datavorous/entrance-exam-dataset huggingface dataset:
The interactions of ligands with protein,

Options:
A usually result in the activation of the proteins
B are relatively rare in biological systems
C are usually irreversible
D are usually transient

The correct answer is:
are usually transient

Solution:
The ligand-protein binding refers to the induced fit theory. The structural adaptations occur when a ligand binds to a protein involving a conformational change in the protein. This change alters the binding site to be more complementary to the ligand.

Topic: Chemistry, Biomolecules, NEET
The interatomic distances in H2 and Cl2 molecules are 74 and 198 pm respectively. the bond length of HCL

Options:
A 124 pm
B 248 pm
C 272 pm
D 136 pm

The correct answer is:
136 pm

Solution:
Interatomic distance in hydrogen H2 molecule LH2=74 pm and interatomic distance in chlorine Cl2 molecule LCl2=198 pm . Atoms linked by one Ï -bond, covalent radius r=Interatomic distance2 . Therefore covalent radius of hydrogen rH=LH22=742=37 pm and that of chlorine rCl=LCl22=1982=99 pm
As H 2 g + Cl 2 g â2HCL g
Therefore bond length of HCl molecule LHCl
=rH+rCl=37+99=136 pm.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The interdependent evolution of flowering plants and pollinating insects together is known as

Options:
A predation.
B co-evolution.
C commensalism.
D proto-cooperation.

The correct answer is:
co-evolution.

Solution:
When two different species reciprocally affect other developing a mutualistically beneficial relationship and cause evolution is called as co-evolution. This generally occurs when the two participating species have close ecological interactions. Classic example been that of Fig plant and Wasp, where the fig provides egglaying site for the insect and in return the insect helps in pollination.

Topic: Biology, Organisms and Populations, NEET
The interference pattern is observed at P due to superimposition of two rays coming out from a source S as shown in the figure. The value of l for which maxima is obtained at P is ( R is a perfect reflecting surface): -

Options:
A l=2n-1λ23-1
B l=2n-1λ3-1
C l=2n-1λ342-3
D l=2nλ3-1

The correct answer is:
l=2n-1λ342-3

Solution:
x cos30°=l

x32=l

x=23l

2l3+2l3+λ2-2l=nλ

2l23-1=2n-1λ2

2l2-3=2n-1λ32

l=2n-1λ342-3

Topic: Physics, Wave Optics, NEET
The interference pattern is observed at P due to the superposition of two rays coming out from a source S as shown in the diagram. The value of l for which maxima is obtained at P is, (Given that R is a perfect reflecting surface)

Options:
A lâ¡ = 2 n λ 3 - 1
B lâ¡ = 2 n - 1 Î» 3 4 2 - 3
C lâ¡ = 2 n - 1 λ 3 - 1
D lâ¡ = 2 n - 1 λ 2 3 - 1

The correct answer is:
l⁡ = 2 n - 1 λ 3 4 2 - 3

Solution:
Î x = 4 l 3 + Î» 2 - 2 l = n Î» for construction interference.
Î»2 is added due to reflection from denser medium.
â l = n Î» - Î» 2 4 3 - 2 = 3 2 n - 1 Î» 4 2 - 3

Topic: Physics, Wave Optics, NEET
The interference pattern is obtained with two coherent light sources of intensity ratio, n. In the interference pattern, the ratio, Imax−IminImax+Imin will be

Options:
A nn+1
B 2nn+1
C nn+12
D 2nn+12

The correct answer is:
2nn+1

Solution:
I1I2=n.

Imax=I1+I22=n+12I2.

Imin=I1-I22=n-12I2.

Imax-IminImax+Imin=4n2n+1=2nn+1

Topic: Physics, Wave Optics, NEET, NEET 2016 (Phase 2)
The intermediate carbocation formed in the reactions of HI, HBr and HCl with propene is the same and the bond energy of HCl, HBr and HI is 430.5 kJ mol–1, 363.7 kJ mol–1and 296.8 kJ mol–1respectively. What will be the order of reactivity of these halogen acids?

