datavorous/entrance-exam-dataset huggingface dataset:
The correct option for a redox couple is

Options:
A Both the reduced and oxidized forms involve same element
B Cathode and anode together
C Both are oxidised forms involving same element
D Both are reduced forms involving same element

The correct answer is:
Both the reduced and oxidized forms involve same element

Solution:
A redox couple is defined as having together the oxidised and reduced forms of a substance taking part in an oxidation or reduction half reaction.

Topic: Chemistry, Redox Reactions, NEET, NEET 2023 (Manipur)
The correct option for free expansion of an ideal gas under adiabatic condition is

Options:
A q=0,ΔT<0 and w>0
B q<0,ΔT=0 and w=0
C q>0,ΔT>0 and w>0
D q=0,ΔT=0 and w=0

The correct answer is:
q=0,ΔT=0 and w=0

Solution:
Free expansion of ideal gas

Pext=0

wpv=0

q=0 (adiabatic process)

ΔE=q+w

ΔE=0

ΔE=nCVΔT=0

q=0, ΔT=0, w=0

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2020 (Phase 1)
The correct option for the number of body centred unit cells in all 14 types of Bravais lattice unit cells is:

Options:
A 7
B 5
C 2
D 3

The correct answer is:
3

Solution:
Body-centred unit cell is possible in following crystal system:-
a) Cubic
b) Tetragonal
c) Orthorhombic

Thus, three body-centred unit cells are possible among 14 types of Bravais lattice.

Topic: Chemistry, Solid State, NEET, NEET 2021
The correct option for the rate law that corresponds to overall first order reaction is

Options:
A Rate $=k[A]^{1 / 2}[B]^2$
B Rate $=k[A]^{-1 / 2}[\mathrm{~B}]^{3 / 2}$
C Rate $=k[A]^0[\mathrm{~B}]^2$
D Rate $=k[A][B]$

The correct answer is:
Rate $=k[A]^{-1 / 2}[\mathrm{~B}]^{3 / 2}$

Solution:
Rate $=k[A]^{-\frac{1}{2}}[B]^{\frac{3}{2}}$
So, order of reaction $=-\frac{1}{2}+\frac{3}{2}=1$

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2023 (Manipur)
The correct option for the value of vapour pressure of a solution at 45ºC with benzene to octane in molar ratio 3 : 2 is:

[At 45°C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]

Options:
A 160 mm of Hg
B 168 mm of Hg
C 336 mm of Hg
D 350 mm of Hg

The correct answer is:
336 mm of Hg

Solution:
Given: PB°=280 mm of Hg, Po°=420 mm of Hg

XB=35 and Xo=25

We know that according to Raoult's Law

PT=XBPB°+XoPo°=35×280+25×420=336 mm of Hg

Topic: Chemistry, Solutions, NEET, NEET 2021
The correct option in which the density of argon (Atomic mass $=40$ ) is highest

Options:
A $0^{\circ} \mathrm{C}, 4 \mathrm{~atm}$
B $273^{\circ} \mathrm{C}, 4$ atm
C STP
D $0^{\circ} \mathrm{C}, 2 \mathrm{~atm}$

The correct answer is:
$0^{\circ} \mathrm{C}, 4 \mathrm{~atm}$

Solution:
$\mathrm{PV}=\mathrm{nRT}$
$\begin{aligned} P & =\frac{n}{V} R T \\ P & =\frac{d}{M} R T\end{aligned}$
$P=\frac{d}{M} R T$
$\frac{P M}{R T}=d$

Topic: Chemistry, States of Matter, NEET, NEET 2023 (Manipur)
The correct option representing a Freundlich adsorption isotherm is

Options:
A $\frac{x}{m}=k p^{0.3}$
B $\frac{x}{m}=k p^{2.5}$
C $\frac{x}{m}=k p^{-0.5}$
D $\frac{x}{m}=k p^{-1}$

The correct answer is:
$\frac{x}{m}=k p^{0.3}$

Solution:
According to Freundlich adsorption isotherm,
$$
\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kp}^{1 / \mathrm{n}}
$$
where, $\frac{\mathrm{x}}{\mathrm{m}}=$ amount of the gas adsorbed per unit mass of adsorbent $\mathrm{p}=$ pressure $\mathrm{k}$ and $\mathrm{n}=$ constants.
The value of $\mathrm{n}$ lies in between 0 to 1 .
Thus, $\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kp}^{0.3}$ and option (1) is correct.

