon contains 8 electrons in its valence shell, 2 electrons are used in bonding with two fluorine atoms and 6 electrons remain unbonded, i.e., 3 lone pairs of electrons.

4 XeF4: Xenon contains 8 electrons in its valence shell, 4 electrons are used in bonding with four fluorine atoms and 4 electrons remain unbonded, i.e., 2 lone pairs of electrons.

5 XeOF4: Xenon contains 8 electrons in its valence shell, 4 electrons are used in bonding with four fluorine atoms and 2 electrons are used in bonding with one oxygen atom and 2 electrons remain unbonded, i.e., 1 lone pair of electrons.

Hence, the pair of xenon compounds, which have a same number of lone pairs of electrons on the central atom is XeO3 and XeF6.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), AP EAMCET, AP EAMCET 2018 (25 Apr Shift 1)
The pair that contains two P-H bonds in each of the oxoacids is:

Options:
A H3PO3 and H3PO2
B H4P2O5 and H3PO3
C H4P2O5 and H4P2O6
D H3PO2 and H4P2O5

The correct answer is:
H3PO2 and H4P2O5

Solution:
First form structures of all the oxy acids of phophorous.

Then identify the bond of P-H:

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2019 (10 Jan Shift 2)
The pair that does not require calcination is

Options:
A ZnO and Fe2O3.xH2O
B ZnO and MgO
C ZnCO3 and CaO
D Fe2O3 and CaCO3.MgCO3

The correct answer is:
ZnO and MgO

Solution:
Calcination is required for hydroxide carbonate and hydrated ores. So, calcination for ZnO and MgO will not occur.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The pair that does not require calcination is

Options:
A ZnO and MgO
B Fe2O3 and CaCO3.MgCO3
C ZnO and Fe2O3.xH2O
D ZnCO3 and CaO

The correct answer is:
ZnO and MgO

Solution:
Calcination is required for carbonate and hydrated oxide ores. ZnO and MgO both are oxides, so they donot require calcination.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
The pair that has similar atomic radii is:

Options:
A Mo and W
B Sc and Ni
C Mn and Re
D Ti and Hf

The correct answer is:
Mo and W

Solution:
Among the given pairs Mo and W have the same atomic radius of 139 pm and this is due to Lanthanoid contraction.
This happens due to the poor screening ability of 4f electrons. The radius of other pairs are:
Sc (162 pm) and Ni (124 pm)
Mo (139 pm) and W (139 pm)
Mn 140 pmand Re 135 pm
Ti170 pm and Hf 159 pm.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2019 (12 Apr Shift 2)
The pairs having same dimensional formula

Options:
A Angular momentum, torque
B Torque, work
C Plank's constant, Boltzmann's constant
D Gas constant, pressure

The correct answer is:
Torque, work

Solution:
We know,

Work done = Force×Displacement

Torque = force×perpendicular distance between force and axis of rotation

Dimension of Work= [ML2T-2]

Dimension of Torque =ML2T-2

Here it is clear that, dimension of work and dimension of torque are equal.

Topic: Physics, Units and Dimensions, JEE Main
The pairs of species of oxygen and their magnetic behaviours are noted below. Which of the following presents the correct description?

Options:
A $\mathrm{O}_2^{-}, \mathrm{O}_2^{2-}$ - Both diamagnetic
B $\mathrm{O}^{+}, \mathrm{O}_2^{2-}$ - Both paramagnetic
C $\mathrm{O}_2^{+}, \mathrm{O}_2$ - Both paramagnetic
D $\mathrm{O}, \mathrm{O}_2^{2-}$ - Both paramagnetic

The correct answer is:
$\mathrm{O}_2^{+}, \mathrm{O}_2$ - Both paramagnetic

Solution:
The molecular orbital configurations of $\mathrm{O}_2^{+}, \mathrm{O}_2^{-}, \mathrm{O}_2^{2-}$ and $\mathrm{O}_2$ are
$$
\begin{aligned}
& \mathrm{O}_2^{+}=\sigma 1 s^2, \sigma 1 s^2, \sigma 2 s^2, \sigma 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \\
& \pi 2 p_x^1 \approx \pi 2 p_y^0 \\
& \mathrm{O}_2^{-}=\sigma 1 s^2, \sigma \cdot \% 1 s^2, \sigma 2 s^2, \sigma 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \\
& \approx \pi 2 p_y^2, \pi 2 p_x^2 \approx \pi 2 p_y^1 \\
& \mathrm{O}_2^{2-}=\sigma 1 s^2, \sigma \cdot s^2, \sigma 2 s^2, \sigma 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \\
& \approx \pi 2 p_y^2, \pi 2 p_x^2 \approx \pi 2 p_y^2 . \\
& \mathrm{O}_2=\sigma 1 s^2, \sigma 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \\
& \approx \pi 2 p_y^2, \pi 2 p_x^1 \approx \pi 2 p_y^1
\end{aligned}
$$
And the electronic configuration of $\mathrm{O}$ and $\mathrm{O}^{+}$ are
$$
\begin{array}{r}
\mathrm{O}=1 s^2, 2 s^2, 2 p_x^2, 2 p_y^1, 2 p_z^1 \\
\mathrm{O}^{+}=1 s^2, 2 s^2, 2 p_x^1, 2 p_y^1, 2 p_z^1
\end{array}
$$
As $\mathrm{O}_2^{+}, \mathrm{O}_2, \mathrm{O}_2^{-}, \mathrm{O}$ and $\mathrm{O}^{+}$have unpaired electrons, hence are paramagnetic.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2011 (Mains)
The pairs of species of oxygen and their magnetic behaviours are noted below. Which of the following presents the correct description?