Options:
A Bond dissociation energy is least for HI and maximum for HCl, therefore, order of reactivity will be HI > HBr > HCl.
B Bond dissociation energy is maximum for HI and minimum for HCl, therefore, order of reactivity will be HI < HBr < HCl.
C Bond dissociation energy is least for HI and maximum for HCl, therefore, order of reactivity will be HI = HBr = HCl.
D Bond dissociation energy is least for HCl and maximum for HBr, therefore, order of reactivity will be HCl > HI > HBr.

The correct answer is:
Bond dissociation energy is least for HI and maximum for HCl, therefore, order of reactivity will be HI > HBr > HCl.

Solution:
Correct Option is : (A)
Bond dissociation energy is least for HI and maximum for HCl, therefore, order of reactivity will be HI > HBr > HCl.

Topic: Chemistry, Hydrocarbons, NEET
The intermediate compound 'X' in the following chemical reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, NEET, NEET 2021
The intermediate involves in reaction of phenol with chloroform and sodium hydroxide to give o- hydroxy benzaldehyde is

Options:
A Dichloro carbene
B Trichloro carbene
C Chlorine atoms
D Chlorine molecules

The correct answer is:
Dichloro carbene

Solution:
This is Reimer-Tiemann reaction where the electrophile is dichlorocarbene (:CCl2) generated from chloroform by the action of a base.
OH-+CHCl3 â HOH+ CCl3-âCl-+:CCl2

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
The intermediate product 'X' of following synthesis is indentified as

→Dil.H2SO4,Δ 2,6-Dinitroaniline

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, NEET
The internal and external diameters of a hollow cylinder are measured with the help of vernier callipers. Their values are 4.23±0.01cm and 3.87±0.01cm respectively. The thickness of the wall of the cylinder is

Options:
A 0.36 Â± 0.02 cm
B 0.18 Â± 0.02cm
C 0.36 Â± 0.01cm
D 0.18 Â± 0.01cm

The correct answer is:
0.18 Â± 0.02cm

Solution:
t=D1−D22=4.23−3.872=0.18 cm
Δt=ΔD1+ΔD2
=0.01+0.01=0.02 cm
⇒t=(0.18±0.02) cm

Topic: Physics, Units and Dimensions, NEET
The internal and external diameters of a hollow cylinder are measured with the help of vernier callipers. Their values are 4.23±0.01cm and 3.87±0.01cm respectively. The thickness of the wall of the cylinder is

Options:
A 0.36 Â± 0.02 cm
B 0.18 Â± 0.02cm
C 0.36 Â± 0.01cm
D 0.18 Â± 0.01cm

The correct answer is:
0.18 Â± 0.01cm

Solution:
t=D1−D22=4.23−3.872=0.18 cm
Δt=ΔD1+ΔD22
= 0.01+0.01 2 =0.01 cm
⇒t=(0.18±0.01) cm

Topic: Physics, Units and Dimensions, NEET
The internal energy change in a system that has absorbed $2 \mathrm{kcal}$ of heat and done $500 \mathrm{~J}$ of work is

Options:
A $8900 \mathrm{~J}$
B $6400 \mathrm{~J}$
C $5400 \mathrm{~J}$
D $7900 \mathrm{~J}$

The correct answer is:
$7900 \mathrm{~J}$

Solution:
Key Idea Heat given to a system ( $\Delta Q$ ) is equal to the sum of increase in the internal energy $(\Delta \mathrm{u})$ and the work done $(\Delta W)$ by the system against the surrounding and $1 \mathrm{cal}=4.2 \mathrm{~J}$.
According to first law of thermodynamics
$$
\begin{aligned}
\Delta \mathrm{U} & =\mathrm{Q}-\mathrm{W} \\
& =2 \times 4.2 \times 1000-500 \\
& =8400-500 \\
& =7900 \mathrm{~J}
\end{aligned}
$$

