NEET Chemistry - MCQ Practice Questions

Boiling points of hydrides depend on intermolecular forces, which vary based on molecular mass and hydrogen bonding strength.

Step 1: Identify Group 16 Hydrides and Their Boiling Points

Group 16 elements (chalcogens) form hydrides with the general formula H₂X. Let's list their boiling points:

Step 2: Analyze Intermolecular Forces

H₂O has exceptional boiling point due to strong hydrogen bonding between oxygen (highly electronegative) and hydrogen. However, S, Se, and Te are less electronegative, so H₂S, H₂Se, and H₂Te exhibit only weak dipole-dipole interactions and London dispersion forces:

Among H₂S, H₂Se, and H₂Te, molecular mass increases down the group, strengthening London dispersion forces. H₂S has the smallest molecular mass (34 g/mol) among these three, resulting in the weakest intermolecular forces.

The correct answer is (B) H₂S with a boiling point of -60°C, the lowest among the group.

At 25°C (298 K), (2.303RT/F) = (2.303 × 8.314 × 298)/96485 ≈ 0.0592 V. This is commonly used in half-cell potential calculations.

For a first-order reaction, the relationship between time, concentration, and rate constant follows a logarithmic decay pattern that connects to the half-life.

Step 1: Apply the First-Order Rate Law

For a first-order reaction, we use the integrated rate equation that relates remaining concentration to time.

Since the reaction is 75% complete, 25% of the reactant remains, so $\frac{[A{]}_t}{[A{]}_0}=0.25$ or $\frac{[A{]}_0}{[A{]}_t}=4$

Step 2: Calculate Half-Life Using the Half-Life Formula

The half-life for a first-order reaction is independent of initial concentration and is given by:

The half-life of the reaction is 22.5 minutes.

The answer is (B) 22.5 minutes.

At equilibrium, the chemical potential of a substance is the same in all phases present. This is the condition for phase equilibrium.

Initial: PA = 2, PB = 1. Change: -1.5 for A, -1.5 for B, +1.5 for C and D. Equilibrium: PA = 0.5, PB = -0.5 (invalid). Using Kp: 4 = (PC × PD)/(0.5 × PB). Since stoichiometry is 1:1:1:1, PC = PD = 1.5 atm.

Understanding:

We must determine the correct IUPAC name of the complex ion $[\text{Co}({\text{NH}}_3{)}_4{\text{Cl}}_2{]}^{+}$.

Step 1: Determine the oxidation state of cobalt.

Let the oxidation state of Co be $x$. Each ${\text{NH}}_3$ is neutral and each ${\text{Cl}}^{-}$ carries $-1$ charge:

So cobalt is in the $+3$ oxidation state.

Step 2: Name the ligands in alphabetical order.

According to 2005 IUPAC recommendations, anionic ligands ending in "-ide" replace the old "-o" suffix. Chloride becomes chlorido and ammonia becomes ammine. Alphabetically, "ammine" (a) comes before "chlorido" (c), so the order is: tetraammine then dichlorido.

Step 3: Assemble the full name.

The metal is cobalt in oxidation state $+3$, giving cobalt(III). The complete IUPAC name is tetraamminedichloridocobalt(III) ion.

Answer:

The correct IUPAC name is tetraamminedichloridocobalt(III) ion.

Quick Tip:

Under 2005 IUPAC rules, anionic ligands are named as "chlorido", "bromido", etc. (not the older "chloro", "bromo"). NEET questions increasingly use this updated nomenclature.

Understanding:

We must find the coordination number (CN) and oxidation state of Pt in $[\text{Pt}(\text{en}{)}_2{\text{Cl}}_2{]}^{2+}$.

Step 1: Calculate coordination number.

Each en ligand donates 2 donor atoms, and each Cl donates 1:

Step 2: Calculate oxidation state of Pt.

Let oxidation state of Pt be $x$. en is neutral, each ${\text{Cl}}^{-}$ is $-1$:

Answer:

The coordination number is 6 and the oxidation state of platinum is $+4$.