Options:
A $\mathrm{O}_2^{-}, \mathrm{O}_2^{2-}$ - Both diamagnetic
B $\mathrm{O}^{+}, \mathrm{O}_2^{2-}$ - Both paramagnetic
C $\mathrm{O}_2^{+}, \mathrm{O}_2$ - Both paramagnetic
D $\mathrm{O}, \mathrm{O}_2^{2-}$ - Both paramagnetic

The correct answer is:
$\mathrm{O}_2^{+}, \mathrm{O}_2$ - Both paramagnetic

Solution:
The molecular orbital configurations of $\mathrm{O}_2^{+}, \mathrm{O}_2^{-}, \mathrm{O}_2^{2-}$ and $\mathrm{O}_2$ are
$$
\begin{aligned}
& \mathrm{O}_2^{+}=\sigma 1 s^2, \sigma 1 s^2, \sigma 2 s^2, \sigma 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \\
& \pi 2 p_x^1 \approx \pi 2 p_y^0 \\
& \mathrm{O}_2^{-}=\sigma 1 s^2, \sigma \cdot \% 1 s^2, \sigma 2 s^2, \sigma 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \\
& \approx \pi 2 p_y^2, \pi 2 p_x^2 \approx \pi 2 p_y^1 \\
& \mathrm{O}_2^{2-}=\sigma 1 s^2, \sigma \cdot s^2, \sigma 2 s^2, \sigma 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \\
& \approx \pi 2 p_y^2, \pi 2 p_x^2 \approx \pi 2 p_y^2 . \\
& \mathrm{O}_2=\sigma 1 s^2, \sigma 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \\
& \approx \pi 2 p_y^2, \pi 2 p_x^1 \approx \pi 2 p_y^1
\end{aligned}
$$
And the electronic configuration of $\mathrm{O}$ and $\mathrm{O}^{+}$ are
$$
\begin{array}{r}
\mathrm{O}=1 s^2, 2 s^2, 2 p_x^2, 2 p_y^1, 2 p_z^1 \\
\mathrm{O}^{+}=1 s^2, 2 s^2, 2 p_x^1, 2 p_y^1, 2 p_z^1
\end{array}
$$
As $\mathrm{O}_2^{+}, \mathrm{O}_2, \mathrm{O}_2^{-}, \mathrm{O}$ and $\mathrm{O}^{+}$have unpaired electrons, hence are paramagnetic.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The parabola having its focus at $(3,2)$ and directrix along the y-axis has its vertex at

Options:
A $(2,2)$
B $\left(\frac{3}{2}, 2\right)$
C $\left(\frac{1}{2}, 2\right)$
D $\left(\frac{2}{3}, 2\right)$

The correct answer is:
$\left(\frac{3}{2}, 2\right)$

Solution:
Vertex of the parabola is a point which lies on the axis of the parabola, which is a line $\perp$ to the directrix through the focus, i.e., $\mathrm{y}=$ 2 and equidistant from the focus and directrix $x=0$, so that the vertex is $\left(\frac{3}{2}, 2\right)$

Topic: Mathematics, Parabola, JEE Main
The parabola with directrix $x+2 y-1=0$ and focus $(1,0)$ is

Options:
A $4 x^2-4 x y+y^2-8 x+4 y+4=0$
B $4 x^2+4 x y+y^2-8 x+4 y+4=0$
C $4 x^2+5 x y+y^2+8 x-4 y+4=0$
D $4 x^2-4 x y+y^2-8 x-4 y+4=0$

The correct answer is:
$4 x^2-4 x y+y^2-8 x+4 y+4=0$

Solution:
Let $P(x, y)$ be any point on the parabola
By definition of Parabola $P M=P S$
$\Rightarrow \quad \frac{x+2 y-1}{\sqrt{1+4}}=\sqrt{(x-1)^2+y^2}$
On squaring both sides, we get
$\begin{gathered}
x^2+4 y^2+1+4 x y-4 y-2 x \\
=5\left(x^2+1-2 x+y^2\right) \\
\Rightarrow \quad 4 x^2+y^2-8 x+4 y-4 x y+4=0
\end{gathered}$

Topic: Mathematics, Parabola, JEE Main
The parabola with directrix \(x+2 y-1=0\) and focus \((1,0)\) is

Options:
A \(4 x^2-4 x y+y^2-8 x+4 y+4=0\)
B \(4 x^2+4 x y+y^2-8 x+4 y+4=0\)
C \(4 x^2+4 x y+y^2+8 x-4 y+4=0\)
D \(4 x^2-4 x y+y^2-8 x-4 y+4=0\)

The correct answer is:
\(4 x^2-4 x y+y^2-8 x+4 y+4=0\)

Solution:
Equation of parabola having directrix \(x+2 y-1=0\) and focus \((1,0)\) is
\(\begin{aligned}
& \frac{|x+2 y-1|}{\sqrt{1+4}}=\sqrt{(x-1)^2+(y-0)^2} \\
\Rightarrow \quad & (x+2 y-1)^2=5\left[(x-1)^2+y^2\right] \\
\Rightarrow \quad & x^2+4 y^2+1+4 x y-2 x-4 y=5 \\
& \quad\left(x^2+y^2-2 x+1\right) \\
\Rightarrow \quad & 4 x^2-4 x y+y^2-8 x+4 y+4=0
\end{aligned}\)
Hence, option (a) is correct.

Topic: Mathematics, Parabola, JEE Main