AIEEE 2006 Chemistry Question Paper with Answer and Solution

(A) Given the equation: ${x^p}{y^q} = {(x + y)^{p + q}}$
Taking the natural logarithm on both sides:
$\ln({x^p}{y^q}) = \ln({(x + y)^{p + q}})$
$p\ln x + q\ln y = (p + q)\ln(x + y)$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(p\ln x + q\ln y) = \frac{d}{dx}((p + q)\ln(x + y))$
$\frac{p}{x} + \frac{q}{y}\frac{dy}{dx} = \frac{p + q}{x + y}(1 + \frac{dy}{dx})$
Multiply by $xy(x + y)$ to clear denominators:
$py(x + y) + qx(x + y)\frac{dy}{dx} = x(p + q)y + x(p + q)x\frac{dy}{dx}$
$pxy + py^2 + qx^2\frac{dy}{dx} + qxy\frac{dy}{dx} = pxy + qxy + (px^2 + qx^2)\frac{dy}{dx}$
Rearranging terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx}(qx^2 + qxy - px^2 - qx^2) = pxy + qxy - pxy - py^2$
$\frac{dy}{dx}(qxy - px^2) = qxy - py^2$
$\frac{dy}{dx} = \frac{qxy - py^2}{qx^2 - pxy} = \frac{y(qx - py)}{x(qx - py)}$
$\frac{dy}{dx} = \frac{y}{x}$

(B) The resistance at temperature $t$ is given by the formula $R_t = R_0(1 + \alpha \Delta T)$,where $R_0$ is the resistance at $0 \ ^oC$ and $\Delta T$ is the temperature change.
Given: $R_1 = 100 \ \Omega$ at $t_1 = 100 \ ^oC$,$R_2 = 200 \ \Omega$ at $t_2 = T$,and $\alpha = 0.005 \ ^oC^{-1}$.
Using the relation $R_t = R_{ref}[1 + \alpha(t - t_{ref})]$,we can write:
$100 = R_0(1 + 0.005 \times 100) \implies 100 = R_0(1 + 0.5) \implies 100 = 1.5 R_0 \implies R_0 = \frac{100}{1.5} = \frac{200}{3} \ \Omega$.
Now,for $R_2 = 200 \ \Omega$ at temperature $T$:
$200 = R_0(1 + 0.005 \times T) = \frac{200}{3}(1 + 0.005T)$.
Dividing both sides by $200$:
$1 = \frac{1}{3}(1 + 0.005T) \implies 3 = 1 + 0.005T \implies 2 = 0.005T$.
$T = \frac{2}{0.005} = \frac{2000}{5} = 400 \ ^oC$.

(C) Given: Initial temperature $\theta_1 = 100\, ^oC$,Initial resistance $R_{\theta_1} = 100\, \Omega$,Final resistance $R_{\theta_2} = 200\, \Omega$,Temperature coefficient $\alpha = 0.005\, (^oC)^{-1}$.
The formula relating resistance and temperature is $R_{\theta_2} = R_{\theta_1} [1 + \alpha (\theta_2 - \theta_1)]$.
Rearranging the formula to solve for $\theta_2$:
$R_{\theta_2} - R_{\theta_1} = \alpha R_{\theta_1} (\theta_2 - \theta_1)$
$\frac{R_{\theta_2} - R_{\theta_1}}{\alpha R_{\theta_1}} = \theta_2 - \theta_1$
$\theta_2 = \frac{R_{\theta_2} - R_{\theta_1}}{\alpha R_{\theta_1}} + \theta_1$
Substituting the values:
$\theta_2 = \frac{200 - 100}{0.005 \times 100} + 100$
$\theta_2 = \frac{100}{0.5} + 100$
$\theta_2 = 200 + 100 = 300\, ^oC$.

(D) Given the quadratic equation $x^2 + px + q = 0$ with roots $\alpha = \tan 30^\circ$ and $\beta = \tan 15^\circ$.
From the relation between roots and coefficients,we have:
Sum of roots: $\alpha + \beta = -p$
Product of roots: $\alpha \beta = q$
We know $\tan 30^\circ = \frac{1}{\sqrt{3}}$ and $\tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} = \frac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{(\sqrt{3}-1)^2}{3-1} = \frac{4-2\sqrt{3}}{2} = 2 - \sqrt{3}$.
Thus,$p = -(\alpha + \beta) = -(\frac{1}{\sqrt{3}} + 2 - \sqrt{3}) = -(\frac{1 + 2\sqrt{3} - 3}{\sqrt{3}}) = -(\frac{2\sqrt{3} - 2}{\sqrt{3}}) = \frac{2 - 2\sqrt{3}}{\sqrt{3}}$.
And $q = \alpha \beta = \frac{1}{\sqrt{3}}(2 - \sqrt{3}) = \frac{2}{\sqrt{3}} - 1$.
Now,calculate $2 + q - p$:
$2 + (\frac{2}{\sqrt{3}} - 1) - (\frac{2 - 2\sqrt{3}}{\sqrt{3}}) = 1 + \frac{2}{\sqrt{3}} - \frac{2}{\sqrt{3}} + \frac{2\sqrt{3}}{\sqrt{3}} = 1 + 2 = 3$.

(C) The center of the circle is the intersection point of the two diameters.
Solving $3x - 4y = 7$ and $2x - 3y = 5$:
Multiply the first by $2$ and the second by $3$: $6x - 8y = 14$ and $6x - 9y = 15$.
Subtracting the equations: $(6x - 8y) - (6x - 9y) = 14 - 15 \implies y = -1$.
Substituting $y = -1$ into $3x - 4y = 7$: $3x - 4(-1) = 7 \implies 3x + 4 = 7 \implies 3x = 3 \implies x = 1$.
So,the center $(h, k)$ is $(1, -1)$.
Given the area of the circle is $49\pi$,we have $\pi r^2 = 49\pi \implies r^2 = 49 \implies r = 7$.
The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.
$(x - 1)^2 + (y + 1)^2 = 7^2$.
$x^2 - 2x + 1 + y^2 + 2y + 1 = 49$.
$x^2 + y^2 - 2x + 2y + 2 = 49$.
$x^2 + y^2 - 2x + 2y - 47 = 0$.

(B) Let the intercepts of the line on the $x$-axis and $y$-axis be $a$ and $b$ respectively. The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$.
Since the point $A(3, 4)$ bisects the segment intercepted between the axes,the coordinates of $A$ are $(\frac{a}{2}, \frac{b}{2})$.
Equating the coordinates,we get $\frac{a}{2} = 3 \Rightarrow a = 6$ and $\frac{b}{2} = 4 \Rightarrow b = 8$.
Substituting these values into the intercept form equation: $\frac{x}{6} + \frac{y}{8} = 1$.
Multiplying by $24$,we get $4x + 3y = 24$.

