datavorous/entrance-exam-dataset huggingface dataset:
The electric field strength in N C-1 that is required to just prevent a water drop carrying a charge 16Ã10-19 C from falling under gravity is
(g=9.8 m s-2 , the mass of water drop =0.0016 g)

Options:
A 9.8×10-16
B 9.8×1016
C 9.8×10-13
D 9.8×1013

The correct answer is:
9.8×1013

Solution:
Charge on drop =16×10-19 C
Mass of the drop =0.0016 g
=16×10-4 g
=16×10-7kg
g=9.8 m s-2
In balance position,
mg=qE
Where, E is electric field strength.
E=mgq=16×10-7×9.816×10-19
=9.8×1013 N C-1.

Topic: Physics, Electrostatics, NEET
The electric potential at a point in free space due to a charge $Q$ coulomb is $Q \times 10^{11} \mathrm{~V}$. The electric field at that point is

Options:
A $4 \pi \varepsilon_0 Q \times 10^{22} \mathrm{~V} / \mathrm{m}$
B $12 \pi \varepsilon_0 \mathrm{Q} \times 10^{20} \mathrm{~V} / \mathrm{m}$
C $4 \pi \varepsilon_0 Q \times 10^{20} \mathrm{~V} / \mathrm{m}$
D $12 \pi \varepsilon_0 Q \times 10^{22} \mathrm{~V} / \mathrm{m}$

The correct answer is:
$4 \pi \varepsilon_0 Q \times 10^{22} \mathrm{~V} / \mathrm{m}$

Solution:
Key Idea : Search for the relations of electric potential and electric field at a particular point. At any point, electric potential due to charge $Q$ is
$$
V=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r}
$$
where $r$ is the distance of observation point from the charge.
At the same point, electric field is
$$
E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2}
$$
Combining Eqs. (i) and (ii), we have
$$
\begin{aligned}
E & =\frac{4 \pi \varepsilon_0 V^2}{Q}=\frac{4 \pi \varepsilon_0 \times\left(Q \times 10^{11}\right)^2}{Q} \\
& =4 \pi \varepsilon_0 \mathrm{Q} \times 10^{22} \mathrm{~V} / \mathrm{m}
\end{aligned}
$$

Topic: Physics, Electrostatics, NEET, NEET 2008 (Screening)
The electric potential at a point $(x, y, z)$ is given by
$$
V=-x^2 y-x z^3+4
$$
The electric field $\vec{E}$ at that point is

Options:
A $\vec{E}=\hat{i}\left(2 x y+z^3\right)+\hat{j} x^2+\hat{k} 3 x z^2$
B $\vec{E}=\hat{i} 2 x y+\hat{j}\left(x^2+y^2\right)+\hat{k}\left(3 x z-y^2\right)$
C $\vec{E}=\hat{i} z^3+\hat{j} x y z+\hat{k} z^2$
D $\vec{E}=\hat{i}\left(2 x y-z^3\right)+\hat{j} x y^2+\hat{k} z^2 x$

The correct answer is:
$\vec{E}=\hat{i}\left(2 x y+z^3\right)+\hat{j} x^2+\hat{k} 3 x z^2$

Solution:
Key Idea Electric field at a point is equal to the negative gradient of the electrostatic potential at that point.
Potential gradient relates with electric field according to the following relation $E=\frac{-d V}{d r}$
$$
\begin{aligned}
\vec{E} & =-\frac{\partial V}{\partial r}=\left[-\frac{\partial V}{\partial x} \hat{i}-\frac{\partial V}{\partial y} \hat{j}-\frac{\partial V}{\partial x} \hat{k}\right] \\
& =\left[\hat{i}\left(2 x y+z^3\right)+\hat{j} x^2+\hat{k} 3 x z^2\right]
\end{aligned}
$$

Topic: Physics, Electrostatics, NEET, NEET 2009 (Screening)
The electric potential at point A is 20 V and B is -20 V.The work done by an external force in moving electron slowly from B to A is