Topic: Chemistry, Surface Chemistry, NEET, NEET 2019 (Odisha)
The correct order for the rate of $\alpha, \beta$-dehydrohalogenation for the following compounds is

Options:
A (iii) $ < $ (ii) $ < $ (i)
B (ii) < (iii) < (i)
C (i) $ < $ (ii) $ < $ (iii)
D (ii) $ < $ (i) $ < $ (iii)

The correct answer is:
(ii) < (iii) < (i)

Solution:
- More is the stability of alkene formed, more will be the rate of $\alpha-\beta$ dehydrohalogenation

Topic: Chemistry, Hydrocarbons, NEET, NEET 2023 (Manipur)
The correct order for the wavelengths of absorption in the visible region for the following is: Ni NO 2 6 4− , Ni NH 3 6 2+ , Ni H 2 O 6 2+

Options:
A [ Ni(H 2 O ) 6 ] 2+ ≈ [ Ni(NH 3 ) 6 ] 2+ ≈ [ Ni(NO 2 ) 6 ] 4−
B [ Ni(NH 3 ) 6 ] 2+ > ​ [Ni(NO 2 ) 6 ] 4− > [ Ni(H 2 O) 6 ] 2+
C [ Ni(H 2 O ) 6 ] 2+ > ​ [ Ni(NO 2 ) 6 ] 4− > [ Ni(NH 3 ) 6 ] 2+
D [ Ni(H 2 O ) 6 ] 2+ > [ Ni(NH 3 ) 6 ] 2+ > [ Ni(NO 2 ) 6 ] 4â

The correct answer is:
[ Ni(H 2 O ) 6 ] 2+ > [ Ni(NH 3 ) 6 ] 2+ > [ Ni(NO 2 ) 6 ] 4−

Solution:
The order of ligands in the spectrochemtextcal series is

H 2 O<N H 3 <N O 2 â

Thus, energy absorbed will be in the order :

[Niâ ( H 2 O) 6 ] 2+ <ââ [Niââ (N H 3 ) 6 ] 2+ <ââ [Niâ (N O 2 ) 6 ] 4â

Since Î»âââ 1 E ( E= hc Î» ) the wavelength of light absorbed will be in the order :

[Niâ ( H 2 O) 6 ] 2+> [Niâ(N H 3 ) 6 ] 2+>â [Niâ (N O 2 ) 6 ] 4â

Topic: Chemistry, Coordination Compounds, NEET
The correct order in which oxygen-oxygen bond dissociation energy increases is

Options:
A H2O2<O2<O3
B H2O2<O3<O2
C O2<O3<H2O2
D O2<H2O2<O3

The correct answer is:
H2O2<O3<O2

Solution:
H2O2 has a peroxide bond with a bond order of 1 while in O3 the bond order is 1.5 and in O2 it is 2. Hence, the bond dissociation energy order is H2O2<O3<O2

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The correct order in which the $\mathrm{O}-\mathrm{O}$ bond length increases in the following is:

Options:
A $\mathrm{O}_2 < \mathrm{H}_2 \mathrm{O}_2 < \mathrm{O}_3$
B $\mathrm{O}_3 < \mathrm{H}_2 \mathrm{O}_2 < \mathrm{O}_2$
C $\mathrm{H}_2 \mathrm{O}_2 < \mathrm{O}_2 < \mathrm{O}_2$
D $\mathrm{O}_2 < \mathrm{O}_3 < \mathrm{H}_2 \mathrm{O}_2$

The correct answer is:
$\mathrm{O}_2 < \mathrm{O}_3 < \mathrm{H}_2 \mathrm{O}_2$

Solution:
The bond Length decreases with increase of bond order which is the multiplicity of bond. If $\mathrm{BO}$ is 1 its a single bond; 2 for double bond and 3 for triple bond.
The bond length of $\mathrm{O}-\mathrm{O}$ in $\mathrm{H}_2 \mathrm{O}_2$ is 147.5pm, in $\mathrm{O}_3$ is $128 \mathrm{pm}$ and in $\mathrm{O}_2$ it is $121 \mathrm{pm}$.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2005
The correct order in which the O-O bond length increases in the following is-

