Q.Predict in which of the following, entropy increases/decreases:

Step 1 — The tool: molecular orbital bond order

where $N_b$ = electrons in bonding MOs and $N_a$ = electrons in antibonding MOs. For these second-row diatomics the filling order (for $\geq 14$ electrons, i.e. $\mathrm{O_2}$ and beyond) is

Step 2 — Work out each species

(1) NO — total electrons $= 7 + 8 = 15$

Up to $\mathrm{N_2}$-like filling we get $N_b = 10$, $N_a = 5$ (one lone electron sits in a $\pi^*$ orbital — which is why NO is paramagnetic and readily loses that antibonding electron to form $\mathrm{NO^+}$):

(2) CO — total electrons $= 6 + 8 = 14$ (isoelectronic with $\mathrm{N_2}$)

A triple bond — consistent with CO's very high bond dissociation enthalpy.

(3) $\mathrm{O_2^{-}}$ (superoxide) — total electrons $= 16 + 1 = 17$

The extra electron goes into an antibonding $\pi^*$ orbital, so $N_b = 10$, $N_a = 7$:

Adding electrons to antibonding orbitals weakens and lengthens the bond — hence superoxide's longer O–O bond than $\mathrm{O_2}$.

(4) $\mathrm{O_2}$ — total electrons $= 8 + 8 = 16$

The two unpaired $\pi^*$ electrons explain $\mathrm{O_2}$'s paramagnetism — MO theory's great success.

Step 3 — Assemble the match

That is exactly option (A).

ANSWER: A

Step 1 — Decide the direction of the shift.

Initially all four species are $1\,\text{M}$, so

Since $Q = 1 < K = 100$, the reaction proceeds forward (left → right) to reach equilibrium.

Step 2 — ICE table. Let $x$ mol L$^{-1}$ of A react.

Step 3 — Apply the equilibrium constant.

Step 4 — Take the (positive) square root.

Step 5 — Equilibrium concentration of D.

Check: $[\ce{A}]=[\ce{B}]=1-0.818=0.182$, so $K = \dfrac{(1.818)^2}{(0.182)^2} = \left(\dfrac{1.818}{0.182}\right)^2 = (9.99)^2 \approx 100$. ✓

(Note the distractor: $0.182$ is the equilibrium concentration of A or B, and $0.818$ is the extent $x$ — not $[\ce{D}]$.)

ANSWER: B

Step 1 — Establish the ground rules for a galvanic cell.

A galvanic (voltaic) cell converts the energy of a spontaneous redox reaction into electrical energy.

• At the anode, oxidation occurs: $\mathrm{M \rightarrow M^{n+} + n e^-}$. Electrons are produced here and pile up, so the anode becomes the NEGATIVE terminal.
• At the cathode, reduction occurs: $\mathrm{M^{n+} + n e^- \rightarrow M}$. Electrons are consumed here, so the cathode is electron-deficient — the POSITIVE terminal.
• Electrons therefore flow through the external wire from anode → cathode.
• Conventional current is defined opposite to electron flow, so in the external circuit it flows cathode → anode.

(Note the contrast with an electrolytic cell, where an external supply forces the reaction and the anode is positive — this sign flip is the classic trap in this question.)

Step 2 — Evaluate each statement.

1. Current flows from cathode to anode. ✓ CORRECT. Electrons go anode → cathode externally; conventional current, being opposite, goes cathode → anode in the external circuit. This is precisely what makes the cathode the $+$ terminal that current "comes out of".
2. Anode is positive terminal. ✗ WRONG. In a galvanic cell the anode is the site of oxidation, accumulates electrons and is the negative terminal. (Only in an electrolytic cell is the anode positive.) (c) If $E_{cell} < 0$, then it is spontaneous reaction. ✗ WRONG. Gibbs energy and cell potential are linked by

   $\Delta G = -nFE_{cell}$

   A reaction is spontaneous only when $\Delta G < 0$, which requires

   $-nFE_{cell} < 0 \qquad\Rightarrow\qquad E_{cell} > 0$

   (as $n$ and $F$ are positive). So spontaneity demands a positive $E_{cell}$; a negative $E_{cell}$ means the reaction is non-spontaneous (it would need electrolysis). The statement has the sign exactly backwards.