Topic: Physics, Thermodynamics, NEET, NEET 2009 (Screening)
The internal resistance of a $2.1 \mathrm{~V}$ cell which gives a current of $0.2 \mathrm{~A}$ through a resistance of $10 \Omega$ is

Options:
A $0.2 \Omega$
B $0.5 \Omega$
C $0.8 \Omega$
D $1.0 \Omega$

The correct answer is:
$0.5 \Omega$

Solution:
\(\begin{aligned} & I=\frac{E}{R+r} \\ & 0.2 \times(10+r)=2.1 \\ & 10+r=\frac{2.1}{2} \times 10 \\ & r=10.5-10=0.5 \Omega\end{aligned}\)

Topic: Physics, Current Electricity, NEET, NEET 2013 (All India)
The international treaty of Montreal Protocol became effective in the year of

Options:
A 1989
B 1999
C 1889
D 1987

The correct answer is:
1989

Solution:
The Montreal Protocol is an international treaty designed to protect the ozone layer by phasing out the production of numerous substances that are responsible for ozone depletion. It was agreed on 16th September 1987, and entered into force on 1st January 1989.

Topic: Biology, Environmental Issues, NEET
The internodal segments of tobacco stems the callus proliferated only if it contains ______(i)_____ in addition to ____(ii)____.

Options:
A (i) Yeast extract ; (ii) coconut milk
B (i) Coconut milk; (ii) Auxin
C (i) Yeast extract ; (ii) Cytokinin
D (i) coconut milk.; (ii) Gibberellic acid

The correct answer is:
(i) Coconut milk; (ii) Auxin

Solution:
Coconut milk contains cytokinin. Both Auxin and Cytokinin help in the proliferation of Callus. If auxin is more root will while if cytokinin is more shoot will form.

Topic: Biology, Plant - Growth and Development, NEET
The interstitial cells of the testes synthesize and secrete:

Options:
A Androgens
B Luteinizing hormone
C Aldosterone
D Interstitial cell-stimulating hormone

The correct answer is:
Androgens

Solution:
The regions outside the seminiferous tubules called interstitial spaces contain small blood vessels and interstitial cells or Leydig cells. Leydig cells synthesise and secrete testicular hormones called androgens.

Topic: Biology, Human Reproduction, NEET
The intine of pollen grains is composed of

Options:
A Lipid and protein
B Cellulose and pectin
C Lignin and cutin
D Pectin and lignin

The correct answer is:
Cellulose and pectin

Solution:
The inner intine is thin, delicate structure and is made up of cellulose and pectose.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The intrameiotic interphase is called:

Options:
A Interphase
B Cytokinesis
C Interkinesis
D Karyokinesis

The correct answer is:
Interkinesis

Solution:
The stage between two meiotic divisions is called interkinesis. It is generally short-lived and is followed by prophase-II, much simpler prophase than prophase-I of meiosis-I.

Topic: Biology, Cell Cycle and Cell Division, NEET
The intrinsic rate of natural increase for Indian human population according to the 2011 census was

Options:
A 0.0205
B 0.0147
C 0.0652
D 0.0351

The correct answer is:
0.0147

Solution:
The intrinsic rate of natural increase (r) is the difference between birth rate and death rate. According to the 2011 census, India has a birth rate of 21.8 per 1000 and death rate 7.1 per 1000 individuals. Thus, r = 21.8 - 7.1 = 14.7 per 1000, which is 0.0147.

Topic: Biology, Organisms and Populations, NEET
The introduction of T-DNA into plants involves :

Options:
A Altering the pH of the soil, then heat-shocking the plants
B Exposing the plants to cold for a brief period
C Allowing the plant roots to stand in water
D Infection of the plant by Agrobacterium tumefaciens

The correct answer is:
Infection of the plant by Agrobacterium tumefaciens

Solution:
Ti plasmid of Agrobacterium tumefaciens: Agrobacterium tumefaciens, a pathogen of several dicot plants is able to deliver a piece of DNA known as ‘T-DNA’ to transform normal plant cells into a tumor and direct these tumor cells to produce the chemicals required by the pathogen. This bacterium invades plants at the site of wound, transforming them and nearby cells to form a tumor called crown gall.