Options:
A O2<O3<H2O2
B H2O2<O3<O2
C O3<O2<H2O2
D O2<H2O2<O3

The correct answer is:
O2<O3<H2O2

Solution:
The increasing order is
Due to resonance in O3, the O-O bond length will be in between O=O and O-O

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The correct order of 2nd ionization energy for the elements carbon, nitrogen, oxygen and fluorine is

Options:
A F>O>N>C
B C>N>O>F
C O>F>N>C
D O>N>F>C

The correct answer is:
O>F>N>C

Solution:
The second ionisation energy means the energy required to remove the electron from the corresponding monovalent cation of the respective atom. For IE2, electronic. Configuration of C+,N+,O+ and F+ -ions which are as follows.
C+=2s2,2p1
N+=2s2,2p2
O+=2s2,2p3
F+=2s2,2p4
âµâO+ is half - filed configuration, obviously it is most stable and greater ionization energy than F+ as well as
N+ . The ionisation energy of an element depends upon its size as. IEâ 1 Size of atom/ ion smaller the atom/ion,
larger mill be IE.
Hence, correct order will be, O>F>N>C

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
The correct order of acid strength is:

Options:
A $\mathrm{HClO}_4 < \mathrm{HClO}_3 < \mathrm{HClO}_2 < \mathrm{HClO}$
B $\mathrm{HClO} < \mathrm{HClO}_2 < \mathrm{HClO}_3 < \mathrm{HClO}_4$
C $\mathrm{HClO}_4 < \mathrm{HClO}^2 < \mathrm{HClO}_2 < \mathrm{HClO}_3$
D $\mathrm{HClO}_2 < \mathrm{HClO}_3 < \mathrm{HClO}_4 < \mathrm{HClO}$

The correct answer is:
$\mathrm{HClO} < \mathrm{HClO}_2 < \mathrm{HClO}_3 < \mathrm{HClO}_4$

Solution:
Oxygen is more electronegative than chlorine. With an increase in the number of $\mathrm{O}$ atoms attached to $\mathrm{Cl}$, more electrons are pulled away from $\mathrm{O}-\mathrm{H}$ bond and weaker becomes the $\mathrm{O}-\mathrm{H}$ bond. This increases the acid strength. The decreasing order of acidic strength is $\mathrm{HClO} < \mathrm{HClO}_2 < $ $\mathrm{HClO}_3 < \mathrm{HClO}_4$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET, NEET 2005
The correct order of acidic strength

Options:
A Cl 2 O 7  >  SO 3  >  P 4 O 10
B CO2>N2O5>SO3
C K2>CaO>MgO
D Na2O>MgO>Al2O3

The correct answer is:
Cl 2 O 7  >  SO 3  >  P 4 O 10

Solution:
Acidic nature of oxide depends on Non-metallic nature of element, it increases with increase in non-metallic character. Non-metallic nature decrease in the order Cl>S>P .

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
The correct order of acidity for the following compounds is

Options:
A I > II > III > IV
B III > I > II > IV
C III > IV > II > I
D I > III > IV > II

The correct answer is:
I > II > III > IV

Solution:
Due to ortho-effect, (I) and (II) are stronger acid than (III) and (IV). Due to two ortho hydroxyl groups in (I), it is stronger acid than (II). (III) is a stronger acid than (IV) because at m-position, -OH group cannot exert its +R effect but can only exert its –I effect while at p-position, -OH group exerts its strong +R effect. Thus, the correct order of acidity is
I > II > III > IV.

Topic: Chemistry, General Organic Chemistry, NEET
The correct order of acidity for the following compounds is

Options:
A I > III > IV > II > V
B III > IV > II > I > V
C III > I > II > IV > V
D I > II > III > V > IV

The correct answer is:
I > II > III > V > IV

Solution:
Acidic nature ∝ electron withdrawing group (-I, -R) ∝ ortho effect
Ortho substituted benzoic acid is always more acidic than its other substituted isomers as here steric inhibition of resonance occurs

Topic: Chemistry, General Organic Chemistry, NEET
The correct order of atomic radii in group 13 elements is

Options:
A B<Ga<Al<Tl<In
B B<Al<Ga<In<Tl
C B<Al<In<Ga<Tl
D B<Ga<Al<In<Tl

The correct answer is:
B<Ga<Al<Tl<In

Solution:
In group 13, due to inert electron pair effect of Ga, the size order will be

Al>Ga

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET, NEET 2018
The correct order of atomic radii in group 13 elements is

Options:
A B < Ga < Al < Tl < ln
B B < Al < Ga < ln < Tl
C B < Al < ln < Ga < Tl
D B < Ga < Al < ln < Tl

The correct answer is:
B < Ga < Al < ln < Tl

Solution:
Ga is little smaller or have slightly lower atomic radius due to poor Screening effect by inner 3d electrons.

