IIT JEE 1999 Chemistry Question Paper with Answer and Solution

(B) Consider a gas-filled compartment moving with acceleration $a$ in the horizontal direction.
In the non-inertial frame of reference of the compartment,a fictitious force acts on the gas molecules in the direction opposite to the acceleration (i.e.,towards the rear side).
This causes the gas molecules to crowd towards the rear side of the compartment.
As a result,the density and consequently the pressure of the gas increase at the rear side.
Conversely,the density and pressure decrease at the front side of the compartment.
Therefore,the pressure is lower in the front side.

(D) Oxygen is a diatomic gas,so its internal energy for $n_1 = 2$ moles is given by $U_1 = n_1 \times \frac{f_1}{2} RT$,where $f_1 = 5$ (degrees of freedom for diatomic gas).
$U_1 = 2 \times \frac{5}{2} RT = 5 RT$.
Argon is a monoatomic gas,so its internal energy for $n_2 = 4$ moles is given by $U_2 = n_2 \times \frac{f_2}{2} RT$,where $f_2 = 3$ (degrees of freedom for monoatomic gas).
$U_2 = 4 \times \frac{3}{2} RT = 6 RT$.
The total internal energy of the system is $U_{total} = U_1 + U_2 = 5 RT + 6 RT = 11 RT$.

(C) The chemical reaction between calcium phosphide $(Ca_3P_2)$ and water is as follows:
$Ca_3P_2 + 6H_2O \to 2PH_3 + 3Ca(OH)_2$
According to the stoichiometry of the balanced chemical equation,$1 \text{ mole}$ of $Ca_3P_2$ reacts with $6 \text{ moles}$ of $H_2O$ to produce $2 \text{ moles}$ of phosphine $(PH_3)$ and $3 \text{ moles}$ of calcium hydroxide $(Ca(OH)_2)$.
Therefore,the correct option is $C$.

(D) To determine the hybridization of the carbon atoms,we number the chain: $C_1H_2 = C_2H - C_3H_2 - C_4H_2 - C_5 \equiv C_6H$.
$C_1$ is $sp^2$ hybridized.
$C_2$ is $sp^2$ hybridized.
$C_3$ is $sp^3$ hybridized.
$C_4$ is $sp^3$ hybridized.
$C_5$ is $sp$ hybridized.
$C_6$ is $sp$ hybridized.
The bond between $C_2$ and $C_3$ involves an $sp^2$ hybridized carbon and an $sp^3$ hybridized carbon.
Therefore,the bond type is $sp^2 - sp^3$.

(D) The bond order $(B.O.)$ is inversely proportional to the bond length.
For $CO$,the bond order is $3$.
For $CO_2$ $(O=C=O)$,the bond order is $2$.
For $CO_3^{2-}$,the resonance hybrid results in a bond order of $1.33$.
Since bond length increases as bond order decreases,the order of increasing bond length is $CO < CO_2 < CO_3^{2-}$.
Therefore,option $(D)$ is correct.

(D) The ideal gas equation is $PV = nRT$.
Real gases approach ideal behavior under conditions of high temperature and low pressure,where the intermolecular forces are negligible and the volume of the gas particles is insignificant compared to the total volume of the container.

(A) The reaction is $3X_{(g)} + Y_{(g)} \rightleftharpoons X_3Y_{(g)}$.
Since the number of moles of gaseous reactants $(4 \ mol)$ is different from the number of moles of gaseous products $(1 \ mol)$,the equilibrium position is affected by pressure.
Since chemical reactions involve enthalpy changes,the equilibrium constant and the position of equilibrium are affected by temperature.
$A$ catalyst only increases the rate of the forward and backward reactions equally and does not affect the equilibrium position or the amount of product at equilibrium.
Therefore,the amount of $X_3Y$ at equilibrium is affected by temperature and pressure.

(D) The correct answer is $D$.
$A$ buffer solution is prepared by mixing a weak acid with its conjugate base (salt of a strong base) or a weak base with its conjugate acid (salt of a strong acid).
$1$. Sodium acetate $(CH_3COONa)$ and acetic acid $(CH_3COOH)$ form an acidic buffer because acetic acid is a weak acid and sodium acetate is its salt with a strong base $(NaOH)$.
$2$. Ammonia $(NH_3)$ and ammonium chloride $(NH_4Cl)$ form a basic buffer because ammonia is a weak base and ammonium chloride is its salt with a strong acid $(HCl)$.
$3$. Sodium acetate and hydrochloric acid $(HCl)$ react to form acetic acid and sodium chloride,which does not constitute a buffer system in the required proportions.
Therefore,both $A$ and $C$ are correct.

(A) $HCl$ is a strong acid,so its $pH$ is the lowest $(pH \approx 1)$.
$NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$,so it undergoes cationic hydrolysis,resulting in an acidic solution $(pH < 7)$.
$NaCl$ is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$,so it is neutral $(pH = 7)$.
$NaCN$ is a salt of a weak acid $(HCN)$ and a strong base $(NaOH)$,so it undergoes anionic hydrolysis,resulting in a basic solution $(pH > 7)$.
Therefore,the order of increasing $pH$ is: $HCl < NH_4Cl < NaCl < NaCN$.