(D) Given function is $f(x) = \frac{x}{2} + \frac{2}{x}$.
First,find the derivative $f'(x)$:
$f'(x) = \frac{1}{2} - \frac{2}{x^2}$.
For local maxima or minima,set $f'(x) = 0$:
$\frac{1}{2} - \frac{2}{x^2} = 0 \implies \frac{1}{2} = \frac{2}{x^2} \implies x^2 = 4 \implies x = \pm 2$.
Now,find the second derivative $f''(x)$:
$f''(x) = \frac{d}{dx}(\frac{1}{2} - 2x^{-2}) = 0 - 2(-2)x^{-3} = \frac{4}{x^3}$.
Check the sign of $f''(x)$ at critical points:
At $x = 2$,$f''(2) = \frac{4}{2^3} = \frac{4}{8} = \frac{1}{2} > 0$.
Since $f''(2) > 0$,the function has a local minimum at $x = 2$.
At $x = -2$,$f''(-2) = \frac{4}{(-2)^3} = \frac{4}{-8} = -\frac{1}{2} < 0$.
Since $f''(-2) < 0$,the function has a local maximum at $x = -2$.

(D) Let $AB$ be a chord of the circle $C$ and $M(h, k)$ be the midpoint of $AB$.
Since the chord $AB$ subtends an angle of $\frac{2\pi}{3}$ at the center $O(0,0)$,the triangle $\triangle OAB$ is an isosceles triangle with $OA = OB = 3$.
The line segment $OM$ is the perpendicular bisector of the chord $AB$.
In the right-angled triangle $\triangle OAM$,the angle $\angle AOM = \frac{1}{2} \angle AOB = \frac{1}{2} \times \frac{2\pi}{3} = \frac{\pi}{3} = 60^{\circ}$.
In $\triangle OAM$,we have $OM = OA \cos(60^{\circ})$.
Since $M(h, k)$ is the midpoint,the distance $OM = \sqrt{h^2 + k^2}$.
Thus,$\sqrt{h^2 + k^2} = 3 \cos(60^{\circ}) = 3 \times \frac{1}{2} = \frac{3}{2}$.
Squaring both sides,we get $h^2 + k^2 = \frac{9}{4}$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 = \frac{9}{4}$.

(B) Let $I = \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx \quad \dots(i)$
Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,we get:
$I = \int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{9-(9-x)} + \sqrt{9-x}} \, dx$
$I = \int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} \, dx \quad \dots(ii)$
Adding equations $(i)$ and $(ii)$:
$2I = \int_{3}^{6} \frac{\sqrt{x} + \sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} \, dx$
$2I = \int_{3}^{6} 1 \, dx$
$2I = [x]_{3}^{6} = 6 - 3 = 3$
$I = \frac{3}{2}$

(D) Let the axis be $nn'$. The axis passes through $A$ and is parallel to the diagonal $BD$.
The distance of point $A$ from the axis is $0$. Thus,the moment of inertia of the mass at $A$ is $0$.
The distance of point $B$ from the axis is $d_B = \frac{l}{\sqrt{2}}$.
The distance of point $D$ from the axis is $d_D = \frac{l}{\sqrt{2}}$.
The distance of point $C$ from the axis is $d_C = \sqrt{2}l$ (since $C$ is at the opposite corner of the square from $A$).
The total moment of inertia $I$ is given by:
$I = m(d_A)^2 + m(d_B)^2 + m(d_D)^2 + m(d_C)^2$
$I = m(0)^2 + m\left(\frac{l}{\sqrt{2}}\right)^2 + m\left(\frac{l}{\sqrt{2}}\right)^2 + m(\sqrt{2}l)^2$
$I = 0 + m\left(\frac{l^2}{2}\right) + m\left(\frac{l^2}{2}\right) + 2ml^2$
$I = ml^2 + 2ml^2 = 3ml^2$

(B) The resonant frequencies of a string fixed at both ends are given by $f_n = n \times f_1$,where $f_1 = \frac{v}{2L}$ is the fundamental frequency and $n$ is an integer $(n = 1, 2, 3, \dots)$.
Given two consecutive resonant frequencies are $f_n = 315 \ Hz$ and $f_{n+1} = 420 \ Hz$.
We know that $f_{n+1} - f_n = (n+1)f_1 - nf_1 = f_1$.
Therefore,the difference between any two consecutive resonant frequencies is equal to the fundamental frequency $f_1$.
$f_1 = 420 \ Hz - 315 \ Hz = 105 \ Hz$.
Since $105 \ Hz$ is the fundamental frequency,it is also the lowest resonant frequency for the string.

(D) In one mole of $Mg_{3}(PO_{4})_{2}$,there are $8 \ mol$ of oxygen atoms.
Let $M$ be the number of moles of $Mg_{3}(PO_{4})_{2}$.
The number of moles of oxygen atoms is given by $M \times 8$.
According to the question,$M \times 8 = 0.25 \ mol$.
Therefore,$M = \frac{0.25}{8} = 0.03125 \ mol$.
This can be expressed as $3.125 \times 10^{-2} \ mol$.

(B) The formula for the angular momentum $(L)$ of an electron in the $n^{th}$ orbit is given by Bohr's postulate:
$L = mvr = \frac{nh}{2\pi}$
Given that the orbit number $n = 5$,we substitute this value into the formula:
$L = \frac{5h}{2\pi}$
Calculating the value:
$L = 2.5 \frac{h}{\pi}$
Therefore,the correct option is $B$.

(A) The percentage error in velocity is given as $0.001\%$.
$\frac{\Delta V}{V} \times 100 = 0.001$
$\Delta V = \frac{0.001 \times 300}{100} = 3 \times 10^{-3} \ ms^{-1}$
According to Heisenberg's uncertainty principle:
$\Delta x \cdot m \Delta V \geq \frac{h}{4 \pi}$
$\Delta x = \frac{h}{4 \pi m \Delta V}$
Substituting the values ($h = 6.63 \times 10^{-34} \ J \cdot s$,$m = 9.1 \times 10^{-31} \ kg$,$\Delta V = 3 \times 10^{-3} \ ms^{-1}$):
$\Delta x = \frac{6.63 \times 10^{-34}}{4 \times 3.14159 \times 9.1 \times 10^{-31} \times 3 \times 10^{-3}}$
$\Delta x \approx 1.92 \times 10^{-2} \ m$

(C) Isoelectronic species are those that have the same number of electrons.
For $K^{+}$ (atomic number $19$): Number of electrons $= 19 - 1 = 18$.
For $Cl^{-}$ (atomic number $17$): Number of electrons $= 17 + 1 = 18$.
For $Ca^{2+}$ (atomic number $20$): Number of electrons $= 20 - 2 = 18$.
For $Sc^{3+}$ (atomic number $21$): Number of electrons $= 21 - 3 = 18$.
Since all these ions have $18$ electrons,they are isoelectronic.

(B) For alkali metals $(Group \ 1)$,reactivity increases down the group because the ionization enthalpy decreases,making it easier to lose the valence electron.
For halogens $(Group \ 17)$,reactivity decreases down the group because the electron gain enthalpy becomes less negative,making it harder to gain an electron.
Therefore,in alkali metals,reactivity increases,and in halogens,it decreases with an increase in atomic number down the group.

