NEET 2020 Chemistry Question Paper with Answer and Solution

(C) Cationic detergents are quaternary ammonium salts of amines with acetates,chlorides,or bromides as anions.
They are called cationic detergents because the cationic part of these molecules contains a long hydrocarbon chain and a positive charge on the $N$ atom.
For example,$CH_3(CH_2)_{15}N^+(CH_3)_3Br^-$ (cetyltrimethylammonium bromide) is a cationic detergent.

(B) The correct matches are:
$a. CO_{(g)} + H_{2(g)} \rightarrow iii. \text{Synthesis gas}$
$b. \text{Temporary hardness of water} \rightarrow i. Mg(HCO_3)_2 + Ca(HCO_3)_2$
$c. B_2H_6 \rightarrow ii. \text{An electron deficient hydride}$
$d. H_2O_2 \rightarrow iv. \text{Non-planar structure}$
Thus, the correct sequence is $a-iii, b-i, c-ii, d-iv$.

(A) The tertiary butyl carbocation $( (CH_3)_3C^+ )$ has $9$ $\alpha$-hydrogens,whereas the secondary butyl carbocation $( CH_3CH^+CH_2CH_3 )$ has $5$ $\alpha$-hydrogens.
Stability of carbocations is directly proportional to the number of hyperconjugative structures,which in turn is determined by the number of $\alpha$-hydrogens.
Since the tertiary butyl carbocation has more $\alpha$-hydrogens,it is more stable due to hyperconjugation.

(D) In $CH_4$,the oxidation number of $H$ is $+1$. Let the oxidation number of $C$ be $x$. So,$x + 4(+1) = 0$,which gives $x = -4$.
In $CCl_4$,the oxidation number of $Cl$ is $-1$. Let the oxidation number of $C$ be $y$. So,$y + 4(-1) = 0$,which gives $y = +4$.
Therefore,the oxidation number of carbon changes from $-4$ to $+4$.

(B) For $He_{2}$,the total number of electrons is $4$.
The molecular orbital configuration is $\sigma_{1s}^{2} \sigma_{1s}^{*2}$.
Bond Order $(B.O.) = \frac{1}{2} [N_{b} - N_{a}] = \frac{1}{2} [2 - 2] = 0$.
Since the bond order is $0$,the molecule $He_{2}$ does not exist.

(D) The Wurtz reaction is primarily used for the preparation of symmetrical alkanes containing an even number of carbon atoms.
$n-$Butane $(C_4H_{10})$ is symmetrical and can be prepared from ethyl bromide $(C_2H_5Br)$.
$n-$Hexane $(C_6H_{14})$ is symmetrical and can be prepared from propyl bromide $(C_3H_7Br)$.
$2,3-$Dimethylbutane is symmetrical and can be prepared from isopropyl bromide $(CH_3CH(Br)CH_3)$.
$n-$Heptane $(C_7H_{16})$ is an unsymmetrical alkane with an odd number of carbon atoms. Attempting to synthesize it via the Wurtz reaction would result in a mixture of products (e.g.,$n-$hexane,$n-$heptane,and $n-$octane) due to the coupling of different alkyl halides,leading to a poor yield.

(B) For the neutral atom ${ }_{71}^{175} Lu$:
The atomic number $(Z)$ is $71$,which represents the number of protons $(p^+)$.
Since the atom is neutral,the number of electrons $(e^-)$ is equal to the number of protons,so $e^- = 71$.
The mass number $(A)$ is $175$. The number of neutrons $(n^0)$ is calculated as $A - Z = 175 - 71 = 104$.
Therefore,the number of protons,neutrons,and electrons are $71, 104$,and $71$ respectively.

(A) molecule has a zero dipole moment if it is symmetrical and the bond dipoles cancel each other out.
$BF_3$ (trigonal planar) has a net dipole moment of $0$.
$BeF_2$ (linear) has a net dipole moment of $0$.
$CO_2$ (linear) has a net dipole moment of $0$.
$1,4-$dichlorobenzene (para-substituted) has a net dipole moment of $0$ due to symmetry.
Therefore,the set containing $BF_3, BeF_2, CO_2,$ and $1,4-$dichlorobenzene has a zero dipole moment.