Elements	B	Ga	Al	In	TI

Atomic radii (pm)	85	135	143	167	170

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
The correct order of boiling point is:

Options:
A NH3<HF<H2O<H2O2
B NH3<HF<H2O2<H2O
C NH3<H2O<HF<H2O2
D HF<NH3<H2O<H2O2

The correct answer is:
NH3<HF<H2O<H2O2

Solution:
NH3<HF<H2O<H2O2 Boiling point order
Number of hydrogen bonding increases boiling point increases, NH3 has weakest H-bond

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
The correct order of bond angles in the following compounds/species is :

Options:
A $\mathrm{H}_2 \mathrm{O} < \mathrm{NH}_3 < \stackrel{+}{\mathrm{NH}_4} < \mathrm{CO}_2$
B $\mathrm{H}_2 \mathrm{O} < \stackrel{+}{\mathrm{NH}_4} < \mathrm{NH}_3 < \mathrm{CO}_2$
C $\mathrm{H}_2 \mathrm{O} < \stackrel{+}{\mathrm{NH}_4}=\mathrm{NH}_3 < \mathrm{CO}_2$
D $\mathrm{CO}_2 < \mathrm{NH}_3 < \mathrm{H}_2 \mathrm{O} < \stackrel{+}{\mathrm{NH}_4}$

The correct answer is:
$\mathrm{H}_2 \mathrm{O} < \mathrm{NH}_3 < \stackrel{+}{\mathrm{NH}_4} < \mathrm{CO}_2$

Solution:
According to VSEPR theory, bond angle decreases with increase in lone pair, for molecules of same hybridisation.
With increase in s-character, bond angle increase.
$\therefore$ order of bond angle $=$

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2022 (Phase 2)
The correct order of ClâO bond lengths in ClO â , ClO 2 â , ClO 3 â , ClO 4 â is

Options:
A ClO − < ClO 2 − < ClO 3 − < ClO 4 −
B ClO 4 − < ClO 3 − < ClO 2 − < ClO −
C ClO 3 − < ClO 4 − < ClO 2 − < ClO −
D ClO 4 − = ClO 3 − < ClO 2 − < ClO −

The correct answer is:
ClO 4 − < ClO 3 − < ClO 2 − < ClO −

Solution:
Bond length is inversely proportional to bond order. Greater the delocalization of Ïe-(-ve charge) shorter the bond length.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The correct order of decreasing second ionisation enthalpy of Ti (22), V (23), Cr (24) and Mn (25) is-

Options:
A Mn > Cr > Ti > V
B Ti > V > Cr > Mn
C Cr > Mn > V > Ti
D V > Mn > Cr > Ti

The correct answer is:
Cr > Mn > V > Ti

Solution:
Electronic configuration of the given elements are
Mn → 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 5 4 s 2
Cr → 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 5 4 s 1
Ti → 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 2 4 s 2
V → 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 3 4 s 2
In general, ionization potential ( both 1st and 2nd ) increases form left to right across the period due to increase in effective nuclear charge. On this basis, the second IP values should exhibit the trend :
Mn > Cr > V > Ti
But the actual observed order is :
Cr > Mn > V > Ti
Practically, only chromium is exceptional and rest others show the normal trend. This exceptional behaviour of chromium is due to the stable configuration ( 3d5 ) that it achieves after the loss of first electron.

Topic: Chemistry, d and f Block Elements, NEET
The correct order of decreasing second ionisation enthalpy of Ti(22), V (23), Cr (24) and $\mathrm{Mn}(25)$ is

Options:
A $\mathrm{Cr}>\mathrm{Mn}>\mathrm{V}>$ Ti
B V $>$ Mn $>$ Cr $>$ Ti
C Mn $>$ Cr $>$ Ti $>$ V
D Ti $>$ V $>$ Cr $>$ Mn