Options:
A -6.4Ã10-18 J
B +6.4Ã10-18Â J
C 4Ã10-20Â J
D -4Ã10-20Â J

The correct answer is:
-6.4Ã10-18 J

Solution:
W=qVf-Vi=-eVA-VB
âW=-1.6Ã10-1920--20
âW=-1.6Ã10-19Ã40
âW=-6.4Ã10-18 J

Topic: Physics, Electrostatics, NEET
The electric potential between a proton and an electron is given by V = V 0 ln r r 0 where r0 is a constant. Assuming Bohrâs model to be applicable, write variation of rn with n, n being the principal quantum number.

Options:
A r n â n
B r n ∝ 1 n
C r n ∝ n 2
D r n ∝ 1 n 2

The correct answer is:
r n ∝ n

Solution:
U = eV = eV0 ln (rr0)
&verbar;F&verbar;=&verbar;- dU dr&verbar;=eV0r
This force provides necessary centripetal force.
Hence, mv2r=eV0r
Or v=eV0m ... (i)
Moreover, mvr=nh2π ... (ii)
Dividing Eq. (ii) by (i), we have
mr=nh2πmeV0 or r n ∝ n

Topic: Physics, Atomic Physics, NEET
The electric potential between a proton and an electron is given by V = V 0 ln r r 0 where r0 is a constant. Assuming Bohr's model to be applicable, write a variation of rn with n, n being the principal quantum number.

Options:
A rn∝n
B rn∝1n
C rn∝n2
D rn∝1n2

The correct answer is:
rn∝n

Solution:
U=eV=eV0 lnrr0
F=-dUdr=eV0r
This force provides the necessary centripetal force.
Hence
mv2r=eV0r
or v=eV0m ...(1)
Moreover,
mvr=nh2π ...(2)
Dividing Eq. (2) by Eq. (1), we have
mr=nh2πmeV0 or r n ∝ n

Topic: Physics, Atomic Physics, NEET
The electric potential V at any point x, y, z in space is given by V=4x2 V. The electric field E (in Vm) at the point 1, 0, 2 is

Options:
A +8 in x direction
B 8 in -x direction
C 16 in +x direction
D 16 in -x direction

The correct answer is:
8 in -x direction

Solution:
E=-dVdx=-ddx4x2, E=-8x

Topic: Physics, Electrostatics, NEET
The electrical conductivity of a semiconductor increases when electromagnetic radiation having a wavelength shorter than 2480 nm is incident on it. The bandgap (in eV) for the semiconductor is

Options:
A 0.9
B 0.7
C 0.5
D 1.1

The correct answer is:
0.5

Solution:
λmin=2480 nm=24800 Ao
Energy (in eV) =12375λ
E⁡=1237524800 eV ⇒E⁡=0.5 eV

Topic: Physics, Semiconductors, NEET
The electrical conductivity of a semiconductor increases when electromagnetic radiation of a wavelength shorter than 2480 nm is incident on it. The bandgap (in eV) for the semiconductor is

Options:
A 0.9
B 0.7
C 0.5
D 0.1

The correct answer is:
0.5

Solution:
Band gap is the energy gap between conduction and valence band
For excitation of e- from valence band to conduction band
Ephtonâ¥Eg
hCÎ»â¥Eg
E g = 1240 eV Î» (nm)
E g = 1240 eV 2480 = 0.5 eV

Topic: Physics, Semiconductors, NEET
The electrical conductivity of semiconductor increases when electromagnetic radiation of wavelength shorter than 24800 â« is incident on it. The band gap for the semiconductor is

Options:
A 0.9eV
B 0.7eV
C 0.5eV
D 1.1eV

The correct answer is:
0.5eV

Solution:
ÎÎ= band gape energy
Î»= maximum wave length
â=hcÎ»=12400Î» eV
âE=1240024800=0.5 eV

Topic: Physics, Semiconductors, NEET
The electrical resistivity of a semiconductor

Options:
A increases with temperature
B decreases with temperature
C increases at low temperature and then decreases
D does not change with temperature

The correct answer is:
decreases with temperature

Solution:
Resistivity of semiconductors decreases with increase in temperature because at higher temperature, the number of current carriers increases.