When the bacterium contacts a damaged plant cell, it delivers a T-DNA (transferred DNA) fragment from plasmid into host cell that integrates at a random position in plant cells chromosome. The tumor inducing (Ti) plasmid of A. tumefaciens has now been modified into a cloning vector which is no more pathogenic to the plants but is still able to use the mechanism to deliver genes of interest.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2015 (Phase 2)
The inverse square law in electrostatics is
for the force between an electron and a proton. The (1/r) dependence of | | F can be understood in quantum theory as being due to the fact that the ‘particle’ of light (photon) is massless. If photons had a mass mp, force would be modified to
. What is the estimated change in the ground state energy of a hydrogen atom if the proton mass (mp) were 10-6times the mass of an electron?

Options:
A 2.72 λrBeV
B 0.0272 λrBeV
C 27.2 λrBeV
D 0.272 λrBeV

The correct answer is:
27.2 λrBeV

Solution:
Correct Option is : (C)
27.2 λrBeV

Topic: Chemistry, Atomic Physics, NEET
The ionic product of Ni(OH)2 is 2.0 Ã 10-15. The molar solubility of Ni(OH)2 in 0.10 M NaOH is-

Options:
A 3.2 × 10-12
B 2.0 Ã 10-13
C â4.34 × 10-12
D â0.58 × 10-4

The correct answer is:
2.0 × 10-13

Solution:
Ni(OH)2 â Ni2+ + 2OH-
s 2s
Total [OH-] = 2s + 0.1
Ionic product = [s] [2s + 0.1]2
âS (0.01) =2Ã10-15
0.01 s = 2 Ã 10-15
s = 2 Ã 10-13

Topic: Chemistry, Ionic Equilibrium, NEET
The ionic radii of A+ and B- ions are 0.98×10-10m and 1.81×10-10 m . The coordination number of each ion in AB is

Options:
A 6
B 4
C 8
D 2

The correct answer is:
6

Solution:
Radius ratio of A+B-=0.98×10-10 m1.81×10-10 m=0.981.81=0.541
If the radius ratio is between 0.414 and 0.732 then Co-ordination number is 6.

Topic: Chemistry, Solid State, NEET, NEET 2016 (Phase 1)
The ionisation constant of ammonium hydroxide is $1.77 \times 10^{-5}$ at $298 \mathrm{~K}$. Hydrolysis constant of ammonium chloride is

Options:
A $5.65 \times 10^{-10}$
B $6.50 \times 10^{-12}$
C $5.65 \times 10^{-13}$
D $5.65 \times 10^{-12}$

The correct answer is:
$5.65 \times 10^{-10}$

Solution:
Given, $\mathrm{K}_{\mathrm{a}}\left(\mathrm{NH}_4 \mathrm{OH}\right)=1.77 \times 10^{-5}$
$$
\begin{aligned}
& \mathrm{NH}_4 \mathrm{OH} \rightleftharpoons \mathrm{NH}_4^{+}+\mathrm{OH}^{-} \\
& \mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_4 \mathrm{OH}\right]}=1.77 \times 10^{-5}
\end{aligned}
$$
Hydrolysis of $\mathrm{NH}_4 \mathrm{Cl}$ takes place as,
$$
\mathrm{NH}_4 \mathrm{Cl}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NH}_4 \mathrm{OH}+\mathrm{HCl}
$$
or $\mathrm{NH}_4^{+}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NH}_4 \mathrm{OH}+\mathrm{H}^{+}$
Hydrolysis constant
$$
, \mathrm{K}_{\mathrm{h}}=\frac{\left[\mathrm{NH}_4 \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{NH}_4^{+}\right]} \ldots
$$
or
$$
\mathrm{K}_{\mathrm{h}}=\frac{\left[\mathrm{NH}_4 \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}^{-}\right]}
$$
From Eqs (i) and (iii)
$$
\begin{aligned}
\mathrm{K}_{\mathrm{h}} & =\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}}} \quad\left[\because\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=\mathrm{K}_{\mathrm{w}}\right] \\
& =\frac{10^{-14}}{1.77 \times 10^{-5}} \\
& =5.65 \times 10^{-10}
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2009 (Screening)
The ionisation energy of 10 time ionised sodium atom is

