NEET Chemistry - MCQ Practice Questions

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.

Understanding:

We need to identify the pair where both species have the same molecular geometry.

Formula:

Molecular geometry is determined by VSEPR theory based on the number of bond pairs and lone pairs:

Step 1: Analyse Option A — ${\text{NH}}_3$ and ${\text{BF}}_3$.

${\text{NH}}_3$: 3 bond pairs + 1 lone pair $\Rightarrow$ trigonal pyramidal.

${\text{BF}}_3$: 3 bond pairs + 0 lone pairs $\Rightarrow$ trigonal planar.

Different geometries.

Step 2: Analyse Option B — ${\text{H}}_2\text{O}$ and ${\text{CO}}_2$.

${\text{H}}_2\text{O}$: 2 bond pairs + 2 lone pairs $\Rightarrow$ bent (V-shaped).

${\text{CO}}_2$: 2 double bonds + 0 lone pairs $\Rightarrow$ linear.

Different geometries.

Step 3: Analyse Option C — ${\text{PCl}}_3$ and ${\text{SO}}_3^{2-}$.

${\text{PCl}}_3$: P has 3 bond pairs + 1 lone pair $\Rightarrow$ trigonal pyramidal.

${\text{SO}}_3^{2-}$: S has 3 bond pairs + 1 lone pair $\Rightarrow$ trigonal pyramidal.

Both are trigonal pyramidal.

Step 4: Analyse Option D — ${\text{SF}}_4$ and ${\text{XeF}}_4$.

${\text{SF}}_4$: 4 bond pairs + 1 lone pair ($s{p}^{3}d$) $\Rightarrow$ see-saw (seesaw) shape.

${\text{XeF}}_4$: 4 bond pairs + 2 lone pairs ($s{p}^3{d}^2$) $\Rightarrow$ square planar.

Different geometries.

Answer:

${\text{PCl}}_3$ and ${\text{SO}}_3^{2-}$ both have 3 bond pairs and 1 lone pair on the central atom, giving both a trigonal pyramidal geometry.

Quick Tip:

When checking if two molecules share geometry, always count both bond pairs AND lone pairs. Two molecules with the same bond-pair count but different lone-pair counts will have different shapes even if they have the same number of atoms.