Quick Tip:

Always multiply the number of each ligand by its denticity to get the coordination number. Bidentate ligands like en, ox, and bipy each contribute 2 to the CN.

Understanding:

We must identify which complex can exist as non-superimposable mirror images (enantiomers), i.e., shows optical isomerism.

Step 1: Screen each complex for a plane of symmetry.

Step 2: Analyse $[\text{Co}(\text{en}{)}_3{]}^{3+}$.

This is a tris(bidentate) octahedral complex. With three bidentate en ligands wrapped around the metal, the complex belongs to the $D_3$ point group, which has no plane of symmetry, centre of inversion, or improper rotation axis. It therefore exists as $\Delta$ (right-handed) and $\Lambda$ (left-handed) enantiomers.

Answer:

$[\text{Co}(\text{en}{)}_3{]}^{3+}$ exhibits optical isomerism because it is a chiral, non-superimposable tris-bidentate octahedral complex.

Quick Tip:

Tris-bidentate octahedral complexes (e.g., $[\text{M}(\text{en}{)}_3{]}^{n+}$, $[\text{M}(\text{ox}{)}_3{]}^{n-}$) are the classic examples of optical isomerism in coordination chemistry.

Understanding:

We must calculate the CFSE for $[\text{Fe}({\text{H}}_2\text{O}{)}_6{]}^{2+}$.

Step 1: Write the high-spin $d^6$ electron configuration.

For high spin $d^6$: electrons fill $t_{2g}$ and $e_g$ with maximum spin:

(4 electrons in $t_{2g}$, 2 electrons in $e_g$)

Step 2: Calculate CFSE.

Answer:

The CFSE for high-spin $[\text{Fe}({\text{H}}_2\text{O}{)}_6{]}^{2+}$ is $-0.4{\Delta }_o$.

Quick Tip:

For high-spin $d^6$: remember the configuration is $t_{2g}^4{e}_g^2$, not $t_{2g}^6$. A common error is to assume low-spin, which gives $t_{2g}^6$ and CFSE $=-2.4{\Delta }_o$.

Understanding:

We must identify which pair are linkage isomers — complexes with the same molecular formula where an ambidentate ligand is bonded through different donor atoms.

Step 1: Check each option.

Step 2: Confirm.

Linkage isomerism requires an ambidentate ligand (${\text{NO}}_2^{-}$, ${\text{SCN}}^{-}$, ${\text{CN}}^{-}$) bonded through different atoms. Only Option A meets this definition.

Answer:

The pair in Option A are linkage isomers, as ${\text{NO}}_2^{-}$ coordinates through N in one and through O in the other.

Quick Tip:

The four common types of isomerism to distinguish: linkage, ionisation, coordination, and hydrate. Always check whether the same ambidentate ligand is involved before calling a pair "linkage isomers".

Understanding:

We must find the number of unpaired electrons in the tetrahedral complex $[{\text{MnBr}}_4{]}^{2-}$.

Step 1: Determine spin state.

In a tetrahedral field, ${\Delta }_t\approx \frac{4}{9}{\Delta }_o$, which is small. Weak-field ligands like ${\text{Br}}^{-}$ always give high-spin tetrahedral complexes.

Step 2: Fill $d^5$ in high-spin tetrahedral.

In a tetrahedral field, the $d$ orbitals split into $e$ (lower, 2 orbitals) and $t_2$ (upper, 3 orbitals). For high-spin $d^5$:

All five electrons are unpaired (one in each $d$ orbital by Hund's rule).

Step 3: Count unpaired electrons.

Answer:

The number of unpaired electrons in $[{\text{MnBr}}_4{]}^{2-}$ is 5.

Quick Tip:

All tetrahedral complexes are effectively high-spin because ${\Delta }_t$ is only $\frac{4}{9}{\Delta }_o$, too small to force pairing. Low-spin tetrahedral complexes are essentially unknown.

Understanding:

We must find the number of geometrical isomers for a square planar complex of type $[\text{MA}\text{B}\text{C}\text{D}]$ — four different monodentate ligands.

Step 1: Apply the general rule.