(C) According to Molecular Orbital Theory $(MOT)$,we determine the presence of unpaired electrons by filling the molecular orbitals:
$1$. For $N_2^+$ ($13$ electrons): $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^1$. It has $1$ unpaired electron.
$2$. For $O_2$ ($16$ electrons): $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*1} \pi_{2p_y}^{*1}$. It has $2$ unpaired electrons.
$3$. For $O_2^{2-}$ ($18$ electrons): $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*2} \pi_{2p_y}^{*2}$. All electrons are paired.
$4$. For $B_2$ ($10$ electrons): $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p_x}^1 \pi_{2p_y}^1$. It has $2$ unpaired electrons.
Therefore,$O_2^{2-}$ is the only species without unpaired electrons.

(D) In $XeF_4$,the geometry is square planar with all $Xe-F$ bonds being equal due to symmetry.
In $BF_4^-$ and $SiF_4$,the geometry is tetrahedral,where all four bonds are equivalent.
In $SF_4$,the sulfur atom undergoes $sp^3d$ hybridization with one lone pair,resulting in a see-saw geometry.
Due to the presence of the lone pair,the axial and equatorial $S-F$ bonds have different bond lengths and bond angles,making them unequal.

(C) For an ideal gas expanding in an isolated system,the internal energy change $\Delta U = q + w = 0$. Since $q = 0$ (isolated system),$w = 0$.
For an ideal gas,$\Delta U = nC_v\Delta T = 0$,which implies $\Delta T = 0$ or $T_f = T_i$ for both processes.
However,if the question implies expansion against external pressure where work is done,in an isolated system $(q=0)$,$\Delta U = w$.
Since $w_{rev} < w_{irrev}$ (more work done in reversible expansion),the temperature drop is greater in the reversible process.
Therefore,$(T_f)_{rev} < (T_f)_{irrev}$.

(C) The standard enthalpy of formation of $CH_4$ is given by the reaction:
$C_{(s)} + 2H_{2(g)} \rightarrow CH_{4(g)}$
To calculate the average $C-H$ bond energy,we need to consider the atomization of the reactants:
$1.$ Enthalpy of sublimation of carbon: $C_{(s)} \rightarrow C_{(g)}$
$2.$ Dissociation energy of hydrogen: $2H_{2(g)} \rightarrow 4H_{(g)}$
By using Hess's Law,we combine these with the enthalpy of formation to find the total bond dissociation energy of $CH_4$,which is then divided by $4$ to get the average $C-H$ bond energy.
Therefore,the dissociation energy of $H_2$ and the enthalpy of sublimation of carbon are required.

(A) The formation reaction for $ICl_{(g)}$ is: $\frac{1}{2} I_{2(s)} + \frac{1}{2} Cl_{2(g)} \rightarrow ICl_{(g)}$
The enthalpy of formation is calculated using the bond dissociation energies and sublimation energy:
$\Delta H_f^o = [\frac{1}{2} \Delta H_{sub}(I_2) + \frac{1}{2} BE(I-I) + \frac{1}{2} BE(Cl-Cl)] - BE(I-Cl)$
Substituting the given values:
$\Delta H_f^o = [\frac{1}{2}(62.76) + \frac{1}{2}(151.0) + \frac{1}{2}(242.3)] - 211.3$
$\Delta H_f^o = [31.38 + 75.5 + 121.15] - 211.3$
$\Delta H_f^o = 228.03 - 211.3 = 16.73 \ kJ \ mol^{-1}$
Rounding to one decimal place,we get $+16.8 \ kJ \ mol^{-1}$.

(D) The chemical equation for the formation of $CO$ is:
$C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$
From the first law of thermodynamics,we know that $\Delta H = \Delta U + \Delta n_g RT$.
Rearranging the equation,we get $\Delta H - \Delta U = \Delta n_g RT$.
Here,$\Delta n_g$ is the change in the number of moles of gaseous species:
$\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting the values:
$\Delta H - \Delta U = \frac{1}{2} \times 8.314 \times 298 = 1238.78 \ J \ mol^{-1}$.

(C) The dissociation reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
Initial moles: $1 \quad 0 \quad 0$
Moles at equilibrium: $(1 - x) \quad x \quad x$
Total moles at equilibrium = $(1 - x) + x + x = 1 + x$
Partial pressure of a gas is given by: $P_{gas} = \text{mole fraction} \times P_{total}$
Mole fraction of $PCl_3 = \frac{x}{1 + x}$
Therefore,$P_{PCl_3} = \left( \frac{x}{1 + x} \right) P$

(C) For the reaction $SO_{3(g)} \rightleftharpoons SO_{2(g)} + \frac{1}{2} O_{2(g)}$,the equilibrium constant is $K_c = \frac{[SO_2][O_2]^{1/2}}{[SO_3]} = 4.9 \times 10^{-2}$.
To obtain the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$,we reverse the original reaction and multiply it by $2$.
If a reaction is reversed,the new equilibrium constant becomes $1/K_c$. If it is multiplied by a factor $n$,the new equilibrium constant becomes $(K_c)^n$.
Therefore,for the new reaction,$K_c' = \frac{1}{(K_c)^2} = \frac{1}{(4.9 \times 10^{-2})^2}$.
$K_c' = \frac{1}{24.01 \times 10^{-4}} = \frac{10^4}{24.01} \approx 416.5$.

(D) In aqueous solution,alkali metal ions are hydrated. The extent of hydration is inversely proportional to the ionic size. Smaller ions have higher charge density and thus get more hydrated,resulting in a larger hydrated radius.
Since ionic mobility is inversely proportional to the size of the hydrated ion,the hydrated radius order is $Li^{+} > Na^{+} > K^{+} > Rb^{+}$.
Therefore,the ionic mobility order is $Li^{+} < Na^{+} < K^{+} < Rb^{+}$.
Thus,the ionic mobility is maximum for $Rb^{+}$.

(A) $1$. In cyclic compounds containing a double bond,the double bond is given priority for numbering.
$2$. The carbon atoms of the double bond are assigned positions $1$ and $2$ such that the substituents get the lowest possible locants.
$3$. Starting from the carbon attached to the chlorine atom as position $1$,we move towards the double bond to assign position $2$ to the other carbon of the double bond.
$4$. This gives the bromine atom at position $3$.
$5$. Thus,the correct name is $3-$bromo$-1-$chlorocyclohexene.

(B) The stability of free radicals is determined by resonance and inductive effects.
Resonance provides greater stabilization than the inductive effect.
$(C_6H_5)_3 \dot{C}$ is the most stable due to resonance with three phenyl rings.
$(C_6H_5)_2 \dot{C}H$ is the next most stable due to resonance with two phenyl rings.
$(CH_3)_3 \dot{C}$ (tertiary alkyl radical) is more stable than $(CH_3)_2 \dot{C}H$ (secondary alkyl radical) due to the greater inductive effect and hyperconjugation.
Thus,the increasing order of stability is: $(CH_3)_2 \dot{C}H < (CH_3)_3 \dot{C} < (C_6H_5)_2 \dot{C}H < (C_6H_5)_3 \dot{C}$.