(D) The correct option is $(D)$. All the given reactions are endothermic because they proceed by the absorption of heat from the surroundings.
$1$. Conversion of graphite to diamond: $C_{\text{graphite}} \rightarrow C_{\text{diamond}}$ $(\Delta H > 0)$
$2$. Decomposition of water: $2H_2O_{(l)} \rightarrow 2H_{2(g)} + O_{2(g)}$ $(\Delta H > 0)$
$3$. Dehydrogenation of ethane: $C_2H_{6(g)} \rightarrow C_2H_{4(g)} + H_{2(g)}$ $(\Delta H > 0)$
Since all these processes require energy input,they are endothermic.

(A) $1$. For $S_8$: Since it is an elemental form of sulphur,the oxidation number is $0$.
$2$. For $S_2F_2$: Let the oxidation number of $S$ be $x$. Since the oxidation number of $F$ is $-1$,we have $2x + 2(-1) = 0$,which gives $2x = 2$,so $x = +1$.
$3$. For $H_2S$: Let the oxidation number of $S$ be $x$. Since the oxidation number of $H$ is $+1$,we have $2(+1) + x = 0$,which gives $x = -2$.
Therefore,the oxidation numbers are $0, +1, -2$.

(D) The ionic radius depends on the effective nuclear charge and the number of shells.
$(A)$ $Ti^{4+}$ $(Z=22)$ and $Mn^{7+}$ $(Z=25)$ are isoelectronic ($18$ electrons). As nuclear charge increases,size decreases,so $Ti^{4+} > Mn^{7+}$.
$(B)$ $^{35}Cl^{-}$ and $^{37}Cl^{-}$ are isotopes. Isotopes have the same number of electrons and protons,so their ionic radii are equal.
$(C)$ $K^{+}$ and $Cl^{-}$ are isoelectronic ($18$ electrons). $K^{+}$ has a higher nuclear charge $(Z=19)$ than $Cl^{-}$ $(Z=17)$,so $K^{+} < Cl^{-}$.
$(D)$ $P^{3+}$ and $P^{5+}$ are ions of the same element. $P^{5+}$ has lost more electrons than $P^{3+}$,resulting in a higher effective nuclear charge per electron,making $P^{3+} > P^{5+}$.
Thus,the correct statement is $(D)$.

(C) $Ca_3P_2 + 6H_2O \rightarrow 2PH_3 + 3Ca(OH)_2$
According to the balanced chemical equation,$1 \text{ mole}$ of calcium phosphide $(Ca_3P_2)$ reacts with excess water to produce $2 \text{ moles}$ of phosphine $(PH_3)$ and $3 \text{ moles}$ of calcium hydroxide $(Ca(OH)_2)$.

(C) The chemical reaction between calcium phosphide $(Ca_3P_2)$ and water is given by the equation:
$Ca_3P_2 + 6H_2O \to 3Ca(OH)_2 + 2PH_3$
From the stoichiometry of the balanced equation,$1 \text{ mole}$ of $Ca_3P_2$ reacts with $6 \text{ moles}$ of $H_2O$ to produce $2 \text{ moles}$ of phosphine $(PH_3)$ and $3 \text{ moles}$ of calcium hydroxide $(Ca(OH)_2)$.

(B) The correct order is $BCl_3 > PCl_3 > AsCl_3 > BiCl_3$.
$BCl_3$ has a trigonal planar geometry with a bond angle of $120^{\circ}$ due to $sp^2$ hybridization.
In the case of $PCl_3$,$AsCl_3$,and $BiCl_3$,the central atom is $sp^3$ hybridized with one lone pair,resulting in a pyramidal geometry.
As we move down the group from $P$ to $Bi$,the electronegativity of the central atom decreases,causing the bond pairs to move further away from the central atom.
Consequently,the bond angle decreases as we move down the group.

(C) $D-(+)-$ tartaric acid is named based on its configuration relative to $D-(+)$ glyceraldehyde and its positive optical rotation.

(A) The hydration of $1$-butyne $(CH_3-CH_2-C \equiv CH)$ in the presence of $HgSO_4$ and $H_2SO_4$ follows Markovnikov's rule.
Water adds across the triple bond to form an unstable enol intermediate $(CH_3-CH_2-C(OH)=CH_2)$.
This enol then tautomerizes to form the more stable ketone,$2$-butanone $(CH_3-CH_2-C(=O)-CH_3)$.

(D) The formula for normality is $\text{Normality} = \text{Molarity} \times \text{n-factor}$.
For phosphorous acid $(H_3PO_3)$,the structure contains two $P-OH$ bonds,making it a dibasic acid.
Therefore,the n-factor (basicity) of $H_3PO_3$ is $2$.
$\text{Normality} = 0.3 \ M \times 2 = 0.6 \ N$.

(A) The hydration of terminal alkynes like $but-1-yne$ in the presence of $HgSO_4$ and $H_2SO_4$ follows Markovnikov's rule.
Step $1$: The alkyne reacts with water to form an unstable enol intermediate: $CH_3-CH_2-C \equiv CH + H_2O \xrightarrow{HgSO_4/H_2SO_4} [CH_3-CH_2-C(OH)=CH_2]$.
Step $2$: The enol undergoes tautomerization to form a stable ketone: $[CH_3-CH_2-C(OH)=CH_2] ightarrow CH_3-CH_2-COCH_3$ $(Butan-2-one)$.
Therefore,the correct product is $Butan-2-one$.