For a square planar complex with four different ligands ($\text{MABCD}$), the number of geometrical isomers is 3. This is because we can place each ligand in a fixed position (say top) and arrange the remaining three, but after removing duplicates from mirror equivalence, exactly 3 distinct geometric arrangements emerge.

Step 2: Enumerate them.

Label positions as 1 (top), 2 (right), 3 (bottom), 4 (left). Fix A at position 1. The three isomers arise from which ligand is placed trans to A:

1. A trans to B (→ C and D are trans to each other)

2. A trans to C (→ B and D are trans to each other)

3. A trans to D (→ B and C are trans to each other)

These three arrangements are all distinct. Note these are geometric, not optical, isomers (though some may also be optically active).

Answer:

There are 3 geometrical isomers for the square planar complex $[\text{Pt}({\text{NH}}_3)(\text{py})(\text{Cl})(\text{Br})]$.

Quick Tip:

For square planar $[\text{MABCD}]$: geometric isomers = 3. For $[{\text{MA}}_2\text{BC}]$: 2 isomers (cis/trans). For $[{\text{MA}}_2{\text{B}}_2]$: 2 isomers (cis/trans). Memorise these counts.

Understanding:

We must identify which complex releases the most free ${\text{Cl}}^{-}$ ions in solution immediately, giving the most precipitate with ${\text{AgNO}}_3$. The question asks which gives a precipitate "immediately" — all options with ionisable chloride will, but the question implies the one that precipitates all three ${\text{Cl}}^{-}$ ions is most definitive. However, since all A, B, C do precipitate, the question specifically tests how many ${\text{Cl}}^{-}$ are outside the coordination sphere.

Step 1: Count ionisable (outer sphere) chloride ions for each complex.

Step 2: Identify the correct answer.

The complex that ionises to give the maximum free ${\text{Cl}}^{-}$ immediately (3 ions) is $[\text{Co}({\text{NH}}_3{)}_6]{\text{Cl}}_3$. This is the classic Werner's experiment result: it gives a white precipitate (AgCl) with 3 equivalents of AgNO${}_3$.

Answer:

$[\text{Co}({\text{NH}}_3{)}_6]{\text{Cl}}_3$ gives 3 free ${\text{Cl}}^{-}$ ions in solution, producing the most AgCl precipitate immediately.

Quick Tip:

This is Werner's classic molar conductance / AgNO${}_3$ test. Chlorides inside the coordination sphere do NOT precipitate with AgNO${}_3$ immediately — only outer-sphere (ionic) chlorides do.

Understanding:

We must determine the hybridisation and geometry of $[\text{Ni}(\text{CN}{)}_4{]}^{2-}$.

Step 1: Determine the $d^8$ configuration under strong field.

Free ${\text{Ni}}^{2+}$ ([$d^8$]): electron configuration is $3d^8$. With strong-field ${\text{CN}}^{-}$, the two unpaired electrons in the $3d$ orbitals are forced to pair:

This vacates one $3d$ orbital.

Step 2: Determine hybridisation.

The vacant $3d$ orbital, along with $4s$ and two $4p$ orbitals, hybridise to form four $ds{p}^2$ hybrid orbitals:

This gives a square planar geometry.

Step 3: Contrast with weak-field.

$[\text{Ni}(\text{Cl}{)}_4{]}^{2-}$ (weak field) is $s{p}^3$ and tetrahedral (2 unpaired electrons, paramagnetic). $[\text{Ni}(\text{CN}{)}_4{]}^{2-}$ is diamagnetic.

Answer:

$[\text{Ni}(\text{CN}{)}_4{]}^{2-}$ has $ds{p}^2$ hybridisation and a square planar geometry.

Quick Tip:

$d^8$ metal ions with strong-field ligands (e.g., ${\text{Ni}}^{2+}$, ${\text{Pd}}^{2+}$, ${\text{Pt}}^{2+}$) always adopt square planar geometry with $ds{p}^2$ hybridisation.

Understanding:

We must identify the correct statement about the trans-effect in square planar complexes.

Step 1: Define the trans-effect.