Topic: Chemistry, Solid State, NEET
The electrode potentials for
$\begin{aligned}
& \mathrm{Cu}^{2+}(a q)+e^{-} \longrightarrow \mathrm{Cu}^{+}(a q) \\
& \text { and } \mathrm{Cu}^{+}(a q)+e^{-} \longrightarrow \mathrm{Cu}(s) \\
& \text { are }+0.15 \mathrm{~V} \text { and }+0.50 \mathrm{~V} \text { respectively. The value } \\
& \text { of } E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ} \text { will be }
\end{aligned}$

Options:
A $0.325 \mathrm{~V}$
B $0.650 \mathrm{~V}$
C $0.150 \mathrm{~V}$
D $0.500 \mathrm{~V}$

The correct answer is:
$0.325 \mathrm{~V}$

Solution:
$$
\begin{aligned}
& \mathrm{Cu}^{2+}+e^{-} \quad \rightarrow \mathrm{Cu}^{+} ; \\
& E_1^{\circ}=0.15 \mathrm{~V} ; \Delta G_1^{\circ}=-n_1 E_1^{\circ} F \\
& \mathrm{Cu}^{+}+e^{-} \rightarrow \mathrm{Cu} ; \\
& E_2^{\circ}=0.50 \mathrm{~V} ; \Delta G_2^{\circ}=-n_2 E_2^{\circ} F \\
& \hline \mathrm{Cu}^{2+}+2 e^{-} \quad \rightarrow \mathrm{Cu} ; E^{\circ}=? ; \Delta G^{\circ}=-n E^{\circ} F \\
& \Delta G^{\circ}=\Delta G_1^{\circ}+\Delta G_2^{\circ} \\
& \text { or }-2 E^{\circ} F=-1 F \times 0.15+(-1 F \times 0.50) \\
& \text { or }-2 E^{\circ} \mathrm{F}=-0.15 \mathrm{~F}-0.50 \mathrm{~F} \\
& \text { or }-2 F E^{\circ}=-F(0.15+0.50) \\
& \therefore \quad E^{\circ}= \frac{0.65}{2}=0.325 \mathrm{~V}
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, NEET, NEET 2011 (Screening)
The electrolyte ratio of solution conductivity of [CoCl2(NH3)4]+Cl–is:

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

The correct answer is:
1:1

Solution:
Correct Option is : 1
1:1

Topic: Chemistry, Coordination Compounds, NEET
The electrolyte ratio of solution conductivity of [CoCl(NH3)5]2+2Cl–is:

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

The correct answer is:
1:2

Solution:
Correct Option is : 2
1:2

Topic: Chemistry, Coordination Compounds, NEET
The electrolytes usually used in the electroplating of gold and silver, respectively, are

Options:
A [Au(OH)4]- and [Ag(OH)2]-
B [Au(CN)2]- and [AgCl2]-
C [Au(NH3)2]+ and [Ag(CN)2]-
D [Au(CN)2]- and [Ag(CN)2]-

The correct answer is:
[Au(CN)2]- and [Ag(CN)2]-

Solution:
Amine complexes of gold are not stable, so, they cannot be used as a solvent.Moreover, in electroplating with silver, silver cyanide is used as electrolyte because the cyanide ion, CN- , reacts with silver ion, Ag+ to form the complex ion Ag(CN)2- . This produces a silver and more adherent silver plating.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
The electrolytic reduction of nitrobenzene in strongly acidic medium produces:

Options:
A p-aminophenol
B Azoxybenzene
C Azobenzene
D Aniline

The correct answer is:
p-aminophenol

Solution:
The electrolytic reduction of nitrobenzene in an acidic medium gives phenyl hydroxyl amine. Since the medium is acidic so phenyl hydroxylamine undergoes rearrangement reaction and gives p-aminophenol.

