NEET Chemistry - MCQ Practice Questions

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.

Understanding:

We need to find the equilibrium constant $K$ from the standard EMF.

Formula:

The relation between $E_{\text{cell}}^{\circ }$ and $K$ is:

Step 1: Rearrange for $\log K$

Step 2: Find $K$

Using $\frac{RT}{F}=0.0257\text{ V}$:

Answer:

The equilibrium constant is approximately $1.57\times 10^{37}$.

Quick Tip:

When using the $\ln$ form with $\frac{RT}{F}=0.0257\text{ V}$, note that $e^{85.6}=10^{85.6/2.303}=10^{37.16}$, giving $\approx 1.57\times 10^{37}$.

Understanding:

We need to calculate molar conductivity from conductivity and concentration.

Formula:

where $c$ must be in ${\text{mol cm}}^{-3}$ (i.e., ${\text{mol mL}}^{-1}\times 10^{-3}$, or convert using $1\text{ L}=1000{\text{ cm}}^3$).

Step 1: Convert concentration to ${\text{mol cm}}^{-3}$

Step 2: Calculate ${\Lambda }_m$

Answer:

The molar conductivity of the $\text{KCl}$ solution is $141{\text{ S cm}}^2{\text{ mol}}^{-1}$.

Quick Tip:

A very common error is forgetting to convert $c$ from ${\text{mol L}}^{-1}$ to ${\text{mol cm}}^{-3}$. Always divide by $1000$ when $\kappa$ is in ${\text{S cm}}^{-1}$.

Understanding:

We must calculate the standard cell EMF and assess spontaneity.

Formula:

A positive $E_{\text{cell}}^{\circ }$ indicates a spontaneous reaction ($\Delta G^{\circ }<0$).

Step 1: Calculate $E_{\text{cell}}^{\circ }$

Step 2: Assess spontaneity

Since $\Delta G^{\circ }<0$, the reaction is spontaneous.

Answer:

The standard EMF is $+1.10\text{ V}$ and the cell reaction is spontaneous.

Quick Tip:

Always use $E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }$ (both as reduction potentials). Never reverse the sign of the cathode.

Understanding:

We need to identify the anode reaction when aqueous ${\text{CuSO}}_4$ is electrolysed with copper electrodes.

Step 1: Identify the anode process

At the anode, oxidation occurs. When the electrode itself is made of an active (reactive) metal like copper, the electrode dissolves preferentially over the oxidation of water, because the oxidation potential of copper is more favourable than that of water.

Step 2: Write the anode reaction

Step 3: Identify other electrode reactions

At the cathode, ${\text{Cu}}^{2+}$ ions from solution are reduced and deposited:

This is the principle used in electrorefining of copper.

Answer:

At the anode, copper dissolves into the solution as ${\text{Cu}}^{2+}$ ions.

Quick Tip:

With inert electrodes (like Pt or graphite) and aqueous ${\text{CuSO}}_4$, the anode would evolve ${\text{O}}_2$. With active copper electrodes, the metal itself oxidises — this distinction is a classic NEET trap.

Understanding:

We must apply the Nernst equation to find the electrode potential at non-standard concentration.

Formula:

For the reduction half-reaction ${\text{Zn}}^{2+}+2e^{-}\to \text{Zn}$:

Step 1: Substitute values

Step 2: Round the result

Answer:

The electrode potential for ${\text{Zn}}^{2+}/\text{Zn}$ at $[{\text{Zn}}^{2+}]=0.001\text{ M}$ is $-0.849\text{ V}$.

Quick Tip:

For a reduction half-cell, decreasing the ion concentration makes the potential more negative. The Nernst equation for a reduction is $E=E^{\circ }-\frac{0.0592}{n}\log Q$, where $Q=\frac{1}{[{\text{Zn}}^{2+}]}$ for this half-reaction.

Understanding:

We need to find the charge (in Faradays) needed to deposit $27\text{ g}$ of $\text{Al}$.

Formula:

From Faraday's first law:

where $n_e$ is the number of electrons per atom and $F=96500{\text{ C mol}}^{-1}$.