The trans-effect is a kinetic phenomenon: a ligand already coordinated to a metal labilises (weakens and facilitates substitution of) the ligand situated trans to it. It is observed most prominently in square planar complexes, especially of ${\text{Pt}}^{2+}$.

Step 2: Evaluate each option.

Answer:

The trans-effect is a kinetic phenomenon in which a coordinated ligand labilises the ligand trans to it during substitution reactions in square planar complexes.

Quick Tip:

Distinguish: trans-effect = kinetic (rate of substitution); trans-influence = thermodynamic (ground-state bond weakening). NEET questions often test this distinction.

Understanding:

We need the enthalpy change for the formation of 2 mol of ${\text{NH}}_3$ from its elements.

Formula:

Step 1: Apply Hess's Law

The standard enthalpy of formation of elements in their standard state is zero, so:

Answer:

The enthalpy change for the given reaction is $-92{\text{ kJ mol}}^{-1}$.

Quick Tip:

The standard enthalpy of formation is always per mole of product formed. When 2 moles are produced, multiply by 2. A common error is to use $-46$ directly without scaling.

Understanding:

We need $\Delta H_f^{\circ }$ for ${\text{CH}}_4(g)$, i.e., the enthalpy change for:

Formula:

By Hess's Law:

Step 1: Construct the target equation

Add equation 1 and $2\times$ equation 2, then subtract equation 3:

Step 2: Verify the equation

Using $\Delta H_{\text{rxn}}=\sum \Delta H_f^{\circ }(\text{products})-\sum \Delta H_f^{\circ }(\text{reactants})$ for combustion:

Answer:

The standard enthalpy of formation of ${\text{CH}}_4(g)$ is $-75{\text{ kJ mol}}^{-1}$.

Quick Tip:

To find $\Delta H_f^{\circ }$ of a compound using combustion data, rearrange the combustion equation algebraically. Reversing equation 3 changes the sign from $-890$ to $+890$.

Understanding:

We need $\Delta S_{\text{vap}}$ for vaporisation of water at its boiling point.

Formula:

At equilibrium (constant $T$ and $P$), $\Delta G=0$, so:

Step 1: Substitute values

Answer:

The entropy change of vaporisation is approximately $0.109{\text{ kJ mol}}^{-1}{\text{K}}^{-1}$.

Quick Tip:

Trouton's rule states that $\Delta S_{\text{vap}}\approx 88{\text{ J mol}}^{-1}{\text{K}}^{-1}$ for most non-associated liquids. Water's value ($\approx 109{\text{ J mol}}^{-1}{\text{K}}^{-1}$) is higher due to hydrogen bonding.

Understanding:

We need the temperature above which $\Delta G>0$ (reaction becomes non-spontaneous).

Formula:

The crossover temperature (where $\Delta G=0$) is:

Step 1: Identify the spontaneity pattern

Since both $\Delta H<0$ and $\Delta S<0$, the $-T\Delta S$ term is positive (adding to $\Delta G$). The reaction is spontaneous at low $T$ but becomes non-spontaneous at high $T$.

Step 2: Find the crossover temperature

Step 3: Interpret

Below $2500\text{ K}$: $\Delta G<0$ (spontaneous). Above $2500\text{ K}$: $\Delta G>0$ (non-spontaneous).

Answer:

The reaction becomes non-spontaneous above approximately $2500\text{ K}$.

Quick Tip:

When $\Delta H$ and $\Delta S$ have the same sign, there is always a crossover temperature. Always convert $\Delta H$ to joules (or $\Delta S$ to kJ) before dividing to keep units consistent.

Understanding:

We use bond enthalpies to find $\Delta H$ for the reaction.

Formula:

Step 1: Bonds broken (reactants)

Step 2: Bonds formed (products)

Two moles of $\text{HCl}$ are formed, each with one $\text{H–Cl}$ bond:

Step 3: Calculate $\Delta H$

Answer:

The enthalpy change for the reaction is $-184{\text{ kJ mol}}^{-1}$.

Quick Tip:

Bond breaking is always endothermic (positive contribution) and bond forming is always exothermic (negative contribution). A common error is reversing this sign convention.