(D) The given reaction is a Hofmann elimination,which occurs when a quaternary ammonium salt is heated with a base like $OH^-$.
In Hofmann elimination,the base abstracts a proton from the $\beta$-carbon that is least sterically hindered,leading to the formation of the least substituted alkene as the major product.
In the provided substrate,the quaternary nitrogen is attached to a methyl group,an ethyl group,a $n$-butyl group,and a cyclohexenyl ring.
Comparing the $\beta$-hydrogens available:
$1$. $\beta_1$ on the ethyl group leads to $CH_2=CH_2$ (ethene).
$2$. $\beta_2$ on the $n$-butyl group leads to $CH_2=CHCH_2CH_3$ ($1$-butene).
$3$. $\beta_3$ and $\beta_4$ on the ring lead to cyclic alkenes.
According to the Hofmann rule,the least substituted alkene,which is $CH_2=CH_2$ (ethene),is formed as the major product because the ethyl group is the least sterically hindered group attached to the nitrogen.

(A) In $2-$fluoroethanol,the Gauche conformation is unexpectedly more stable than the Anti conformation due to the formation of an intramolecular hydrogen bond between the fluorine atom $(F^{\delta -})$ and the hydroxyl hydrogen atom $(H^{\delta +})$.
The Eclipse conformation is the least stable due to high torsional strain and steric repulsion.
Therefore,the increasing order of stability is: $Eclipse < Anti < Gauche$.

(D) Dipole-dipole interactions occur between polar molecules that possess permanent dipoles.
In this interaction,the positive end of one polar molecule is attracted to the negative end of another polar molecule.
$KCl$ and water involve ion-dipole interactions.
Benzene and carbon tetrachloride are non-polar,involving London dispersion forces.
Benzene and ethanol involve dipole-induced dipole interactions.
Acetonitrile $(CH_{3}CN)$ and acetone $(CH_{3})_{2}CO$ are both polar molecules,and therefore,the major interaction between them is dipole-dipole interaction.

(C) In the reaction $2HI + H_2SO_4 \rightarrow I_2 + SO_2 + 2H_2O$,the oxidation state of sulfur $(S)$ decreases from $+6$ in $H_2SO_4$ to $+4$ in $SO_2$,indicating reduction.
Simultaneously,the oxidation state of iodine $(I)$ increases from $-1$ in $HI$ to $0$ in $I_2$,indicating oxidation.
Since $H_2SO_4$ causes the oxidation of $HI$ while being reduced itself,it acts as an oxidizing agent.

(B) The acidity of oxoacids depends on the stability of their conjugate bases.
$HNO_3$ (nitric acid) forms the nitrate ion $(NO_3^-)$ as its conjugate base,which is stabilized by resonance.
$HNO_2$ (nitrous acid) forms the nitrite ion $(NO_2^-)$ as its conjugate base.
Since the conjugate base of $HNO_3$ is more resonance-stabilized than that of $HNO_2$,$HNO_3$ is a stronger acid than $HNO_2$.
Other options are incorrect because:
$HClO_4$ is a stronger acid than $HClO_3$ due to higher oxidation state of $Cl$.
$H_2SO_3$ is a stronger acid than $H_3PO_3$.
$HCl$ is a stronger acid than $HF$ in aqueous medium due to the high bond dissociation energy of $H-F$.

(A) The first ionization enthalpy generally increases across a period from left to right due to an increase in effective nuclear charge.
$B$ (Boron,group $13$) has the lowest ionization enthalpy among the given elements.
Comparing $P$ (Phosphorus,group $15$) and $S$ (Sulfur,group $16$),$P$ has a higher ionization enthalpy than $S$ because $P$ has a stable half-filled $p$-orbital configuration $(3s^2 3p^3)$.
$F$ (Fluorine,group $17$) has the highest ionization enthalpy due to its small size and high effective nuclear charge.
Thus,the order is $B < S < P < F$.

(B) Hypochlorous acid $(HOCl)$ undergoes disproportionation reaction upon heating or standing.
The reaction is given by:
$3 HOCl \rightarrow 2 HCl + HClO_3$
In this reaction,the chlorine atom in $HOCl$ (oxidation state $+1$) is simultaneously oxidized to $HClO_3$ (oxidation state $+5$) and reduced to $HCl$ (oxidation state $-1$).

(B) $1$. Identify the oxidation state of the central metal atom $Co$: Let the oxidation state be $x$. The complex is $[Co(NO_2)(NH_3)_5]Cl_2$. The charge on $NO_2^-$ is $-1$,$NH_3$ is $0$,and $Cl^-$ is $-1$. Thus,$x + (-1) + 5(0) + 2(-1) = 0$,which gives $x - 3 = 0$,so $x = +3$.
$2$. Name the ligands: There are $5$ ammonia ligands $(ammine)$ and $1$ nitro ligand $(nitrito-N)$.
$3$. Assemble the name: Ligands are named in alphabetical order: $ammine$ comes before $nitrito-N$. So,it is pentaammine.
$4$. Combine: The name is pentaamminenitrito-$N$-cobalt$(III)$ chloride.

(A) According to Fajan's rule,the covalent character of a compound increases with an increase in the oxidation state of the metal cation.
In the $+4$ oxidation state,the metal ion has a smaller size and a higher charge density,which leads to greater polarizing power.
Consequently,$MCl_4$ exhibits more covalent character,whereas $MCl_2$ exhibits more ionic character.
Therefore,$MCl_2$ is more ionic than $MCl_4$.

(D) In $Fe(CO)_5$,the $Fe-C$ bond is formed by the donation of a lone pair of electrons from the $CO$ ligand into the empty $d$-orbitals of the $Fe$ atom,which creates a $\sigma$-bond.
Additionally,there is back-bonding where electrons from the filled $d$-orbitals of the $Fe$ atom are donated into the empty antibonding $\pi^*$-orbitals of the $CO$ ligand,which creates a $\pi$-bond.
Therefore,the $Fe-C$ bond possesses both $\sigma$ and $\pi$ characters.

(C) The photoelectric effect is an instantaneous process. Experimental observations confirm that there is no measurable time lag between the incidence of a photon and the emission of a photoelectron. The time taken for this process is approximately of the order of $10^{-10} \ s$.

(C) The magnetic flux linked with the coil is given by $\phi = 10t^2 - 50t + 250$.
According to Faraday's law of electromagnetic induction,the induced emf $e$ is given by $e = -\frac{d\phi}{dt}$.
Differentiating $\phi$ with respect to $t$:
$\frac{d\phi}{dt} = \frac{d}{dt}(10t^2 - 50t + 250) = 20t - 50$.
Therefore,$e = -(20t - 50) = 50 - 20t$.
At $t = 3 \ s$,the induced emf is:
$e = 50 - 20(3) = 50 - 60 = -10 \ V$.