Rearranged:

Step 1: Find moles of $\text{Al}$

Step 2: Determine electrons required

The cathode half-reaction is:

So $1\text{ mol}$ of $\text{Al}$ requires $3\text{ mol}$ of electrons.

Step 3: Convert to Faradays

Answer:

Depositing $27\text{ g}$ ($1\text{ mol}$) of $\text{Al}$ requires $3$ Faradays of charge.

Quick Tip:

Faradays required $=$ moles of substance $\times$ $n$ (electrons per ion). For ${\text{Al}}^{3+}$, $n=3$, so $1\text{ mol}$ always needs $3\text{ F}$.

Understanding:

We need to derive ${\Lambda }_m^{\circ }({\text{NH}}_4\text{OH})$ using Kohlrausch's law and the given strong electrolytes.

Step 1: Write individual ionic contributions

By Kohlrausch's law:

Step 2: Combine to get ${\Lambda }_m^{\circ }({\text{NH}}_4\text{OH})$

The target is:

Subtracting ${\Lambda }_m^{\circ }(\text{NaCl})$ from the sum of the first two:

Answer:

The correct expression is ${\Lambda }_m^{\circ }({\text{NH}}_4\text{OH})={\Lambda }_m^{\circ }({\text{NH}}_4\text{Cl})+{\Lambda }_m^{\circ }(\text{NaOH})-{\Lambda }_m^{\circ }(\text{NaCl})$.

Quick Tip:

The Hess's law analogy works here — add the equations that give the ions you need and cancel the unwanted ions by subtracting an equation that contains them.

Understanding:

We must calculate $\Delta G^{\circ }$ for the Daniell cell reaction.

Formula:

Step 1: Substitute values

Step 2: Confirm sign

Since $E_{\text{cell}}^{\circ }>0$, $\Delta G^{\circ }<0$, confirming a spontaneous reaction.

Answer:

The standard Gibbs energy change for the reaction is $-212.3{\text{ kJ mol}}^{-1}$.

Quick Tip:

The unit check: ${\text{C mol}}^{-1}\times \text{V}=\text{C}\times {\text{J C}}^{-1}\times {\text{mol}}^{-1}={\text{J mol}}^{-1}$. Divide by $1000$ to get ${\text{kJ mol}}^{-1}$.

Understanding:

We must use the cell constant to find the conductivity of $\text{NaOH}$ solution.

Formula:

The cell constant $G^{\ast }=\kappa \times R$, so:

and then:

Step 1: Calculate the cell constant

Step 2: Calculate ${\kappa }_{\text{NaOH}}$

Answer:

The conductivity of $0.01\text{ M}$ $\text{NaOH}$ solution is $7.05\times 10^{-4}{\text{ S cm}}^{-1}$.

Quick Tip:

The cell constant $G^{\ast }$ (in ${\text{cm}}^{-1}$) is a property of the cell geometry alone — it is the same regardless of the solution. Calibrate it with a known solution ($\text{KCl}$), then apply it to the unknown.

Understanding:

We need to identify the compound with the highest lattice energy among the given options.

Formula:

Lattice energy is given by the Born–Landé equation (qualitatively):

where $z^{+}$ and $z^{-}$ are the charges on the cation and anion respectively, and $r^{+}+r^{-}$ is the interionic distance.

Step 1: Identify ionic charges and sizes.

Step 2: Compare $\text{MgO}$ and $\text{CaO}$.

Both have charge product = 4, but ${\text{Mg}}^{2+}$ (ionic radius $\approx 0.72\text{A˚}$) is smaller than ${\text{Ca}}^{2+}$ (ionic radius $\approx 1.00\text{A˚}$). The smaller interionic distance in $\text{MgO}$ gives it a higher lattice energy.

Answer:

$\text{MgO}$ has the highest lattice energy due to the combination of high ionic charges ($+2$ and $-2$) and the small ionic radii of ${\text{Mg}}^{2+}$ and ${\text{O}}^{2-}$.

Quick Tip:

When comparing lattice energies, always check the charge product first — doubly charged ions dominate. Among those with equal charges, the smaller the ions, the greater the lattice energy.

