NEET Chemistry - MCQ Practice Questions

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.