Thus, under a strongly acidic medium, nitrobenzene on electrolytic reduction gives p-aminophenol.

Topic: Chemistry, Amines, NEET, NEET 2015 (Phase 1)
The electrolytic reduction of nitrobenzene in strongly acidic medium produces

Options:
A Azoxybenzene
B Aniline
C Azobenzene
D p-Aminophenol

The correct answer is:
p-Aminophenol

Solution:
The electrolytic reduction of nitrobenzene in strongly acidic medium produces p-Aminophenol as follows

Topic: Chemistry, Amines, NEET
The electromagnetic radiation which has the smallest wavelength are

Options:
A X-rays
B Gamma rays
C Ultraviolet rays
D Microwaves

The correct answer is:
Gamma rays

Solution:
Among the given options, $\gamma$-rays has the smallest wavelength.

Topic: Physics, Electromagnetic Waves, NEET, NEET 2024 (Re-NEET)
The electromagnetic waves travel in free space with the velocity of

Options:
A Sound
B Light
C Greater than that of light
D Greater than that of sound

The correct answer is:
Light

Solution:
Electromagnetic waves travel in free space or vacuum with the velocity of light 3×108ms-1.

Topic: Physics, Electromagnetic Waves, NEET
The electron concentration in an undefined n-type semiconductor is the same as hole concentration in a undefined p-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them.

Options:
A current in n-type = current in p- type
B current in n- type < current in p- type.
C current in n- type > current in p- type.
D No current will flow in p-type, current will only flow in n-type.

The correct answer is:
current in n- type > current in p- type.

Solution:
Electron's effective mass is smaller than holes therefore mobility of electrons is higher than holes and for equal electric field drift velocity of the electron will be greater compared to holes.

I=neAvdI=neAμE

∵ μe>μhole so, vde>vdhole∴Ie>Ihole

Topic: Physics, Semiconductors, NEET, NEET 2021
The electron density of intrinsic semiconductor at room temperature is 1016 m-3. When doped with a trivalent impurity, the electron density is decreased to 1014 m-3 at the same temperature. The majority carrier density is

Options:
A 1016 m-3
B 1018 m-3
C 1021 m-3
D 1020 m-3

The correct answer is:
1018 m-3

Solution:
ne×nh=ni2
ni=1016 m-3
ne=1014 m-3 given
⇒ nh=ni2ne=101621014=10321014=1018 m -3
Given semiconductor is doped with a trivalent impurity, therefore majority charge carriers are holes.

Topic: Physics, Semiconductors, NEET
The electron energy in hydrogen atom is given by - En= ( –2.18 × 10–18) / n2J. (i)Calculate the energy required to remove an electron completely from the n = 2 orbit. (ii)What is the longest wavelength of light in cm that can be used to cause this transition?

Options:
A (i) 7.58 × 10–19J and (ii)9852Å
B (i) 7.58 × 10–19J and (ii)3647Å
C (i) 5.45 × 10–19J and (ii)6587Å
D (i) 5.45 × 10–19J and (ii)3647Å

The correct answer is:
(i) 5.45 × 10–19J and (ii)3647Å

Solution:
Correct Option is : (D)
(i) 5.45 × 10–19J and (ii)3647Å

Topic: Chemistry, Structure of Atom, NEET
The electron gain enthalpy (in kJ/mol ) of fluorine, chlorine, bromine and iodine, respectively, are

Options:
A – 333, – 349, – 325 and – 296
B – 349, – 333, – 325 and – 296
C – 333, – 325, – 349 and – 296
D – 296, – 325, – 333 and – 349

The correct answer is:
– 333, – 349, – 325 and – 296

Solution:
Normally electron gain enthalpy increases with electro negativity but chlorine has higher electron gain enthalpy than fluorine (exception) due easier addition of electron in Cl as in F addition of electron is tough due to more electronic density based electronic repulsion.
Cl>F>Br>I(Electron affinity order)