(C) Since the boxes are rigid and thermally isolated from the surroundings,the heat lost by the hotter gas (helium) equals the heat gained by the colder gas (nitrogen).
For helium (monatomic gas),the molar heat capacity at constant volume is $C_{V,He} = \frac{3R}{2}$.
For nitrogen (diatomic gas),the molar heat capacity at constant volume is $C_{V,N_2} = \frac{5R}{2}$.
Let $T_f$ be the final equilibrium temperature.
Heat lost by helium = Heat gained by nitrogen
$n_{He} C_{V,He} (T_{initial,He} - T_f) = n_{N_2} C_{V,N_2} (T_f - T_{initial,N_2})$
$1 \times \frac{3R}{2} \times (\frac{7}{3} T_0 - T_f) = 1 \times \frac{5R}{2} \times (T_f - T_0)$
Canceling $\frac{R}{2}$ from both sides:
$3(\frac{7}{3} T_0 - T_f) = 5(T_f - T_0)$
$7 T_0 - 3 T_f = 5 T_f - 5 T_0$
$12 T_0 = 8 T_f$
$T_f = \frac{12}{8} T_0 = \frac{3}{2} T_0$

(A) When the circuit is connected for a long time,the inductor acts as a short circuit (ideal inductor). The steady-state current $I_0$ is given by $I_0 = \frac{E}{R} = \frac{100 \, V}{100 \, \Omega} = 1 \, A$.
When the battery is disconnected and points $A$ and $B$ are short-circuited,the circuit becomes an $LR$ decay circuit. The current at time $t$ is given by $I(t) = I_0 e^{-t/\tau}$,where $\tau$ is the time constant.
The time constant $\tau = \frac{L}{R} = \frac{100 \times 10^{-3} \, H}{100 \, \Omega} = 10^{-3} \, s = 1 \, ms$.
Given $t = 1 \, ms = 10^{-3} \, s$,the current is $I = 1 \times e^{-(10^{-3})/(10^{-3})} = 1 \times e^{-1} = \frac{1}{e} \, A$.

(C) The total power radiated by the sun is given by the Stefan-Boltzmann law: $P = \sigma T^4 (4 \pi R^2)$.
This energy spreads out over a spherical surface of radius $r$ as it travels from the sun to the earth. The intensity $I$ (power per unit area) at a distance $r$ from the sun is given by: $I = \frac{P}{4 \pi r^2} = \frac{\sigma T^4 (4 \pi R^2)}{4 \pi r^2} = \frac{\sigma R^2 T^4}{r^2}$.
The earth acts as a disc of radius $r_0$ intercepting this radiation. The area of this disc is $A = \pi r_0^2$.
Therefore,the total radiant power incident on the earth is $Q = I \times A = \left( \frac{\sigma R^2 T^4}{r^2} \right) \times (\pi r_0^2) = \frac{\pi r_0^2 R^2 \sigma T^4}{r^2}$.

(A) For an adiabatic process,the work done on the gas is given by $W = \frac{nR \Delta T}{\gamma - 1}$.
Given: $W = 146 \, kJ = 146 \times 10^3 \, J$,$n = 1 \, kmol = 10^3 \, mol$,$\Delta T = 7 \, K$,and $R = 8.3 \, J \, mol^{-1} \, K^{-1}$.
Substituting the values into the formula:
$146 \times 10^3 = \frac{10^3 \times 8.3 \times 7}{\gamma - 1}$
$146 = \frac{58.1}{\gamma - 1}$
$\gamma - 1 = \frac{58.1}{146} \approx 0.3979 \approx 0.4$
$\gamma = 1 + 0.4 = 1.4$.
Since $\gamma = 1.4$ for a diatomic gas,the gas is diatomic.

(C) When two conductors are connected by a conducting wire,charge flows until their potentials become equal,i.e.,$V_1 = V_2$.
Since the potential of a spherical conductor is given by $V = K \frac{Q}{r}$,we have $K \frac{Q_1}{r_1} = K \frac{Q_2}{r_2}$,which implies $\frac{Q_1}{r_1} = \frac{Q_2}{r_2}$ or $\frac{Q_1}{Q_2} = \frac{r_1}{r_2}$.
The electric field at the surface of a spherical conductor is $E = K \frac{Q}{r^2}$.
Therefore,the ratio of the electric fields is $\frac{E_1}{E_2} = \frac{K Q_1 / r_1^2}{K Q_2 / r_2^2} = \frac{Q_1}{Q_2} \times \frac{r_2^2}{r_1^2}$.
Substituting $\frac{Q_1}{Q_2} = \frac{r_1}{r_2}$,we get $\frac{E_1}{E_2} = \frac{r_1}{r_2} \times \frac{r_2^2}{r_1^2} = \frac{r_2}{r_1}$.
Given $r_1 = 1 \ mm$ and $r_2 = 2 \ mm$,the ratio is $\frac{E_1}{E_2} = \frac{2}{1} = 2 : 1$.

(A) Ferromagnetic materials (like $N_1$) are strongly attracted by a magnet.
Paramagnetic materials (like $N_2$) are weakly attracted by a magnet.
Diamagnetic materials (like $N_3$) are weakly repelled by a magnet.
Therefore,when a magnet is brought close to them,it will attract $N_1$ strongly,$N_2$ weakly,and repel $N_3$ weakly.

(C) Since the boxes are in thermal contact and isolated from the surroundings,the heat lost by one gas equals the heat gained by the other.
$\Delta Q_A + \Delta Q_B = 0$
$\Delta U_A + \Delta U_B = 0$
$n_A C_{vA} (T_f - T_A) + n_B C_{vB} (T_f - T_B) = 0$
For nitrogen (diatomic),$C_{vA} = \frac{5}{2} R$. For helium (monatomic),$C_{vB} = \frac{3}{2} R$.
Given $n_A = 1, n_B = 1, T_A = T_0, T_B = \frac{7}{3} T_0$.
$1 \times \frac{5}{2} R (T_f - T_0) + 1 \times \frac{3}{2} R (T_f - \frac{7}{3} T_0) = 0$
Dividing by $\frac{1}{2} R$:
$5(T_f - T_0) + 3(T_f - \frac{7}{3} T_0) = 0$
$5 T_f - 5 T_0 + 3 T_f - 7 T_0 = 0$
$8 T_f = 12 T_0$
$T_f = \frac{12}{8} T_0 = \frac{3}{2} T_0$

(B) In a series $LCR$ circuit at resonance,the inductive reactance $X_L$ equals the capacitive reactance $X_C$,i.e.,$X_L = X_C = \frac{1}{\omega C}$.
The current $I$ in the circuit is given by $I = \frac{V_R}{R}$.
Given $V_R = 100\,V$ and $R = 1\,k\Omega = 1000\,\Omega$,we have $I = \frac{100}{1000} = 0.1\,A$.
At resonance,the voltage across the inductor $V_L$ is given by $V_L = I \times X_L$.
Since $X_L = X_C = \frac{1}{\omega C}$,we have $V_L = I \times \frac{1}{\omega C}$.
Substituting the given values: $V_L = 0.1 \times \frac{1}{200 \times 2 \times 10^{-6}}$.
$V_L = 0.1 \times \frac{1}{400 \times 10^{-6}} = 0.1 \times \frac{10^6}{400} = 0.1 \times 2500 = 250\,V$.
Thus,the voltage across $L$ is $250\,V$.

(A) Ferromagnetic substances are strongly magnetized in the direction of the external magnetic field,leading to a strong attraction.
Paramagnetic substances are weakly magnetized in the direction of the external magnetic field,leading to a weak attraction.
Diamagnetic substances are weakly magnetized in the direction opposite to the external magnetic field,leading to a weak repulsion.
Therefore,when a magnet is brought close to these needles,it will attract $N_1$ strongly,attract $N_2$ weakly,and repel $N_3$ weakly.

(C) The formula for terminal velocity $v$ is given by $v = \frac{2}{9 \eta} r^2 g (\rho - \sigma)$,where $r$ is the radius of the sphere,$\rho$ is the density of the sphere,$\sigma$ is the density of the liquid,and $\eta$ is the coefficient of viscosity.
Since $r$,$g$,and $\eta$ are constant for both spheres in the same liquid,we have $v \propto (\rho - \sigma)$.
For the gold sphere: $v_g = k (\rho_g - \sigma)$,where $k = \frac{2 r^2 g}{9 \eta}$.
$0.2 = k (19.5 \times 10^3 - 1.5 \times 10^3) = k (18 \times 10^3)$.
For the silver sphere: $v_s = k (\rho_s - \sigma) = k (10.5 \times 10^3 - 1.5 \times 10^3) = k (9 \times 10^3)$.
Taking the ratio: $\frac{v_s}{v_g} = \frac{9 \times 10^3}{18 \times 10^3} = \frac{1}{2}$.
Therefore,$v_s = \frac{v_g}{2} = \frac{0.2}{2} = 0.1\, m/s$.

(C) The terminal velocity $V_T$ of a sphere of radius $r$ and density $\rho$ falling through a liquid of density $\sigma$ and viscosity $\eta$ is given by: $V_T = \frac{2r^2}{9\eta}(\rho - \sigma)g$.
Since $r$,$\eta$,and $\sigma$ are the same for both spheres,$V_T \propto (\rho - \sigma)$.
Let $V_{Au}$ and $V_{Ag}$ be the terminal velocities of gold and silver spheres respectively.
$\frac{V_{Ag}}{V_{Au}} = \frac{\rho_{Ag} - \sigma}{\rho_{Au} - \sigma}$.
Given: $\rho_{Au} = 19.5 \times 10^3 \, kg/m^3$,$\rho_{Ag} = 10.5 \times 10^3 \, kg/m^3$,$\sigma = 1.5 \times 10^3 \, kg/m^3$,and $V_{Au} = 0.2 \, m/s$.
$V_{Ag} = 0.2 \times \frac{10.5 - 1.5}{19.5 - 1.5} = 0.2 \times \frac{9}{18} = 0.2 \times 0.5 = 0.1 \, m/s$.

(A) Given: Molarity $(M) = 2.05 \ M$,Density $(d) = 1.02 \ g/mL$,Molar mass of acetic acid $(CH_3COOH) = 60 \ g/mol$.
Molality $(m)$ is calculated using the formula: $m = \frac{1000 \times M}{1000 \times d - M \times M_{solute}}$.
Substituting the values: $m = \frac{1000 \times 2.05}{1000 \times 1.02 - 2.05 \times 60}$.
$m = \frac{2050}{1020 - 123} = \frac{2050}{897} \approx 2.28 \ mol \ kg^{-1}$.

(B) As we move down group-$15$,the electronegativity of the central atom decreases.
Due to the decrease in electronegativity,the bond pairs of electrons move farther away from the central atom.
This results in a decrease in the $bp-bp$ (bond pair-bond pair) repulsion.
Consequently,the bond angle decreases from $NH_3$ to $SbH_3$.

(B) The cell reactions are:
$(I) \ Ag \rightarrow Ag^{+} + e^-$,$E^o = -0.800\, V$
$(II) \ Ag + I^{-} \rightarrow AgI + e^-$,$E^o = 0.152\, V$
Subtracting $(I)$ from $(II)$ gives the reaction for the solubility product:
$AgI \rightarrow Ag^{+} + I^{-}$
$E^o_{cell} = E^o_{(II)} - E^o_{(I)} = 0.152 - (-0.800) = -0.952\, V$
Using the relation $E^o_{cell} = \frac{0.059}{n} \log K_{sp}$ at $25\, ^oC$:
$-0.952 = \frac{0.059}{1} \log K_{sp}$
$\log K_{sp} = -\frac{0.952}{0.059} = -16.135 \approx -16.13$

(C) The rate of the $S_N2$ reaction is directly proportional to the nucleophilicity of the attacking nucleophile.
Nucleophilicity is inversely related to the strength of the corresponding conjugate acid.
The order of acidity of the conjugate acids is: $CH_3COOH (AcOH) > C_6H_5OH (PhOH) > H_2O > CH_3OH$.
Therefore,the order of basicity (nucleophilicity) is the reverse: $CH_3COO^{-} < C_6H_5O^{-} < HO^{-} < CH_3O^{-}$.
Thus,the decreasing order of the rate of reaction is: $D > C > A > B$.

(B) face-centred cubic $(FCC)$ unit cell contains $4$ atoms per unit cell.
The volume of a single atom is given by the formula $V = \frac{4}{3} \pi r^3$.
Therefore,the total volume of atoms in the unit cell is $4 \times \frac{4}{3} \pi r^3 = \frac{16}{3} \pi r^3$.

(A) According to Langmuir's model of adsorption,the rate of adsorption is proportional to the pressure of the gas $(p)$ and the number of vacant sites on the surface.
The mass of gas striking a given area of surface is directly proportional to the pressure of the gas $(p)$ because the frequency of molecular collisions with the surface increases linearly with pressure.
Therefore,the correct statement is that the mass of gas striking a given area of surface is proportional to the pressure of the gas.

(B) Moles of glucose $(n_{glucose}) = \frac{18 \ g}{180 \ g/mol} = 0.1 \ mol$.
Moles of water $(n_{water}) = \frac{178.2 \ g}{18 \ g/mol} = 9.9 \ mol$.
Mole fraction of water $(\chi_{water}) = \frac{n_{water}}{n_{water} + n_{glucose}} = \frac{9.9}{9.9 + 0.1} = \frac{9.9}{10} = 0.99$.
The vapour pressure of pure water at $100^\circ C$ is $760 \ Torr$.
According to Raoult's law,the vapour pressure of the solution $(P_{sol}) = \chi_{water} \times P^\circ_{water}$.
$P_{sol} = 0.99 \times 760 \ Torr = 752.4 \ Torr$.

(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity of a weak electrolyte like acetic acid $(CH_3COOH)$ at infinite dilution can be calculated using the molar conductivities of strong electrolytes.
The expression is:
$\Lambda_{CH_3COOH}^o = \Lambda_{CH_3COONa}^o + \Lambda_{HCl}^o - \Lambda_{NaCl}^o$
Given that we have $\Lambda_{HCl}^o$,we need $\Lambda_{CH_3COONa}^o$ and $\Lambda_{NaCl}^o$ to complete the calculation. However,if the question implies we are starting from a set of values,the missing component to relate these to acetic acid is $\Lambda_{CH_3COONa}^o$ and $\Lambda_{NaCl}^o$. Based on the provided options,$\Lambda_{NaCl}^o$ is the standard additional value required to cancel out the ions not present in the target electrolyte.

(B) Step $1$: Calculate the cell constant $(G^* = l/a)$.
Given for $0.1 \, M$ solution: $R = 100 \, \Omega$ and $\kappa = 1.29 \, S \, m^{-1}$.
$G^* = \kappa \times R = 1.29 \, S \, m^{-1} \times 100 \, \Omega = 129 \, m^{-1}$.
Step $2$: Calculate conductivity $(\kappa)$ for $0.2 \, M$ solution.
Given $R = 520 \, \Omega$ for $0.2 \, M$ solution.
$\kappa = \frac{G^*}{R} = \frac{129 \, m^{-1}}{520 \, \Omega} \approx 0.248 \, S \, m^{-1}$.
Step $3$: Calculate molar conductivity $(\Lambda_m)$.
$\Lambda_m = \frac{\kappa}{C} = \frac{0.248 \, S \, m^{-1}}{0.2 \, mol \, L^{-1}} = \frac{0.248 \, S \, m^{-1}}{0.2 \times 10^3 \, mol \, m^{-3}} = 1.24 \times 10^{-3} \, S \, m^2 \, mol^{-1}$.
Converting to the required units $(\times 10^{-4} \, S \, m^2 \, mol^{-1})$:
$1.24 \times 10^{-3} = 12.4 \times 10^{-4} \, S \, m^2 \, mol^{-1}$.

(A) Since the reaction is $2nd$ order with respect to $CO$,the rate law is given as:
$r = k[CO]^2$
Let the initial concentration of $CO$ be $a$,so $[CO] = a$.
$r_1 = k(a)^2 = ka^2$
When the concentration is doubled,$[CO] = 2a$.
Therefore,$r_2 = k(2a)^2 = 4ka^2$.
Comparing the two rates,$r_2 = 4r_1$.
Thus,the rate of reaction increases by a factor of $4$.

(C) In the Arrhenius equation $k = A e^{-E_a / RT}$,$E_a$ represents the activation energy.
Activation energy is defined as the minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to the threshold energy,allowing them to react and form products.
Therefore,it is the energy barrier that colliding molecules must overcome to react.

(C) The nuclear transformation is ${}_{92}^{238}U \rightarrow {}_{92}^{234}U + \text{emissions}$.
First,the emission of one $\alpha$-particle $({}_{2}^{4}He)$ results in: ${}_{92}^{238}U \rightarrow {}_{90}^{234}Th + {}_{2}^{4}He$.
To reach the final product ${}_{92}^{234}U$,the ${}_{90}^{234}Th$ must undergo beta decay: ${}_{90}^{234}Th \rightarrow {}_{91}^{234}Pa + {}_{-1}^{0}e$ and ${}_{91}^{234}Pa \rightarrow {}_{92}^{234}U + {}_{-1}^{0}e$.
Thus,the total emissions are one $\alpha$-particle and two $\beta^-$-particles.

(B) Step $(i)$: $NO_{(g)} + Br_{2(g)} \rightleftharpoons NOBr_{2(g)}$ (Fast equilibrium)
Step $(ii)$: $NOBr_{2(g)} + NO_{(g)} \longrightarrow 2NOBr_{(g)}$ (Slow,rate-determining step)
The rate law is determined by the slow step: $\text{Rate} = k[NOBr_2][NO]$.
Since $NOBr_2$ is an intermediate,we express its concentration using the equilibrium constant $K_C$ from step $(i)$: $K_C = \frac{[NOBr_2]}{[NO][Br_2]}$.
Therefore,$[NOBr_2] = K_C[NO][Br_2]$.
Substituting this into the rate law: $\text{Rate} = k \cdot K_C[NO][Br_2][NO] = k'[NO]^2[Br_2]$.
The order of the reaction with respect to $NO_{(g)}$ is $2$.

(D) The atomic number of Nickel $(Ni)$ is $28$. Its electronic configuration is $[Ar] 3d^8 4s^2$.
In the complex $[NiX_4]^{-2}$,let the oxidation state of $Ni$ be $y$.
$y + 4(-1) = -2 \implies y = +2$.
So,$Ni^{+2}$ has the configuration $[Ar] 3d^8$.
In the $3d$ orbital,there are $8$ electrons. According to Hund's rule,these are distributed as $3$ paired and $2$ unpaired electrons.
Since the complex is paramagnetic and the ligand $X^-$ is a weak field ligand (as it forms a tetrahedral complex),no pairing of electrons occurs in the $3d$ orbitals.
Thus,there are $2$ unpaired electrons.
The hybridization involved is $sp^3$,which corresponds to a tetrahedral geometry.

(B) The main reason for lanthanoid contraction is the poor shielding of the outer shell electrons by $4f$ electrons.
Because the $4f$ orbitals have a diffuse shape,their shielding effect is imperfect.
Consequently,the effective nuclear charge experienced by the outer electrons increases,causing the nucleus to attract the outer shell electrons more strongly,which leads to a decrease in atomic and ionic radii across the lanthanoid series.

(A) $EDTA$ is a hexadentate ligand,meaning it has $6$ donor atoms ($4$ oxygen atoms and $2$ nitrogen atoms) that can coordinate to a central metal ion.
Since an octahedral complex requires a coordination number of $6$,a single molecule of $EDTA$ is sufficient to satisfy all $6$ coordination sites of the $Ca^{2+}$ ion.
Therefore,$1$ molecule of $EDTA$ is required.

(C) The electronic configuration of $Ni$ $(Z = 28)$ is $[Ar] 3d^8 4s^2$.
For $Ni^{2+}$,the configuration is $[Ar] 3d^8$.
In $3d^8$,there are $2$ unpaired electrons.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ \mu_B$,where $n$ is the number of unpaired electrons.
Substituting $n = 2$: $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ \mu_B$.
Thus,the closest value is $2.84 \ \mu_B$.

(D) Grignard reagents $(RMgX)$ are strong bases and react with compounds containing active hydrogen atoms (like alcohols) to form alkanes.
$C_{6}H_{5}MgBr + CH_{3}OH \longrightarrow C_{6}H_{6} + CH_{3}OMgBr$
Here,phenyl magnesium bromide reacts with methanol to produce benzene and methoxymagnesium bromide $(Mg(OMe)Br)$.

(D) Fluorobenzene is prepared by the Balz-Schiemann reaction.
In this process,aniline is first converted to benzene diazonium chloride using $NaNO_2$ and $HCl$ at $0-5^{\circ}C$.
Then,it is treated with fluoroboric acid $(HBF_4)$ to form benzene diazonium tetrafluoroborate $(C_6H_5N_2^+BF_4^-)$.
Finally,heating this salt leads to the formation of fluorobenzene,$BF_3$,and $N_2$ gas.

(B) The reaction follows the $E2$ mechanism. In an $E2$ elimination,the leaving group $(-Br)$ and the $\beta$-hydrogen atom must be in an anti-periplanar (trans) orientation to allow for the formation of the double bond. In trans-$2-$phenyl$-1-$bromocyclopentane,the $\beta$-hydrogen at the $C-3$ position is anti to the $-Br$ group at the $C-1$ position. Therefore,the elimination occurs between $C-1$ and $C-3$,resulting in the formation of $3-$phenylcyclopentene.

(B) The reaction involves a nucleophilic substitution of the $-CH_2Cl$ group with the $CN^-$ nucleophile from $NaCN$ in a polar aprotic solvent $(DMF)$.
This reaction proceeds via the $S_N2$ mechanism.
The $-CH_2Cl$ group is a primary alkyl halide,which is highly reactive towards $S_N2$ substitution.
The $-I$ group attached to the benzene ring is an aryl halide,which is much less reactive towards nucleophilic substitution under these conditions.
Therefore,the $CN^-$ nucleophile selectively attacks the $-CH_2Cl$ group to form the $-CH_2CN$ group,while the $-I$ group remains unaffected.
The major product is $3$-iodophenylacetonitrile.

(B) The reaction involves electrophilic addition of $HBr$ to the vinyl ether.
$H^{+}$ from $HBr$ adds to the terminal carbon $(CH_2)$ to form a resonance-stabilized carbocation:
$CH_2=CH-OCH_3 + H^{+} \rightarrow CH_3-C^{+}H-OCH_3 \leftrightarrow CH_3-CH=O^{+}-CH_3$
The $Br^{-}$ ion then attacks the carbocation to form the product:
$CH_3-C^{+}H-OCH_3 + Br^{-} \rightarrow CH_3-CHBr-OCH_3$

(B) The iodoform test is given by alcohols containing the $CH_3-CH(OH)-$ group or carbonyl compounds containing the $CH_3-CO-$ group.
$C_6H_5-CH(OH)-CH_3$ ($1$-phenylethanol) contains the $CH_3-CH(OH)-$ group and thus gives a positive iodoform test.
The reaction is:
$C_6H_5-CH(OH)-CH_3 + 4I_2 + 6NaOH \rightarrow CHI_3 + C_6H_5-COONa + 5NaI + 5H_2O$

(C) When phenols are treated with bromine water,the $-OH$ group strongly activates the benzene ring,leading to electrophilic substitution at all available ortho and para positions.
For $m$-cresol ($3$-methylphenol),the positions ortho and para to the $-OH$ group are available for substitution.
Specifically,the positions ortho to the $-OH$ group (at $C2$ and $C6$) and the position para to the $-OH$ group (at $C4$) are all activated,leading to the formation of $2,4,6-$tribromo$-3-$methylphenol.
Therefore,$m$-cresol gives a tribromo derivative.

(D) The given reaction is the $Reimer-Tiemann$ reaction.
In this reaction,phenol reacts with chloroform $(CHCl_3)$ in the presence of aqueous sodium hydroxide $(NaOH)$ to form salicylaldehyde.
The mechanism involves the formation of dichlorocarbene $(:\,CCl_2)$ as the reactive intermediate,which acts as an electrophile.
Therefore,the correct option is $D$.

(A) The rate of nucleophilic addition of $HCN$ to carbonyl compounds depends on two factors:
$1$. Steric hindrance: As the size of the groups attached to the carbonyl carbon increases,the rate of nucleophilic attack decreases.
$2$. Electronic effects: Electron-donating groups (like alkyl groups) decrease the electrophilicity of the carbonyl carbon,while electron-withdrawing groups (like phenyl rings via resonance) also decrease the reactivity compared to aliphatic counterparts.
Comparing the given compounds:
$(A)\ HCHO$ (Formaldehyde): No alkyl groups,least steric hindrance,most reactive.
$(B)\ CH_3COCH_3$ (Acetone): Two methyl groups,more steric hindrance than $HCHO$.
$(C)\ PhCOCH_3$ (Acetophenone): One phenyl group and one methyl group. The phenyl group provides resonance stabilization to the carbonyl carbon,reducing its electrophilicity.
$(D)\ PhCOPh$ (Benzophenone): Two phenyl groups,highest steric hindrance and significant resonance stabilization,making it the least reactive.
Thus,the reactivity order is $HCHO > CH_3COCH_3 > PhCOCH_3 > PhCOPh$.
The increasing order is $D < C < B < A$.

(A) The acid strength of carboxylic acids depends on the stability of the conjugate base (carboxylate ion).
Electron-withdrawing groups $(EWG)$ increase acidity by stabilizing the negative charge,while electron-donating groups $(EDG)$ decrease acidity by destabilizing it.
The compounds are:
$(A) CH_3CO_2H$ (Acetic acid)
$(B) MeOCH_2CO_2H$ (Methoxyacetic acid,$-OCH_3$ is $EWG$)
$(C) CF_3CO_2H$ (Trifluoroacetic acid,$-CF_3$ is a very strong $EWG$)
$(D) (CH_3)_2CHCO_2H$ (Isobutyric acid,isopropyl group is $EDG$)
Comparing the effects:
$(D)$ has an isopropyl group (strongest $EDG$),making it the least acidic.
$(A)$ has a methyl group (weak $EDG$).
$(B)$ has a methoxy group $(EWG)$.
$(C)$ has three fluorine atoms (strongest $EWG$).
The order of increasing acid strength is $D < A < B < C$.

(B) Anomers of glucose are cyclic diastereomers (epimers) that differ in configuration at the anomeric carbon,which is $C-1$.
They exist in two forms,$\alpha-$ and $\beta-$form,depending on the orientation of the hydroxyl group at $C-1$ relative to the $CH_2OH$ group.

(A) $DNA$ contains two types of nitrogenous bases: purines and pyrimidines.
The purines are adenine $(A)$ and guanine $(G)$.
The pyrimidines present in $DNA$ are cytosine $(C)$ and thymine $(T)$.
In $RNA$,thymine is replaced by uracil $(U)$.
Therefore,the pyrimidine bases in $DNA$ are cytosine and thymine.

(D) In $Fe(CO)_5$,the $Fe-C$ bond is formed by the donation of a lone pair of electrons from the $C$ atom of $CO$ to the empty $d$-orbital of $Fe$,which creates a $\sigma$-bond.
Additionally,there is back-donation of electrons from the filled $d$-orbitals of $Fe$ into the empty antibonding $\pi^*$-orbitals of $CO$,which creates a $\pi$-bond.
Therefore,the $Fe-C$ bond possesses both $\sigma$ and $\pi$ characters.

(D) $1$. Identify the ligands: There are $5$ ammine $(NH_3)$ ligands and $1$ nitrito$-N$ $(NO_2^-)$ ligand.
$2$. Determine the oxidation state of the central metal atom $(Co)$: Let the oxidation state be $x$. The charge on $NH_3$ is $0$,$NO_2^-$ is $-1$,and $Cl^-$ is $-1$. The total charge of the complex is $0$. Thus,$x + 5(0) + 1(-1) + 2(-1) = 0$,which gives $x - 3 = 0$,so $x = +3$.
$3$. Name the complex: The ligands are named in alphabetical order (ammine before nitrito$-N$). The metal is cobalt followed by its oxidation state in Roman numerals in parentheses. The counter ion is chloride.
$4$. The correct name is Pentaammine nitrito$-N$-cobalt$(III)$ chloride.