(d) Cathode is positive terminal. ✓ CORRECT. Reduction consumes electrons at the cathode, leaving it at higher (positive) potential.

Step 3 — Collect the correct statements.

Correct: (a) and (d) only.

ANSWER: C

1. The Arrhenius equation

Here $A$ is the Arrhenius (frequency / pre-exponential) factor — physically, the collision frequency with correct orientation — and $e^{-E_a/RT}$ is the fraction of collisions that carry at least the activation energy.

2. The infinite-temperature limit

Therefore

Interpretation: at infinitely high temperature every collision is energetic enough to cross the barrier, so the energy barrier stops mattering and the rate constant saturates at the collision frequency $A$. This is why $A$ is often described as "the rate constant at infinite temperature".

3. Reading the numerical answer

The question quotes the value $6.0\times10^{5}\ \mathrm{s^{-1}}$ and asks for $A$ in that limit. Since $k \to A$ with the exponential factor equal to unity, the required value is simply

Note also that $A$ must have the same units as $k$ (here $\mathrm{s^{-1}}$, a first-order reaction), because $e^{-E_a/RT}$ is dimensionless.

4. Why the other options are impossible

• (A) $6\times10^{5}\,e^{-E_a/300R}$ and (B) $e^{-E_a/300R}$ still contain an exponential term. But in the $T\to\infty$ limit the exponential is 1, so no such factor can appear in the answer. (Multiplying $k$ by $e^{-E_a/RT}$ rather than dividing is, in any case, the wrong direction.)
• (C) $\dfrac{6\times10^{5}}{300}$ divides by temperature — the Arrhenius equation has no such linear division; $T$ appears only inside the exponent.

ANSWER: D

Concept: Entropy increases with disorder/randomness; solid < liquid < gas, and dissolved/dispersed states have higher entropy than ordered solids.

(A) Rusting of iron: 4Fe(s) + 3O2(g) -> 2Fe2O3(s). Gas is consumed, so entropy DECREASES too - but this is a chemical change usually cited for its enthalpy; still, gas -> solid does lower S.

(B) Melting of ice: solid -> liquid, entropy INCREASES.

(C) Vaporisation of camphor: solid -> gas (sublimation), entropy INCREASES greatly.

(D) Crystallisation of sucrose from solution: randomly dispersed solute molecules become an ordered crystal lattice - entropy clearly DECREASES.

The process asked for is the one where the system becomes more ordered, i.e. crystallisation of sucrose from solution (the standard textbook example of an entropy decrease).

ANSWER: D

The entire question hinges on the reactivity series of metals — a list that ranks metals from most reactive (like potassium) to least reactive (like gold and platinum). A chemical reaction occurs in a displacement scenario only if the metal doing the displacing is above the metal being displaced in this series. If the metal is below, no reaction happens.

Let’s examine each case one by one.

1. Option (A): Gold ornaments washed with dilute HCl

   Gold is one of the least reactive metals — it lies far below hydrogen in the reactivity series. Dilute HCl can only react with metals that are above hydrogen (like zinc, iron, magnesium), because those metals can displace hydrogen from the acid. Gold cannot. So no reaction occurs.

2. Option (B): $\text{ZnSO}_{4(\text{aq})}$ placed in a copper vessel

   Here, the vessel is made of copper, and the solution contains zinc sulfate. For a reaction to happen, copper would have to displace zinc from its sulfate. But copper is less reactive than zinc (zinc is above copper in the series). A less reactive metal cannot displace a more reactive one. So no reaction.

3. Option (C): $\text{AgNO}_3$ solution stirred with a copper spoon

   Copper is more reactive than silver (copper is above silver in the reactivity series). So copper can displace silver from silver nitrate:

You would see a greyish-black deposit of silver on the copper spoon, and the solution turns blue due to copper(II) nitrate. This reaction is possible.