The correct answer is:
$\mathrm{Cr}>\mathrm{Mn}>\mathrm{V}>$ Ti

Solution:
Key Idea : The amount of energy required to remove an electron from a unipositive ion is called second IP which generally increases in a period from left to right.
The second ionisation energies of the first transition series increase almost regularly with increase in atomic number. However, the value for $\mathrm{Cr}$ is sufficiently higher than those of its neighbour, ie, (Mn). This is due to stable configuration of $\mathrm{Cr}^{+}$( $3 d^5$ exactly half filled).
Note : The halffilled and completely filled configurations are more stable.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET, NEET 2008 (Screening)
The correct order of dipole moments for molecules $\mathrm{NH}_3, \mathrm{H}_2 \mathrm{~S}, \mathrm{CH}_4$ and $\mathrm{HF}$, is

Options:
A $\mathrm{NH}_3>\mathrm{HF}>\mathrm{CH}_4>\mathrm{H}_2 \mathrm{~S}$
B $\mathrm{HF}>\mathrm{NH}_3>\mathrm{H}_2 \mathrm{~S}>\mathrm{CH}_4$
C $\mathrm{CH}_4>\mathrm{H}_2 \mathrm{~S}>\mathrm{NH}_3>\mathrm{HF}$
D $\mathrm{H}_2 \mathrm{~S}>\mathrm{NH}_3>\mathrm{HF}>\mathrm{CH}_4$

The correct answer is:
$\mathrm{HF}>\mathrm{NH}_3>\mathrm{H}_2 \mathrm{~S}>\mathrm{CH}_4$

Solution:
Fluorine is the most electronegative element. $\mathrm{H}-\mathrm{F}$ bond has maximum bond polarity. Order of dipole moment is

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2023 (Manipur)
The CORRECT order of duration of various phases of the cardiac cycle is

Options:
A Atrial systole < Ventricular systole < Joint diastole < Ventricular diastole < Atrial diastole
B Atrial systole < Ventricular diastole < Joint diastole < Ventricular systole < Atrial diastole
C Atrial diastole < Ventricular diastole < Joint diastole < Ventricular systole < Atrial systole
D Joint diastole < Ventricular diastole < Atrial diastole

The correct answer is:
Atrial systole < Ventricular systole < Joint diastole < Ventricular diastole < Atrial diastole

Solution:
The action potential generated by the SA node stimulates both the atria to contract, which is known as the atrial systole. This action potential is then transmitted to the ventricles by the AV node and the Bundle of His, which makes the ventricular musculature to contract. Both the atria and ventricles then are in a state of relaxation called the joint diastole which is then followed by the ventricular relaxation, when the blood from the ventricles flow into the pulmonary artery and aorta. The aortic diastole takes place at the end of the cycle wherein the blood from the atria are pushed into the ventricles due to the opening of the tricuspid and bicuspid valves.
A rhythmic contraction and relaxation of heart during one heartbeat is called the cardiac cycle. The time required for one cardiac cycle is 0.8sec.

Topic: Biology, Body Fluids and Circulation, NEET
The correct order of energies of molecular orbitals of N2 molecule, is:

Options:
A (σ1s)<(σ*1s)< (σ2s)< (σ2s*)<(π2px)=(π2py)<(π*2px)=(π*2py)<(σ2pz)<(σ*2pz)
B (σ1s)<(σ*1s)< (σ2s)< (σ2s*)<(π2px)=(π2py)<σ2pz<(π*2px)=(π*2py)<(σ*2pz)
C (σ1s)<(σ*1s)< (σ2s)< (σ2s*)<σ2pz<(π2px)=(π2py)<(π*2px)=(π*2py)<(σ*2pz)
D (σ1s)<(σ*1s)< (σ2s)< (σ2s*)<σ2pz<(σ*2pz)<(π2px)=(π2py)<(π*2px)=(π*2py)

The correct answer is:
(σ1s)<(σ*1s)< (σ2s)< (σ2s*)<(π2px)=(π2py)<σ2pz<(π*2px)=(π*2py)<(σ*2pz)

Solution:
The molecular orbitals filled with electrons according Afbaue, Pauli's exclusion and Hund's rules.

(σ1s)<(σ*1s)< (σ2s)< (σ2s*)<(π2px)=(π2py)<σ2pz<(π*2px)=(π*2py)<(σ*2pz)

The important characteristic feature of this order is that the energy of σ2pz molecular orbital is higher than that of π2px and π2py molecular orbitals, it is due to 2s-2p mixing is allowed in case of nitrogen molecule.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2023 (All India)
The correct order of first ionization enthalpy for the given four element is :

Options:
A C $ < $ N $ < $ F $ < \mathrm{O}$
B $\mathrm{C} < \mathrm{N} < \mathrm{O} < \mathrm{F}$
C $\mathrm{C} < \mathrm{O} < \mathrm{N} < \mathrm{F}$
D $\mathrm{C} < $ F $ < \mathrm{N} < \mathrm{O}$

The correct answer is:
$\mathrm{C} < \mathrm{O} < \mathrm{N} < \mathrm{F}$

Solution:
$\begin{array}{|c|c|c|c|c|}\hline \text {Explanation: } & \mathrm{C} & \mathrm{N} & \mathrm{O} & \mathrm{F} \\ \hline \text{(Half-filled) } & 2 s^2 2 p^2 & 2 s^2 2 p^3 & 2 s^2 2 p^4 & 2 s^2 2 p^5 \\ \hline \end{array}$
Across the period $Z_{\text {eff }}$ increases but nitrogen has half-filled orbital which makes it more stable.
$\therefore$ Order of $\mathrm{IE}_1=\mathrm{C} < \mathrm{O} < \mathrm{N} < \mathrm{F}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET, NEET 2022 (Phase 2)
The correct order of fossils from lowest cranial capacity to highest cranial capacity is

Options:
A Java man → Homo habilis → Neanderthal man
B Homo habilis → Java man → Neanderthal man
C Neanderthal man → Homo habilis → Java man
D Homo habilis → Neanderthal man → Java man

The correct answer is:
Homo habilis → Java man → Neanderthal man

Solution:
The first hominid was called Homo habilisand his cranial capacity was between 650-800 cc. Homo erectus whose fossils were discovered in Java had a large brain with a cranial capacity of around 900 cc. The Neanderthal man had a cranial capacity of 1400 cc and lived near east and central Asia.

Topic: Biology, Evolution, NEET
The correct order of increasing basic nature for the bases NH3, CH3NH2 and CH32NH in aqueous solutions is:

Options:
A CH3 NH2<NH3<CH32NH
B NH3<CH3NH2<CH32NH
C CH3NH2<CH32NH<NH3
D CH32NH<NH3<CH3NH2

The correct answer is:
NH3<CH3NH2<CH32NH

Solution:
If the alkyl group is bigger than CH3 group, there is steric hindrance to H-bonding and hence + I effect predomtextnates over H-bonding. 2>3>1 . There is an important role of the inductive effect, solvation effect and steric hindrance of alkyl group, which decides the basic strength of alkyl amine in the aqueous state.
CH32 NH>CH3NH2>NH3

Topic: Chemistry, General Organic Chemistry, NEET
The correct order of increasing bond angle in the following species is

Options:
A ClO2-<Cl2O<ClO2
B Cl2O<ClO2<ClO2-
C ClO2<Cl2O<ClO2-
D Cl2O<ClO2-<ClO2

The correct answer is:
Cl2O<ClO2-<ClO2

Solution:
Here the correct order of bond angle is Cl2O110.9,ClO2-111,ClO2117.6

(Cl is less electronegative than O)

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The correct order of increasing bond angles in the following species is

Options:
A $\mathrm{Cl}_2 \mathrm{O} < \mathrm{ClO}_2 < \mathrm{ClO}_2^{-}$
B $\mathrm{ClO}_2 < \mathrm{Cl}_2 \mathrm{O} < \mathrm{ClO}_2^{-}$
C $\mathrm{Cl}_2 \mathrm{O} < \mathrm{ClO}_2^{-} < \mathrm{ClO}_2$
D $\mathrm{ClO}_2^{-} < \mathrm{Cl}_2 \mathrm{O} < \mathrm{ClO}_2$

The correct answer is:
$\mathrm{ClO}_2^{-} < \mathrm{Cl}_2 \mathrm{O} < \mathrm{ClO}_2$

Solution:
Key Idea As the number of lone pairs of electrons increases, bond angle decreases due to repulsion between lp-lp. Moreover, as the electronegativity of central atom decreases, bond angle decreases.
Hence, the order of bond angle is
( $\mathrm{Cl}$ is less electronegative as compared to $\mathrm{O}$.)

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2010 (Screening)
The correct order of increasing bond angles in the following triatomic species is

Options:
A $\mathrm{NO}_2^{-} < \mathrm{NO}_2^{+} < \mathrm{NO}_2$
B $\mathrm{NO}_2^{-} < \mathrm{NO}_2 < \mathrm{NO}_2^{+}$
C $\mathrm{NO}_2^{+} < \mathrm{NO}_2 < \mathrm{NO}_2^{-}$
D $\mathrm{NO}_2^{+} < \mathrm{NO}_2^{-} < \mathrm{NO}_2$

The correct answer is:
$\mathrm{NO}_2^{-} < \mathrm{NO}_2 < \mathrm{NO}_2^{+}$

Solution:
Key Idea : As the number of lone pair of electrons increases, bond angle decreases.
$\mathrm{NO}_2^{+}$ion is isoelectronic with $\mathrm{CO}_2$ molecule. It is a linear ion and its central atom $\left(\mathrm{N}^{+}\right)$ undergoes $s p$-hybridisation, hence bond angle is $180^{\circ}$.
In $\mathrm{NO}_2^{-}$ion, $\mathrm{N}$-atom undergoes $s p^2$ hybridisation. The angle between hybrid orbital should be $120^{\circ}$ but one lone pair of electrons is lying on $\mathrm{N}$-atom, hence bond angle decreases to $115^{\circ}$.
In $\mathrm{NO}_2$ molecule, $\mathrm{N}$-atom has one unpaired electron in $s p^2$-hybrid orbital. The bond angle should be $120^{\circ}$ but actually it is $132^{\circ}$. It may be due to one unpaired electron in $s p^2$-hybrid orbital.
Therefore, the increasing order of bond angles is :
$$
\underset{\left(115^{\circ}\right)}{\mathrm{NO}_2^{-}} < \underset{\left(132^{\circ}\right)}{\mathrm{NO}_2} < \underset{\left(180^{\circ}\right)}{\mathrm{NO}_{+}^{+}}
$$

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2008 (Screening)
The correct order of increasing bond length of
$\mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{O}, \mathrm{C}-\mathrm{C}$ and $\mathrm{C}=\mathrm{C}$ is

Options:
A $\mathrm{C}-\mathrm{C} < \mathrm{C}=\mathrm{C} < \mathrm{C}-\mathrm{O} < \mathrm{C}-\mathrm{H}$
B $\mathrm{C}-\mathrm{O} < \mathrm{C}-\mathrm{H} < \mathrm{C}-\mathrm{C} < \mathrm{C}=\mathrm{C}$
C $\mathrm{C}-\mathrm{H} < \mathrm{C}-\mathrm{O} < \mathrm{C}-\mathrm{C} < \mathrm{C}=\mathrm{C}$
D $\mathrm{C}-\mathrm{H} < \mathrm{C}=\mathrm{C} < \mathrm{C}-\mathrm{O} < \mathrm{C}-\mathrm{C}$

The correct answer is:
$\mathrm{C}-\mathrm{H} < \mathrm{C}=\mathrm{C} < \mathrm{C}-\mathrm{O} < \mathrm{C}-\mathrm{C}$

Solution:
$$
\begin{aligned}
& \mathrm{C}-\mathrm{H}: 0.109 \mathrm{~nm} \\
& \mathrm{C}=\mathrm{C}: 0.134 \mathrm{~nm} \\
& \mathrm{C}-\mathrm{O}: 0.143 \mathrm{~nm} \\
& \mathrm{C}-\mathrm{C}: 0.154 \mathrm{~nm}
\end{aligned}
$$
$\therefore$ Bond length order is
$$
\mathrm{C}-\mathrm{H} < \mathrm{C}=\mathrm{C} < \mathrm{C}-\mathrm{O} < \mathrm{C}-\mathrm{C}
$$

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2011 (Screening)
The correct order of increasing bond length of C-H,C-O,C-C and C=C is

Options:
A C-H<C=C<C-O<C-C
B C-C<C=C<C-O<C-H
C C-O<C-H<C-C<C=C
D C-H<C-O<C-C<C=C

The correct answer is:
C-H<C=C<C-O<C-C

Solution:
TheC−Hbond is the shortest due to the small size of both atoms & large electronegativity difference.
The double bonds are shorter than single bonds.
HenceC=C<C−OandC−C
Due to electronegativity difference betweenCandO
C−O<C−C
Increasing order of bond length is
C-H107pm<C=C134pm<C-O141pm<C-C154pm

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET