AIEEE 2009 Chemistry Question Paper with Answer and Solution

(D) The energy of a photon emitted during a transition between energy levels $n_i$ and $n_f$ is given by $\Delta E = 13.6 Z^2 (\frac{1}{n_f^2} - \frac{1}{n_i^2}) \text{ eV}$.
Ultraviolet $(UV)$ radiation corresponds to higher energy transitions (Lyman series),while infrared $(IR)$ radiation corresponds to lower energy transitions (Paschen,Brackett,or Pfund series).
Since the transition $4 \to 3$ results in $UV$ radiation,and $IR$ radiation requires significantly lower energy transitions,we look for a transition where the energy gap is smaller than that of $4 \to 3$.
The energy difference decreases as $n$ increases. Among the given options,the transition $5 \to 4$ involves the smallest energy change,which corresponds to the infrared region of the electromagnetic spectrum.

(A) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 6.93 \ min$,so $k = \frac{0.693}{6.93} = 0.1 \ min^{-1}$.
The integrated rate equation is $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.
For $99\%$ completion,$[A]_0 = 100$ and $[A]_t = 100 - 99 = 1$.
Substituting the values: $t = \frac{2.303}{0.1} \log \frac{100}{1}$.
$t = 23.03 \times \log(10^2) = 23.03 \times 2 = 46.06 \ min$.

(D) Linkage isomerism occurs in coordination compounds containing ambidentate ligands,which can coordinate through two different donor atoms.
In the pair $[Pd(PPh_3)_2(NCS)_2]$ and $[Pd(PPh_3)_2(SCN)_2]$,the ligand $NCS^-$ (isothiocyanate) coordinates through the nitrogen atom,while $SCN^-$ (thiocyanate) coordinates through the sulfur atom.
Therefore,this pair exhibits linkage isomerism.

(D) Let the roots of the equation $x^2 - 6x + a = 0$ be $\alpha$ and $4\beta$,and the roots of the equation $x^2 - cx + 6 = 0$ be $\alpha$ and $3\beta$,where $\alpha$ is the common root.
From the sum and product of roots:
For the first equation: $\alpha + 4\beta = 6$ and $4\alpha\beta = a$.
For the second equation: $\alpha + 3\beta = c$ and $3\alpha\beta = 6$.
From $3\alpha\beta = 6$,we get $\alpha\beta = 2$,so $\beta = \frac{2}{\alpha}$.
Substitute $\beta = \frac{2}{\alpha}$ into the first equation: $\alpha + 4(\frac{2}{\alpha}) = 6$.
$\alpha + \frac{8}{\alpha} = 6 \Rightarrow \alpha^2 - 6\alpha + 8 = 0$.
Solving the quadratic equation: $(\alpha - 2)(\alpha - 4) = 0$.
So,$\alpha = 2$ or $\alpha = 4$.
If $\alpha = 2$,then $\beta = 1$. The roots of the first equation are $2$ and $4(1) = 4$. The roots of the second equation are $2$ and $3(1) = 3$. This satisfies the condition.
If $\alpha = 4$,then $\beta = 0.5$,which is not an integer. Since the problem states the other roots are integers,$\alpha = 2$ is the correct common root.

(A) The de Broglie wavelength $(\lambda)$ is given by the formula: $\lambda = \frac{h}{mv}$
Given:
$h = 6.63 \times 10^{-34} \ J \ s$
$m = 1.67 \times 10^{-27} \ kg$
$v = 1.0 \times 10^3 \ m \ s^{-1}$
Substituting the values:
$\lambda = \frac{6.63 \times 10^{-34}}{1.67 \times 10^{-27} \times 1.0 \times 10^3} \ m$
$\lambda = \frac{6.63}{1.67} \times 10^{-34 + 27 - 3} \ m$
$\lambda \approx 3.97 \times 10^{-10} \ m$
Since $1 \ nm = 10^{-9} \ m$,we have:
$\lambda \approx 0.397 \times 10^{-9} \ m \approx 0.40 \ nm$