4. Option (D): Conc. $\text{HNO}_3$ stored in a platinum vessel Platinum is one of the most unreactive metals — it is below almost everything, including hydrogen. Concentrated nitric acid is a strong oxidizer, but platinum does not react with it under normal conditions. In fact, platinum is often used to store or handle concentrated acids because of its inertness. No reaction.

The relationship between enthalpy change ($\Delta H$) and internal energy change ($\Delta U$) for a reaction is given by:

where $\Delta n_g$ is the change in the number of moles of gaseous substances (products minus reactants). This comes from the definition $H = U + PV$, and for ideal gases $PV = nRT$, so at constant temperature the $PV$ term changes only when the number of gas moles changes.

For $\Delta H > \Delta U$, we need $\Delta n_g > 0$ — the reaction must produce more gas moles than it consumes. Let’s check each option.

1. Option (A): $\text{N}_{2(g)} + \text{O}_{2(g)} \longrightarrow 2\text{NO}_{(g)}$

   • Reactant gas moles: $1 + 1 = 2$
   • Product gas moles: $2$
   • $\Delta n_g = 2 - 2 = 0$
   • So $\Delta H = \Delta U$. Not greater.

2. Option (B): $\text{N}_{2(g)} + 3\text{H}_{2(g)} \longrightarrow 2\text{NH}_{3(g)}$

   • Reactant gas moles: $1 + 3 = 4$
   • Product gas moles: $2$
   • $\Delta n_g = 2 - 4 = -2$
   • So $\Delta H < \Delta U$. Not greater.

3. Option (C): $\text{CaCO}_{3(s)} \longrightarrow \text{CaO}_{(s)} + \text{CO}_{2(g)}$

   • Reactant gas moles: $0$ (both solids)
   • Product gas moles: $1$ (only $\text{CO}_2$ is a gas)
   • $\Delta n_g = 1 - 0 = +1$
   • So $\Delta H = \Delta U + RT$, meaning $\Delta H > \Delta U$. This fits.

4. Option (D): $\text{CH}_{4(g)} + 2\text{O}_{2(g)} \longrightarrow \text{CO}_{2(g)} + 2\text{H}_2\text{O}_{(l)}$

   • Reactant gas moles: $1 + 2 = 3$
   • Product gas moles: $1$ (only $\text{CO}_2$ is a gas; water is liquid)
   • $\Delta n_g = 1 - 3 = -2$
   • So $\Delta H < \Delta U$. Not greater.

The key to this question lies in understanding the classification of crystalline solids based on the type of bonding and structural units. Network crystalline solids (also called covalent network solids) are held together by a continuous framework of covalent bonds extending throughout the crystal. This gives them very high melting points and extreme hardness. In contrast, molecular solids (like iodine or ice) have discrete molecules held by weak intermolecular forces, and ionic solids (like NaCl) consist of ions held by electrostatic attraction.

Let’s examine each option:

1. Option (A): $I_2$ (Iodine)

   Iodine forms diatomic molecules held together in the solid state by weak van der Waals forces. These are discrete $I_2$ units, not a continuous covalent network. It is a molecular solid, not a network solid.

2. Option (B): NaCl (Sodium chloride)

   NaCl is the classic example of an ionic solid. It consists of $Na^+$ and $Cl^-$ ions arranged in a lattice, held by ionic bonds. While the structure is extended, the bonding is electrostatic, not covalent. So it is not a network covalent solid.

3. Option (C): AlN (Aluminium nitride)

   AlN is a compound where each Al atom is covalently bonded to four N atoms in a tetrahedral arrangement, and each N is similarly bonded to four Al atoms. This forms a three-dimensional network of covalent bonds, analogous to diamond or silicon carbide. It is a network covalent solid with a high melting point (over 2000°C) and great hardness. This matches the definition perfectly.

4. Option (D): Ice (Solid $H_2O$)

   In ice, water molecules are held together by hydrogen bonds, which are intermolecular forces (though stronger than van der Waals forces). The molecules themselves remain intact; there is no continuous covalent network. Ice is a molecular solid (specifically, a hydrogen-bonded molecular solid).