(C) Paper chromatography is based on the principle of partition chromatography.
In this technique,the substances are distributed or partitioned between two liquid phases.
The stationary phase is the water molecules held in the pores of the filter paper,while the mobile phase is a solvent or mixture of solvents that moves over the paper.

(A-D) Let us verify the $IUPAC$ names for elements with $Z > 100$:
$1$. Unnilunium $(101)$ is Mendelevium $(Md)$. This is correct.
$2$. Unniltrium $(103)$ is Lawrencium $(Lr)$. This is correct.
$3$. Unnilhexium $(106)$ is Seaborgium $(Sg)$. This is correct.
$4$. Ununnilium $(110)$ is Darmstadtium $(Ds)$. This is correct.
Wait,re-evaluating the systematic nomenclature:
Unnilunium $(101)$ = $Unu$ (Mendelevium).
Unniltrium $(103)$ = $Unt$ (Lawrencium).
Unnilhexium $(106)$ = $Unh$ (Seaborgium).
Ununnilium $(110)$ = $Uun$ (Darmstadtium).
All given matches are actually correct. However,if the question implies an error in the provided options,let us re-check the systematic names.
Actually,all matches provided in the table are correct according to $IUPAC$ nomenclature. Since the question asks to identify the incorrect match and all are correct,there might be a typo in the question source. Given the standard curriculum,all these are correct matches.

(B) $NaOH$ is a strong electrolyte,so it dissociates completely:
$NaOH_{(aq)} \to Na^+_{(aq)} + OH^-_{(aq)}$
$[OH^-] = 0.1 \ M$
For $Ni(OH)_2$:
$Ni(OH)_{2(s)} \rightleftharpoons Ni^{2+}_{(aq)} + 2OH^-_{(aq)}$
Let the solubility of $Ni(OH)_2$ in the presence of $0.1 \ M \ NaOH$ be $S' \ M$.
Then,$[Ni^{2+}] = S'$ and $[OH^-] = (0.1 + 2S') \ M$.
Since $S'$ is very small,we can approximate $(0.1 + 2S') \approx 0.1$.
$K_{sp} = [Ni^{2+}][OH^-]^2$
$2 \times 10^{-15} = (S')(0.1)^2$
$2 \times 10^{-15} = S' \times 0.01$
$S' = \frac{2 \times 10^{-15}}{10^{-2}} = 2 \times 10^{-13} \ M$.

(C) When $HCl$ gas is passed through the solution,the concentration of $Cl^{-}$ ions increases significantly due to the dissociation of $HCl$.
According to the common ion effect,the ionic product of $NaCl$ $([Na^{+}][Cl^{-}])$ exceeds its solubility product $(K_{sp})$.
Since $NaCl$ has a lower solubility compared to $MgCl_{2}$ and $CaCl_{2}$ in the presence of high $Cl^{-}$ concentration,$NaCl$ crystallizes out of the solution.
$MgCl_{2}$ and $CaCl_{2}$ remain in the solution as they are more soluble.

(B) For free expansion of an ideal gas,the external pressure $P_{ext} = 0$.
Since $w = -P_{ext} \Delta V$,it follows that $w = 0$.
For an adiabatic process,the heat exchange $q = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$.
Substituting the values,$\Delta U = 0 + 0 = 0$.
For an ideal gas,internal energy change $\Delta U = n C_{v,m} \Delta T$.
Since $\Delta U = 0$,it implies $\Delta T = 0$.
Therefore,the correct conditions are $q = 0, \Delta T = 0, w = 0$.

(A) Let us analyze each statement:
$(a)$ Incorrect: Solid $CO_{2(s)}$ (dry ice) is used as a refrigerant,not gaseous $CO_{2(g)}$.
$(b)$ Incorrect: The structure of $C_{60}$ (Buckminsterfullerene) contains $20$ six-membered rings and $12$ five-membered rings,not $12$ six-membered and $20$ five-membered.
$(c)$ Correct: $ZSM-5$ is a shape-selective catalyst used in the petrochemical industry to convert alcohols directly into gasoline.
$(d)$ Correct: Carbon monoxide $(CO)$ is a colorless,odorless,and neutral gas.
Therefore,statements $(c)$ and $(d)$ are correct.