Topic: Chemistry, Thermodynamics (C), NEET
The electron in a hydrogen atom makes a transition from n=n1 to n=n2 state. The time period of the electron in the initial state n1is eight times that in the final state n2. The possible values of n1and n2 are

Options:
A n1=8,n2=1
B n1=4,n2=2
C n1=2,n2=4
D n1=1,n2=8

The correct answer is:
n1=4,n2=2

Solution:
In a hydrogen atom, the time period is given by
T∝n3
T1T2=n1n23⇒81=n1n23
⇒n1n2=21
Thus, n1=4 and n2=2

Topic: Physics, Atomic Physics, NEET
The electron in a hydrogen atom makes a transition n1ân2 where n1and n2 are the principal quantum numbers of the two states. Assume the Bohr model to be valid. The time period of the electron in the initial state is 8 times that in the final state. The possible values of n1 and n2 are

Options:
A n1=6,n2=3
B n1=8, n2=2
C n1=n2=1
D n1=8, n2=1

The correct answer is:
n1=6,n2=3

Solution:
As T=2πrv or V=nh2πmr
But r∝n2∴T∝n3
or T1T2=n1n23,As T1=8T2
∴ n1n23=8,n1n2=2
Therefore, in given values n1=6, n2=3

Topic: Physics, Atomic Physics, NEET
The electron was shown experimentally to have wave properties by

Options:
A de Broglie
B Davisson and Germer
C N. Bohr
D Schrodinger

The correct answer is:
Davisson and Germer

Solution:
Wave nature of electron was shown by Davisson and Germer. Davission and germer demonstrated the physical reality of the wave nature of electrons by showing that a beam of electrons could also diffracted by crystals just like light of x - rays.

Topic: Chemistry, Structure of Atom, NEET
The electronegativity difference between $\mathrm{N}$ and $\mathrm{F}$ is greater that that between $\mathrm{N}$ and $\mathrm{H}$ yet the dipole moment of $\mathrm{NH}_3$ (1.5 D) is larger than that of $\mathrm{NF}_3(0.2 \mathrm{D})$. This is because:

Options:
A In \(\mathrm{NH}_3\) as well as in \(\mathrm{NF}_3\) the atomic dipole and bond dipole are in the same direction
B In \(\mathrm{NH}_3\) the atomic dipole and bond dipole are in the same direction whereas in \(\mathrm{NF}_3\) these are in opposite directions
C In \(\mathrm{NH}_3\) as well as \(\mathrm{NF}_3\) the atomic dipole and bond dipole are in opposite directions
D In \(\mathrm{NH}_3\) the atomic dipole and bond dipole are in the opposite directions whereas in \(\mathrm{NF}_3\) these are in the same direction

The correct answer is:
In \(\mathrm{NH}_3\) as well as \(\mathrm{NF}_3\) the atomic dipole and bond dipole are in opposite directions

Solution:
The electronegativity of $\mathrm{N}$ is greater than that of $\mathrm{H}_{\text {in }} \mathrm{NH}_3$, but in $\mathrm{NF}_3$, the electronegativity of $\mathrm{F}$ is greater than that of $\mathrm{N}$. In $\mathrm{NH}_3$, the atomic dipole and bond dipole are in the same direction, whereas in $\mathrm{NF}_3$, these are in opposite directions.
Related Theory
The ammonia molecule has a permanent electric dipole moment equal to 1. $47 \mathrm{D}$.
A Caution
The larger the difference in electronegativity, the larger the dipole moment.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2006
The electronic appliances that are not in use will eventually form part of

Options:
A Radioactive wastes
B Chemical weed
C e-wastes
D Mist

The correct answer is:
e-wastes

Solution:
All the waste collected of the electronic use origin are collectively called as e-waste. Radioactive waste is generated by radiation emitted substances generally observed in nuclear power plant. Chemical weed can be considered chemicals used in controlling growth of weed. Mist is a type of dispersion where droplets of water suspend in air due to warm moist air gets cooled suddenly.

