NEET Chemistry - MCQ Practice Questions

Understanding:

We find the enthalpy of hydration using the Born–Haber cycle relationship for dissolution.

Formula:

The enthalpy of solution is related to lattice and hydration enthalpies by:

Step 1: Solve for $\Delta H_{\text{hydration}}$

Step 2: Verify the sign

Hydration is always exothermic (water molecules stabilise the ions), so $\Delta H_{\text{hydration}}<0$. The result $-784{\text{ kJ mol}}^{-1}$ is consistent.

Answer:

The enthalpy of hydration of $\text{NaCl}$ is $-784{\text{ kJ mol}}^{-1}$.

Quick Tip:

Hydration enthalpy is always negative. If the magnitude of hydration enthalpy exceeds the lattice enthalpy, the salt dissolves with a negative $\Delta H_{\text{solution}}$ (exothermic dissolution). For $\text{NaCl}$, they are nearly equal, giving a near-zero $\Delta H_{\text{solution}}$.

Understanding:

We need to identify which alkyl bromide undergoes ${\text{S}}_{\text{N}}1$ substitution most readily.

Formula:

The rate of ${\text{S}}_{\text{N}}1$ depends on the stability of the carbocation intermediate:

Carbocation stability order:

Step 1: Classify each substrate

Step 2: Identify the most reactive

The $3^{\circ }$ carbocation formed from $({\text{CH}}_3{)}_3\text{CBr}$ is stabilised by three alkyl groups donating electron density via hyperconjugation and induction, making it the most stable and easiest to form.

Answer:

The compound $({\text{CH}}_3{)}_3\text{CBr}$ undergoes ${\text{S}}_{\text{N}}1$ reaction most readily.

Quick Tip:

Always look for the most substituted carbon bearing the leaving group when predicting ${\text{S}}_{\text{N}}1$ reactivity.

Understanding:

We need to identify the final product of aldol condensation of ${\text{CH}}_3\text{CHO}$ followed by dehydration.

Formula:

Aldol condensation:

Step 1: Aldol addition

Two molecules of ${\text{CH}}_3\text{CHO}$ react in the presence of dilute base. The $\alpha$-carbon of one molecule attacks the carbonyl carbon of another:

This gives $3$-hydroxybutanal (aldol product).

Step 2: Dehydration

On heating, the $\beta$-hydroxy aldehyde loses water to form an $\alpha ,\beta$-unsaturated aldehyde:

The product is but-2-enal (crotonaldehyde).

Answer:

The final product of aldol condensation followed by dehydration of acetaldehyde is but-2-enal.

Quick Tip:

The aldol addition product is a $\beta$-hydroxy carbonyl compound; dehydration always removes ${\text{H}}_2\text{O}$ from the $\alpha$ and $\beta$ positions to give the conjugated $\alpha ,\beta$-unsaturated product.

Understanding:

We need to count all $\sigma$ and $\pi$ bonds in ${\text{CH}}_3{\text{CH}}_2\text{C}\equiv \text{CH}$.

Formula:

For any bond:

Step 1: Draw the structure

The connectivity of ${\text{CH}}_3-{\text{CH}}_2-\text{C}\equiv \text{CH}$ gives:

Step 2: Add up

Answer:

But-1-yne contains $9$ sigma bonds and $2$ pi bonds.

Quick Tip:

A quick formula: for ${\text{C}}_n{\text{H}}_m$ with one triple bond, $\sigma =(n-1)+m$ carbon-carbon and carbon-hydrogen $\sigma$ bonds; count $\pi$ bonds separately from the degree of unsaturation.

Understanding:

We need to identify the reagent that converts ${\text{RCONH}}_2$ to ${\text{RNH}}_2$, specifically with loss of one carbon.

Formula:

Hofmann bromamide degradation:

Step 1: Analyse each option

${\text{LiAlH}}_4$ reduces ${\text{RCONH}}_2$ to ${\text{RCH}}_2{\text{NH}}_2$ — the carbon count is retained, giving a primary amine with the same number of carbons.

Step 2: Identify the degradation reaction

In the Hofmann bromamide reaction, ${\text{Br}}_2/\text{NaOH}$ acts on the amide nitrogen, forming an isocyanate intermediate which hydrolyses. The carbonyl carbon is lost as ${\text{CO}}_2$:

The product ${\text{RNH}}_2$ has one carbon fewer than the starting amide.

Step 3: Eliminate other options

${\text{HNO}}_2$ reacts with primary amines (not amides to give a degraded amine), and $\text{Sn}/\text{HCl}$ reduces nitro groups. Neither removes a carbon.

Answer:

The Hofmann bromamide reaction using ${\text{Br}}_2/\text{NaOH}$ converts a primary amide to a primary amine with one fewer carbon.

Quick Tip:

Whenever you see a reaction that decreases the carbon chain by one while converting an amide to an amine, it is always the Hofmann degradation.

Understanding:

We need to assign the correct IUPAC name to ${\text{CH}}_3{\text{CH(OH)CH(CH}}_3){\text{CH}}_2\text{Cl}$.

Formula:

IUPAC rules:

1. Identify the longest continuous carbon chain containing the principal characteristic group ($-\text{OH}$ has higher seniority than $-\text{Cl}$).

2. Number from the end that gives the $-\text{OH}$ group the lowest locant.

Step 1: Identify the parent chain

The longest continuous chain is 4 carbons: $\text{C1}-\text{C2}-\text{C3}-\text{C4}$.

Step 2: Number to give $-\text{OH}$ the lowest locant

Numbering from the ${\text{CH}}_3$ end gives $\text{OH}$ at position $2$ and $\text{Cl}$ at position $4$.

Numbering from the ${\text{CH}}_2\text{Cl}$ end gives $\text{OH}$ at position $3$ — higher, so not preferred.

Step 3: Name substituents

Full name (substituents in alphabetical order): $4$-chloro-$3$-methylbutan-$2$-ol.

Answer:

The IUPAC name of the given compound is $4$-chloro-$3$-methylbutan-$2$-ol.

Quick Tip:

Always number the chain to give the principal characteristic group ($-\text{OH}$ over $-\text{Cl}$) the lowest possible locant, not the halogen.

Understanding:

We need to apply Markovnikov's rule to the addition of $\text{HBr}$ to ${\text{CH}}_3\text{CH}={\text{CH}}_2$ (propene).

Formula:

Markovnikov's rule (mechanistic statement):

The nucleophile (${\text{Br}}^{-}$) then attacks the carbocation.

Step 1: Protonation step

${\text{H}}^{+}$ from $\text{HBr}$ adds to $\text{C1}$ (which has more H atoms):

or to $\text{C2}$:

The $2^{\circ }$ carbocation is more stable, so ${\text{H}}^{+}$ goes to $\text{C1}$.

Step 2: Attack by ${\text{Br}}^{-}$

${\text{Br}}^{-}$ attacks the $2^{\circ }$ carbocation at $\text{C2}$:

Major product: $2$-bromopropane.

Answer:

According to Markovnikov's rule, bromine adds to the secondary carbon $\text{C2}$, giving $2$-bromopropane as the major product.

Quick Tip:

Markovnikov's rule in one line: the electrophile (${\text{H}}^{+}$) goes to the carbon that already has more hydrogens, because this generates the more stable (more substituted) carbocation.

Understanding:

We need to arrange phenol, ethanol, and water in order of decreasing ${\text{pK}}_{\text{a}}$ (i.e., increasing acidity).

Formula:

Step 1: Recall approximate ${\text{pK}}_{\text{a}}$ values

Step 2: Explain the trend

Phenol is most acidic because its conjugate base (phenoxide ion) is stabilised by resonance with the aromatic ring — the negative charge is delocalised over the ring. Water is slightly more acidic than ethanol because the alkyl group in ethanol donates electron density (induction), destabilising the ethoxide ion. Ethanol's conjugate base (ethoxide) is less stable than hydroxide, so ethanol is the weakest acid.

Step 3: Arrange

Answer:

The correct order is ethanol $>$ water $>$ phenol in terms of ${\text{pK}}_{\text{a}}$.

Quick Tip:

Resonance stabilisation of the conjugate base is a much stronger effect than induction; phenoxide's resonance makes phenol thousands of times more acidic than alcohols.

Understanding:

We need to identify the product of catalytic hydrogenation of a nitrile ($\text{R-CN}$) with ${\text{H}}_2$ over nickel.

Formula:

Step 1: Analyse the reaction

A nitrile contains a $\text{C}\equiv \text{N}$ triple bond (one $\sigma$ + two $\pi$ bonds). Catalytic hydrogenation with $2$ moles of ${\text{H}}_2$ reduces the $\text{C}\equiv \text{N}$ completely:

Step 2: Eliminate other options

Answer:

The catalytic hydrogenation of a nitrile produces a primary amine.

Quick Tip:

Reduction of a nitrile always gives a primary amine with the same number of carbons as the nitrile — useful for synthesising amines from halides via $\text{R-X}\to \text{R-CN}\to {\text{R-CH}}_2{\text{NH}}_2$.

Understanding:

We need to identify the pair that are enantiomers (non-superimposable mirror images of each other).

Formula:

Enantiomers are stereoisomers that are non-superimposable mirror images, differing only in the configuration at every chiral centre:

Step 1: Analyse option A

$(R)\text{-}2\text{-bromobutane}$ and $(S)\text{-}2\text{-bromobutane}$ have the same molecular formula ${\text{C}}_4{\text{H}}_9\text{Br}$, the same connectivity, and opposite configurations ($R$ vs $S$) at the single chiral centre (C2). They are non-superimposable mirror images — the definition of enantiomers.

Step 2: Analyse other options

Answer:

$(R)\text{-}2\text{-bromobutane}$ and $(S)\text{-}2\text{-bromobutane}$ are enantiomers.

Quick Tip:

Enantiomers must have the same molecular formula and connectivity but opposite $(R/S)$ configuration at every chiral centre. If any substituent differs, they are not enantiomers.

Understanding:

We need to explain why aniline is less basic than cyclohexylamine (an aliphatic amine).

Formula:

Step 1: Lone pair in cyclohexylamine

In cyclohexylamine, nitrogen is ${\text{sp}}^3$ hybridised. Its lone pair is in a pure ${\text{sp}}^3$ orbital and is completely available to accept a proton.

Step 2: Lone pair in aniline

In aniline, the nitrogen lone pair overlaps with the $\pi$ system of the benzene ring (resonance):

This delocalisation reduces the electron density on nitrogen, making it much less available for protonation. The conjugate acid (anilinium ion) loses this resonance stabilisation, so aniline does not gain as much from protonation.

Step 3: Eliminate option D

Although nitrogen in aniline has partial ${\text{sp}}^2$ character and ${\text{sp}}^2$ orbitals are more electronegative, the dominant reason is resonance delocalisation, not orbital electronegativity. Option D is partially true but not the best or primary explanation taught at this level.

Answer:

Aniline is less basic because resonance delocalisation of the nitrogen lone pair into the benzene ring reduces its availability for accepting a proton.

Quick Tip:

Any time $-{\text{NH}}_2$ is directly attached to an aromatic ring, expect reduced basicity due to resonance. The same effect makes aromatic amines much weaker bases than aliphatic amines.

Understanding:

We need to identify why the $\text{H}-\text{O}-\text{H}$ bond angle ($104.5°$) is less than the ideal tetrahedral angle of $109.5°$.

Formula:

VSEPR theory ranks repulsions as:

where lp = lone pair and bp = bond pair.

Step 1: Determine the electron geometry of water.

Oxygen in ${\text{H}}_2\text{O}$ has 2 bond pairs (to the two H atoms) and 2 lone pairs, giving a tetrahedral electron geometry (4 electron domains) but a bent molecular shape.

Step 2: Apply VSEPR repulsion hierarchy.

Each lone pair occupies more space than a bond pair because lone pairs are held by only one nucleus and spread out more. The two lone pairs exert stronger repulsion on the two $\text{O}-\text{H}$ bond pairs than bond pairs would exert on each other. This lp–bp repulsion pushes the bond pairs closer together, compressing the $\text{H}-\text{O}-\text{H}$ angle below $109.5°$ to approximately $104.5°$.

Step 3: Eliminate the other options.

Option A incorrectly identifies the dominant repulsion type — lp–lp repulsion (between the two lone pairs on O) is the largest, but it is the lp–bp repulsion acting on the bond pairs that compresses the $\text{H}-\text{O}-\text{H}$ angle. Option C has the wrong electron count. Option D misidentifies the hybridisation of oxygen.

Answer:

The bond angle is compressed because the two lone pairs on oxygen exert stronger lone pair–bond pair repulsion on the $\text{O}-\text{H}$ bonds than bond pair–bond pair repulsion would, pushing the bond pairs together.

Quick Tip:

Always compare ${\text{H}}_2\text{O}$ ($104.5°$), ${\text{NH}}_3$ ($107°$), and ${\text{CH}}_4$ ($109.5°$) — each additional lone pair reduces the bond angle by roughly $2-3°$.

Understanding:

We need to compare bond lengths in the oxygen species ${\text{O}}_2^{2-}$, ${\text{O}}_2^{-}$, ${\text{O}}_2$, and ${\text{O}}_2^{+}$ using Molecular Orbital Theory.

Formula:

Bond order is related to bond length and bond strength as:

Higher bond order $\Rightarrow$ shorter and stronger bond.

Step 1: Determine the bond orders using MO theory.

For ${\text{O}}_2$ (16 electrons), the MO configuration gives:

Step 2: Adjust for each ion.

Step 3: Rank bond orders and bond lengths.

Since bond length decreases as bond order increases:

Answer:

${\text{O}}_2^{+}$ has the highest bond order of $2.5$ and therefore the shortest bond length among the four species.

Quick Tip:

In MO theory, cations of ${\text{O}}_2$ are formed by removing antibonding electrons, which increases bond order and decreases bond length.

Understanding:

We need to determine the hybridisation and molecular geometry of ${\text{XeF}}_4$.

Formula:

The number of hybrid orbitals equals the total number of electron domains (bond pairs + lone pairs) around the central atom:

Step 1: Count electron domains on Xe in ${\text{XeF}}_4$.

Xe has 8 valence electrons. Each F contributes one bond, using one electron from Xe. Four F atoms use 4 of Xe's electrons in bonding. Remaining electrons on Xe: $8-4=4$, forming 2 lone pairs.

Step 2: Identify hybridisation.

6 electron domains require 6 hybrid orbitals:

The electron geometry is octahedral.

Step 3: Determine molecular shape.

With 4 bond pairs and 2 lone pairs in an octahedral electron geometry, the 2 lone pairs occupy axial positions (opposite each other) to minimise lp–lp repulsion. The 4 F atoms are in the equatorial plane, giving a square planar molecular shape.

Step 4: Compare with other options.

$s{p}^3$ gives only 4 domains (tetrahedral). $s{p}^{3}d$ gives 5 domains (trigonal bipyramidal parent). Only $s{p}^3{d}^2$ correctly accounts for 6 domains, and with 2 lone pairs placed axially, the molecular shape is square planar, not octahedral.

Answer:

${\text{XeF}}_4$ has $s{p}^3{d}^2$ hybridisation and a square planar molecular geometry.

Quick Tip:

For noble gas compounds, always count lone pairs carefully. ${\text{XeF}}_2$ ($s{p}^{3}d$, linear), ${\text{XeF}}_4$ ($s{p}^3{d}^2$, square planar), and ${\text{XeF}}_6$ ($s{p}^3{d}^3$, distorted octahedral) form a useful comparison set.

Understanding:

We need to identify the most stable Lewis structure of ${\text{CO}}_2$ by evaluating formal charges on each atom.

Formula:

Formal charge is calculated as:

The most stable structure has formal charges closest to zero on all atoms.

Step 1: Evaluate the double-bond structure ($\text{O}=\text{C}=\text{O}$).

For carbon (4 valence electrons, 0 lone pairs, 8 bonding electrons):

For each oxygen (6 valence electrons, 4 lone pair electrons, 4 bonding electrons):

All formal charges are zero — this is the most stable structure.

Step 2: Evaluate the single-bond structure ($\text{O}-\text{C}-\text{O}$).

For carbon: ${\text{FC}}_{\text{C}}=4-0-\frac{4}{2}=+2$

For each oxygen: ${\text{FC}}_{\text{O}}=6-6-\frac{2}{2}=-1$

Large formal charges indicate high instability.

Step 3: Evaluate structures with one triple bond and one single bond.

These give non-zero formal charges on C and O (e.g., $+1$ on C and $-1$ on an O), which are less favourable than the all-zero structure.

Answer:

The structure with two double bonds ($\text{O}=\text{C}=\text{O}$) is the most stable because all formal charges are zero.

Quick Tip:

Always prefer Lewis structures where formal charges are zero or where any negative formal charge sits on the more electronegative atom.

Understanding:

We need to count the number of $\sigma$ and $\pi$ bonds in allene, ${\text{H}}_2\text{C}=\text{C}={\text{CH}}_2$.

Formula:

For any bond:

Step 1: Identify all bonds in allene ${\text{H}}_2\text{C}=\text{C}={\text{CH}}_2$.

The molecule has:

Step 2: Count $\sigma$ bonds.

Each $\text{C}-\text{H}$ bond contributes 1 $\sigma$: $4\times 1=4\sigma$ bonds.

Each $\text{C}=\text{C}$ bond contributes 1 $\sigma$: $2\times 1=2\sigma$ bonds.

Step 3: Count $\pi$ bonds.

Each $\text{C}=\text{C}$ bond contributes 1 $\pi$: $2\times 1=2\pi$ bonds.

Answer:

Allene has 6 sigma bonds and 2 pi bonds.

Quick Tip:

In allene, the two $\pi$ bonds are perpendicular to each other because the central carbon is $sp$ hybridised, which makes the two terminal ${\text{CH}}_2$ planes mutually perpendicular — a key stereochemical feature.

Understanding:

We need to arrange ${\text{BF}}_3$, ${\text{NF}}_3$, and ${\text{NH}}_3$ in increasing order of dipole moment.

Formula:

Dipole moment depends on molecular geometry and the vector sum of individual bond dipoles:

For a symmetric molecule, individual bond dipoles cancel; for asymmetric molecules, they add partially.

Step 1: Analyse ${\text{BF}}_3$.

${\text{BF}}_3$ is trigonal planar ($s{p}^2$ hybridised, no lone pair on B). The three $\text{B}-\text{F}$ bond dipoles are symmetrically arranged at $120°$ and cancel completely.

Step 2: Analyse ${\text{NF}}_3$.

${\text{NF}}_3$ is pyramidal ($s{p}^3$, one lone pair on N). F is more electronegative than N, so each $\text{N}-\text{F}$ bond dipole points away from N (toward F). The lone pair dipole on N points upward (away from the base). The lone pair contribution partially opposes the bond dipoles, resulting in a small net dipole moment.

Step 3: Analyse ${\text{NH}}_3$.

${\text{NH}}_3$ is also pyramidal ($s{p}^3$, one lone pair on N). H is less electronegative than N, so each $\text{N}-\text{H}$ bond dipole points toward N. The lone pair dipole on N also points in the same direction (away from N toward the three H side), so the lone pair and bond dipoles reinforce each other, giving a larger net dipole moment.

Step 4: Arrange in increasing order.

Answer:

The correct increasing order of dipole moment is ${\text{BF}}_3<{\text{NF}}_3<{\text{NH}}_3$.

Quick Tip:

The key contrast is ${\text{NF}}_3$ vs ${\text{NH}}_3$: in ${\text{NF}}_3$ the bond dipoles and lone pair dipole partially cancel; in ${\text{NH}}_3$ they add up. This makes ${\text{NH}}_3$ have a much larger dipole despite both being pyramidal.

Understanding:

We need to find the bond order of $\text{NO}$ using Molecular Orbital (MO) Theory.

Formula:

Step 1: Fill the MO diagram for $\text{NO}$ (15 electrons).

The MO filling order for a heteronuclear diatomic like $\text{NO}$ (using the scheme valid for $Z\le 8$):

Step 2: Count bonding and antibonding electrons.

Step 3: Calculate bond order.

Answer:

The bond order of $\text{NO}$ is $2.5$, corresponding to a bond between a double and a triple bond, and $\text{NO}$ is paramagnetic (one unpaired electron in ${\pi }^{\ast }$).

Quick Tip:

When $\text{NO}$ loses an electron to form ${\text{NO}}^{+}$, the antibonding ${\pi }^{\ast }$ electron is removed, raising the bond order to $3$ (isoelectronic with ${\text{N}}_2$). This makes ${\text{NO}}^{+}$ more stable and diamagnetic.

Understanding:

We need to explain why ${\text{PCl}}_5$ is a stable molecule but ${\text{NCl}}_5$ cannot be formed.

Formula:

For a central atom to form 5 bonds (expanded octet), it must have empty low-energy $d$ orbitals available in its valence shell to accommodate more than 8 electrons:

Step 1: Examine phosphorus.

Phosphorus belongs to the third period. Its valence shell is $n=3$, which contains $3s$, $3p$, and $3d$ orbitals. The $3d$ orbitals are vacant and energetically accessible, so P can use them to form additional bonds beyond the octet. This allows ${\text{PCl}}_5$ to form with P in an $s{p}^{3}d$ hybridised state (5 electron domains).

Step 2: Examine nitrogen.

Nitrogen belongs to the second period. Its valence shell is $n=2$, which contains only $2s$ and $2p$ orbitals. There are no $d$ orbitals in the $n=2$ shell. Nitrogen is strictly limited to an octet (maximum 4 bonds), making ${\text{NCl}}_5$ impossible to form.

Step 3: Eliminate incorrect options.

Electronegativity does not prevent bonding with Cl. Atomic radius of N is smaller than P, not larger — and a smaller radius makes it harder, not easier, to accommodate more atoms. Nitrogen forms predominantly covalent, not ionic, bonds with Cl.

Answer:

${\text{PCl}}_5$ exists because phosphorus has vacant $3d$ orbitals available for expanded octet formation, while nitrogen in the second period has no $d$ orbitals in its valence shell and cannot exceed an octet.

Quick Tip:

This is a recurring NEET question type. Second-period elements (C, N, O, F) can never exhibit expanded octets because $n=2$ has no $d$ subshell. Third-period and beyond can.

Understanding:

We need to compare $\text{C}-\text{O}$ bond lengths in $\text{CO}$, ${\text{CO}}_2$, and ${\text{CO}}_3^{2-}$, given that higher bond order gives shorter bond length.

Formula:

Step 1: Determine bond order in $\text{CO}$.

$\text{CO}$ has a triple bond ($\text{C}\equiv \text{O}$).

Step 2: Determine bond order in ${\text{CO}}_2$.

${\text{CO}}_2$ has the structure $\text{O}=\text{C}=\text{O}$ with two double bonds.

Step 3: Determine bond order in ${\text{CO}}_3^{2-}$.

${\text{CO}}_3^{2-}$ has resonance among three equivalent structures, delocalising the bonds. The bond order is:

Step 4: Rank bond lengths.

Since bond length decreases with increasing bond order:

Answer:

The increasing order of $\text{C}-\text{O}$ bond length is $\text{CO}<{\text{CO}}_2<{\text{CO}}_3^{2-}$.

Quick Tip:

Approximate $\text{C}-\text{O}$ bond lengths: $\text{CO}\approx 113\text{pm}$, ${\text{CO}}_2\approx 116\text{pm}$, ${\text{CO}}_3^{2-}\approx 129\text{pm}$. Remembering the trend (triple $<$ double $<$ fractional) is enough for NEET.