Options:
A 13.611 eV
B 13.6112 eV
C 13.6 Ã 112 eV
D 13.6 eV

The correct answer is:
13.6 Ã 112 eV

Solution:
The energy of nth orbit of a hydrogen-like atom is,
En = -13.6 Z2n2
Here, Z=11 for Na atom. 10 electrons are removed already. For the last electron to be removed n=1.
∴En=-13.6 ×(11)2(1)2 eV
= -13.6 ×(11)2 eV

Topic: Physics, Atomic Physics, NEET
The ionisation energy of H-atoms is 13.6 ev. What will be the ionisation energy of He+ and Li2+ ions?

Options:
A 54.4 ev, 122.4 ev
B 5.44 ev, 12.24 ev
C 6.82 ev, 13.6 ev
D 13.4 ev, 122.4 ev

The correct answer is:
54.4 ev, 122.4 ev

Solution:
Ionization energy of H-atom = 13.6 ev
Ionization energy of He+=Z2× Ionization energy of hydrogen atom
=(2)2×13.6=54.4 ev
Ionization energy of Li2+=Z2× Ionization energy of H-atom
=(3)2×13.6=122.4 ev.

Topic: Chemistry, Structure of Atom, NEET
The ionisation potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. The spectral lines emitted by hydrogen atoms according to Bohr's theory will be

Options:
A One
B Two
C Three
D Four

The correct answer is:
Three

Solution:
Energy in excited state =-13.6+12.1=-1.5 eV
∴ -13.6n2=-1.5
∴ n=13.61.5=3
Number of spectral lines
=n(n-1)2=3(3-1)2=3

Topic: Physics, Atomic Physics, NEET
The ionization constant of acetic acid is 1.74 × 10-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

Options:
A 0.009, 0.00093 M, 3.03
B 0.008, 0.042 M, 2.22
C 0.009, 0.041 M, 2.22
D 0.0089, 0.00093 M, 3.8

The correct answer is:
0.009, 0.00093 M, 3.03

Solution:
Correct Option is : (A)
0.009, 0.00093 M, 3.03

Topic: Chemistry, Ionic Equilibrium, NEET
The ionization constant of benzoic acid is 6.46 × 10-5and Ksp for silver benzoate is 2.5 × 10-13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Options:
A 10 times
B 3.317 times
C 3.326 times
D 10000 times

The correct answer is:
3.317 times

Solution:
Correct Option is : (B)
3.317 times

Topic: Chemistry, Ionic Equilibrium, NEET
The ionization constant of chloroacetic acid is 1.35 × 10-3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?

Options:
A pH of acid = 1.94, pH of sodium salt solution = 7.94
B pH of acid = 2.75, pH of sodium salt solution = 7.10
C pH of acid = 2.95, pH of sodium salt solution = 6.90
D pH of acid = 3.05, pH of sodium salt solution = 6.80

The correct answer is:
pH of acid = 1.94, pH of sodium salt solution = 7.94

Solution:
Correct Option is : (A)
pH of acid = 1.94, pH of sodium salt solution = 7.94

Topic: Chemistry, Ionic Equilibrium, NEET
The ionization constant of dimethylamine is 5.4 × 10-4. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?

Options:
A α = 0.164 , 0.54% ionized
B α = 0.024, 2.0% ionized
C α = 0.025, 25.0% ionized
D α = 0.022, 1.1% ionized

The correct answer is:
α = 0.164 , 0.54% ionized

Solution:
Correct Option is : (A)
α = 0.164 , 0.54% ionized

Topic: Chemistry, Ionic Equilibrium, NEET
The ionization constant of HF is 3.2 × 10-4. Calculate the degree of dissociation of HF in its 0.02 M solution. Calculate the concentration of all species present (H₃O⁺, F⁻, and HF) in the solution and its pH.

Options:
A Degree of dissociation = 0.017, [H₃O⁺] = 0.037 M, [F⁻] = 0.017 M, [HF] = 1.963 M, pH = 0.71
B Degree of dissociation = 0.032, [H₃O⁺] = 0.066 M, [F⁻] = 0.032 M, [HF] = 1.968 M, pH = 2.62
C Degree of dissociation = 0.020, [H₃O⁺] = 0.040 M, [F⁻] = 0.020 M, [HF] = 1.980 M, pH = 0.69
D Degree of dissociation = 0.025, [H₃O⁺] = 0.049 M, [F⁻] = 0.025 M, [HF] = 1.975 M, pH = 0.70

The correct answer is:
Degree of dissociation = 0.020, [H₃O⁺] = 0.040 M, [F⁻] = 0.020 M, [HF] = 1.980 M, pH = 0.69

Solution:
Correct Option is : (C)
Degree of dissociation = 0.020, [H₃O⁺] = 0.040 M, [F⁻] = 0.020 M, [HF] = 1.980 M, pH = 0.69

Topic: Chemistry, Ionic Equilibrium, NEET
The ionization constant of \(\mathrm{HF}, \mathrm{HCOOH}\) and \(\mathrm{HCN}\) at \(298 \mathrm{~K}\) are \(6.8 \times 10^{-4}, 1.8 \times 10^{-4}\) and \(4.8 \times 10^{-9}\) respectively. Calculate the ionization constants of the corresponding conjugate base.

Options:
A \(6.8 \times 10^{-4}, 1.8 \times 10^{-4}, 4.8 \times 10^{-9}\)
B \(3.2 \times 10^{-11}, 2.2 \times 10^{-11}, 2.1 \times 10^{-5}\)
C \(1.47 \times 10^{-10}, 1.56 \times 10^{-10}, 2.08 \times 10^{-5}\)
D \(1.1 \times 10^{-14}, 1.3 \times 10^{-14}, 3.2 \times 10^{-9}\)

The correct answer is:
\(3.2 \times 10^{-11}, 2.2 \times 10^{-11}, 2.1 \times 10^{-5}\)

Solution:
Correct Option is : (B)
\(3.2 \times 10^{-11}, 2.2 \times 10^{-11}, 2.1 \times 10^{-5}\)

Topic: Chemistry, Ionic Equilibrium, NEET
The ionization constant of nitrous acid is 4.5 × 10-4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

Options:
A pH = 8.00, Degree of hydrolysis (α) = 0.0000023
B pH = 3.50, Degree of hydrolysis (α) = 0.0000009
C pH = 3.20, Degree of hydrolysis (α) = 0.0000012
D pH = 3.10, Degree of hydrolysis (α) = 0.0000012

The correct answer is:
pH = 8.00, Degree of hydrolysis (α) = 0.0000023

Solution:
Correct Option is : (A)
pH = 8.00, Degree of hydrolysis (α) = 0.0000023

Topic: Chemistry, Ionic Equilibrium, NEET
The ionization constant of phenol is 1.0 × 10-10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01 M in sodium phenolate?

