IIT JEE 1997 Chemistry Question Paper with Answer and Solution

(B) It was in $1913$ that Neils Bohr put forth the stability of the atom and with the help of Planck's quantum theory explained the reason for spectral lines.
Bohr first made use of quantum theory to explain the structure of atoms and proposed that the energy of electrons in an atom is quantized.

(A) $1$. $BF_3$ has a trigonal planar geometry with $sp^2$ hybridization,but it is non-polar due to the cancellation of dipole moments.
$2$. $SiF_4$ has a tetrahedral geometry with $sp^3$ hybridization and is non-polar.
$3$. $HClO_2$ has a central $Cl$ atom with $sp^3$ hybridization.
$4$. $H_2CO_3$ (carbonic acid) has a central carbon atom with $sp^2$ hybridization. Due to the asymmetric arrangement of atoms and the difference in electronegativity between $C$,$O$,and $H$,the molecule is polar.

(B) The compressibility factor $Z$ is defined as $Z = \frac{PV}{nRT}$.
For an ideal gas,the equation of state is $PV = nRT$.
Substituting this into the expression for $Z$,we get $Z = \frac{nRT}{nRT} = 1$.
Therefore,the compressibility factor for an ideal gas is $1$.

(B) For an ideal gas,the equation of state is $PV = nRT$.
By definition,the compressibility factor $Z$ is given by the ratio $Z = \frac{PV}{nRT}$.
Substituting $PV = nRT$ into the expression for $Z$,we get $Z = \frac{nRT}{nRT} = 1$.
Therefore,the compressibility factor for an ideal gas is $1$.

(D) The dissociation of the salt $M_2X_3$ is given by:
$M_2X_3(s) \rightleftharpoons 2M^{3+}(aq) + 3X^{2-}(aq)$
If the solubility is $y \ mol \ dm^{-3}$,then the concentration of $M^{3+}$ is $2y \ mol \ dm^{-3}$ and the concentration of $X^{2-}$ is $3y \ mol \ dm^{-3}$.
The solubility product $K_{sp}$ is defined as:
$K_{sp} = [M^{3+}]^2 [X^{2-}]^3$
Substituting the values:
$K_{sp} = (2y)^2 \times (3y)^3$
$K_{sp} = (4y^2) \times (27y^3)$
$K_{sp} = 108y^5 \ mol^5 \ dm^{-15}$
Therefore,the correct option is $(D)$.

(D) The standard molar enthalpy of formation of a compound is defined as the enthalpy change when $1 \ mol$ of the compound is formed from its constituent elements in their most stable states at $298 \ K$ and $1 \ bar$ pressure.
For $CO_2$,the reaction is: $C(\text{graphite}) + O_2(g) \rightarrow CO_2(g)$.
This reaction also represents the combustion of carbon (graphite) in oxygen to form $CO_2$.
Therefore,the standard molar enthalpy of formation of $CO_2$ is equal to the standard molar enthalpy of combustion of carbon (graphite).

(B) The correct answer is $(B)$.
$1$. The first ionisation potential $(IE_1)$ of $Al$ $([Ne] 3s^2 3p^1)$ is less than $Mg$ $([Ne] 3s^2)$ because $Mg$ has a stable fully-filled $s$-orbital.
$2$. The second ionisation potential $(IE_2)$ of $Na$ $([Ne] 3s^1 \rightarrow [Ne])$ is much higher than that of $Mg$ $([Ne] 3s^2 \rightarrow [Ne] 3s^1)$ because the second electron in $Na$ is removed from a stable noble gas core $([Ne])$.
$3$. $IE_1$ of $Na$ is less than $Mg$ due to higher effective nuclear charge and stable configuration of $Mg$.
$4$. The third ionisation potential $(IE_3)$ of $Mg$ is higher than $Al$ because the third electron in $Mg$ is removed from a stable noble gas core $([Ne])$,whereas in $Al$ it is removed from the $3s^2$ subshell.

(D) The critical temperature of a substance depends on the strength of intermolecular forces.
$H_2O$ is a polar molecule with a permanent dipole moment $(1.84 \ D)$,which leads to strong dipole-dipole interactions and hydrogen bonding.
In contrast,$O_2$ is a non-polar molecule with a dipole moment of $0 \ D$,exhibiting only weak London dispersion forces.
Therefore,$H_2O$ has a higher critical temperature than $O_2$.

(B) In alkaline earth metals,as the atomic number increases,the atomic size increases.
For hydroxides,the lattice energy decreases more rapidly than the hydration energy as we move down the group from $Be$ to $Ba$.
This leads to an increase in the solubility of their hydroxides.
Conversely,properties like ionisation energy $(IE)$,electronegativity,and the solubility of their sulphates decrease as the atomic number increases.

(C) The electronic configurations are as follows:
$1$. $KO_2$ contains the superoxide ion $O_2^-$. The molecular orbital configuration of $O_2^-$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. It has one unpaired electron.
$2$. $NO_2^-$ has $5 + 6(2) + 1 = 18$ valence electrons. All electrons are paired.
$3$. $BaO_2$ contains the peroxide ion $O_2^{2-}$. The configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. All electrons are paired.
$4$. $NO_2^+$ has $5 + 6(2) - 1 = 16$ valence electrons. All electrons are paired.
Therefore,only $KO_2$ contains an unpaired electron.

(A) Compound $A$ is methyl hydrogen tartrate (where the top group is $COOCH_3$ and the bottom is $COOH$).
Compound $B$ is also methyl hydrogen tartrate (where the top group is $COOH$ and the bottom is $COOCH_3$).
By rotating compound $B$ by $180^{\circ}$ in the plane of the paper,the $COOH$ group moves to the top and the $COOCH_3$ group moves to the bottom,while the $H$ and $OH$ groups on the chiral carbons remain in their relative positions.
After rotation,the structure of $B$ becomes identical to the structure of $A$.
Therefore,$A$ and $B$ are identical.

(D) $2$-methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$.
Mono-chlorination gives the following products:
$1.$ $ClCH_2-CH(CH_3)-CH_2-CH_3$ (Chiral center at $C2$,$1$ pair of enantiomers)
$2.$ $CH_3-CCl(CH_3)-CH_2-CH_3$ (Achiral)
$3.$ $CH_3-CH(CH_3)-CHCl-CH_3$ (Chiral center at $C3$,$1$ pair of enantiomers)
$4.$ $CH_3-CH(CH_3)-CH_2-CH_2Cl$ (Achiral)
Total enantiomeric pairs = $1 + 1 = 2$.

(D) The structure of $2$-methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$.
Monochlorination can occur at four different types of hydrogen atoms:
$1$. At $C-1$ (terminal methyl group): $Cl-CH_2-CH(CH_3)-CH_2-CH_3$. This molecule has a chiral center at $C-2$,so it exists as a pair of enantiomers.
$2$. At $C-2$: $CH_3-CCl(CH_3)-CH_2-CH_3$. This molecule is achiral.
$3$. At $C-3$: $CH_3-CH(CH_3)-CHCl-CH_3$. This molecule has a chiral center at $C-3$,so it exists as a pair of enantiomers.
$4$. At $C-4$: $CH_3-CH(CH_3)-CH_2-CH_2Cl$. This molecule is achiral.
Thus,there are $2$ chiral products,each forming an enantiomeric pair. Therefore,the total number of enantiomeric pairs is $2$.

(B) The addition of $HCl$ to propene follows Markovnikov's rule,even in the presence of peroxide.
Peroxide effect (Kharasch effect) is only applicable to $HBr$ and not to $HCl$ or $HI$.
In the reaction,the electrophile $H^+$ attacks the double bond to form the more stable secondary carbocation,$CH_3-CH^{+}-CH_3$,as the reaction intermediate.
Therefore,the correct intermediate is $CH_3-CH^{+}-CH_3$.

(D) The debromination of $meso-2,3-dibromobutane$ proceeds via an $E2$ mechanism,which is stereospecific.
In the $meso$ form,the two bromine atoms are anti to each other in the staggered conformation.
Elimination of $Br_2$ from this conformation leads to the formation of $trans-2-butene$.
$trans-2-butene$ is more stable than $cis-2-butene$ due to reduced steric hindrance between the methyl groups.

(C) Nitration of aromatic compounds is an electrophilic aromatic substitution reaction where the rate-determining step is the formation of the sigma complex (arenium ion).
Since the $C-H$ or $C-D$ bond cleavage occurs after the rate-determining step,there is no primary kinetic isotope effect observed.
Therefore,the rate of nitration of benzene $(C_6H_6)$ and hexadeuterobenzene $(C_6D_6)$ is essentially the same.
Statement $C$ is false because the rates are equal,not greater.

(A) In the nitration of benzene,the mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$ is used.
Here,$H_2SO_4$ acts as a stronger acid and protonates $HNO_3$ to generate the electrophile $NO_2^+$.
The reaction is as follows:
$HNO_3 + H_2SO_4 \rightleftharpoons H_2NO_3^+ + HSO_4^-$
$H_2NO_3^+ \rightleftharpoons H_2O + NO_2^+$
Since $HNO_3$ accepts a proton from $H_2SO_4$,it acts as a base in this reaction.

(C) In $VOSO_4$,the oxidation state of Vanadium is $V^{+4}$.
The electronic configuration of $V^{+4}$ is $[Ar] 3d^1 4s^0$.
Since it has $1$ unpaired electron,it is paramagnetic.
Due to the presence of an unpaired electron and $d-d$ transition,it is a coloured compound.

(A) The balanced redox reaction in an acidic medium is:
$3MnO_4^- + 5FeC_2O_4 + 24H^+ \to 3Mn^{2+} + 5Fe^{3+} + 10CO_2 + 12H_2O$
From the stoichiometry of the balanced equation,$5 \ mol$ of $FeC_2O_4$ reacts with $3 \ mol$ of $KMnO_4$.
Therefore,$1 \ mol$ of $FeC_2O_4$ will react with $3/5 \ mol$ of $KMnO_4$.

(A) Given $g(x) = \int_0^x \cos^4 t \, dt$.
We need to evaluate $g(x + \pi) = \int_0^{x + \pi} \cos^4 t \, dt$.
Using the property of definite integrals,we can split the interval:
$g(x + \pi) = \int_0^{\pi} \cos^4 t \, dt + \int_{\pi}^{x + \pi} \cos^4 t \, dt$.
By definition,the first integral is $g(\pi)$.
For the second integral,let $t = u + \pi$,then $dt = du$. When $t = \pi$,$u = 0$,and when $t = x + \pi$,$u = x$.
Since $\cos^4(u + \pi) = (-\cos u)^4 = \cos^4 u$,the second integral becomes $\int_0^x \cos^4 u \, du = g(x)$.
Therefore,$g(x + \pi) = g(\pi) + g(x)$.

(C) Given $pK_b = 10.83$ for $F^-$.
The base dissociation constant $K_b = 10^{-pK_b} = 10^{-10.83} = 1.479 \times 10^{-11} \approx 1.48 \times 10^{-11}$.
At $25\,^{\circ}C$,the ionic product of water $K_w = 10^{-14}$.
The relationship between $K_a$ and $K_b$ for a conjugate acid-base pair is $K_a \times K_b = K_w$.
Therefore,$K_a = \frac{K_w}{K_b} = \frac{10^{-14}}{1.48 \times 10^{-11}} = 6.756 \times 10^{-4} \approx 6.75 \times 10^{-4}$.

(B) The molar heat capacity at constant pressure is defined as $C_P = (\frac{\delta H}{\delta T})_P$.
At phase equilibrium between ice and water,the temperature remains constant even when heat is added,meaning $\delta T = 0$.
Therefore,$C_P = \frac{\delta H}{0} = \infty$.

(B) According to Wien's displacement law,the product of the absolute temperature $T$ and the wavelength $\lambda_m$ corresponding to maximum intensity is a constant:
$T \lambda_m = b$
This implies $T \propto \frac{1}{\lambda_m}$.
Given:
$\lambda_S = 510 \ nm$ (wavelength for the Sun)
$\lambda_N = 350 \ nm$ (wavelength for the North Star)
Therefore,the ratio of the surface temperature of the Sun $(T_S)$ to the North Star $(T_N)$ is:
$\frac{T_S}{T_N} = \frac{\lambda_N}{\lambda_S}$
Substituting the values:
$\frac{T_S}{T_N} = \frac{350}{510} \approx 0.686 \approx 0.69$.

(C) The ideal gas equation is given by $PV = nRT$,where $P$ is pressure,$V$ is volume,$n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the absolute temperature.
For the $O_2$ gas vessel:
$P_1 = P, V_1 = V, n_1 = 1, T_1 = T$
So,$PV = (1)RT = RT$ --- (Equation $1$)
For the $He$ gas vessel:
$P_2 = ?, V_2 = V, n_2 = 1, T_2 = 2T$
Using the ideal gas equation: $P_2 V = n_2 R T_2$
$P_2 V = (1)R(2T) = 2RT$ --- (Equation $2$)
Dividing Equation $2$ by Equation $1$:
$P_2 V / PV = 2RT / RT$
$P_2 / P = 2$
$P_2 = 2P$
Therefore,the pressure of the $He$ gas is $2P$.

(C) The energy of an $X$-ray photon is given by the relation $E = \frac{hc}{\lambda}$.
Given $\lambda = 0.021 \, nm = 0.21 \, \mathring{A}$.
Using the approximation $E \approx \frac{12400 \, eV \cdot \mathring{A}}{\lambda (\text{in } \mathring{A})}$,we get:
$E = \frac{12400}{0.21} \, eV \approx 59047 \, eV$.
Since the $K_{\alpha}$ line corresponds to the transition from the $L$-shell to the $K$-shell,the energy difference is $\Delta E = E_L - E_K = E_{photon} \approx 59 \, keV$.

(C) We are given $I_1 = \int_{1-k}^{k} x f\{x(1-x)\} dx$ and $I_2 = \int_{1-k}^{k} f\{x(1-x)\} dx$.
Using the property $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a+b-x) dx$,we apply it to $I_1$ with $a = 1-k$ and $b = k$,so $a+b = 1$.
$I_1 = \int_{1-k}^{k} (1-x) f\{(1-x)(1-(1-x))\} dx$
$I_1 = \int_{1-k}^{k} (1-x) f\{(1-x)x\} dx$
$I_1 = \int_{1-k}^{k} f\{x(1-x)\} dx - \int_{1-k}^{k} x f\{x(1-x)\} dx$
$I_1 = I_2 - I_1$
$2I_1 = I_2$
Therefore,$\frac{I_1}{I_2} = \frac{1}{2}$.

(B) Let us analyze the chemical composition of the given ores:
$1$. Calamine is $ZnCO_3$ and siderite is $FeCO_3$. Both are carbonates. This statement is correct.
$2$. Argentite is $Ag_2S$ (a sulphide) and cuprite is $Cu_2O$ (an oxide). Since both are not oxides,this statement is incorrect.
$3$. Zinc blende is $ZnS$ and pyrites (iron pyrites) is $FeS_2$. Both are sulphides. This statement is correct.
$4$. Malachite is $CuCO_3 \cdot Cu(OH)_2$ and azurite is $2CuCO_3 \cdot Cu(OH)_2$. Both are ores of copper. This statement is correct.
Therefore,the incorrect statement is $B$.

(B) The chemical formulas for the ores are as follows:
$1$. Calamine is $ZnCO_{3}$ and siderite is $FeCO_{3}$ (both are carbonates).
$2$. Argentite is $Ag_{2}S$ (sulphide) and cuprite is $Cu_{2}O$ (oxide). Thus,the statement that both are oxides is incorrect.
$3$. Zinc blende is $ZnS$ and iron pyrites is $FeS_{2}$ (both are sulphides).
$4$. Malachite is $CuCO_{3} \cdot Cu(OH)_{2}$ and azurite is $Cu_{3}(CO_{3})_{2}(OH)_{2}$ (both are copper ores).
Therefore,the incorrect statement is $B$.

(C) In $K_2Cr_2O_7$,the oxidation state of $Cr$ is $+6$. $Cr^{+6} = [Ar] 3d^0$.
In $(NH_4)_2[TiCl_6]$,the oxidation state of $Ti$ is $+4$. $Ti^{+4} = [Ar] 3d^0$.
In $VOSO_4$,the oxidation state of $V$ is $+4$. $V^{+4} = [Ar] 3d^1$.
In $K_3[Cu(CN)_4]$,the oxidation state of $Cu$ is $+1$. $Cu^{+1} = [Ar] 3d^{10}$.
In $VOSO_4$,the $V^{+4}$ configuration is $d^1$,which contains one unpaired $d$-electron,making it paramagnetic and coloured.
All other compounds listed have no unpaired $d$-electrons,making them diamagnetic in nature.

(C) The ideal gas equation is given by $PV = nRT$.
Since the vessels are identical,the volume $V$ is the same for both gases.
For the first vessel containing $O_2$ gas: $P_1 V = n_1 R T_1$,where $n_1 = 1 \, mole$,$T_1 = T$,and $P_1 = P$.
So,$PV = (1)RT \Rightarrow PV = RT$.
For the second vessel containing $He$ gas: $P_2 V = n_2 R T_2$,where $n_2 = 1 \, mole$,$T_2 = 2T$,and $P_2$ is the unknown pressure.
So,$P_2 V = (1)R(2T) \Rightarrow P_2 V = 2RT$.
Dividing the two equations: $\frac{P_2 V}{PV} = \frac{2RT}{RT}$.
This simplifies to $\frac{P_2}{P} = 2$,which gives $P_2 = 2P$.

(B) $1$. Analyze the circuit: The battery is connected to a $100 \, \Omega$ resistor in series with a parallel combination of two branches.
$2$. Branch $1$ contains diode $D_1$ (forward biased) and a $150 \, \Omega$ resistor. Total resistance of this branch = $R_{D1} + R_1 = 50 \, \Omega + 150 \, \Omega = 200 \, \Omega$.
$3$. Branch $2$ contains diode $D_2$ (reverse biased). Since it has infinite backward resistance, no current flows through this branch.
$4$. The total resistance of the circuit is the sum of the $100 \, \Omega$ resistor and the equivalent resistance of the parallel branches. Since only Branch $1$ is active, the total resistance $R_{eq} = 100 \, \Omega + 200 \, \Omega = 300 \, \Omega$.
$5$. The current $I$ through the $100 \, \Omega$ resistor is given by Ohm's Law: $I = \frac{V}{R_{eq}} = \frac{6 \, V}{300 \, \Omega} = 0.02 \, A$.

(B) The structure of $2$-methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$.
Monochlorination can occur at four different types of hydrogen atoms:
$1$. At $C_1$ (terminal methyl group): $CH_2Cl-CH(CH_3)-CH_2-CH_3$. This product has a chiral center at $C_2$,so it exists as a pair of enantiomers ($R$ and $S$).
$2$. At $C_2$: $CH_3-CCl(CH_3)-CH_2-CH_3$. This product has a chiral center at $C_2$,so it exists as a pair of enantiomers ($R$ and $S$).
$3$. At $C_3$: $CH_3-CH(CH_3)-CHCl-CH_3$. This product has a chiral center at $C_3$,so it exists as a pair of enantiomers ($R$ and $S$).
$4$. At $C_4$ (terminal methyl group): $CH_3-CH(CH_3)-CH_2-CH_2Cl$. This product is achiral.
Thus,there are $3$ products that exist as enantiomeric pairs. Therefore,the number of enantiomeric pairs is $3$.

(D) The time taken for the plate to fall through a distance $h$ under gravity is given by $h = \frac{1}{2} g t^2$.
Given $h = 10 \, cm = 0.1 \, m$ and $g = 9.8 \, m/s^2$ (or $10 \, m/s^2$ for simplicity).
Using $g = 10 \, m/s^2$,we have $0.1 = \frac{1}{2} \times 10 \times t^2$.
$0.1 = 5 t^2 \implies t^2 = 0.02 \implies t = \sqrt{0.02} = \sqrt{\frac{2}{100}} = \frac{\sqrt{2}}{10} \, s$.
In this time $t$,the tuning fork completes $n = 8$ oscillations.
The frequency $f$ is given by $f = \frac{n}{t}$.
$f = \frac{8}{\frac{\sqrt{2}}{10}} = \frac{80}{\sqrt{2}} = 40 \sqrt{2} \approx 40 \times 1.414 = 56.56 \, Hz$.
Rounding to the nearest provided option,the frequency is $56 \, Hz$.

(B) According to Wien's displacement law,the product of the wavelength corresponding to maximum intensity $(\lambda_{\max})$ and the absolute temperature $(T)$ of a black body is constant.
$\lambda_{\max} T = b$ (constant)
Therefore,$T \propto \frac{1}{\lambda_{\max}}$.
Let $T_S$ and $\lambda_S$ be the temperature and wavelength of the sun,and $T_N$ and $\lambda_N$ be the temperature and wavelength of the north star.
$\frac{T_S}{T_N} = \frac{(\lambda_N)_{\max}}{(\lambda_S)_{\max}}$
Given $(\lambda_S)_{\max} = 510 \ nm$ and $(\lambda_N)_{\max} = 350 \ nm$.
$\frac{T_S}{T_N} = \frac{350}{510} = \frac{35}{51} \approx 0.686 \approx 0.69$.

(B) The molar heat capacity $(C)$ is defined as $C = \frac{dQ}{dT}$.
When water is in equilibrium with ice at a constant pressure,the phase transition occurs at a constant temperature (the melting point).
Since the temperature remains constant during the phase change,the change in temperature $\Delta T = 0$.
However,there is heat exchange $(\Delta Q \neq 0)$ as the substance absorbs or releases latent heat.
Therefore,$C = \frac{\Delta Q}{\Delta T} = \frac{\Delta Q}{0} \rightarrow \infty$.
Thus,the molar heat capacity is infinite. Option $B$ is correct.