Topic: Biology, Environmental Issues, NEET
The electronic configuration of an element is 1s2, 2s2, 2p6, 3s2 3p3. What is the atomic number of the element which is just below the above element in the periodic table ?

Options:
A 34
B 49
C 33
D 31

The correct answer is:
33

Solution:
Given electronic configuration of element is 1s2, 2s2, 2p6, 3s2 3p3
Z=15; 7N ,15P, 33As

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
The electronic configuration of an element is 1s 2 ,  2s 2 ,  2p 6 ,  3s 2 ,  3p 3 . What is the atomic number of the element, which is just below the above element in the periodic table?

Options:
A 33
B 34
C 31
D 49

The correct answer is:
33

Solution:
The electronic configuration of valence shell of element below the given element is 4s24p3 . But before the filling of 4p-orbitals start, the electrons enters the 3d-orbitals, which can accommodate 10 electrons. Hence, the electronic configuration of the element, which is just below the given element must be 1s2,2s22p6,3s23p6,4s23d10,4p3 . So its atomic number would be 33.

Topic: Chemistry, Structure of Atom, NEET
The electronic configuration of nitrogen is 1s2,2s22px12py12pz1 and not 1s2,2s22px22py12pz0 . This is explained by

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

The correct answer is:
Hund's rule

Solution:
Hund’s rule states that pairing of electrons in the orbitals of a subshell (orbitals of equal energy) starts when each of them is singly filled.

Topic: Chemistry, Structure of Atom, NEET
The electronic configuration of the valence shell of Cu is 3d¹⁰4s¹ and not 3d⁹4s². How is this configuration explained?

Options:
A The extra stability gained by having a half-filled d-subshell (4d¹⁰) and a filled s-subshell (4s¹) outweighs the energy required to promote an electron from 4s to 3d. This leads to the observed configuration.
B The extra stability gained by having a half-filled d-subshell (3d¹⁰) and a filled s-subshell (4s¹) outweighs the energy required to promote an electron from 4s to 4d. This leads to the observed configuration.
C The extra stability gained by having a half-filled d-subshell (3d9) and a filled s-subshell (4s¹) outweighs the energy required to promote an electron from 4s to 3d. This leads to the observed configuration.
D The extra stability gained by having a half-filled d-subshell (3d¹⁰) and a filled s-subshell (4s¹) outweighs the energy required to promote an electron from 4s to 3d. This leads to the observed configuration.

The correct answer is:
The extra stability gained by having a half-filled d-subshell (3d¹⁰) and a filled s-subshell (4s¹) outweighs the energy required to promote an electron from 4s to 3d. This leads to the observed configuration.

Solution:
Correct Option is : (D)
The extra stability gained by having a half-filled d-subshell (3d¹⁰) and a filled s-subshell (4s¹) outweighs the energy required to promote an electron from 4s to 3d. This leads to the observed configuration.

Topic: Chemistry, Structure of Atom, NEET
The electronic configuration with the highest ionization enthalpy is:

Options:
A Ar 3d10 4s2 4p3
B Ne3s2Â 3p1
C Ne 3s2 3p3
D NeÂ 3s2Â 3p2

The correct answer is:
Ne 3s2 3p3

Solution:
(i) Ionisation energy decrease with increasing atomic radius.
(ii) Half filled and completely filled electronic configuration are extra stable.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
The electronic configurations of bivalent europium and trivalent cerium are respectively
(atomic number: Xe=54,Ce=58,Eu=63 )

Options:
A [Xe]4f76s2 and [Xe]4f26s2
B [Xe]4f7 and [Xe]4f1
C [Xe]4f4 and [Xe]4f9
D [Xe]4f2 and [Xe]4f7

The correct answer is:
[Xe]4f7 and [Xe]4f1

Solution:
In ground state, europium and cerium has following electronic configuration
63Eu⇒[Xe]4f76s2
58Ce⇒[Xe]4f15d16s2
Bivalent means Eu lost two electrons and trivalent means Ce lost three electrons so.
Eu2+→[Xe]4f7
Ce3+:[Xe]4f1