(C) Let $L = \mathop {\lim }\limits_{x \to 0} \frac{{x\tan 2x - 2x\tan x}}{{{{(1 - \cos 2x)}^2}}}$
Using the identity $1 - \cos 2x = 2\sin^2 x$,we get:
$L = \mathop {\lim }\limits_{x \to 0} \frac{{x(\tan 2x - 2\tan x)}}{{{{(2\sin^2 x)}^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{x(\tan 2x - 2\tan x)}}{{4\sin^4 x}}$
Using the Taylor series expansion $\tan \theta = \theta + \frac{\theta^3}{3} + \dots$ and $\sin x \approx x$ as $x \to 0$:
$L = \mathop {\lim }\limits_{x \to 0} \frac{{x \left[ (2x + \frac{(2x)^3}{3} + \dots) - 2(x + \frac{x^3}{3} + \dots) \right]}}{{4x^4}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{x \left[ 2x + \frac{8x^3}{3} - 2x - \frac{2x^3}{3} \right]}}{{4x^4}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{x \cdot (2x^3)}}{{4x^4}} = \mathop {\lim }\limits_{x \to 0} \frac{{2x^4}}{{4x^4}} = \frac{2}{4} = \frac{1}{2}$.

(A) Let $f(x) = x^2 - 2ax + a^2 + a - 3$.
For the roots to be real and less than $3$,we must satisfy three conditions:
$1$. Discriminant $D \geq 0$:
$D = (-2a)^2 - 4(1)(a^2 + a - 3) = 4a^2 - 4a^2 - 4a + 12 = 12 - 4a \geq 0 \implies a \leq 3$.
$2$. Vertex $x_v < 3$:
$x_v = -b/(2a) = 2a/2 = a < 3$.
$3$. $f(3) > 0$:
$f(3) = 3^2 - 2a(3) + a^2 + a - 3 = 9 - 6a + a^2 + a - 3 = a^2 - 5a + 6 > 0$.
$(a - 2)(a - 3) > 0 \implies a < 2$ or $a > 3$.
Combining all conditions: $a \leq 3$ $AND$ $a < 3$ $AND$ ($a < 2$ or $a > 3$).
The intersection is $a < 2$.

(D) Let the point on the $x$-axis be $(h, 0)$. The equation of the chord with midpoint $(h, 0)$ is given by $T = S_1$.
The equation of the circle is $x^{2} + y^{2} - px - qy = 0$.
The equation of the chord is $xh + y(0) - \frac{p}{2}(x + h) - \frac{q}{2}(y + 0) = h^{2} + 0^{2} - ph - q(0)$.
$xh - \frac{p}{2}(x + h) - \frac{qy}{2} = h^{2} - ph$.
Since this chord passes through $(p, q)$,we substitute $x = p$ and $y = q$:
$ph - \frac{p}{2}(p + h) - \frac{q^{2}}{2} = h^{2} - ph$.
Multiplying by $2$: $2ph - p^{2} - ph - q^{2} = 2h^{2} - 2ph$.
$2h^{2} - 3ph + p^{2} + q^{2} = 0$.
Since there are two distinct chords,the quadratic equation in $h$ must have two distinct real roots,so the discriminant $D > 0$.
$D = (-3p)^{2} - 4(2)(p^{2} + q^{2}) > 0$.
$9p^{2} - 8p^{2} - 8q^{2} > 0$.
$p^{2} - 8q^{2} > 0 \Rightarrow p^{2} > 8q^{2}$.

(D) The internal energy $U$ of $n$ moles of an ideal gas is given by $U = \frac{f}{2} nRT$,where $f$ is the number of degrees of freedom.
For $O_2$ (diatomic gas),the number of degrees of freedom $f_1 = 5$ (neglecting vibrational modes).
Internal energy of $2$ moles of $O_2$ is $U_1 = \frac{5}{2} \times 2 \times RT = 5 RT$.
For $Ar$ (monatomic gas),the number of degrees of freedom $f_2 = 3$.
Internal energy of $4$ moles of $Ar$ is $U_2 = \frac{3}{2} \times 4 \times RT = 6 RT$.
The total internal energy of the system is $U_{total} = U_1 + U_2 = 5 RT + 6 RT = 11 RT$.

(A) The energy of an orbital is determined by the $(n + l)$ rule.
$(i)$ $n = 4, l = 1 \implies n + l = 5$ ($4p$ orbital).
$(ii)$ $n = 4, l = 0 \implies n + l = 4$ ($4s$ orbital).
$(iii)$ $n = 3, l = 2 \implies n + l = 5$ ($3d$ orbital).
$(iv)$ $n = 3, l = 1 \implies n + l = 4$ ($3p$ orbital).
For orbitals with the same $(n + l)$ value,the one with the lower $n$ value has lower energy.
Comparing $(ii)$ $(n=4)$ and $(iv)$ $(n=3)$,$(iv)$ has lower energy.
Comparing $(i)$ $(n=4)$ and $(iii)$ $(n=3)$,$(iii)$ has lower energy.
Thus,the order of increasing energy is $(iv) < (ii) < (iii) < (i)$.