Understanding:

We need to find the oxidation state of sulphur in ${\text{H}}_2{\text{S}}_2{\text{O}}_7$.

Formula:

Using the rule that the sum of all oxidation states in a neutral compound equals zero:

Step 1: Assign known oxidation states.

Step 2: Set up and solve the equation.

Answer:

The oxidation state of sulphur in ${\text{H}}_2{\text{S}}_2{\text{O}}_7$ is $+6$.

Quick Tip:

In all oxoacids of sulphur in its highest oxidation form (like ${\text{H}}_2{\text{SO}}_4$, ${\text{H}}_2{\text{S}}_2{\text{O}}_7$, ${\text{H}}_2{\text{S}}_2{\text{O}}_8$), the oxidation state of S is $+6$. Pyrosulphuric acid is simply ${\text{H}}_2{\text{SO}}_4+{\text{SO}}_3$.

Understanding:

We need to identify the correct reason for fluorine's anomalous behaviour among halogens.

Step 1: Recall key properties of fluorine.

Fluorine is the smallest halogen and belongs to Period 2. Its valence shell is $n=2$, which contains only $2s$ and $2p$ subshells — there are no $d$-orbitals available in the second period.

Step 2: Consequences of no $d$-orbitals.

Step 3: Evaluate other options.

Answer:

Fluorine's anomalous behaviour arises from the absence of $d$-orbitals in its valence shell, which limits its covalency and contributes to its role as the strongest oxidising halogen.

Quick Tip:

The absence of $d$-orbitals is the single most important reason for the anomalous behaviour of all second-period elements (F, O, N) compared to their heavier congeners.

Understanding:

We need to identify the products of the reaction of ${\text{Cl}}_2$ with excess $\text{NaOH}$ at $70°\text{C}$.

Step 1: Recall the temperature-dependent disproportionation of ${\text{Cl}}_2$ with $\text{NaOH}$.

The reaction of ${\text{Cl}}_2$ with $\text{NaOH}$ proceeds differently depending on temperature:

Products: sodium chloride ($\text{NaCl}$) and sodium hypochlorite ($\text{NaOCl}$).

Products: sodium chloride ($\text{NaCl}$) and sodium chlorate (${\text{NaClO}}_3$).

Step 2: Identify what forms at $70°\text{C}$.

At higher temperatures, $\text{NaOCl}$ (formed initially) undergoes further disproportionation to give ${\text{NaClO}}_3$:

So the final products at $70°\text{C}$ are $\text{NaCl}$ and ${\text{NaClO}}_3$.

Answer:

When ${\text{Cl}}_2$ reacts with excess $\text{NaOH}$ at $70°\text{C}$, the products are $\text{NaCl}$ and ${\text{NaClO}}_3$.

Quick Tip:

A simple memory aid: **Cold NaOH → hypochlorite (OCl${}^{-}$); Hot NaOH → chlorate (ClO${}_3^{-}$)**. This pattern holds for all halogens reacting with alkali.

Understanding:

We need to arrange the sulphates of alkaline earth metals ($\text{Be, Mg, Ca, Sr, Ba}$) in increasing order of solubility in water.

Step 1: Recall the trend in solubility of Group 2 sulphates.

As we go down Group 2 (Be → Ba), the cationic size increases. For sulphates, the hydration enthalpy decreases more sharply than the lattice enthalpy as cation size increases. This means the net energy released on dissolving decreases, reducing solubility.

Step 2: Establish the trend.

In terms of increasing solubility (lowest to highest):

${\text{BaSO}}_4$ is practically insoluble (used as a barium meal in X-rays), while ${\text{BeSO}}_4$ and ${\text{MgSO}}_4$ (Epsom salt) are readily soluble.

Answer:

The correct increasing order of solubility of alkaline earth metal sulphates is:

Quick Tip:

For Group 2 salts: sulphates and carbonates become less soluble down the group (hydration enthalpy falls faster than lattice enthalpy). Hydroxides and fluorides show the opposite trend — more soluble down the group.