(B) Given,velocity $v = 600 \, m/s$ and percentage error $= 0.005 \%$.
$\Delta v = \frac{0.005}{100} \times 600 = 0.03 \, m/s = 3 \times 10^{-2} \, m/s$.
According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$,where $\Delta p = m \Delta v$.
$\Delta x = \frac{h}{4 \pi m \Delta v}$.
Substituting the values: $\Delta x = \frac{6.6 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 3 \times 10^{-2}}$.
$\Delta x = \frac{6.6 \times 10^{-34}}{34.2264 \times 10^{-33}} \approx 0.1928 \times 10^{-1} \, m = 1.928 \times 10^{-3} \, m$.

(A) Given,for reaction:
$(I) \quad H_2O_{(\ell)} \rightarrow H^{+}_{(aq)} + OH^{-}_{(aq)} \quad \Delta H_r = 57.32 \, kJ$
$(II) \quad H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(\ell)} \quad \Delta H_r = -286.20 \, kJ$
For reaction $(I)$:
$\Delta H_r = \Delta H_f^o(H^{+}, aq) + \Delta H_f^o(OH^{-}, aq) - \Delta H_f^o(H_2O, \ell)$
Since $\Delta H_f^o(H^{+}, aq) = 0$,we have:
$57.32 = 0 + \Delta H_f^o(OH^{-}, aq) - \Delta H_f^o(H_2O, \ell) \quad \dots (III)$
For reaction $(II)$:
$\Delta H_f^o(H_2O, \ell) = -286.20 \, kJ$
Substituting this value in equation $(III)$:
$57.32 = \Delta H_f^o(OH^{-}, aq) - (-286.20)$
$\Delta H_f^o(OH^{-}, aq) = 57.32 - 286.20$
$\Delta H_f^o(OH^{-}, aq) = -228.88 \, kJ$

(A) The dissolution of $Na_2CO_3$ in water is given by: $Na_2CO_3 \rightarrow 2Na^{+} + CO_3^{2-}$.
Given the concentration of $Na_2CO_3$ is $1.0 \times 10^{-4} \ M$,the concentration of carbonate ions is $[CO_3^{2-}] = 1.0 \times 10^{-4} \ M$.
For the precipitation of $BaCO_3$ to begin,the ionic product must exceed the solubility product constant $(K_{sp})$.
The condition for the start of precipitation is: $[Ba^{2+}][CO_3^{2-}] = K_{sp}(BaCO_3)$.
Substituting the given values: $[Ba^{2+}] \times (1.0 \times 10^{-4}) = 5.1 \times 10^{-9}$.
Solving for $[Ba^{2+}]$: $[Ba^{2+}] = \frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}} = 5.1 \times 10^{-5} \ M$.

(A) The structure of neopentane is $CH_3-C(CH_3)_2-CH_3$.
To determine the $IUPAC$ name,we identify the longest carbon chain,which contains $3$ carbon atoms,making it a propane derivative.
There are two methyl groups attached to the second carbon atom.
Therefore,the $IUPAC$ name is $2,2-$dimethylpropane.

(B) The stability of carbanions is determined by the electronic effects of the attached groups.
Electron-withdrawing groups ($-I$ effect) stabilize the negative charge,while electron-donating groups ($+I$ effect) destabilize it.
$1$. $\overline{C}Cl_3$: The three $Cl$ atoms exert a strong $-I$ effect,significantly stabilizing the negative charge.
$2$. $C_6H_5\overline{C}H_2$: The phenyl group provides stability through $-I$ effect and resonance (delocalization of electrons).
$3$. $(CH_3)_2\overline{C}H$: This is a secondary carbanion with two electron-donating $CH_3$ groups ($+I$ effect),which destabilize the negative charge.
$4$. $(CH_3)_3\overline{C}$: This is a tertiary carbanion with three electron-donating $CH_3$ groups ($+I$ effect),making it the least stable.
Thus,the order of decreasing stability is: $\overline{C}Cl_3 > C_6H_5\overline{C}H_2 > (CH_3)_2\overline{C}H > (CH_3)_3\overline{C}$.