(A) The atomic number of nitrogen is $Z = 7$. The electronic configuration is $1s^2, 2s^2, 2p^3$.
According to Hund's rule of maximum multiplicity,for a given electron configuration,the term with the maximum multiplicity has the lowest energy.
Therefore,the three electrons in the $2p$ orbitals must occupy separate orbitals with parallel spins.
This corresponds to the configuration $1s^2, 2s^2, 2p_x^1, 2p_y^1, 2p_z^1$,which is represented by option $A$.

(A) The central atom $S$ in $H_2S$ has $6$ valence electrons. It forms $2$ bond pairs with $H$ atoms and has $2$ lone pairs of electrons.
According to $VSEPR$ theory,the presence of $2$ lone pairs causes the molecule to adopt a bent or angular geometry.
Due to the difference in electronegativity between $S$ and $H$ and the angular shape,the bond dipoles do not cancel each other out.
Therefore,$H_2S$ has a non-zero dipole moment.

(B) $HCl$ is a strong acid. In its $0.1 \ M$ solution,$[H^{+}] = 0.1 \ M$,hence $pH = 1$.
$NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$,so it undergoes cationic hydrolysis to form an acidic solution $(pH < 7)$.
$NaCl$ is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$,so it does not undergo hydrolysis,resulting in a neutral solution $(pH = 7)$.
$NaCN$ is a salt of a weak acid $(HCN)$ and a strong base $(NaOH)$,so it undergoes anionic hydrolysis to form an alkaline solution $(pH > 7)$.
Therefore,the $pH$ increases in the order: $HCl < NH_4Cl < NaCl < NaCN$.

(C) In resonance structures,the nitrogen atom $(N)$ in the nitro group has a valency of $4$ (forming $4$ bonds).
If $N$ forms $5$ bonds,it would exceed its octet,which is not possible for second-period elements like nitrogen.
Structure $(C)$ shows the nitrogen atom forming $5$ bonds (one double bond with the ring,two double bonds with oxygens,and one single bond with the other oxygen),which violates the octet rule.
Therefore,it is the most unlikely representation.

(C) Since this is a head-on elastic collision between two identical spheres,their linear velocities will be exchanged. Sphere $A$ comes to rest,and sphere $B$ moves with a linear velocity $v$.
Since the surface is frictionless,there is no torque acting on the spheres about their centers of mass. Therefore,their angular velocities remain unchanged.
Thus,$\omega_A = \omega$ and $\omega_B = 0$.
Consequently,after the collision,sphere $A$ will only perform rotational motion (without linear motion),and sphere $B$ will only perform linear motion (without rotation).

(B) The function is given by $f(x) = (x^2 - 1) |(x - 1)(x - 2)| + \cos(|x|)$.
First,consider the term $|(x - 1)(x - 2)|$. This term is non-differentiable at the roots of the quadratic expression inside the absolute value,which are $x = 1$ and $x = 2$.
However,we must check if the factor $(x^2 - 1)$ compensates for this non-differentiability.
At $x = 1$,$f(x) = (x^2 - 1) |(x - 1)(x - 2)| + \cos(|x|)$. The term $(x^2 - 1)$ has a factor of $(x - 1)$,so $(x^2 - 1)|x - 1| = (x + 1)(x - 1)|x - 1|$. Since $\lim_{x \to 1} (x - 1)|x - 1| = 0$ and the derivative of $(x - 1)|x - 1|$ at $x = 1$ is $0$,the function is differentiable at $x = 1$.
At $x = 2$,the term $|x - 2|$ is not differentiable. The expression is $f(x) = (x^2 - 1)|x - 1||x - 2| + \cos(|x|)$. Since $(x^2 - 1)|x - 1|$ is non-zero at $x = 2$ (specifically,$(2^2 - 1)|2 - 1| = 3 \neq 0$),the product $3|x - 2|$ is not differentiable at $x = 2$.
Finally,consider $\cos(|x|)$. This is $\cos(x)$ for $x \ge 0$ and $\cos(-x) = \cos(x)$ for $x < 0$. The derivative of $\cos(|x|)$ at $x = 0$ is $\lim_{h \to 0^+} \frac{\cos(h) - 1}{h} = 0$ and $\lim_{h \to 0^-} \frac{\cos(-h) - 1}{h} = 0$. Thus,$\cos(|x|)$ is differentiable at $x = 0$.
Therefore,the function is not differentiable at $x = 2$.

(C) We need to evaluate $I = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} [2 \sin x] \,dx$.
For $x \in [\frac{\pi}{2}, \frac{3\pi}{2}]$,$\sin x$ ranges from $1$ to $-1$,so $2 \sin x$ ranges from $2$ to $-2$.
We break the integral based on the values of $[2 \sin x]$:
$1) \text{ For } x \in [\frac{\pi}{2}, \frac{5\pi}{6}], \frac{1}{2} \leq \sin x \leq 1 \Rightarrow 1 \leq 2 \sin x \leq 2$. Thus $[2 \sin x] = 1$ for $x \in [\frac{\pi}{2}, \frac{5\pi}{6})$ and $2$ at $x = \frac{\pi}{2}$.
$2) \text{ For } x \in [\frac{5\pi}{6}, \pi], 0 \leq \sin x \leq \frac{1}{2} \Rightarrow 0 \leq 2 \sin x \leq 1$. Thus $[2 \sin x] = 0$ for $x \in [\frac{5\pi}{6}, \pi)$.
$3) \text{ For } x \in [\pi, \frac{7\pi}{6}], -\frac{1}{2} \leq \sin x \leq 0 \Rightarrow -1 \leq 2 \sin x \leq 0$. Thus $[2 \sin x] = -1$ for $x \in [\pi, \frac{7\pi}{6})$.
$4) \text{ For } x \in [\frac{7\pi}{6}, \frac{3\pi}{2}], -1 \leq \sin x \leq -\frac{1}{2} \Rightarrow -2 \leq 2 \sin x \leq -1$. Thus $[2 \sin x] = -2$ for $x \in [\frac{7\pi}{6}, \frac{3\pi}{2})$.
$I = \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} (1) \,dx + \int_{\frac{5\pi}{6}}^{\pi} (0) \,dx + \int_{\pi}^{\frac{7\pi}{6}} (-1) \,dx + \int_{\frac{7\pi}{6}}^{\frac{3\pi}{2}} (-2) \,dx$
$I = (\frac{5\pi}{6} - \frac{\pi}{2}) + 0 - (\frac{7\pi}{6} - \pi) - 2(\frac{3\pi}{2} - \frac{7\pi}{6})$
$I = \frac{\pi}{3} - \frac{\pi}{6} - 2(\frac{9\pi - 7\pi}{6}) = \frac{\pi}{6} - 2(\frac{2\pi}{6}) = \frac{\pi}{6} - \frac{2\pi}{3} = \frac{\pi - 4\pi}{6} = -\frac{3\pi}{6} = -\frac{\pi}{2}$.

(A) According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases.
If the $(n+l)$ values are the same,the orbital with the lower value of $n$ has lower energy.
Calculating $(n+l)$ values for each:
$(i)$ $n=4, l=1 \implies n+l=5$
$(ii)$ $n=4, l=0 \implies n+l=4$
$(iii)$ $n=3, l=2 \implies n+l=5$
$(iv)$ $n=3, l=1 \implies n+l=4$
Comparing the values:
For $(n+l)=4$: $(iv)$ has $n=3$ and $(ii)$ has $n=4$. Thus,$iv < ii$.
For $(n+l)=5$: $(iii)$ has $n=3$ and $(i)$ has $n=4$. Thus,$iii < i$.
Combining these,the order of increasing energy is: $iv < ii < iii < i$.

(A) According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases. If two orbitals have the same $(n+l)$ value,the one with the lower $n$ value has lower energy.
Calculating $(n+l)$ for each:
$(i) n = 4, l = 1 \implies n+l = 4+1 = 5$
$(ii) n = 4, l = 0 \implies n+l = 4+0 = 4$
$(iii) n = 3, l = 2 \implies n+l = 3+2 = 5$
$(iv) n = 3, l = 1 \implies n+l = 3+1 = 4$
Comparing the values:
For $(ii)$ and $(iv)$,$(n+l) = 4$. Since $(iv)$ has a lower $n$ value $(3 < 4)$,$(iv) < (ii)$.
For $(i)$ and $(iii)$,$(n+l) = 5$. Since $(iii)$ has a lower $n$ value $(3 < 4)$,$(iii) < (i)$.
Combining these,the order of increasing energy is $(iv) < (ii) < (iii) < (i)$.

(C) The chemical reaction between calcium phosphide and water is represented as:
$Ca_3P_2 + 6H_2O \rightarrow 2PH_3 + 3Ca(OH)_2$
From the stoichiometry of the balanced equation,$1 \text{ mole}$ of $Ca_3P_2$ reacts with $6 \text{ moles}$ of $H_2O$ to produce $2 \text{ moles}$ of phosphine $(PH_3)$ and $3 \text{ moles}$ of calcium hydroxide $(Ca(OH)_2)$.
Therefore,one mole of calcium phosphide yields $2 \text{ moles}$ of phosphine.

(C) Energy is released in a nuclear reaction if the total binding energy of the products is greater than the total binding energy of the reactants. This corresponds to an increase in the binding energy per nucleon.
For nuclear fission,a heavy nucleus splits into lighter nuclei with higher binding energy per nucleon. For nuclear fusion,lighter nuclei combine to form a heavier nucleus with higher binding energy per nucleon.
Looking at the graph:
- $Z$ has a mass number of $30$ and a binding energy per nucleon of $5.0 \text{ MeV}$.
- $Y$ has a mass number of $60$ and a binding energy per nucleon of $8.5 \text{ MeV}$.
- $X$ has a mass number of $90$ and a binding energy per nucleon of $8.0 \text{ MeV}$.
- $W$ has a mass number of $120$ and a binding energy per nucleon of $7.5 \text{ MeV}$.
In the process $W \to 2Y$:
- Reactant $W$ $(A=120)$ has a total binding energy of $120 \times 7.5 = 900 \text{ MeV}$.
- Products $2Y$ $(2 \times 60 = 120)$ have a total binding energy of $2 \times (60 \times 8.5) = 1020 \text{ MeV}$.
- Since the total binding energy of the products $(1020 \text{ MeV})$ is greater than that of the reactant $(900 \text{ MeV})$,energy is released.

(C) Energy is released in a nuclear process if the total binding energy of the products is greater than the total binding energy of the reactants.
From the graph:
For $Z$: Mass number $A = 30$,$BE/A = 5.0 \text{ MeV}$
For $Y$: Mass number $A = 60$,$BE/A = 8.5 \text{ MeV}$
For $X$: Mass number $A = 90$,$BE/A = 8.0 \text{ MeV}$
For $W$: Mass number $A = 120$,$BE/A = 7.5 \text{ MeV}$
Let's evaluate option $(c)$: $W \to 2Y$
Total $BE$ of reactant $W = 120 \times 7.5 = 900 \text{ MeV}$
Total $BE$ of products $2Y = 2 \times (60 \times 8.5) = 2 \times 510 = 1020 \text{ MeV}$
Since the total $BE$ of the products $(1020 \text{ MeV})$ is greater than the total $BE$ of the reactant $(900 \text{ MeV})$,energy is released in this process.

(A) In a hexagonal close-packed $(HCP)$ structure,each atom is surrounded by $12$ nearest neighbors.
Specifically,an atom in a given layer is in contact with $6$ atoms in its own layer,$3$ atoms in the layer above,and $3$ atoms in the layer below,totaling $6 + 3 + 3 = 12$.

(C) Energy is released in a nuclear process if the total binding energy of the products is greater than the total binding energy of the reactants.
From the graph, we identify the mass numbers $(A)$ and binding energy per nucleon $(BE/A)$ for the nuclei:
For $W$: $A = 120, BE/A = 7.5 \, MeV \implies \text{Total } BE = 120 \times 7.5 = 900 \, MeV$.
For $Y$: $A = 60, BE/A = 8.5 \, MeV \implies \text{Total } BE = 60 \times 8.5 = 510 \, MeV$.
Now, let us check option $(c)$: $W \to 2Y$.
Total $BE$ of reactants $= 900 \, MeV$.
Total $BE$ of products $= 2 \times (510) = 1020 \, MeV$.
Since the total binding energy of the products $(1020 \, MeV)$ is greater than that of the reactants $(900 \, MeV)$, energy is released in this process.

(C) According to the law of conservation of linear momentum,since the initial particle is at rest,the total momentum before and after decay must be zero.
$0 = m_1 \vec{v}_1 + m_2 \vec{v}_2 \Rightarrow m_1 \vec{v}_1 = -m_2 \vec{v}_2$.
Taking the magnitude of the momenta,we have $|p_1| = |p_2| = p$,where $p = m_1 v_1 = m_2 v_2$.
The de-Broglie wavelength is given by $\lambda = h/p$.
Since both particles have the same magnitude of momentum $p$,their de-Broglie wavelengths are $\lambda_1 = h/p$ and $\lambda_2 = h/p$.
Therefore,the ratio $\lambda_1/\lambda_2 = (h/p) / (h/p) = 1.0$.

(D) The thermal decomposition of ammonium dichromate is an exothermic reaction.
The chemical equation for the reaction is:
$(NH_4)_2Cr_2O_7 \rightarrow N_2 + Cr_2O_3 + 4H_2O$
As shown in the equation,the gas evolved is nitrogen $(N_2)$.

(C) Given that the half-life of $X$ equals the mean life of $Y$: $(T_{1/2})_X = (\tau)_Y$.
We know that $(T_{1/2})_X = \frac{0.693}{\lambda_X}$ and $(\tau)_Y = \frac{1}{\lambda_Y}$.
Equating these, we get $\frac{0.693}{\lambda_X} = \frac{1}{\lambda_Y}$, which implies $\lambda_X = 0.693 \lambda_Y$.
Since $0.693 < 1$, it follows that $\lambda_X < \lambda_Y$.
The rate of decay is given by $R = \lambda N$.
Initially, the number of atoms $N$ is the same for both elements. Since $\lambda_Y > \lambda_X$, the decay rate $R_Y = \lambda_Y N$ will be greater than $R_X = \lambda_X N$.
Therefore, $Y$ will decay at a faster rate than $X$.

(D) The total internal energy $U$ of a gas mixture is the sum of the internal energies of its individual components: $U = U_1 + U_2$.
For oxygen $(O_2)$,which is a diatomic gas,the degrees of freedom $f_1 = 5$ (neglecting vibrational modes). Thus,$C_{v1} = \frac{f_1}{2}R = \frac{5}{2}R$.
For argon $(Ar)$,which is a monatomic gas,the degrees of freedom $f_2 = 3$. Thus,$C_{v2} = \frac{f_2}{2}R = \frac{3}{2}R$.
Given $n_1 = 2$ moles of oxygen and $n_2 = 4$ moles of argon:
$U = n_1 C_{v1} T + n_2 C_{v2} T$
$U = 2 \times (\frac{5}{2}R)T + 4 \times (\frac{3}{2}R)T$
$U = 5RT + 6RT = 11RT$.

(C) Energy is released in a nuclear process when the total binding energy $(B.E.)$ of the products is greater than the total binding energy of the reactants.
From the graph, we can identify the mass number $(A)$ and binding energy per nucleon $(B.E./A)$ for each nucleus:
For $W: A = 120, B.E./A = 7.5 \, MeV \implies \text{Total } B.E. = 120 \times 7.5 = 900 \, MeV$
For $Y: A = 60, B.E./A = 8.5 \, MeV \implies \text{Total } B.E. = 60 \times 8.5 = 510 \, MeV$
For $X: A = 90, B.E./A = 8.0 \, MeV \implies \text{Total } B.E. = 90 \times 8.0 = 720 \, MeV$
For $Z: A = 30, B.E./A = 5.0 \, MeV \implies \text{Total } B.E. = 30 \times 5.0 = 150 \, MeV$
Now, let us check the options:
$(A)$ $Y \to 2Z$: Reactant $B.E. = 510 \, MeV$, Product $B.E. = 2 \times 150 = 300 \, MeV$. (Energy absorbed)
$(B)$ $W \to X + Z$: Reactant $B.E. = 900 \, MeV$, Product $B.E. = 720 + 150 = 870 \, MeV$. (Energy absorbed)
$(C)$ $W \to 2Y$: Reactant $B.E. = 900 \, MeV$, Product $B.E. = 2 \times 510 = 1020 \, MeV$. (Energy released as $1020 > 900$)
$(D)$ $X \to Y + Z$: Reactant $B.E. = 720 \, MeV$, Product $B.E. = 510 + 150 = 660 \, MeV$. (Energy absorbed)
Therefore, the correct process is $W \to 2Y$.

(C) Energy is released in a nuclear reaction if the total binding energy of the products is greater than the total binding energy of the reactants.
From the graph:
For nucleus $W$: Mass number $A = 120$,$BE/A = 7.5 \, MeV$.
For nucleus $Y$: Mass number $A = 60$,$BE/A = 8.5 \, MeV$.
For nucleus $Z$: Mass number $A = 30$,$BE/A = 5.0 \, MeV$.
Let us check option $(c)$: $W \to 2Y$
Total $BE$ of reactant $W = 120 \times 7.5 = 900 \, MeV$.
Total $BE$ of products $= 2 \times (60 \times 8.5) = 1020 \, MeV$.
Since the total $BE$ of products $(1020 \, MeV)$ is greater than the total $BE$ of the reactant $(900 \, MeV)$,energy is released in this process.

(A) According to the $(n+l)$ rule,the higher the value of $(n+l)$,the higher is the energy of the orbital.
If the $(n+l)$ values are the same,the orbital with the lower value of $n$ has lower energy.
Calculating $(n+l)$ values:
$(i) n=4, l=1 \implies n+l = 5$
$(ii) n=4, l=0 \implies n+l = 4$
$(iii) n=3, l=2 \implies n+l = 5$
$(iv) n=3, l=1 \implies n+l = 4$
Comparing the values:
For $(iv)$ and $(ii)$,both have $(n+l) = 4$. Since $n=3 < n=4$,$(iv) < (ii)$.
For $(i)$ and $(iii)$,both have $(n+l) = 5$. Since $n=3 < n=4$,$(iii) < (i)$.
Combining these,the order of increasing energy is $(iv) < (ii) < (iii) < (i)$.

(B) The structure of the dichromate dianion $(Cr_2O_7^{2-})$ consists of two $CrO_4$ tetrahedra sharing a common oxygen atom.
In this structure,there are $6$ terminal $Cr-O$ bonds (three on each chromium atom) and $2$ bridging $Cr-O-Cr$ bonds.
The $6$ terminal $Cr-O$ bonds are equivalent due to resonance,while the $2$ bridging bonds are different from the terminal ones.
Therefore,$6$ $Cr-O$ bonds are equivalent.

(C) In the commercial electrochemical process for aluminium extraction,known as the Hall-$H$éroult process,the electrolyte used is a molten mixture of $Al_2O_3$ (alumina),$Na_3AlF_6$ (cryolite),and $CaF_2$ (fluorspar).
Cryolite lowers the melting point of the mixture and increases its electrical conductivity.

(D) The thermal decomposition of ammonium dichromate is given by the following reaction:
$(NH_4)_2Cr_2O_7 \to N_2 + Cr_2O_3 + 4H_2O$
As shown in the equation,the gas evolved is nitrogen $(N_2)$.

(D) The reaction of ammonia $(NH_3)$ with hypochlorite anion $(OCl^-)$ is a method for the preparation of hydrazine $(N_2H_4)$.
The balanced chemical equation is:
$2NH_3 + OCl^- \to N_2H_4 + Cl^- + H_2O$
However,in the presence of excess ammonia and specific conditions,the reaction can also produce ammonium chloride $(NH_4Cl)$ as a byproduct.
Therefore,both $NH_4Cl$ and $N_2H_4$ can be formed.

(D) $SbCl_5$ is a strong Lewis acid. It reacts with the $C-Cl$ bond of $1-$chloro$-1-$phenylethane to abstract the chloride ion,forming a stable benzylic carbocation intermediate: $Ph-CH(CH_3)-Cl + SbCl_5 \rightarrow [Ph-CH^+(CH_3)] + [SbCl_6]^-$.
Since the carbocation is planar,the chloride ion can attack from either side,leading to the formation of a racemic mixture of the $(d)$ and $(l)$ forms.

(D) The freezing point of a solution is the temperature at which the vapour pressure of the solution becomes equal to the vapour pressure of the pure solid solvent.
At the freezing point,only the solvent molecules solidify,while the solute molecules remain in the liquid phase.
Furthermore,the addition of a non-volatile solute lowers the vapour pressure of the solvent,which leads to a depression in the freezing point.
Therefore,both statements $A$ and $B$ are correct.

(B) In a hexagonal close packed $(hcp)$ structure,each atom is in contact with $12$ other atoms.
Therefore,the coordination number of a metal crystallizing in an $hcp$ structure is $12$.

(D) $1$. According to the Van't Hoff equation,$\ln K_p = -\frac{\Delta H^o}{RT} + C$. Thus,a plot of $\log \, K_p$ versus $1/T$ is linear. Statement $(a)$ is correct.
$2$. For a first-order reaction,$\ln [X]_t = -kt + \ln [X]_0$. Converting to base $10$,$\log \, [X]_t = -\frac{kt}{2.303} + \log \, [X]_0$. Thus,a plot of $\log \, [X]$ versus time is linear. Statement $(b)$ is correct.
$3$. According to Boyle's law,at constant temperature,$P \propto 1/V$,which implies $P = k(1/V)$. Thus,a plot of $P$ versus $1/V$ is linear. Statement $(c)$ is correct.
$4$. Since $(a)$,$(b)$,and $(c)$ are all correct,the correct option is $(d)$.

(A) The reduction potential represents the tendency to gain electrons. Given the order $Z > Y > X$,$Z$ has the highest tendency to be reduced,followed by $Y$,and then $X$.
An oxidizing agent is a substance that gets reduced itself and oxidizes others. $A$ species with a higher reduction potential can oxidize a species with a lower reduction potential.
Since the reduction potential of $Y > X$,$Y$ can oxidize $X$.
Since the reduction potential of $Z > Y$,$Y$ cannot oxidize $Z$ (in fact,$Z$ would oxidize $Y$).
Therefore,$Y$ will oxidize $X$ but not $Z$.

(A) The thermal decomposition of ammonium dichromate is given by the following equation:
$(NH_4)_2Cr_2O_7 \rightarrow N_2 + 4H_2O + Cr_2O_3$
As shown in the reaction,the gas evolved is nitrogen $(N_2)$.

(D) In $Ni(CO)_4$,the oxidation state of $Ni$ is $0$. The configuration is $3d^8 4s^2$. Due to the strong field ligand $CO$,electrons pair up,resulting in $sp^3$ hybridization,which gives a tetrahedral geometry.
In $Ni(PPh_3)_2Cl_2$,the oxidation state of $Ni$ is $+2$. The configuration is $3d^8$. $PPh_3$ is a bulky ligand,and $Cl^-$ is a weak field ligand. This complex also undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry.
Therefore,both complexes are tetrahedral.

(A) The reaction of an alkyl aryl ether with $HI$ involves the cleavage of the $C-O$ bond.
In the given ether,$C_6H_5-O-CH_2-C_6H_5$,the oxygen is attached to a phenyl group $(C_6H_5-)$ and a benzyl group $(-CH_2-C_6H_5)$.
The $C-O$ bond between the oxygen and the phenyl group is stronger due to partial double bond character (resonance),while the $C-O$ bond between the oxygen and the benzyl group is weaker because the resulting carbocation $(C_6H_5CH_2^+)$ is resonance-stabilized.
Therefore,the $HI$ attacks the benzyl carbon,leading to the formation of benzyl iodide $(C_6H_5CH_2I)$ and phenol $(C_6H_5OH)$.

(B) Acetone $(CH_3-CO-CH_3)$ undergoes keto-enol tautomerism in the presence of $D_2O$.
The $\alpha$-hydrogens are acidic and exchange with deuterium atoms from $D_2O$.
This exchange process continues until all six $\alpha$-hydrogens are replaced by deuterium,resulting in the formation of hexadeuteroacetone $(CD_3-CO-CD_3)$.

(D) The reaction between propionic acid and sodium bicarbonate is an acid-base reaction:
$CH_3CH_2COOH_{(aq)} + NaHCO_{3(aq)} \to CH_3CH_2COONa_{(aq)} + H_2O_{(l)} + CO_{2(g)}$
In this reaction,the carboxylic acid $(R-COOH)$ acts as a proton donor,and the bicarbonate ion $(HCO_3^-)$ acts as a base.
The $HCO_3^-$ ion accepts a proton to form carbonic acid $(H_2CO_3)$,which then decomposes into $H_2O$ and $CO_2$.
Therefore,the carbon atom in the evolved $CO_2$ gas originates from the bicarbonate ion $(NaHCO_3)$.
Thus,the correct option is $(D)$.

(B) The carbylamine test is a characteristic reaction for primary $(1^\circ)$ amines (both aliphatic and aromatic).
$2, 4$-dimethylaniline is a primary aromatic amine $(Ar-NH_2)$,so it reacts with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to form an isocyanide (carbylamine),which has a foul smell.
$N, N$-dimethylaniline is a tertiary $(3^\circ)$ amine.
$N$-methyl-$o$-methylaniline is a secondary $(2^\circ)$ amine.
$p$-methylbenzylamine is a primary amine,but the question specifically refers to the structure provided in the image,which is $2, 4$-dimethylaniline (also known as $2, 4$-xylidine).