(B) Electrode potential is a relative term,meaning it is always measured with respect to a reference electrode.
In electrochemistry,the standard hydrogen electrode is arbitrarily chosen as the reference electrode.
To establish a basis for comparison with all other electrode reactions,its standard electrode potential is defined as $0.00 \ V$ at all temperatures.

(B) The incorrect statement is $B$.
Calamine is $ZnCO_3$ and siderite is $FeCO_3$,which are both carbonates.
Argentite is $Ag_2S$ (sulphide) and cuprite is $Cu_2O$ (oxide),so the statement that both are oxides is incorrect.
Zinc blende is $ZnS$ and pyrites (iron pyrites) is $FeS_2$,which are both sulphides.
Malachite is $Cu(OH)_2 \cdot CuCO_3$ and azurite is $2CuCO_3 \cdot Cu(OH)_2$,which are both ores of copper.

(C) When $6 \ M \ NaOH$ is added in excess to a solution containing $Fe^{3+}$,$Zn^{2+}$,and $Cu^{2+}$:
$1$. $Fe^{3+}$ reacts with $OH^-$ to form a reddish-brown precipitate of $Fe(OH)_3$,which is insoluble in excess $NaOH$.
$2$. $Zn^{2+}$ reacts with $OH^-$ to form $Zn(OH)_2$,which dissolves in excess $NaOH$ to form the soluble complex $[Zn(OH)_4]^{2-}$.
$3$. $Cu^{2+}$ reacts with $OH^-$ to form a blue precipitate of $Cu(OH)_2$,which is insoluble in excess $NaOH$.
However,in the context of standard qualitative analysis,$Fe^{3+}$ is separated as $Fe(OH)_3$ from the others. Given the options,$6 \ M \ NaOH$ is the correct reagent to distinguish $Fe^{3+}$ (as a precipitate) from the amphoteric $Zn^{2+}$ (as a soluble complex).

(B) The leaving group ability is inversely proportional to the basicity of the leaving group. Weaker bases are better leaving groups.
The acidity of the corresponding conjugate acids follows the order: $CF_3SO_3H > CH_3SO_3H > CH_3COOH > CH_3OH$.
Therefore,the basicity order of their conjugate bases is: $CH_3O^- > CH_3COO^- > CH_3SO_3^- > CF_3SO_3^-$.
Thus,the leaving group ability order is $IV > III > I > II$.

(A) Among the given compounds,the reactivity towards nucleophilic attack depends on the electrophilicity of the carbonyl carbon.
$MeCOCl$ (acid chloride) is the most reactive because the $Cl$ atom exerts a strong $-I$ effect and a weak $+R$ effect,making the carbonyl carbon highly electron-deficient.
The general order of reactivity towards nucleophilic acyl substitution is: $MeCOCl > MeCOOCOMe > MeCOOMe > MeCHO$.
Therefore,$MeCOCl$ is the most susceptible to nucleophilic attack.

(D) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$1$. $ICH_2COCH_2CH_3$ contains the $CH_3CO-$ group (after considering the reaction conditions,it acts as a methyl ketone derivative),which reacts with $I_2/NaOH$ to form $CHI_3$ (yellow precipitate).
$2$. $CH_3CH(OH)CH_2CH_3$ is a secondary alcohol with a $CH_3CH(OH)-$ group,which is oxidized to a methyl ketone and then undergoes the iodoform reaction to give $CHI_3$ (yellow precipitate).
Therefore,both $(a)$ and $(c)$ give a yellow precipitate.

(C) $K_3[Cu(CN)_4]$: $Cu$ is in $+1$ oxidation state $(3d^{10})$,so it has no unpaired electrons,making it diamagnetic and colourless.
$(NH_4)_2[TiCl_6]$: $Ti$ is in $+4$ oxidation state $(3d^0)$,so it has no unpaired electrons,making it diamagnetic and colourless.
$VOSO_4$: $V$ is in $+4$ oxidation state $(3d^1)$,so it has one unpaired electron,making it paramagnetic and coloured.
$K_2Cr_2O_7$: $Cr$ is in $+6$ oxidation state $(3d^0)$,so it has no unpaired electrons. It is diamagnetic,but appears coloured due to ligand-to-metal charge transfer $(LMCT)$.