(B) The alkene that exhibits geometrical isomerism is $2-$butene.
$2-$Butene exists as $cis$ and $trans$ isomers.
In the $cis$-isomer,the two methyl groups are on the same side of the double bond,while in the $trans$-isomer,they are on opposite sides.
Due to the restricted rotation around the $C=C$ double bond,it exhibits geometrical isomerism.

(B) The molecule exhibits both geometrical and optical isomerism.
The compound $CH_3-CH=CH-CH(OH)-CH_3$ contains one carbon-carbon double bond $(C=C)$ that can exhibit geometrical $(cis-trans)$ isomerism.
It also contains one chiral carbon atom $(C^*)$ at the $C-4$ position,which can exhibit optical isomerism.
Since the double bond and the chiral center are not symmetrically related in a way that would create meso compounds,the total number of stereoisomers is given by $2^n$,where $n$ is the number of stereogenic centers.
Here,$n = 2$ (one double bond + one chiral center).
Therefore,the total number of stereoisomers = $2^2 = 4$.

(C) We know that $p \leftrightarrow q$ is equivalent to $\sim (p \leftrightarrow \sim q)$.
Let us verify this using a truth table:
$p$ | $q$ | $\sim q$ | $p \leftrightarrow \sim q$ | $\sim (p \leftrightarrow \sim q)$ | $p \leftrightarrow q$
$T$ | $T$ | $F$ | $F$ | $T$ | $T$
$T$ | $F$ | $T$ | $T$ | $F$ | $F$
$F$ | $T$ | $F$ | $T$ | $F$ | $F$
$F$ | $F$ | $T$ | $F$ | $T$ | $T$
From the truth table,the column for $\sim (p \leftrightarrow \sim q)$ is identical to the column for $p \leftrightarrow q$. Thus,Statement $-1$ is True.
Since the truth values of $\sim (p \leftrightarrow \sim q)$ depend on the truth values of $p$ and $q$ (it is $T$ when $p$ and $q$ have the same truth value and $F$ otherwise),it is not a tautology. Thus,Statement $-2$ is False.

(D) The energy of the emitted photon is given by $\Delta E = 13.6 Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}$.
Ultraviolet radiation corresponds to higher energy transitions (Lyman series,$n_f = 1$).
Infrared radiation corresponds to lower energy transitions (Paschen series $n_f = 3$,Brackett series $n_f = 4$,or Pfund series $n_f = 5$).
Comparing the options,the transition $5 \to 4$ results in a smaller energy change compared to the others,which is characteristic of the infrared region (Brackett series).
Therefore,the correct transition is $5 \to 4$.

(D) According to Einstein's photoelectric equation: $E = \phi + K_{max}$.
Here,$E$ is the energy of the incident photon,$\phi$ is the work function,and $K_{max}$ is the maximum kinetic energy of the photoelectrons.
The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Given $hc = 1240 \, eV-nm$ and $\lambda = 400 \, nm$,we have $E = \frac{1240}{400} \, eV = 3.1 \, eV$.
Substituting the values into the photoelectric equation: $3.1 \, eV = \phi + 1.68 \, eV$.
Therefore,the work function $\phi = 3.1 \, eV - 1.68 \, eV = 1.42 \, eV$.

(B) The precipitation of $BaCO_3$ begins when the ionic product exceeds the solubility product constant $(K_{sp})$.
The equilibrium expression is: $K_{sp} = [Ba^{2+}][CO_3^{2-}]$.
Given: $K_{sp} = 5.1 \times 10^{-9}$ and $[CO_3^{2-}] = 1.0 \times 10^{-3} \ M$.
Substituting the values: $5.1 \times 10^{-9} = [Ba^{2+}] \times (1.0 \times 10^{-3})$.
Therefore,$[Ba^{2+}] = \frac{5.1 \times 10^{-9}}{1.0 \times 10^{-3}} = 5.1 \times 10^{-6} \ M$.

(A) For a first-order reaction,the rate constant $K$ is given by $K = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 6.93 \, \text{min}$,so $K = \frac{0.693}{6.93} = 0.1 \, \text{min}^{-1}$.
The time $t$ required for $99 \%$ completion is given by $t = \frac{2.303}{K} \log \frac{[A]_0}{[A]_t}$.
Here,$[A]_0 = 100$ and $[A]_t = 100 - 99 = 1$.
$t = \frac{2.303}{0.1} \log \frac{100}{1} = 23.03 \times \log(10^2) = 23.03 \times 2 = 46.06 \, \text{min}$.

(C) The stability order is: $\bar{C}Cl_3 > C_6H_5\bar{C}H_2 > (CH_3)_2\bar{C}H > (CH_3)_3\bar{C}$.
$1$. $\bar{C}Cl_3$: Most stable due to the strong $-I$ effect of three $Cl$ atoms and the delocalization of the negative charge into the vacant $d$-orbitals of $Cl$.
$2$. $C_6H_5\bar{C}H_2$: Stable due to resonance delocalization of the negative charge into the benzene ring.
$3$. $(CH_3)_2\bar{C}H$: $2^{\circ}$ carbanion,stability decreases due to the electron-donating $+I$ effect of two $CH_3$ groups.
$4$. $(CH_3)_3\bar{C}$: $3^{\circ}$ carbanion,least stable due to the destabilizing $+I$ effect of three $CH_3$ groups.

(D) The given circuit consists of two $NOT$ gates followed by a $NOR$ gate.
Let the inputs be $A$ and $B$.
The outputs of the two $NOT$ gates are $\bar{A}$ and $\bar{B}$.
These are fed into a $NOR$ gate,so the final output $Y$ is given by:
$Y = \overline{\bar{A} + \bar{B}}$
Using De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{(\bar{A})} \cdot \overline{(\bar{B})} = A \cdot B$.
Thus,the circuit acts as an $AND$ gate.
The output $Y$ is high $(1)$ only when both inputs $A$ and $B$ are high $(1)$.
By observing the input waveforms,the output waveform will be high only during the time interval where both $A$ and $B$ are high.

(A) The reaction of phenol with sodium hydroxide $(NaOH)$ and carbon dioxide $(CO_2)$ followed by acidic workup is known as the $Kolbe-Schmitt$ reaction.
In this reaction,phenol is first converted to sodium phenoxide,which then undergoes electrophilic substitution with $CO_2$ to form sodium salicylate.
Upon acidification,this yields $2-hydroxybenzoic$ acid,commonly known as salicylic acid.
The chemical equation is: $C_6H_5OH + NaOH + CO_2$ $\rightarrow C_6H_4(OH)COONa$ $\xrightarrow{H^+} C_6H_4(OH)COOH$.

(D) The principle of conservation of mechanical energy states that the total mechanical energy remains constant.
At the lowest point,the rotational kinetic energy is maximum,and the potential energy of the centre of mass is minimum (taken as zero).
At the highest point,the rotational kinetic energy is zero,and the potential energy of the centre of mass is maximum.
Let $I$ be the moment of inertia of the rod about the axis passing through its end,which is $I = \frac{1}{3}ml^2$.
Using the conservation of energy:
$K_i + U_i = K_f + U_f$
$\frac{1}{2}I\omega^2 + 0 = 0 + mgh$
Substituting the value of $I$:
$\frac{1}{2} \times (\frac{1}{3}ml^2) \times \omega^2 = mgh$
$\frac{1}{6}ml^2\omega^2 = mgh$
Solving for $h$:
$h = \frac{l^2\omega^2}{6g}$

(D) For physisorption,the enthalpy of adsorption $(\Delta H_{adsorption})$ is low and negative.
Physisorption is an exothermic process,meaning heat is released during the adsorption process. Therefore,the enthalpy change is negative,not positive.

(D) According to the principle of conservation of mechanical energy,the initial rotational kinetic energy is converted into the gravitational potential energy of the centre of mass at the maximum height.
$TE_{i} = TE_{f}$
$\frac{1}{2} I \omega^{2} = mgh$
Here,$I$ is the moment of inertia of the rod about the axis passing through its end,which is $I = \frac{1}{3} m l^{2}$.
Substituting this into the equation:
$\frac{1}{2} \times (\frac{1}{3} m l^{2}) \omega^{2} = mgh$
$\frac{1}{6} m l^{2} \omega^{2} = mgh$
Canceling $m$ from both sides:
$h = \frac{1}{6} \frac{l^{2} \omega^{2}}{g}$

(B) According to Einstein's photoelectric equation: $K_{max} = E - W_0$,where $E$ is the energy of the incident photon and $W_0$ is the work function of the metal.
The energy of the incident photon is given by: $E = \frac{hc}{\lambda} = \frac{1240\, eV\, nm}{400\, nm} = 3.1\, eV$.
Given that the maximum kinetic energy $K_{max} = 1.68\, eV$,we can substitute these values into the equation:
$1.68\, eV = 3.1\, eV - W_0$.
Rearranging to solve for the work function $W_0$:
$W_0 = 3.1\, eV - 1.68\, eV = 1.42\, eV$.

(C) When the switch $S$ is closed at $t=0$,the inductor $L$ and resistor $R_2$ form a series circuit connected across the battery of emf $E=12\,V$. The resistor $R_1$ is in parallel with this branch and does not affect the current through the inductor.
The current $i$ in the $L-R_2$ branch at time $t$ is given by:
$i = \frac{E}{R_2} (1 - e^{-R_2 t / L})$
The potential drop across the inductor $L$ is given by $V_L = L \frac{di}{dt}$.
First,find $\frac{di}{dt}$:
$\frac{di}{dt} = \frac{E}{R_2} \cdot \frac{R_2}{L} \cdot e^{-R_2 t / L} = \frac{E}{L} e^{-R_2 t / L}$
Now,calculate $V_L$:
$V_L = L \left( \frac{E}{L} e^{-R_2 t / L} \right) = E e^{-R_2 t / L}$
Given $E = 12\,V$,$R_2 = 2\,\Omega$,and $L = 400\,mH = 0.4\,H$:
$\frac{R_2}{L} = \frac{2}{0.4} = 5\,s^{-1}$
Therefore,$V_L = 12 e^{-5t}\,V$.

(C) Physisorption is an exothermic process because the formation of weak van der Waal's forces between the adsorbate and adsorbent releases energy.
Therefore,the enthalpy of adsorption $(\Delta H_{adsorption})$ is always negative,typically ranging between $-20 \text{ to } -40 \text{ kJ/mol}$.
Option $C$ states that the enthalpy is positive,which is incorrect.

(A) For an $fcc$ unit cell, the relationship between the edge length $a$ and the radius $r$ is given by $4r = \sqrt{2}a$.
Given $a = 361 \ pm$.
Substituting the values: $r = \frac{\sqrt{2} \times 361}{4} = \frac{1.414 \times 361}{4} \approx 127.6 \ pm$.
Rounding to the nearest integer, the radius is $127 \ pm$.

(B) In a mixture of $n$-heptane and ethanol,the intermolecular forces between $n$-heptane and ethanol molecules are weaker than the forces between $n$-heptane$-n$-heptane molecules and ethanol-ethanol molecules.
Due to the presence of hydrogen bonding in pure ethanol,the $n$-heptane molecules disrupt these interactions.
As a result,the total vapor pressure of the solution is higher than that predicted by Raoult's Law.
Therefore,the solution is non-ideal and shows a positive $(+ve)$ deviation from Raoult's Law.

(B) According to Raoult's law,$P_{\text{total}} = P_X^{\circ} X_X + P_Y^{\circ} X_Y.$
For the first case: $X_X = \frac{1}{1+3} = \frac{1}{4}$ and $X_Y = \frac{3}{4}.$
$550 = P_X^{\circ} \times \frac{1}{4} + P_Y^{\circ} \times \frac{3}{4} \implies P_X^{\circ} + 3P_Y^{\circ} = 2200 \dots (i)$
For the second case,total moles of $Y = 3+1 = 4 \ mol.$ Total moles of solution = $1+4 = 5 \ mol.$
$X_X = \frac{1}{5}$ and $X_Y = \frac{4}{5}.$
New vapour pressure = $550 + 10 = 560 \ mm \ Hg.$
$560 = P_X^{\circ} \times \frac{1}{5} + P_Y^{\circ} \times \frac{4}{5} \implies P_X^{\circ} + 4P_Y^{\circ} = 2800 \dots (ii)$
Subtracting equation $(i)$ from $(ii)$:
$(P_X^{\circ} + 4P_Y^{\circ}) - (P_X^{\circ} + 3P_Y^{\circ}) = 2800 - 2200$
$P_Y^{\circ} = 600 \ mm \ Hg.$
Substituting $P_Y^{\circ}$ in equation $(i)$:
$P_X^{\circ} + 3(600) = 2200$
$P_X^{\circ} = 2200 - 1800 = 400 \ mm \ Hg.$
Thus,the vapour pressures are $400 \ mm \ Hg$ and $600 \ mm \ Hg$ respectively.

(C) The reaction is $CH_3OH_{(l)} + \frac{3}{2} O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$.
Standard Gibbs energy change of the reaction is calculated as:
$\Delta G_r^\circ = \Delta G_f^\circ(CO_2, g) + 2 \Delta G_f^\circ(H_2O, l) - \Delta G_f^\circ(CH_3OH, l) - \frac{3}{2} \Delta G_f^\circ(O_2, g)$
Substituting the given values:
$\Delta G_r^\circ = -394.4 + 2(-237.2) - (-166.2) - 0$
$\Delta G_r^\circ = -394.4 - 474.4 + 166.2 = -702.6 \ kJ \ mol^{-1}$.
The efficiency of a fuel cell is given by $\eta = \frac{\Delta G}{\Delta H} \times 100$.
$\eta = \frac{-702.6}{-726} \times 100 \approx 96.77 \% \approx 97 \%$.

(B) The given half-cell reactions are:
$(i) Fe^{3+} + 3e^- \rightarrow Fe, E^{\circ}_1 = -0.036 \ V, n_1 = 3$
$(ii) Fe^{2+} + 2e^- \rightarrow Fe, E^{\circ}_2 = -0.439 \ V, n_2 = 2$
We need to find $E^{\circ}$ for the reaction:
$(iii) Fe^{3+} + e^- \rightarrow Fe^{2+}, n_3 = 1$
This reaction can be obtained by $(i) - (ii)$.
Using the relation $\Delta G^{\circ} = -nFE^{\circ}$,we have:
$n_3 E^{\circ}_3 = n_1 E^{\circ}_1 - n_2 E^{\circ}_2$
$1 \times E^{\circ} = 3 \times (-0.036) - 2 \times (-0.439)$
$E^{\circ} = -0.108 + 0.878$
$E^{\circ} = 0.770 \ V$

(B) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 6.93 \, \text{min}$,so $k = \frac{0.693}{6.93} = 0.1 \, \text{min}^{-1}$.
The integrated rate equation for a first-order reaction is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
For $99 \%$ completion,$[A]_0 = 100$ and $[A]_t = 100 - 99 = 1$.
Substituting the values: $0.1 = \frac{2.303}{t} \log \frac{100}{1}$.
$0.1 = \frac{2.303 \times 2}{t}$.
$t = \frac{4.606}{0.1} = 46.06 \, \text{min}$.

(D) The reaction $XeO_3 + 6HF \rightarrow XeF_6 + 3H_2O$ is not feasible.
$XeF_6$ is highly reactive and undergoes hydrolysis in the presence of water to form $XeO_3$ and $HF$. Therefore,the reverse reaction is not spontaneous under standard conditions.
$XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$

(C) Optical isomerism is exhibited by complexes that lack a plane of symmetry or a center of inversion.
For octahedral complexes of the type $[M(AA)_2a_2]^{n+}$,the cis-isomer is optically active because it lacks a plane of symmetry.
In the given options,$[Co(en)_2(NH_3)_2]^{3+}$ is of the type $[M(AA)_2a_2]^{n+}$.
The cis-form of $[Co(en)_2(NH_3)_2]^{3+}$ exists as a pair of enantiomers (non-superimposable mirror images).
Therefore,the correct option is $C$.

(A) Transition metals in their lower oxidation states generally form basic oxides or hydroxides,whereas those in their higher oxidation states form acidic oxides or hydroxides.
Therefore,the statement that transition metals show basic character in their highest oxidation states is incorrect.
For example:
$\underbrace{MnO, Mn_2O_3}_{(\text{basic}) (+2, +3)} \quad \underbrace{Mn_2O_7}_{(\text{acidic}) (+7)}$

(A) Linkage isomerism occurs in coordination compounds containing ambidentate ligands,which can bind to the central metal atom through two different donor atoms.
In the pair $[Pd(PH_3)_2(NCS)_2]$ and $[Pd(PH_3)_2(SCN)_2]$,the ligand $SCN^-$ is an ambidentate ligand.
It can coordinate through the sulfur atom $(S)$ to form thiocyanato complexes $(M-SCN)$ or through the nitrogen atom $(N)$ to form isothiocyanato complexes $(M-NCS)$.
Therefore,this pair represents linkage isomers.

(B) The chemistry of lanthanoids is dominated by the $+3$ oxidation state.
Most $Ln^{3+}$ ions contain unpaired $f$-electrons,which undergo $f-f$ transitions,making them coloured.
Therefore,the statement that $Ln(III)$ compounds are generally colourless is incorrect.
$La^{3+}$ $(f^0)$ and $Lu^{3+}$ $(f^{14})$ are colourless,but most others are coloured.

(B) The reaction of phenol with $NaOH$ followed by $CO_2$ at $6 \ atm$ and $140 \ ^\circ C$,and subsequent acidification,is known as the Kolbe-Schmitt reaction.
In this reaction,phenol is first converted to sodium phenoxide.
Sodium phenoxide then reacts with $CO_2$ to form sodium salicylate,which upon acidification yields salicylic acid as the major product.

(C) The reaction between a carboxylic acid and an alcohol in the presence of concentrated $H_2SO_4$ is known as esterification.
The product formed is an ester,which is characterized by a fruity smell.
Given that ethanol $(C_2H_5OH)$ is one of the reactants,the other reactant must be a carboxylic acid to produce an ester.
Among the options,$CH_3COOH$ (acetic acid) is the only carboxylic acid.
The reaction is: $CH_3COOH + C_2H_5OH \xrightarrow{conc. H_2SO_4} CH_3COOC_2H_5 + H_2O$.

(C) When $CH_3CHCl_2$ ($1$,$1$-dichloroethane) is heated with aqueous $KOH$,it undergoes nucleophilic substitution to form a geminal diol,$CH_3CH(OH)_2$.
Geminal diols are unstable and lose a water molecule to form acetaldehyde $(CH_3CHO)$:
$CH_3CHCl_2$ $\xrightarrow{aq. KOH} CH_3CH(OH)_2$ $\xrightarrow{-H_2O} CH_3CHO$.

(A) The mechanism of the Cannizzaro reaction involves the nucleophilic attack of the hydroxide ion on the carbonyl carbon to form a di-anion intermediate.
The rate-determining step (slowest step) is the transfer of a hydride ion $(H^-)$ from the di-anion intermediate to the carbonyl carbon of a second molecule of aldehyde.
This hydride transfer results in the formation of a carboxylate ion and an alkoxide ion,which subsequently undergo proton exchange to yield the final products,$PhCH_2OH$ and $PhCOO^-$.

(B) Buna-$N$ is a synthetic rubber formed by the copolymerization of $1,3$-butadiene $(CH_2=CH-CH=CH_2)$ and acrylonitrile $(CH_2=CH-CN)$ in the presence of a peroxide catalyst.

(C) Carbohydrates are polyhydroxy aldehydes or polyhydroxy ketones.This means they contain multiple hydroxyl groups $(-OH)$ and either an aldehyde group $(-CHO)$ or a ketone group $( > C=O)$.