Topic: Chemistry, d and f Block Elements, NEET
The electronic configurations of Eu (Atomic no. 63), Gd (Atomic no. 64) and Tb (Atomic no. 65) are:

Options:
A Xe4f76s2Â ,Â Xe4f86s2Â andÂ Xe4f85d16s2
B Xe4f65d16s2Â ,Â Xe4f75d16s2Â andÂ Xe4f96s2
C Xe4f65d16s2Â ,Â Xe4f75d16s2Â andÂ Xe4f85d16s2
D Xe4f76s2 , Xe4f75d16s2 and Xe4f96s2

The correct answer is:
Xe4f76s2 , Xe4f75d16s2 and Xe4f96s2

Solution:
The electronic configurations of Eu (Atomic no. 63), Gd (Atomic no. 64) and Tb (Atomic no. 65) are:
Eu-Xe4f7, 6s2
Gd-Xe4f7,5d1,6s2
Tb-Xe4f9,6s2

Topic: Chemistry, d and f Block Elements, NEET, NEET 2016 (Phase 1)
The electrons, that are removed from photosystem-II and photosystem-I during z-scheme, must be replaced. This is achieved by electrons released

Options:
A By photosystem-I and NADP+
B By splitting of water and NADPH
C By ATP and photosystem-II
D By splitting of water and photosystem-II

The correct answer is:
By splitting of water and photosystem-II

Solution:
Photosystem II is associated with oxygen-evolving complex - that has 4 Mg ions, one calcium ion and one chlorine ion - that splits water molecules to 4H+ and O2. Hence photosystem II obtain electron by splitting of water.
Photosystem I gets an electron from photosystem II. This Z-scheme is also called cyclic photophosphorylation.

Topic: Biology, Photosynthesis in Higher Plants, NEET
The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is

Options:
A proportional to the square to the root of the distance between the plates.
B linearly proportional to be distance between the plates.
C independent of the distance between the plates.
D inversely proportional to the distance between the plates.

The correct answer is:
independent of the distance between the plates.

Solution:
For isolated capacitor Q= constant.

Now electric field because of one plate of the capacitor is E=σ2∈0=Q2A∈0.
∴ Force on 1 metal plate is
F=QE
F=Qσ2∈0=Q22A∈0
∴ The electrostatic force is independent of the distance between plates.

Topic: Physics, Capacitance, NEET, NEET 2018
The electrostatic potential due to an electric dipole at an equatorial point is

Options:
A maximum
B zero
C minimum
D not defined

The correct answer is:
zero

Solution:
The electrical potential of a dipole is given by,

V=14πε0p.r^r2

At equatorial plane θ=π/2, therefore, electric potential at any point in the equatorial plane of a dipole is zero.

Topic: Physics, Electrostatics, NEET
The element expected to form largest ion to achieve the nearest noble gas configuration is :

Options:
A Na
B O
C F
D N

The correct answer is:
N

Solution:
When we form the ions with the nearest noble gas electronic configuration using given elements, all the ions formed are isoelectronic species(species with same number of electrons).

Among the isoelectronic species, species with more positive charge have small size and species with more negative charge are large size ions.

The ionic size order of ions is

N3->O2->F->Na+

Topic: Chemistry, Structure of Atom, NEET, NEET 2023 (All India)
The element having electronic configuration [Kr]4d10,4f14,5s25p65d2,6s2 belongs to

Options:
A s-block
B p-block
C d-block
D f-block

The correct answer is:
d-block

Solution:
This configuration has incompletely filled d-orbitals, therefore, it is the configuration of a d-block element.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
The element used for welding metals with high melting points is :

Options:
A $\mathrm{Cl}_2$
B $\mathrm{H}_2$
C $\mathrm{Ne}$
D $\mathrm{He}$

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

Solution:
Oxy-hydrogen flame produces a very high temperature which is used for welding and melting of metals with high melting point.

Topic: Chemistry, Hydrogen, NEET, NEET 2022 (Phase 2)