Options:
A 1.4 × 10-8, 3.2 × 10-8
B 2.3 × 10-9, 2.3 × 10-10
C 2.2 × 10-6, 1.0 × 10-8
D 1.0 × 10-10, 1.0 × 10-9

The correct answer is:
2.2 × 10-6, 1.0 × 10-8

Solution:
Correct Option is : (C)
2.2 × 10-6, 1.0 × 10-8

Topic: Chemistry, Ionic Equilibrium, NEET
The ionization constant of propanoic acid is 1.32 × 10-5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

Options:
A α = 0.0016, pH = 3.10, α' = 0.00013
B α = 0.010, pH = 2.00, α' = 0.012
C α = 0.012, pH = 2.12, α' = 0.014
D α = 0.014, pH = 2.14, α' = 0.016

The correct answer is:
α = 0.0016, pH = 3.10, α' = 0.00013

Solution:
Correct Option is : (A)
α = 0.0016, pH = 3.10, α' = 0.00013

Topic: Chemistry, Ionic Equilibrium, NEET
The ionization energy of He+ ion at minimum position is

Options:
A 13.6 eV
B 27.2 eV
C 54.4 eV
D 68.0 eV

The correct answer is:
54.4 eV

Solution:
Energy of helium ions.
En = - 13.6 Z2n2 eV
In minimum position, n=1
For He+, Z=2
E=-13.6 Ã (2)21 eV
E=54.4 eV

Topic: Physics, Atomic Physics, NEET
The ionization energy of He+ is 19.6×10-18Jatom-1. The energy of the first stationary state of Li+2 will be

Options:
A 21.2×10-18J/atom
B 44.10×10-18J/atom
C 63.2×10-18J/atom
D 84.2×10-18J/atom

The correct answer is:
44.10×10-18J/atom

Solution:
E1 for Li+2=E1 for H×Z2Li
E1 for He+=E1 for H×Z2He
Thus,
E1 for Li+2=94×E1 for He+
=94×19.6×10-18
=44.10×10-18J/atom

Topic: Chemistry, Structure of Atom, NEET
The ionization energy of hydrogen atom is 13.6 eV . Hydrogen atoms in the ground state are excited by electromagnetic radiation of energy 12.1 eV . How many spectral lines will be emitted by the hydrogen atoms

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Number of spectral lines =n(n-1)2=3(3-1)2=3

Topic: Physics, Atomic Physics, NEET
The ionization energy of the electron in the hydrogen atom in its ground state is $13.6 \mathrm{eV}$. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between

Options:
A $n=3$ to $n=2$ states
B $n=3$ to $n=1$ states
C $n=2$ to $n=1$ states
D $n=4$ to $n=3$ states

The correct answer is:
$n=4$ to $n=3$ states

Solution:
Key Idea Number of spectral lines abtained due to transition of electron from $\boldsymbol{n}^{\text {th }}$ orbit to lower orbit is $\mathrm{N}=\frac{\mathrm{n}(\mathrm{n}-1)}{2}$ and for maximum wavelength the difference between the orbits of the series should be minimum.
Number of spectral lines $\mathrm{N}=\frac{\mathrm{n}(\mathrm{n}-1)}{2}$
$$
\begin{aligned}
\Rightarrow & \frac{n(n-1)}{2} & =6 \\
\text { or } & n^2-n-12 & =0 \\
\text { or } & (n-4)(n+3) & =0 \\
\text { or } & n & =4
\end{aligned}
$$
Now as the first line of the series has the maximum wavelength, therefore electron jumps from the $4^{\text {th }}$ orbit to the third orbit.

Topic: Physics, Atomic Physics, NEET, NEET 2009 (Screening)
The ionization enthalpies of Li and Na are 520 kJ mol-1 and 495 kJ mol-1 respectively. The energy required to convert all the atoms present in 7 mg of Li vapours and 23 mg of sodium vapours of their respective gaseous cations respectively are

Options:
A 52 J, 49.5 J
B 520 J, 495 J
C 49.5 J, 52 J
D 495 J, 520 J

The correct answer is:
520 J, 495 J

Solution:
Number of moles = weight in gm atomic weight
Number of moles Li =71000Ã7=10-3
Number of moles of Na =231000Ã23Ã=10-3
Amount of energy = moles Ã energy needed for 1 mole.
So, the amounts of energies required for 10-3 mole each of Li and Na are 520 kJ Ã10-3 or 520 J and 495 J respectively.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET