IIT JEE 2004 Chemistry Question Paper with Answer and Solution

(D) The radius of an orbit in a hydrogen-like species is given by the formula: $r_n = 0.529 \times \frac{n^2}{Z} \ \mathring{A}$.
For the first Bohr orbit of the hydrogen atom $(H)$: $n = 1$ and $Z = 1$. Thus,$r_H = 0.529 \times \frac{1^2}{1} = 0.529 \ \mathring{A}$.
Now,checking the options:
For $Be^{3+}$: $Z = 4$ and $n = 2$. Thus,$r = 0.529 \times \frac{2^2}{4} = 0.529 \times \frac{4}{4} = 0.529 \ \mathring{A}$.
Since the radius of $Be^{3+}$ $(n = 2)$ is equal to the radius of the first Bohr orbit of hydrogen,the correct option is $D$.

(B) The molecular orbital configuration for $O_2$ is $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x^2 \equiv \pi 2p_y^2)(\pi^* 2p_x^1 \equiv \pi^* 2p_y^1)$.
Bond order of $O_2 = \frac{10-6}{2} = 2.0$. It has two unpaired electrons,so it is paramagnetic.
The molecular orbital configuration for $O_2^+$ is $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x^2 \equiv \pi 2p_y^2)(\pi^* 2p_x^1 \equiv \pi^* 2p_y^0)$.
Bond order of $O_2^+ = \frac{10-5}{2} = 2.5$.
Since $O_2^+$ has one unpaired electron,it is paramagnetic.
Comparing the values,$2.5 > 2.0$,so the bond order of $O_2^+ > O_2$.

(C) The root mean square velocity of one mole of a monoatomic gas is given by $V_{rms} = \sqrt{\frac{3RT}{M}} \dots (1)$.
The average kinetic energy $(E)$ of one mole of a monoatomic gas is given by $E = \frac{3}{2}RT$.
From this,we can express $RT$ as $RT = \frac{2E}{3} \dots (2)$.
Substituting equation $(2)$ into equation $(1)$:
$V_{rms} = \sqrt{\frac{3}{M} \times \frac{2E}{3}}$
$V_{rms} = \sqrt{\frac{2E}{M}}$.

(B) The salt $NaX$ is formed from a weak acid $HX$ and a strong base $NaOH$. The hydrolysis reaction is: $X^{-} + H_2O ⇌ HX + OH^{-}$.
The hydrolysis constant $K_h$ is given by $K_h = \frac{K_w}{K_a} = \frac{10^{-14}}{10^{-5}} = 10^{-9}$.
The degree of hydrolysis $h$ for a salt of a weak acid and strong base is given by $h = \sqrt{\frac{K_h}{C}}$,where $C = 0.1 \ M$.
$h = \sqrt{\frac{10^{-9}}{0.1}} = \sqrt{10^{-8}} = 10^{-4}$.
To express this as a percentage: $h \times 100 = 10^{-4} \times 100 = 0.01 \%$.

(B) At the boiling point,the process of vaporization is in equilibrium,so $\Delta G = 0$.
Using the relation $\Delta G = \Delta H - T\Delta S$,we get $\Delta H = T\Delta S$.
Given $\Delta H = 30 \ kJ \ mol^{-1} = 30000 \ J \ mol^{-1}$ and $\Delta S = 75 \ J \ mol^{-1} \ K^{-1}$.
$T = \frac{\Delta H}{\Delta S} = \frac{30000 \ J \ mol^{-1}}{75 \ J \ mol^{-1} \ K^{-1}} = 400 \ K$.

(C) For an ideal gas,the enthalpy change $\Delta H$ is given by the formula $\Delta H = n C_p \Delta T$.
Since the process is isothermal,the change in temperature $\Delta T = 0$.
Therefore,$\Delta H = n C_p \times 0 = 0 \ kJ$.

(A) In an alkaline medium,the permanganate ion $(MnO_4^{-})$ acts as an oxidizing agent and oxidizes the iodide ion $(I^{-})$ to the iodate ion $(IO_3^{-})$.
The balanced chemical equation for this reaction is:
$I^{-} + 6OH^{-} + 6MnO_4^{-} \rightarrow IO_3^{-} + 6MnO_4^{2-} + 3H_2O$
Thus,the correct product is $IO_3^{-}$.

(A) The thermal decomposition of ammonium dichromate is given by: $(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} N_2 \uparrow + Cr_2O_3 + 4H_2O$. The gas liberated is nitrogen $(N_2)$.
Similarly,the thermal decomposition of ammonium nitrite is: $NH_4NO_2 \xrightarrow{\Delta} N_2 \uparrow + 2H_2O$.
Thus,both reactions produce nitrogen gas.

(A) To find the empirical formula,we calculate the mole ratio of each element:

Element	Percentage / Atomic Mass	Moles	Simple Ratio
$C$	$54.5 / 12$	$4.54$	$4.54 / 2.27 = 2$
$H$	$9.1 / 1$	$9.1$	$9.1 / 2.27 = 4$
$O$	$36.4 / 16$	$2.27$	$2.27 / 2.27 = 1$

The ratio of $C:H:O$ is $2:4:1$. Therefore,the empirical formula is $C_2H_4O$.

(B) The acidity of a proton depends on the stability of its conjugate base.
$1$. Position $(X)$ is the carboxylic acid group $(-COOH)$,which is inherently more acidic than the ammonium groups $(-NH_3^+)$.
$2$. Between the two ammonium groups $(Y)$ and $(Z)$,the position $(Y)$ is closer to the electron-withdrawing $-COOH$ group. The inductive effect ($-I$ effect) of the $-COOH$ group destabilizes the positive charge on the adjacent $-NH_3^+$ group at $(Y)$ more than at $(Z)$,making the proton at $(Y)$ more acidic than the one at $(Z)$.
$3$. Therefore,the order of acidity is $(X) > (Y) > (Z)$.

(D) The given molecule is a bicyclic system where one ring is attached to an $-NH-$ group (which is an activating group due to the lone pair on nitrogen) and the other ring is attached to a $-C(=O)-$ group (which is a deactivating group).
Electrophilic aromatic substitution,such as bromination using $Br_2/Fe$,occurs preferentially on the more electron-rich ring.
The ring containing the $-NH-$ group is activated,making it more susceptible to electrophilic attack.
The $-NH-$ group is ortho/para directing. Due to steric hindrance from the methyl group at the ortho position,the bromination occurs at the para position relative to the $-NH-$ group.
Thus,the bromine atom will be substituted at the para position of the activated ring.

(C) Acid catalyzed hydration of alkenes follows Markovnikov's rule.
$2-$phenylpropene $(CH_3-C(Ph)=CH_2)$ reacts with $H_3O^{+}$ to form a stable carbocation intermediate.
The proton $(H^{+})$ adds to the terminal carbon $(CH_2)$ to form a tertiary benzylic carbocation $(CH_3-C^{+}(Ph)-CH_3)$.
Water then attacks this carbocation to form $2-$phenylpropan$-2-$ol $(CH_3-C(OH)(Ph)-CH_3)$.

(D) The conversion of an internal alkyne like $2$-hexyne into a $trans$-alkene is achieved by Birch reduction using an alkali metal (like $Li$ or $Na$) in liquid ammonia $(NH_3)$.
This reaction proceeds via a $trans$-vinylic radical anion intermediate,which leads to the formation of the thermodynamically more stable $trans$-alkene product.

(B) The equation of any tangent to the parabola ${y^2} = 4x$ is given by $y = mx + \frac{1}{m}$.
Since the tangent passes through the point $(1, 4)$,we substitute these coordinates into the equation:
$4 = m(1) + \frac{1}{m}$
Multiplying by $m$,we get the quadratic equation:
${m^2} - 4m + 1 = 0$
Let the roots be ${m_1}$ and ${m_2}$. From the properties of quadratic equations:
${m_1} + {m_2} = 4$ and ${m_1}{m_2} = 1$.
The difference between the slopes is given by:
$|{m_1} - {m_2}| = \sqrt{({m_1} + {m_2})^2 - 4{m_1}{m_2}} = \sqrt{4^2 - 4(1)} = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3}$.
If $\theta$ is the angle between the tangents,then:
$\tan \theta = \left| \frac{{m_1} - {m_2}}{1 + {m_1}{m_2}} \right| = \frac{2\sqrt{3}}{1 + 1} = \frac{2\sqrt{3}}{2} = \sqrt{3}$.
Therefore,$\theta = \arctan(\sqrt{3}) = \frac{\pi}{3}$.

(A) Let $b = b_1 i + b_2 j + b_3 k$.
Given $a \times b = j - k$,we have:
$j - k = \begin{vmatrix} i & j & k \\ 1 & 1 & 1 \\ b_1 & b_2 & b_3 \end{vmatrix}$
Expanding the determinant:
$j - k = i(b_3 - b_2) - j(b_3 - b_1) + k(b_2 - b_1)$
Comparing coefficients:
$b_3 - b_2 = 0 \Rightarrow b_3 = b_2$
$-(b_3 - b_1) = 1 \Rightarrow b_1 - b_3 = 1 \Rightarrow b_1 = b_3 + 1$
$b_2 - b_1 = -1$
Substituting $b_3 = b_2$ into $b_1 = b_3 + 1$,we get $b_1 = b_2 + 1$.
Given $a \cdot b = 1$,we have:
$(i + j + k) \cdot (b_1 i + b_2 j + b_3 k) = 1$
$b_1 + b_2 + b_3 = 1$
Substituting $b_1 = b_2 + 1$ and $b_3 = b_2$:
$(b_2 + 1) + b_2 + b_2 = 1$
$3b_2 + 1 = 1$
$3b_2 = 0 \Rightarrow b_2 = 0$
Thus,$b_3 = 0$ and $b_1 = 0 + 1 = 1$.
Therefore,$b = 1i + 0j + 0k = i$.

(C) $1$. In the $P-V$ diagram,the isothermal expansion from $V_1$ to $V_2$ follows a curve with a smaller slope compared to the adiabatic compression from $V_2$ to $V_1$.
$2$. Since the adiabatic curve is steeper than the isothermal curve,the final pressure $P_3$ at volume $V_1$ must be greater than the initial pressure $P_1$ $(P_3 > P_1)$.
$3$. The work done in a cyclic process is represented by the area enclosed by the $P-V$ loop. Since the process is counter-clockwise (expansion at lower pressure,compression at higher pressure),the net work done on the gas is positive,meaning the work done by the gas $(W)$ is negative $(W < 0)$.

(B) Any point on the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$ is given by $(2\lambda+1, 3\lambda-1, 4\lambda+1)$.
Any point on the line $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu$ is given by $(\mu+3, 2\mu+k, \mu)$.
For the lines to intersect,there must exist values of $\lambda$ and $\mu$ such that:
$2\lambda+1 = \mu+3$ $(i)$
$3\lambda-1 = 2\mu+k$ $(ii)$
$4\lambda+1 = \mu$ $(iii)$
Substituting $\mu = 4\lambda+1$ from $(iii)$ into $(i)$:
$2\lambda+1 = (4\lambda+1)+3$
$2\lambda+1 = 4\lambda+4$
$-2\lambda = 3 \Rightarrow \lambda = -\frac{3}{2}$.
Now,find $\mu$ using $(iii)$:
$\mu = 4(-\frac{3}{2})+1 = -6+1 = -5$.
Substitute $\lambda = -\frac{3}{2}$ and $\mu = -5$ into $(ii)$:
$3(-\frac{3}{2})-1 = 2(-5)+k$
$-\frac{9}{2}-1 = -10+k$
$-\frac{11}{2} = -10+k$
$k = 10 - \frac{11}{2} = \frac{20-11}{2} = \frac{9}{2}$.

(B) The curves are $y = ax^2$ and $x = ay^2$ (which is $y = \sqrt{x/a}$ for $y \ge 0$).
To find the intersection point,substitute $y = ax^2$ into $x = ay^2$:
$x = a(ax^2)^2 = a^3x^4$
$x(a^3x^3 - 1) = 0$
Since $a > 0$,the intersection points are $x = 0$ and $x = 1/a$.
The area $A$ is given by:
$A = \int_{0}^{1/a} (\sqrt{x/a} - ax^2) dx = 1$
$\frac{1}{\sqrt{a}} \int_{0}^{1/a} x^{1/2} dx - a \int_{0}^{1/a} x^2 dx = 1$
$\frac{1}{\sqrt{a}} [\frac{2}{3} x^{3/2}]_{0}^{1/a} - a [\frac{x^3}{3}]_{0}^{1/a} = 1$
$\frac{1}{\sqrt{a}} \cdot \frac{2}{3} \cdot (\frac{1}{a})^{3/2} - \frac{a}{3} \cdot (\frac{1}{a})^3 = 1$
$\frac{2}{3a^2} - \frac{1}{3a^2} = 1$
$\frac{1}{3a^2} = 1$
$a^2 = \frac{1}{3} \implies a = \frac{1}{\sqrt{3}}$ (since $a > 0$).

(C) The reactivity of Grignard reagents $(RMgX)$ with carbonyl compounds depends on steric hindrance and the electrophilicity of the carbonyl carbon.
$1.$ Aldehydes are generally more reactive than ketones due to less steric hindrance and higher electrophilicity of the carbonyl carbon. Thus,$(II)$ $CH_3-CHO$ is the most reactive.
$2.$ Among ketones,as the size of the alkyl or aryl groups increases,steric hindrance increases and reactivity decreases. Also,the $+I$ effect of methyl groups and resonance in phenyl groups reduce the positive charge on the carbonyl carbon.
$3.$ $(III)$ $CH_3-CO-CH_3$ (Acetone) is more reactive than $(I)$ $Ph-CO-Ph$ (Benzophenone) because phenyl groups are much bulkier than methyl groups and provide resonance stabilization.
Therefore,the correct order is $II > III > I$.

(D) The electronic configuration of $Ni^{2+}$ is $3d^8$.
In $[NiCl_4]^{2-}$,$Cl^-$ is a weak field ligand,so it does not cause pairing of electrons.
Thus,the hybridization is $sp^3$,which results in a tetrahedral geometry.
In contrast,$[PdCl_4]^{2-}$,$[Ni(CN)_4]^{2-}$,and $[Pd(CN)_4]^{2-}$ involve $4d$ or $5d$ metals or strong field ligands,leading to $dsp^2$ hybridization and square planar geometry.

(D) Given $f(x) = x^3 + bx^2 + cx + d$.
Find the derivative: $f'(x) = 3x^2 + 2bx + c$.
To determine the nature of the function,check the discriminant $D$ of the quadratic $f'(x)$:
$D = (2b)^2 - 4(3)(c) = 4b^2 - 12c = 4(b^2 - 3c)$.
Given $0 < b^2 < c$,we know that $b^2 - 3c < 0$ because $b^2 < c$ implies $b^2 - 3c < c - 3c = -2c < 0$.
Since the leading coefficient of $f'(x)$ is $3 > 0$ and $D < 0$,$f'(x) > 0$ for all $x \in \mathbb{R}$.
Therefore,$f(x)$ is strictly increasing.

(C) $1$. In the $P-V$ diagram,the process $AB$ represents an isothermal expansion from $V_1$ to $V_2$,and the process $BC$ represents an adiabatic compression from $V_2$ back to $V_1$.
$2$. From the graph,it is clear that the final pressure $P_3$ at point $C$ is greater than the initial pressure $P_1$ at point $A$ $(P_3 > P_1)$.
$3$. The work done in an isothermal expansion $(W_{iso})$ is the area under the curve $AB$,which is positive. The work done in an adiabatic compression $(W_{adia})$ is the negative of the area under the curve $BC$,which is negative.
$4$. Since the adiabatic curve $BC$ is steeper than the isothermal curve $AB$,the magnitude of work done during adiabatic compression is greater than the work done during isothermal expansion $(|W_{adia}| > |W_{iso}|)$.
$5$. Therefore,the net work done $W = W_{iso} + W_{adia}$ is negative $(W < 0)$.

(C) $1$. In an isothermal expansion from $V_1$ to $V_2$,the work done by the gas is positive $(W_{iso} > 0)$.
$2$. In an adiabatic compression from $V_2$ to $V_1$,the work done on the gas is negative $(W_{adia} < 0)$.
$3$. The slope of an adiabatic process is $\gamma$ times the slope of an isothermal process. Since the adiabatic curve is steeper,the final pressure $P_3$ at volume $V_1$ after adiabatic compression will be greater than the initial pressure $P_1$ $(P_3 > P_1)$.
$4$. The total work done $W$ is the area enclosed by the cycle. Since the cycle is traversed in an anti-clockwise direction,the net work done is negative $(W < 0)$.

(B) The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}} = \frac{1.5 \times 10^9}{4.5 \times 10^9} = \frac{1}{3}$.
The number of remaining $U^{238}$ nuclei $(N_U)$ is given by $N_U = N_0 \left(\frac{1}{2}\right)^n = N_0 \left(\frac{1}{2}\right)^{1/3}$.
Given $2^{1/3} = 1.26$,we have $N_U = \frac{N_0}{1.26}$.
The number of $Pb$ nuclei formed is $N_{Pb} = N_0 - N_U = N_0 - \frac{N_0}{1.26} = N_0 \left(1 - \frac{1}{1.26}\right) = N_0 \left(\frac{0.26}{1.26}\right)$.
The ratio of the number of nuclei of $Pb$ to $U^{238}$ is $\frac{N_{Pb}}{N_U} = \frac{N_0 (0.26 / 1.26)}{N_0 / 1.26} = 0.26$.

(B) $NaX$ is a salt of a weak acid and a strong base.
For such a salt,the degree of hydrolysis $h$ is given by the formula:
$h = \sqrt{\frac{K_w}{K_a \times C}}$
Given:
$K_w = 10^{-14}$
$K_a = 1 \times 10^{-5}$
$C = 0.1 \text{ } M$
Substituting the values:
$h = \sqrt{\frac{10^{-14}}{10^{-5} \times 0.1}} = \sqrt{\frac{10^{-14}}{10^{-6}}} = \sqrt{10^{-8}} = 10^{-4}$
Percentage hydrolysis $= h \times 100 = 10^{-4} \times 100 = 0.01 \%$

(D) The activity of a radioactive sample follows the law $A = A_0 (1/2)^n$,where $n$ is the number of half-lives.
Given that the activity reduces from $6000\,dps$ to $3000\,dps$ in $140\,days$,the half-life $T_{1/2}$ is $140\,days$.
At $t = 280\,days$,the number of half-lives elapsed is $n = 280 / 140 = 2$.
Let $A_i$ be the initial activity. At $t = 280\,days$,the activity $A_{280} = 6000\,dps$.
Using the formula $A_{280} = A_i (1/2)^n$,we have $6000 = A_i (1/2)^2$.
$6000 = A_i / 4$.
$A_i = 6000 \times 4 = 24000\,dps$.

(B) $NaX$ is a salt of a weak acid and a strong base.
For such a salt,the degree of hydrolysis $h$ is given by the formula:
$h = \sqrt{\frac{K_w}{K_a \times C}}$
Given:
$K_w = 10^{-14}$
$K_a = 1 \times 10^{-5}$
$C = 0.1 \ M$
Substituting the values:
$h = \sqrt{\frac{10^{-14}}{10^{-5} \times 0.1}} = \sqrt{\frac{10^{-14}}{10^{-6}}} = \sqrt{10^{-8}} = 10^{-4}$
Percentage hydrolysis $= h \times 100 = 10^{-4} \times 100 = 0.01 \ \%$.

(D) The radius of the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $r_n = 0.529 \times \frac{n^2}{Z} \ \mathring{A}$.
For the $1^{st}$ orbit of $H$ atom,$n=1$ and $Z=1$,so $r_H = 0.529 \times \frac{1^2}{1} = 0.529 \ \mathring{A}$.
Checking the options:
$A$: For $He^{+} \ (n=2, Z=2)$,$r = 0.529 \times \frac{2^2}{2} = 0.529 \times 2 = 1.058 \ \mathring{A}$.
$B$: For $Li^{2+} \ (n=2, Z=3)$,$r = 0.529 \times \frac{2^2}{3} = 0.529 \times \frac{4}{3} = 0.705 \ \mathring{A}$.
$C$: For $Li^{2+} \ (n=3, Z=3)$,$r = 0.529 \times \frac{3^2}{3} = 0.529 \times 3 = 1.587 \ \mathring{A}$.
$D$: For $Be^{3+} \ (n=2, Z=4)$,$r = 0.529 \times \frac{2^2}{4} = 0.529 \times 1 = 0.529 \ \mathring{A}$.
Thus,the radius of $Be^{3+} \ (n=2)$ is the same as that of the $1^{st}$ orbit of $H$ atom.

(B) When a light ray passes through a prism in the position of minimum deviation,the angle of incidence is equal to the angle of emergence $(i = e)$.
In this specific condition,the refracted ray inside the prism $(QR)$ becomes parallel to the base of the prism.
Since the prism is placed on a horizontal table,its base is horizontal.
Therefore,the refracted ray $QR$ inside the prism must be horizontal.

(D) The hybridization of $[NiCl_4]^{2-}$ is $sp^3$.
The shape of $[NiCl_4]^{2-}$ is tetrahedral.
The oxidation state of $Ni$ in this complex is $+2$. The electronic configuration of $Ni^{2+}$ is $3d^8$.
Since the chloride ion $(Cl^-)$ is a weak field ligand,it does not cause electron pairing in the $3d$ orbitals.
Therefore,one $4s$ orbital and three $4p$ orbitals hybridize to form four $sp^3$ hybrid orbitals,which overlap with the orbitals of the four $Cl^-$ ligands,resulting in a tetrahedral geometry.

(B) When a light ray passes through a prism in the position of minimum deviation,the angle of incidence is equal to the angle of emergence $(i = e)$.
In this condition,the refracted ray inside the prism $(QR)$ becomes parallel to the base of the prism.
Since the prism is placed on a horizontal table,its base is horizontal.
Therefore,the ray $QR$ inside the prism must be horizontal.

(C) $1$. In an isothermal expansion from $V_1$ to $V_2$,the work done by the gas is positive $(W_{iso} > 0)$.
$2$. In an adiabatic compression from $V_2$ back to $V_1$,the work done on the gas is negative $(W_{adia} < 0)$.
$3$. On a $P-V$ diagram,the adiabatic curve is steeper than the isothermal curve. Since the adiabatic compression returns to $V_1$ from a lower pressure point at $V_2$,the final pressure $P_3$ must be greater than the initial pressure $P_1$ $(P_3 > P_1)$.
$4$. The net work done $W = W_{iso} + W_{adia}$ corresponds to the area enclosed by the cycle. Since the cycle is traversed in a counter-clockwise direction,the net work done is negative $(W < 0)$.

(D) The magnetic force acting on a charged particle is given by $\vec{F} = q(\vec{v} \times \vec{B})$. Since the magnetic force is always perpendicular to the velocity of the particle,it does no work on the particle. Therefore,the kinetic energy and the speed of the electron remain constant. Thus,$v = u$.
Using the right-hand rule for the magnetic force on an electron (charge $q = -e$): $\vec{F} = -e(\vec{v} \times \vec{B})$. Given $\vec{v} = u\hat{i}$ and $\vec{B} = -B_0\hat{k}$,we have $\vec{F} = -e(u\hat{i} \times -B_0\hat{k}) = -e(uB_0\hat{j}) = -euB_0\hat{j}$.
Initially,the force is directed along the negative $y-$ axis. This causes the electron to follow a circular path that curves downwards. Consequently,when the electron exits the magnetic field region,its coordinate $y$ will be negative $(y < 0)$.

(C) $1$. In an isothermal expansion from $V_1$ to $V_2$,the work done by the gas is positive $(W_{iso} > 0)$,and the area under the curve represents this work.
$2$. In an adiabatic compression from $V_2$ to $V_1$,the work done on the gas is negative $(W_{adia} < 0)$,and the area under the curve represents this work.
$3$. On a $P-V$ diagram,the adiabatic curve is steeper than the isothermal curve. Since the compression happens from $V_2$ to $V_1$ along the adiabatic path,the final pressure $P_3$ will be higher than the initial pressure $P_1$ $(P_3 > P_1)$.
$4$. The total work done $W$ is the sum of the work done in both processes. Since the area under the adiabatic compression curve (which is negative work) is greater than the area under the isothermal expansion curve (which is positive work),the net work done $W$ is negative $(W < 0)$.

(A) According to Einstein's photoelectric equation,$hv - \phi_0 = eV_0$,where $V_0$ is the stopping potential.
$1$. From the graph,curves $a$ and $b$ intersect the anode potential axis at the same point,which means they have the same stopping potential $(V_0)$. Since the stopping potential depends only on the frequency of the incident radiation,we conclude that $f_a = f_b$.
$2$. The saturation current is directly proportional to the intensity of the incident radiation. From the graph,the saturation current for curve $b$ is higher than that for curve $a$. Therefore,the intensities are different,i.e.,$I_a \neq I_b$.

(B) In the condition of minimum deviation for a prism,the angle of incidence is equal to the angle of emergence $(i = e)$.
As a result,the refracted ray inside the prism $(QR)$ becomes parallel to the base of the prism.
Since the prism is placed on a horizontal table,its base is horizontal.
Therefore,the refracted ray $QR$ inside the prism must be horizontal.

(B) The wavelength of a photon with energy $E$ is given by $\lambda_{\text{photon}} = \frac{hc}{E}$.
The de Broglie wavelength of an electron with kinetic energy $E$ is given by $\lambda_{\text{electron}} = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
Taking the ratio of the two wavelengths:
$\frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = \frac{hc/E}{h/\sqrt{2mE}} = \frac{hc}{E} \cdot \frac{\sqrt{2mE}}{h} = c \sqrt{2m} \cdot \frac{\sqrt{E}}{E} = c \sqrt{2m} \cdot \frac{1}{\sqrt{E}}$.
Therefore,$\frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} \propto \frac{1}{\sqrt{E}}$.

(C) $1$. In the $P-V$ diagram,the isothermal expansion from $V_1$ to $V_2$ follows a curve with a smaller slope compared to the adiabatic compression from $V_2$ to $V_1$.
$2$. Since the adiabatic curve is steeper than the isothermal curve,returning to the same volume $V_1$ results in a final pressure $P_3$ that is greater than the initial pressure $P_1$ $(P_3 > P_1)$.
$3$. The total work done $W$ is the area enclosed by the cycle. Since the process is counter-clockwise (expansion at lower pressure,compression at higher pressure),the net work done on the gas is positive,meaning the work done by the gas is negative $(W < 0)$.

(C) The equation of the parabola is $y^2 = 4x$,which is of the form $y^2 = 4ax$,where $a = 1$.
Any tangent to the parabola $y^2 = 4ax$ is given by $y = mx + \frac{a}{m}$.
Substituting $a = 1$,we get $y = mx + \frac{1}{m}$.
Since the tangent passes through the point $(1, 4)$,we have $4 = m(1) + \frac{1}{m}$.
Multiplying by $m$,we get $4m = m^2 + 1$,or $m^2 - 4m + 1 = 0$.
Let the slopes of the two tangents be $m_1$ and $m_2$. Then $m_1 + m_2 = 4$ and $m_1 m_2 = 1$.
The angle $\theta$ between the tangents is given by $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.
We know that $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2 = 4^2 - 4(1) = 16 - 4 = 12$.
Thus,$|m_1 - m_2| = \sqrt{12} = 2\sqrt{3}$.
Substituting these values,$\tan \theta = |\frac{2\sqrt{3}}{1 + 1}| = \frac{2\sqrt{3}}{2} = \sqrt{3}$.
Therefore,$\theta = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$.

(C) The equation of any tangent to the parabola $y^2=4x$ is $y=mx+\frac{1}{m}$.
Since the tangents pass through the point $(1,4)$,we have $4=m(1)+\frac{1}{m}$.
Multiplying by $m$,we get $m^2-4m+1=0$.
Let the slopes of the two tangents be $m_1$ and $m_2$. Then $m_1+m_2=4$ and $m_1m_2=1$.
The difference between the slopes is $|m_1-m_2| = \sqrt{(m_1+m_2)^2-4m_1m_2} = \sqrt{16-4} = \sqrt{12} = 2\sqrt{3}$.
The angle $\theta$ between the tangents is given by $\tan \theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|$.
Substituting the values,$\tan \theta = \left|\frac{2\sqrt{3}}{1+1}\right| = \frac{2\sqrt{3}}{2} = \sqrt{3}$.
Therefore,$\theta = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$.

(B) The equation of a tangent to the parabola $y^2=4ax$ (where $a=1$) is $y=mx+\frac{a}{m}$,which simplifies to $y=mx+\frac{1}{m}$.
Since the tangent passes through the point $(1,4)$,we substitute these coordinates into the equation:
$4 = m(1) + \frac{1}{m}$
$4 = m + \frac{1}{m}$
$m^2 - 4m + 1 = 0$
Let the slopes of the two tangents be $m_1$ and $m_2$. From the quadratic equation,we have:
$m_1 + m_2 = 4$ and $m_1 m_2 = 1$.
The angle $\theta$ between the two tangents is given by:
$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$
Using the identity $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2$:
$(m_1 - m_2)^2 = (4)^2 - 4(1) = 16 - 4 = 12$
$|m_1 - m_2| = \sqrt{12} = 2\sqrt{3}$
Therefore,$\tan \theta = \left| \frac{2\sqrt{3}}{1 + 1} \right| = \frac{2\sqrt{3}}{2} = \sqrt{3}$.
Since $\tan \theta = \sqrt{3}$,we have $\theta = \frac{\pi}{3}$.

(C) $H_2S_2O_8$ (Peroxodisulfuric acid or Marshall's acid) contains a peroxide linkage ($O - O$ bond).
Its structure is $HO - SO_2 - O - O - SO_2 - OH$.

(C) For isotonic solutions,the osmotic pressure is equal,so the van't Hoff factor multiplied by concentration is equal: $i_1 C_1 = i_2 C_2$.
For glucose (non-electrolyte),$i_2 = 1$,so $i_2 C_2 = 1 \times 0.010 = 0.010 \ M$.
For $Na_2SO_4$,the dissociation is $Na_2SO_4 \rightleftharpoons 2Na^{+} + SO_4^{2-}$.
Let $\alpha$ be the degree of dissociation. The van't Hoff factor $i_1 = 1 + (n-1)\alpha$,where $n=3$.
So,$i_1 = 1 + (3-1)\alpha = 1 + 2\alpha$.
The condition $i_1 C_1 = 0.010$ becomes $(1 + 2\alpha) \times 0.004 = 0.010$.
$1 + 2\alpha = \frac{0.010}{0.004} = 2.5$.
$2\alpha = 1.5$.
$\alpha = 0.75$.
Therefore,the percentage of dissociation is $0.75 \times 100 = 75 \%$.

(B) The adsorption process involves the trapping of gas molecules on a solid surface,which restricts their movement. This leads to a decrease in the randomness of the particles,meaning the entropy $(S)$ decreases. Since $\Delta G = \Delta H - T\Delta S$ and for a spontaneous process $\Delta G < 0$,the decrease in entropy $(\Delta S < 0)$ makes the $-T\Delta S$ term positive. For the process to be spontaneous,the enthalpy change $(\Delta H)$ must be negative,making the process exothermic.

(A) The sodium salt is sodium bicarbonate $(NaHCO_3)$.
When $NaHCO_3$ reacts with $MgCl_2$,it forms magnesium bicarbonate,which is soluble in water at room temperature: $2NaHCO_3 + MgCl_2 \to Mg(HCO_3)_2 + 2NaCl$.
Upon heating,magnesium bicarbonate decomposes to form magnesium carbonate,which is a white precipitate: $Mg(HCO_3)_2 \xrightarrow{\Delta} MgCO_3 \downarrow + H_2O + CO_2 \uparrow$.
Therefore,the anion is $HCO_3^-$.

(B) The central atom $Xe$ has $8$ valence electrons.
In $XeOF_4$,$Xe$ forms $4$ single bonds with $F$ atoms and $1$ double bond with the $O$ atom.
Total electrons involved in bonding $= 4 + 2 = 6$.
Number of lone pair electrons on $Xe = (8 - 6) / 2 = 1$ lone pair.
Thus,the total number of lone pairs on the central atom $Xe$ is $1$.

(B) The cell reaction is: $Zn(s) + Fe^{2+}(aq) \to Zn^{2+}(aq) + Fe(s)$.
Using the Nernst equation: $E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Fe^{2+}]}$.
Given $E_{cell} = 0.2905 \ V$,$n = 2$,$[Zn^{2+}] = 0.01 \ M$,and $[Fe^{2+}] = 0.001 \ M$.
$0.2905 = E^0_{cell} - \frac{0.0591}{2} \log \frac{10^{-2}}{10^{-3}} = E^0_{cell} - 0.02955 \log(10) = E^0_{cell} - 0.02955$.
$E^0_{cell} = 0.2905 + 0.02955 = 0.32005 \approx 0.32 \ V$.
At equilibrium,$E_{cell} = 0$,so $E^0_{cell} = \frac{0.0591}{n} \log K_c$.
$0.32 = \frac{0.0591}{2} \log K_c = 0.02955 \log K_c$.
$\log K_c = \frac{0.32}{0.02955} \approx \frac{0.32}{0.0295}$.
Therefore,$K_c = 10^{\frac{0.32}{0.0295}}$. Note: The provided option $B$ is represented as $10^{0.32/0.0295}$ in standard notation,but based on the provided options,$B$ is the intended choice.

(B) The extraction of $Pb$ (Lead) from its ore $PbS$ (Galena) involves self-reduction.
$2PbO + PbS \xrightarrow{\Delta} 3Pb + SO_2 \uparrow$
$PbSO_4 + PbS \xrightarrow{\Delta} 2Pb + 2SO_2 \uparrow$
The extraction of $Sn$ (Tin) from its ore $SnO_2$ (Cassiterite) involves carbon reduction.
$SnO_2 + 2C \xrightarrow{\Delta} Sn + 2CO$
Therefore,$Pb$ is extracted by self-reduction and $Sn$ is extracted by carbon reduction.

(C) In the complex $Hg[Co(SCN)_4]$,the oxidation state of $Co$ is $+2$.
The electronic configuration of $Co^{2+}$ is $[Ar] 3d^7$.
In a tetrahedral field,$Co^{2+}$ has $3$ unpaired electrons $(n = 3)$.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n + 2)} \ B.M.$
$\mu = \sqrt{3(3 + 2)} = \sqrt{15} \approx 3.87 \ B.M.$

(B) In $CrO_2Cl_2$,the oxidation state of $Cr$ is $x + 2(-2) + 2(-1) = 0$,so $x = +6$. This is the group oxidation state of $Cr$ $(3d^5 4s^1)$.
In $MnO_4^-$,the oxidation state of $Mn$ is $x + 4(-2) = -1$,so $x = +7$. This is the group oxidation state of $Mn$ $(3d^5 4s^2)$.
Both $Cr$ and $Mn$ are in their highest possible oxidation states in these compounds.

(D) The geometry of coordination complexes depends on the nature of the ligand and the central metal ion.
$1$. $[NiCl_4]^{2-}$: $Ni^{2+}$ has a $d^8$ configuration. $Cl^-$ is a weak field ligand,so it does not cause pairing of electrons. The hybridization is $sp^3$,resulting in a tetrahedral geometry.
$2$. $[PdCl_4]^{2-}$,$[Ni(CN)_4]^{2-}$,and $[Pd(CN)_4]^{2-}$: These complexes involve $Pd^{2+}$ or strong field ligands like $CN^-$,which cause pairing of electrons. They undergo $dsp^2$ hybridization,resulting in a square planar geometry.

(C) The reactivity of carbonyl compounds towards nucleophilic addition reactions (such as with Grignard reagents like $PhMgBr$) depends on two main factors:
$1.$ Steric hindrance: As the size and number of groups attached to the carbonyl carbon increase,the approach of the nucleophile becomes more difficult,decreasing reactivity.
$2.$ Electronic effect: Electron-donating groups (like alkyl groups via $+I$ effect) or resonance stabilization (like phenyl groups) decrease the positive charge on the carbonyl carbon,making it less electrophilic.
Comparing the given compounds:
$(II)$ Acetaldehyde $(CH_3-CHO)$ has the least steric hindrance and only one $+I$ group.
$(III)$ Acetone $(CH_3-CO-CH_3)$ has two $+I$ groups and more hindrance than acetaldehyde.
$(I)$ Benzophenone $(Ph-CO-Ph)$ has two bulky phenyl groups and significant resonance,making it the least reactive.
Therefore,the correct order is $(II) > (III) > (I)$.

(D) Tollen's reagent oxidizes compounds containing an aldehyde group,such as glucose.
It also oxidizes $\alpha$-hydroxy ketones that possess a $-COCH_2OH$ group,such as fructose,because they undergo tautomerization to form an aldehyde in an alkaline medium.

(D) When benzamide $(C_6H_5CONH_2)$ is treated with a dehydrating agent like phosphorus oxychloride $(POCl_3)$,it undergoes dehydration to form benzonitrile $(C_6H_5CN)$.
The reaction is as follows:
$C_6H_5CONH_2 \xrightarrow{POCl_3, \Delta} C_6H_5CN + H_2O$
Therefore,the correct option is $(D)$.

(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): $CH_2Cl-CH(CH_3)-CH_2-CH_3$ (achiral).
$2$. At $C-2$ (tertiary carbon): $CH_3-C(Cl)(CH_3)-CH_2-CH_3$ (chiral,exists as $(+)$ and $(-)$ enantiomers).
$3$. At $C-3$ (secondary carbon): $CH_3-CH(CH_3)-CHCl-CH_3$ (chiral,exists as $(+)$ and $(-)$ enantiomers).
$4$. At $C-4$ (terminal methyl): $CH_3-CH(CH_3)-CH_2-CH_2Cl$ (achiral).
Thus,the chiral products are the enantiomers formed by substitution at $C-2$ and $C-3$.
Total number of chiral compounds = $2$ (from $C-2$) + $2$ (from $C-3$) = $4$.

(A) The thermal decomposition of ammonium dichromate $(NH_4)_2Cr_2O_7$ produces nitrogen gas $(N_2)$:
$(NH_4)_2Cr_2O_7 \stackrel{\Delta}{\longrightarrow} N_2 \uparrow + Cr_2O_3 + 4H_2O$
Similarly,ammonium nitrite $(NH_4NO_2)$ on heating also liberates nitrogen gas $(N_2)$:
$NH_4NO_2 \stackrel{\Delta}{\longrightarrow} N_2 \uparrow + 2H_2O$
Thus,the same gas is obtained by heating $NH_4NO_2$.