AIPMT 1999 Chemistry Question Paper with Answer and Solution

(B) The linear velocity $\vec v$ is given by the cross product of angular velocity $\vec \omega$ and position vector $\vec r$:
$\vec v = \vec \omega \times \vec r$
$\vec v = \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & -4 & 1 \\ 5 & -6 & 6 \end{vmatrix}$
Expanding the determinant:
$\vec v = \hat i((-4)(6) - (1)(-6)) - \hat j((3)(6) - (1)(5)) + \hat k((3)(-6) - (-4)(5))$
$\vec v = \hat i(-24 + 6) - \hat j(18 - 5) + \hat k(-18 + 20)$
$\vec v = -18\hat i - 13\hat j + 2\hat k$

(A) The potential difference per unit length,also known as the potential gradient $(k)$,is given by the formula:
$k = \frac{V}{L}$
Where $V$ is the potential difference across the wire and $L$ is the length of the wire.
Given:
$V = 2\, V$
$L = 4\, m$
Substituting the values:
$k = \frac{2}{4} = 0.5\, V/m$
Therefore,the potential difference per unit length is $0.5\, V/m$.

(D) The molar mass of $NH_3$ is $14 + (3 \times 1) = 17 \ g/mol$.
Number of moles of $NH_3 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{4.25 \ g}{17 \ g/mol} = 0.25 \ mol$.
Number of molecules of $NH_3 = 0.25 \times 6.022 \times 10^{23} \approx 1.5055 \times 10^{23}$ molecules.
Each molecule of $NH_3$ contains $4$ atoms ($1$ nitrogen atom and $3$ hydrogen atoms).
Total number of atoms $= 1.5055 \times 10^{23} \times 4 = 6.022 \times 10^{23} \approx 6 \times 10^{23}$.

(D) Arnold $Sommerfeld$ modified $Bohr's$ atomic model in $1916$. He proposed that electrons move in elliptical orbits around the nucleus,in addition to the circular orbits proposed by $Bohr$. This modification helped explain the fine structure of spectral lines in hydrogen-like atoms.

(A) The de-Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.
Given: mass $m = 1 \, g = 10^{-3} \, kg$,velocity $v = 100 \, m/sec$,and Planck's constant $h = 6.63 \times 10^{-34} \, J \cdot s$.
Substituting the values: $\lambda = \frac{6.63 \times 10^{-34} \, J \cdot s}{(10^{-3} \, kg) \times (100 \, m/sec)} = \frac{6.63 \times 10^{-34}}{10^{-1}} = 6.63 \times 10^{-33} \, m$.
Therefore,the correct option is $A$.

(C) According to Heisenberg's uncertainty principle,$\Delta x \times \Delta p \geq \frac{h}{4\pi}$.
Given $\Delta p = 1 \times 10^{-5} \ kg \ m/s$ and $h = 6.62 \times 10^{-34} \ kg \ m^2/s$.
Substituting the values: $\Delta x = \frac{h}{4\pi \times \Delta p}$.
$\Delta x = \frac{6.62 \times 10^{-34}}{4 \times 3.14159 \times 1 \times 10^{-5}}$.
$\Delta x = \frac{6.62 \times 10^{-34}}{12.566 \times 10^{-5}} \approx 5.27 \times 10^{-30} \ m$.

(D) The atomic number of iron $(Fe)$ is $26$.
Following the Aufbau principle,the electrons are filled in the order of increasing energy: $1s, 2s, 2p, 3s, 3p, 4s, 3d$.
Filling $26$ electrons: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$.
Therefore,the correct configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$.

(C) In diborane $(B_2H_6)$,each boron atom is bonded to four hydrogen atoms (two terminal and two bridging).
This results in a tetrahedral geometry around each boron atom.
Therefore,the hybridization of each boron atom is $sp^3$.

(B) $H_2O_2$ has a non-planar structure.
It adopts an open-book configuration to minimize repulsion between the lone pairs of electrons on the oxygen atoms.

(D) The mass of gas can be determined by weighing the container filled with gas and again weighing the container after removing the gas. The difference between the two weights gives the mass of the gas.

(B) The strength of a conjugate base is inversely proportional to the strength of its corresponding acid.
$1$. The conjugate acids of the given bases are: $HCl$ $(Cl^{-})$,$CH_3COOH$ $(CH_3COO^{-})$,$HSO_4^{-}$ $(SO_4^{2-})$,and $HNO_2$ $(NO_2^{-})$.
$2$. Among these acids,$CH_3COOH$ is the weakest acid.
$3$. Since the conjugate base of a weak acid is a strong base,$CH_3COO^{-}$ is the strongest conjugate base among the given options.
$CH_3COOH \rightleftharpoons CH_3COO^{-} + H^{+}$

(D) The dissociation of $CaF_2$ is given by: $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$.
Let the solubility be $s = 2 \times 10^{-4} \, mol/L$.
The solubility product expression is $K_{sp} = [Ca^{2+}][F^-]^2 = (s)(2s)^2 = 4s^3$.
Substituting the value of $s$: $K_{sp} = 4 \times (2 \times 10^{-4})^3$.
$K_{sp} = 4 \times (8 \times 10^{-12}) = 3.2 \times 10^{-11}$.

(A) Given: Concentration of solution $= 0.1 \ M$.
Degree of ionisation $(\alpha) = 2\% = 0.02$.
Ionic product of water $(K_w) = 1 \times 10^{-14}$.
Concentration of $[H^{+}] = C \times \alpha = 0.1 \times 0.02 = 2 \times 10^{-3} \ M$.
Concentration of $[OH^{-}] = \frac{K_w}{[H^{+}]} = \frac{1 \times 10^{-14}}{2 \times 10^{-3}} = 0.5 \times 10^{-11} = 5 \times 10^{-12} \ M$.

(B) In an endothermic reaction,heat is absorbed from the surroundings.
Therefore,the enthalpy change $\Delta H$ is always positive $(+ve)$.

(A) The heat of formation of $SO_2$ is the enthalpy change for the reaction: $S + O_2 \to SO_2$.
Given equations:
$(i) \ S + \frac{3}{2} O_2 \to SO_3 + 2x \ kcal$
$(ii) \ SO_2 + \frac{1}{2} O_2 \to SO_3 + y \ kcal$
To obtain the target reaction,subtract equation $(ii)$ from equation $(i)$:
$(S + \frac{3}{2} O_2) - (SO_2 + \frac{1}{2} O_2) \to (SO_3 - SO_3) + (2x - y) \ kcal$
$S + O_2 - SO_2 \to 2x - y \ kcal$
$S + O_2 \to SO_2 + (2x - y) \ kcal$
Thus,the heat of formation of $SO_2$ is $(2x - y) \ kcal$.

(D) The oxidation state of a metal in a metal carbonyl complex like $[Fe(CO)_5]$ is zero because the ligand $CO$ (carbonyl) is a neutral ligand.
In $[Fe(CO)_5]$,the oxidation state of $Fe$ is calculated as: $x + 5(0) = 0$,which gives $x = 0$.

(D) Let the average oxidation state of $Fe$ be $x$.
In $Fe_3O_4$,the oxidation state of oxygen is $-2$.
The sum of oxidation states of all atoms in a neutral molecule is $0$.
$3x + 4(-2) = 0$
$3x - 8 = 0$
$3x = 8$
$x = \frac{8}{3}$

(B) The correct answer is $(B)$.
Although $F$ has the highest electronegativity,its electron affinity is lower than that of $Cl$ due to its very small atomic size.
In $F$,the incoming electron experiences significant inter-electronic repulsions from the already present electrons in the small $2p$ subshell.
$Cl$,being larger,can accommodate the incoming electron more easily,resulting in a higher energy release (more negative electron gain enthalpy).

(D) The structure of $H_2O_2$ is open book shaped,and thus it is non-planar.

(A) Pencils are not made of lead; they are made from a mixture of clay and graphite.
Therefore,the percentage of lead present in a lead pencil is $0\,\%$.

(C) To find the empirical formula,we calculate the molar ratio of each element:

$C = 40\% \rightarrow 40/12 = 3.33$	$3.33/3.33 = 1$
$H = 13.33\% \rightarrow 13.33/1 = 13.33$	$13.33/3.33 = 4$
$N = 46.67\% \rightarrow 46.67/14 = 3.33$	$3.33/3.33 = 1$

The ratio of $C:H:N$ is $1:4:1$.
Therefore,the empirical formula is $CH_4N$.

(A) The process of decomposition of organic matter in the presence of oxygen (aerobic conditions) without the production of foul-smelling (odoriferous) substances is known as $Decay$. $N_2$ fixation,Nitrification,and Denitrification are specific nitrogen cycle processes that do not describe this general decomposition.

(B) The reaction $CH_4 + Cl_2 \xrightarrow{uv \ light} CH_3Cl + HCl$ involves the replacement of a hydrogen atom in methane $(CH_4)$ by a chlorine atom $(Cl)$.
This type of reaction,where one atom or group of atoms in a molecule is replaced by another atom or group,is known as a substitution reaction.
Specifically,this is a free-radical substitution reaction initiated by $UV$ light.

(C) The hydration of acetylene in the presence of dilute $H_2SO_4$ and $Hg^{2+}$ ions proceeds as follows:
$CH \equiv CH + H_2O \xrightarrow{H_2SO_4, Hg^{2+}} [CH_2 = CH - OH]$
This intermediate,vinyl alcohol,is unstable and undergoes tautomerization (rearrangement) to form a more stable carbonyl compound:
$[CH_2 = CH - OH] \rightarrow CH_3 - CHO$ (Acetaldehyde).

(A) In Friedel-Crafts alkylation,an alkyl group is introduced into the benzene ring by reacting benzene with an alkyl halide in the presence of a Lewis acid catalyst like $AlCl_3$.
For the methylation of benzene,the reactants are benzene $(C_6H_6)$ and methyl chloride $(CH_3Cl)$.
The reaction is: $C_6H_6 + CH_3Cl \xrightarrow{AlCl_3} C_6H_5CH_3 + HCl$.

(A) Given: Molecular weight $(M_w)$ = $200 \ g/mol$.
Since the acid is dibasic,its basicity is $2$.
Equivalent weight $(E)$ = $\frac{M_w}{\text{basicity}} = \frac{200}{2} = 100 \ g/eq$.
Strength $(N)$ = $0.1 \ N$ (decinormal).
Volume $(V)$ = $100 \ mL = 0.1 \ L$.
Using the formula: $N = \frac{w}{E \times V(L)}$,where $w$ is the weight in grams.
$0.1 = \frac{w}{100 \times 0.1}$.
$0.1 = \frac{w}{10}$.
$w = 0.1 \times 10 = 1 \ g$.

(B) $Graphite$ is an allotrope of carbon where each carbon atom is $sp^2$ hybridized,leaving one free electron per atom. These delocalized electrons allow $Graphite$ to conduct electricity,unlike $Diamond$ which has all four valence electrons involved in covalent bonding.

(A) Carboxylic acids are stronger acids than carbonic acid $(H_2CO_3)$.
Therefore,they react with sodium bicarbonate $(NaHCO_3)$ to evolve $CO_2$ gas.
Acetic acid $(CH_3COOH)$ is a carboxylic acid,so it reacts as follows:
$CH_3COOH + NaHCO_3 \to CH_3COONa + H_2O + CO_2 \uparrow$
Phenol is a weaker acid than carbonic acid and does not react with $NaHCO_3$.
$n$-hexanol is an alcohol and does not react with $NaHCO_3$.

(D) Photosynthetic bacteria are prokaryotic organisms. They lack membrane-bound organelles such as chloroplasts. Instead,they possess specialized membranous structures called chromatophores,which contain photosynthetic pigments for the process of photosynthesis.

(B) $Nepenthes$ (Pitcher plant) and Bladderwort $(Utricularia)$ are well-known examples of insectivorous plants.
These plants grow in nitrogen-deficient soil and have evolved modified leaves to trap and digest small insects to fulfill their nitrogen requirements.
$Rafflesia$ is a parasitic plant,and $Viscum$ (mistletoe) is a partial stem parasite,not insectivorous.

(C) The hormone secretin is produced by the $S$-cells located in the mucosal lining of the duodenum. When acidic chyme from the stomach enters the duodenum,it triggers the release of secretin into the bloodstream. This hormone then stimulates the pancreas to secrete bicarbonate-rich pancreatic juice to neutralize the acidity.

(A) Vitamin $K$ is essential for blood clotting. Deficiency of Vitamin $K$ leads to excessive bleeding or hemorrhage. Beri-beri is caused by the deficiency of Vitamin $B_1$ (Thiamine). Therefore,the pair Vitamin $K$ - Beri-beri is mismatched.

(B) Genetic engineering involves the manipulation of genetic material to produce desired traits.
$Plasmids$ are small, circular, double-stranded $DNA$ molecules that are distinct from a cell's chromosomal $DNA$.
They are extensively used as vectors in genetic engineering to carry foreign genes into host cells, making them a fundamental tool in recombinant $DNA$ technology.

(D) Tropical rain forests are the most diverse and highly dense ecosystems with maximum primary productivity (approximately $12000 \text{ kcal/m}^2\text{/yr}$). These regions receive high rainfall and maintain warm temperatures throughout the year,which supports a vast array of flora and fauna.

(D) Secretin is a peptide hormone produced by the $S$-cells of the duodenum,which is the first part of the small intestine.
It is released into the bloodstream in response to the acidity of the chyme entering the duodenum from the stomach.
Its primary function is to stimulate the pancreas to secrete bicarbonate-rich pancreatic juice,which helps neutralize the acidic chyme,and to inhibit the secretion of gastric acid by the stomach.

(C) Gonadotropic hormones (GTHs),which include Luteinizing Hormone $(LH)$ and Follicle Stimulating Hormone $(FSH)$,are secreted by the anterior pituitary gland.
The anterior pituitary gland is also known as the adenohypophysis.
These hormones play a crucial role in regulating the reproductive functions of the gonads (testes and ovaries).

(B) The strength of a conjugate base is inversely proportional to the strength of its corresponding acid.
$1$. $Cl^{-}$ is the conjugate base of $HCl$ (a strong acid).
$2$. $CH_3COO^{-}$ is the conjugate base of $CH_3COOH$ (a weak acid).
$3$. $SO_4^{2-}$ is the conjugate base of $HSO_4^{-}$ (a moderately strong acid).
$4$. $NO_2^{-}$ is the conjugate base of $HNO_2$ (a weak acid).
Comparing the acid strengths: $HCl > HSO_4^{-} > HNO_2 > CH_3COOH$.
Therefore,the order of basicity of their conjugate bases is: $CH_3COO^{-} > NO_2^{-} > SO_4^{2-} > Cl^{-}$.
$CH_3COO^{-}$ is the strongest conjugate base among the given options.

(D) The strength of a conjugate base is inversely proportional to the strength of its corresponding acid.
Stronger acids have weaker conjugate bases,and weaker acids have stronger conjugate bases.
The corresponding acids for the given bases are:
$A) NO_3^-$ comes from $HNO_3$ (Strong acid)
$B) Cl^-$ comes from $HCl$ (Strong acid)
$C) SO_4^{2-}$ comes from $HSO_4^-$ (Stronger acid than $CH_3COOH$)
$D) CH_3COO^-$ comes from $CH_3COOH$ (Weak acid)
Since $CH_3COOH$ is the weakest acid among the given options,its conjugate base $CH_3COO^-$ is the strongest.

(B) unit vector has a magnitude of $1$.
Let the vector be $\vec{A} = 0.5\hat{i} - 0.8\hat{j} + c\hat{k}$.
The magnitude is given by $|\vec{A}| = \sqrt{(0.5)^2 + (-0.8)^2 + c^2} = 1$.
Squaring both sides,we get $(0.5)^2 + (0.8)^2 + c^2 = 1^2$.
$0.25 + 0.64 + c^2 = 1$.
$0.89 + c^2 = 1$.
$c^2 = 1 - 0.89 = 0.11$.
Therefore,$c = \sqrt{0.11}$.

(C) The reactivity of an aromatic ring towards an electrophilic substitution reaction depends on the electron density of the ring.
Groups that donate electrons to the ring (activating groups) increase the electron density,making the ring more susceptible to electrophilic attack.
In the given compounds:
$1$. Benzene: Reference compound.
$2$. Chlorobenzene: $-Cl$ is a deactivating group due to its strong $-I$ effect,which outweighs its $+M$ effect.
$3$. Phenol: The $-OH$ group is a strongly activating group due to its strong $+M$ effect,which significantly increases the electron density at the $o-$ and $p-$ positions.
$4$. Toluene: The $-CH_3$ group is a weakly activating group due to its $+I$ effect and hyperconjugation.
Therefore,phenol is the most reactive towards electrophilic attack.

(C) $o$-Nitrophenol and $p$-nitrophenol can be separated by steam distillation.
$o$-Nitrophenol is steam volatile due to the presence of intramolecular hydrogen bonding,which reduces its intermolecular attraction.
In contrast,$p$-nitrophenol exhibits intermolecular hydrogen bonding,leading to the association of molecules,which makes it less volatile.
Therefore,$o$-nitrophenol distills over with steam,while $p$-nitrophenol remains in the distillation flask.

(A) The angular speed $\omega$ of an object moving in a circle is defined as the angle covered per unit time.
For a complete circle,the angle covered is $2\pi$ radians.
If the time taken to complete one circle is $t$,then the angular speed is given by $\omega = \frac{2\pi}{t}$.
Since both cars complete a circle in the same time $t$,their angular speeds are:
$\omega_1 = \frac{2\pi}{t}$ and $\omega_2 = \frac{2\pi}{t}$.
Therefore,the ratio of the angular speeds is $\frac{\omega_1}{\omega_2} = \frac{2\pi/t}{2\pi/t} = 1 : 1$.

(B) The linear velocity $\vec{v}$ is given by the cross product of angular velocity $\vec{\omega}$ and position vector $\vec{r}$:
$\vec{v} = \vec{\omega} \times \vec{r}$
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -4 & 1 \\ 5 & -6 & 6 \end{vmatrix}$
Expanding the determinant:
$\vec{v} = \hat{i}((-4)(6) - (1)(-6)) - \hat{j}((3)(6) - (1)(5)) + \hat{k}((3)(-6) - (-4)(5))$
$\vec{v} = \hat{i}(-24 + 6) - \hat{j}(18 - 5) + \hat{k}(-18 + 20)$
$\vec{v} = -18\hat{i} - 13\hat{j} + 2\hat{k}$

(B) The relationship between specific conductance $(\kappa)$,cell constant $(G^*)$,and resistance $(R)$ is given by the formula: $\kappa = G^* \times \frac{1}{R}$.
Rearranging the formula to solve for the cell constant: $G^* = \kappa \times R$.
Given values: $\kappa = 0.012 \ \Omega^{-1} \ cm^{-1}$ and $R = 55 \ \Omega$.
Substituting the values: $G^* = 0.012 \ \Omega^{-1} \ cm^{-1} \times 55 \ \Omega = 0.66 \ cm^{-1}$.

(D) The relationship between the speed of light $(c)$,frequency $(f)$,and wavelength $(\lambda)$ is given by the formula: $c = f \lambda$.
Given that the speed of light $c = 3 \times 10^{8}\; m/s$ and the frequency $f = 100\; Hz$.
Rearranging the formula to solve for wavelength: $\lambda = \frac{c}{f}$.
Substituting the values: $\lambda = \frac{3 \times 10^{8}\; m/s}{100\; Hz} = 3 \times 10^{6}\; m$.
Therefore,the correct option is $D$.

(C) The kinetic energy $(KE)$ gained by a charged particle when accelerated through a potential difference $(V)$ is given by the formula: $\Delta KE = qV$.
For a proton,the charge is $q = e$.
Given the potential difference $V = 1\,V$.
Substituting these values,we get: $\Delta KE = e \times 1\,V = 1\,eV$.
Therefore,the kinetic energy of the proton is $1\,eV$.

(D) The strength of a conjugate base is inversely proportional to the strength of its corresponding acid.
Stronger acids have weaker conjugate bases,and weaker acids have stronger conjugate bases.
The corresponding acids for the given bases are:
$NO_3^-$ comes from $HNO_3$ (a strong acid).
$Cl^-$ comes from $HCl$ (a strong acid).
$SO_4^{2-}$ comes from $HSO_4^-$ (a moderately strong acid).
$CH_3COO^-$ comes from $CH_3COOH$ (a weak acid).
Since $CH_3COOH$ is the weakest acid among the given options,its conjugate base $CH_3COO^-$ is the strongest.

(D) Photophosphorylation is the process of synthesizing $ATP$ from $ADP$ and inorganic phosphate $(Pi)$ in the presence of light energy.
This process occurs in the chloroplasts of plants during the light-dependent reactions of photosynthesis.
The reaction can be represented as: $ADP + Pi \xrightarrow{\text{Light energy}} ATP$.

(B) The $Hepatic \text{ } portal \text{ } vein$ is a blood vessel that carries blood from the gastrointestinal tract and spleen to the liver.
It transports nutrient-rich blood, containing absorbed substances like $glucose$ and $amino \text{ } acids$, directly from the digestive tract to the liver for processing, storage, or detoxification before it enters the general systemic circulation.

(B) To determine the number of unpaired $d$-electrons,we look at the electronic configuration of each ion:
$Zn^{+}$ $(Z=30)$: $[Ar] \, 3d^{10} 4s^1$. Number of unpaired $d$-electrons = $0$.
$Fe^{2+}$ $(Z=26)$: $[Ar] \, 3d^6 4s^0$. The $3d^6$ configuration has $4$ unpaired electrons.
$N^{3+}$ $(Z=7)$: $1s^2 2s^2 2p^6$. This ion has no $d$-electrons.
$Cu^{+}$ $(Z=29)$: $[Ar] \, 3d^{10} 4s^0$. Number of unpaired $d$-electrons = $0$.
Comparing these,$Fe^{2+}$ has the most unpaired $d$-electrons $(4)$.

(D) The reaction of potassium ferrocyanide $(K_4[Fe(CN)_6])$ with concentrated sulfuric acid $(H_2SO_4)$ and water $(H_2O)$ is as follows:
$K_4[Fe(CN)_6] + 6H_2SO_4 + 6H_2O \rightarrow 2K_2SO_4 + FeSO_4 + 3(NH_4)_2SO_4 + 6CO$
As shown in the balanced chemical equation,the gas evolved is carbon monoxide $(CO)$.

(D) Purification of aluminium by electrolytic refining is called Hoop's process.
By this method,$99.9\%$ pure aluminium metal is obtained.
The cell used in this method consists of three layers.
In the cell,pure $Al$ acts as the cathode,while the anode is made of impure $Al$.

(D) Among the different allotropes of phosphorus,$White \ phosphorus$ is the most reactive.
This is due to the high angular strain in the $P_4$ molecules,where the bond angle is only $60^\circ$,making it highly unstable and reactive compared to other allotropes like $Red$ or $Black \ phosphorus$.

(D) The anhydride of an acid is formed by the removal of water molecules from the acid.
For nitric acid $(HNO_3)$,the reaction is:
$2HNO_3 \rightarrow N_2O_5 + H_2O$
Thus,$N_2O_5$ is the anhydride of $HNO_3$.

(C) Chlorine can be prepared by the oxidation of hydrochloric acid $(HCl)$ using oxidizing agents like manganese dioxide $(MnO_2)$ or potassium permanganate $(KMnO_4)$.
$1$. Using $MnO_2$: $MnO_2 + 4HCl \to MnCl_2 + 2H_2O + Cl_2$
$2$. Using $KMnO_4$: $2KMnO_4 + 16HCl \to 2KCl + 2MnCl_2 + 8H_2O + 5Cl_2$
Thus,both $MnO_2$ and $KMnO_4$ are used in the preparation of chlorine.

(A) The acidic character of oxides of elements in the same group decreases as we move down the group.
This is because the metallic character increases down the group,which leads to a decrease in the non-metallic character and electronegativity.
Since $N$,$P$,$As$,and $Sb$ belong to Group $15$,the order of acidic strength is $N_2O_5 > P_2O_5 > As_2O_5 > Sb_2O_5$.
Therefore,$N_2O_5$ is the most acidic oxide.

(B) Yellow phosphorus consists of discrete $P_4$ tetrahedral molecules with high angular strain,making it highly reactive.
In contrast,red phosphorus exists as a polymeric chain structure where $P_4$ tetrahedra are linked together.
This polymeric structure reduces the strain and makes it significantly less reactive than yellow phosphorus.

(B) Gammexene,also known as $BHC$ (Benzene hexachloride) or Lindane,is produced by the addition reaction of benzene with chlorine in the presence of sunlight.
The reaction is: $C_6H_6 + 3Cl_2 \xrightarrow{\text{Sunlight}} C_6H_6Cl_6$ $(BHC)$.
Therefore,the correct option is $(b)$.

(C) According to Raoult's law for a non-volatile solute,the relative lowering of vapour pressure is given by:
$\frac{P^0 - P_s}{P^0} = \frac{n}{n + N} \approx \frac{n}{N} = \frac{w \times M}{m \times W}$
Where:
$P^0 = 640 \, mm \, Hg$ (Vapour pressure of pure benzene)
$P_s = 600 \, mm \, Hg$ (Vapour pressure of solution)
$w = 2.175 \, g$ (Mass of solute)
$W = 39.08 \, g$ (Mass of benzene)
$M = 78 \, g/mol$ (Molar mass of benzene,$C_6H_6$)
$m$ = Molar mass of solute
Substituting the values:
$\frac{640 - 600}{640} = \frac{2.175 \times 78}{m \times 39.08}$
$\frac{40}{640} = \frac{169.65}{m \times 39.08}$
$m = \frac{169.65 \times 640}{39.08 \times 40}$
$m = 69.45 \, g/mol$
Thus,the molecular weight of the solid substance is approximately $69.5 \, g/mol$.

(D) The formula for elevation in boiling point is given by: $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{w \times 1000}{M \times W}$,where $w$ is the mass of solute,$M$ is the molar mass,and $W$ is the mass of solvent in grams.
Substituting the values: $\Delta T_b = \frac{K_b \times w \times 1000}{M \times W}$.
Rearranging for $M$: $M = \frac{K_b \times w \times 1000}{\Delta T_b \times W}$.
Given: $K_b = 2.16 \ ^oC \ kg \ mol^{-1}$,$w = 0.15 \ g$,$W = 15 \ g$,and $\Delta T_b = 0.216 \ ^oC$.
$M = \frac{2.16 \times 0.15 \times 1000}{0.216 \times 15} = \frac{324}{3.24} = 100 \ g \ mol^{-1}$.
Thus,the molecular weight is $100$.

(A) For a first order reaction,the half-life period is given by the formula $t_{1/2} = \frac{0.693}{k}$,where $k$ is the rate constant.
Since this expression does not contain the initial concentration $[A]_0$,the half-life period is independent of the initial concentration of the reactant.

(B) The relationship between specific conductance $(K)$,resistance $(R)$,and cell constant $(G^*)$ is given by the formula:
$K = \frac{1}{R} \times G^*$
Rearranging to solve for the cell constant:
$G^* = K \times R$
Given:
$K = 0.012 \ \Omega^{-1} \ cm^{-1}$
$R = 55 \ \Omega$
Calculation:
$G^* = 0.012 \ \Omega^{-1} \ cm^{-1} \times 55 \ \Omega = 0.66 \ cm^{-1}$
Therefore,the cell constant is $0.66 \ cm^{-1}$.

(D) Cassiterite is an ore of $Sn$.
It is also called tinstone and has the molecular formula $SnO_2$.
It is a heavy,metallic,hard tin dioxide that serves as the major ore of tin.
It is colourless when pure,but appears brown or black when iron impurities are present.

(B) To find the number of unpaired $d-$ electrons,we look at the electronic configuration of each ion:
$Zn$ $(Z=30)$: $[Ar] 3d^{10} 4s^2$ ($0$ unpaired electrons).
$Fe^{2+}$ $(Z=26)$: $[Ar] 3d^6 4s^0$. In $3d^6$,the electrons are filled as: $\uparrow\downarrow, \uparrow, \uparrow, \uparrow, \uparrow$. This results in $4$ unpaired electrons.
$Ni^{3+}$ $(Z=28)$: $[Ar] 3d^7 4s^0$. In $3d^7$,the electrons are filled as: $\uparrow\downarrow, \uparrow\downarrow, \uparrow, \uparrow, \uparrow$. This results in $3$ unpaired electrons.
$Cu^{+}$ $(Z=29)$: $[Ar] 3d^{10} 4s^0$ ($0$ unpaired electrons).
Therefore,$Fe^{2+}$ has the maximum number of unpaired $d-$ electrons.

(D) . Bell metal is an alloy primarily composed of copper $(Cu)$ and tin $(Sn)$.
It typically consists of approximately $80\% \ Cu$ and $20\% \ Sn$.

(D) Solution:
$(1)$ In options $B$ and $C$,there are no transition metals,so they are eliminated.
$(2)$ In $CrO_5$ (chromium pentoxide),the oxidation state of $Cr$ is $+6$ due to the presence of peroxo linkages.
$(3)$ In $[Fe(CO)_5]$,carbon monoxide $(CO)$ is a neutral ligand. Since the overall charge of the complex is zero,the oxidation state of $Fe$ is calculated as $x + 5(0) = 0$,which gives $x = 0$.

(A) $o-$ and $p-$ nitrophenols are separated by steam distillation.
$o-$ nitrophenol exhibits intramolecular hydrogen bonding,making it steam volatile.
$p-$ nitrophenol exhibits intermolecular hydrogen bonding,leading to higher molecular association and lower volatility,thus it is not steam volatile.

(A) The reaction between $Benzaldehyde$ $(C_6H_5CHO)$ and concentrated $NaOH$ is a classic example of the $Cannizzaro$ reaction.
Since $Benzaldehyde$ lacks $\alpha$-hydrogen atoms,it undergoes self-oxidation and reduction (disproportionation) in the presence of a strong base.
The products formed are $Benzyl \ alcohol$ $(C_6H_5CH_2OH)$ and $Sodium \ benzoate$ $(C_6H_5COONa)$.

(A) Aldol condensation requires the presence of at least one $\alpha$-hydrogen atom in the aldehyde or ketone.
$HCHO$ (Formaldehyde) does not contain any $\alpha$-carbon atom,and therefore,it lacks $\alpha$-hydrogen atoms.
Consequently,$HCHO$ cannot undergo aldol condensation.

(C) Carboxylic acids are stronger acids than carbonic acid $(H_2CO_3)$.
Therefore,they react with sodium bicarbonate $(NaHCO_3)$ to evolve carbon dioxide $(CO_2)$ gas.
Acetic acid $(CH_3COOH)$ is a carboxylic acid,so it reacts as follows:
$CH_3COOH + NaHCO_3 \to CH_3COONa + H_2O + CO_2 \uparrow$
Phenol and $n-$hexanol are much weaker acids than carbonic acid and do not react with $NaHCO_3$ to evolve $CO_2$.

(C) The reaction is known as the $Hofmann$ bromamide degradation reaction.
In this reaction,an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide to form a primary amine.
The chemical equation is:
$CH_3CONH_2 + Br_2 + 4KOH \xrightarrow{\Delta} CH_3NH_2 + K_2CO_3 + 2KBr + 2H_2O$

(C) The given reaction is the $Stephen$ reduction followed by hydrolysis.
Step $1$: $CH_3-C \equiv N + 2[H] \xrightarrow{SnCl_2/HCl} CH_3-CH=NH$ (Imine,$X$)
Step $2$: $CH_3-CH=NH + H_2O \xrightarrow{\Delta} CH_3-CHO + NH_3$
Here,$Y$ is $CH_3-CHO$ which is Acetaldehyde.

(B) Natural rubber is a linear polymer of $2$-methyl-$1,3$-butadiene,which is commonly known as $Isoprene$.
The polymerization reaction is:
$n CH_2 = C(CH_3) - CH = CH_2 \to -(CH_2 - C(CH_3) = CH - CH_2)_n$ (Natural rubber).

(B)

Sugar	Relative sweetness
$Sucrose$	$100$
$Glucose$	$74$
$Lactose$	$16$
$Fructose$	$173$

Among the given sugars,$Fructose$ has the highest relative sweetness value of $173$.
Therefore,it is the sweetest sugar.
Correct option is $B$.

(A) Given: Resistance $(R)$ = $200 \ \Omega$,Cell constant $(G^*)$ = $1 \ cm^{-1}$,Normality $(N)$ = $0.01 \ N$.
First,calculate the conductivity $(\kappa)$:
$\kappa = \frac{G^*}{R} = \frac{1 \ cm^{-1}}{200 \ \Omega} = 0.005 \ \Omega^{-1} cm^{-1}$.
Next,calculate the equivalent conductance $(\lambda_{eq})$:
$\lambda_{eq} = \frac{\kappa \times 1000}{N} = \frac{0.005 \times 1000}{0.01} = \frac{5}{0.01} = 500 \ \Omega^{-1} cm^2 eq^{-1}$.
This can be expressed as $5 \times 10^2 \ \Omega^{-1} cm^2 eq^{-1}$.

(B) The $Carbylamine$ reaction is used for the preparation of isocyanides (or carbylamines).
In this reaction,a primary amine (aliphatic or aromatic) is heated with chloroform $(CHCl_3)$ and an alcoholic potassium hydroxide $(KOH)$ solution to form an isocyanide (carbylamine),which has a foul smell.
The reaction for the preparation of phenyl isocyanide from aniline is:
$C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$
Therefore,the correct option is $(B)$.