Understanding:

We need to identify which ion has no unpaired electrons (diamagnetic) among the given transition metal ions.

Formula:

An ion is diamagnetic if all its electrons are paired. The magnetic moment is:

where $n$ = number of unpaired electrons. Diamagnetic $\Rightarrow n=0$.

Step 1: Determine electronic configurations.

Step 2: Conclusion.

${\text{Zn}}^{2+}$ with a completely filled $3d^{10}$ configuration has no unpaired electrons and is diamagnetic.

Answer:

${\text{Zn}}^{2+}$ is the only diamagnetic ion among the given options.

Quick Tip:

${\text{Zn}}^{2+}$ ($3d^{10}$), ${\text{Cu}}^{+}$ ($3d^{10}$), ${\text{Sc}}^{3+}$ ($3d^0$), and ${\text{Ti}}^{4+}$ ($3d^0$) are the common diamagnetic transition metal ions — memorise these for NEET.

Understanding:

We need to explain why the $+1$ oxidation state becomes more stable relative to the $+3$ state as we descend Group 13 (B → Tl).

Step 1: Recall the inert pair effect.

In heavier elements of Groups 13–15, the $n{s}^2$ electron pair becomes increasingly reluctant to participate in bond formation. This is called the inert pair effect. As the principal quantum number ($n$) increases, the $ns$ electrons penetrate the core more effectively but are also poorly shielded by $d$ and $f$ electrons, making them more tightly held and less available for bonding.

Step 2: Consequence in Group 13.

Step 3: Evaluate other options.

Answer:

The increasing stability of the $+1$ oxidation state down Group 13 is due to the inert pair effect, where the $n{s}^2$ electrons become increasingly resistant to participation in bonding.

Quick Tip:

The inert pair effect is most pronounced in the 6th period. In Group 14, ${\text{Pb}}^{2+}$ is more stable than ${\text{Pb}}^{4+}$; in Group 15, ${\text{Bi}}^{3+}$ is more stable than ${\text{Bi}}^{5+}$ — same principle.

Understanding:

We need to identify which nitrate salt, on heating, gives both ${\text{NO}}_2$ and ${\text{O}}_2$ as products.

Step 1: Recall the thermal decomposition pattern of metal nitrates.

The products depend on the activity/reactivity of the metal:

Step 2: Distinguish between options A and D.

Both ${\text{Cu(NO}}_3{)}_2$ and ${\text{Pb(NO}}_3{)}_2$ give ${\text{NO}}_2+{\text{O}}_2$ on heating. Among these, ${\text{Pb(NO}}_3{)}_2$ is the most commonly cited NEET example for this reaction and is a standard laboratory preparation of ${\text{NO}}_2$.

The question asks which compound decomposes to give ${\text{NO}}_2$ and ${\text{O}}_2$ — ${\text{Pb(NO}}_3{)}_2$ is the textbook answer for this specific decomposition in NCERT.

Answer:

${\text{Pb(NO}}_3{)}_2$ decomposes on heating to give $\text{PbO}$, ${\text{NO}}_2$, and ${\text{O}}_2$:

Quick Tip:

This reaction is the standard NCERT laboratory method for preparing ${\text{NO}}_2$ gas. Remember: metal nitrates of metals between Mg and Cu in the activity series → metal oxide + ${\text{NO}}_2$ + ${\text{O}}_2$.

Understanding:

We need to give the correct IUPAC name of ${\text{K}}_2[{\text{PtCl}}_4]$.

Step 1: Identify the components of the complex.

Step 2: Determine the oxidation state of Pt.

So platinum is in the $+2$ oxidation state: $\text{Pt(II)}$.

Step 3: Apply IUPAC 2013 naming rules.

Step 4: Check other options.

Answer:

The correct IUPAC name of ${\text{K}}_2[{\text{PtCl}}_4]$ is potassium tetrachloridoplatinate(II).

Quick Tip:

For anionic complexes, the metal name takes the suffix -ate (e.g., platinate, ferrate, cuprate). The 2013 IUPAC rules use -ido for anionic ligands: fluorido, chlorido, bromido, cyanido, hydroxido.