(B) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the formula:
$\Delta_{r}G^{\Theta} = -RT \ln K_{c}$
Given values:
$R = 8.314 \ J \ mol^{-1} \ K^{-1}$
$T = 300 \ K$
$K_{c} = 2 \times 10^{13}$
Substituting these values into the equation:
$\Delta_{r}G^{\Theta} = -8.314 \ J \ mol^{-1} \ K^{-1} \times 300 \ K \times \ln(2 \times 10^{13})$

(D) The formation of methanal $(HCHO)$ during ozonolysis indicates the presence of a terminal double bond $(=CH_2)$ in the alkene.
Among the given options,the structure in option $D$ (allylcyclohexane,which is $cyclohexyl-CH_2-CH=CH_2$) contains a terminal double bond.
Upon ozonolysis,it undergoes cleavage at the double bond to produce methanal $(HCHO)$ and cyclohexylacetaldehyde $(C_6H_{11}-CH_2-CHO)$.

(A) The metal ion described is $K^+$.
$K^+$ ions are essential for activating many enzymes,participating in the oxidation of glucose to produce $ATP$,and along with $Na^+$ ions,they play a crucial role in the transmission of nerve signals.

(A) In the reaction $2 Cl ( g ) \rightarrow Cl _{2}( g )$,two gaseous atoms combine to form one molecule of chlorine gas.
$1$. Enthalpy change $(\Delta_{ r } H)$: Bond formation is an exothermic process,therefore energy is released,making $\Delta_{ r } H < 0$.
$2$. Entropy change $(\Delta_{ r } S)$: The number of moles of gas decreases from $2$ to $1$ $(2 Cl ( g ) \rightarrow 1 Cl _{2}( g ))$. Since the randomness or disorder of the system decreases,$\Delta_{ r } S < 0$.
Thus,both $\Delta_{ r } H < 0$ and $\Delta_{ r } S < 0$.

(C) . $CO$ is a neutral oxide $(ii)$.
$b$. $BaO$ is a basic oxide $(i)$.
$c$. $Al_2O_3$ is an amphoteric oxide $(iv)$.
$d$. $Cl_2O_7$ is an acidic oxide $(iii)$.
Therefore,the correct match is $a-ii, b-i, c-iv, d-iii$.

(D) Step $1$: Calculate the number of moles of each gas.
$n_{N_2} = \frac{7 \ g}{28 \ g \ mol^{-1}} = 0.25 \ mol = \frac{1}{4} \ mol$
$n_{Ar} = \frac{8 \ g}{40 \ g \ mol^{-1}} = 0.2 \ mol = \frac{1}{5} \ mol$
Step $2$: Calculate the mole fraction of $N_2$ $(x_{N_2})$.
$x_{N_2} = \frac{n_{N_2}}{n_{N_2} + n_{Ar}} = \frac{1/4}{1/4 + 1/5} = \frac{0.25}{0.25 + 0.20} = \frac{0.25}{0.45} = \frac{5}{9}$
Step $3$: Calculate the partial pressure of $N_2$ $(p_{N_2})$.
$p_{N_2} = x_{N_2} \times P_{total} = \frac{5}{9} \times 27 \ bar = 5 \times 3 \ bar = 15 \ bar$.

(D) The correct statement is that carboxyhaemoglobin (haemoglobin bound to $CO$) is about $300$ times more stable than oxyhaemoglobin.
Therefore,the statement that it is less stable is incorrect.

(A) The number of atoms is calculated using the formula: $\text{Number of atoms} = \frac{\text{mass}}{\text{molar mass}} \times N_A \times \text{atomicity}$.
$(A)$ For $1 \ g$ of $Li$: $\frac{1}{7} \times N_A \times 1 \approx 0.143 \times N_A$.
$(B)$ For $1 \ g$ of $Ag$: $\frac{1}{108} \times N_A \times 1 \approx 0.009 \times N_A$.
$(C)$ For $1 \ g$ of $Mg$: $\frac{1}{24} \times N_A \times 1 \approx 0.042 \times N_A$.
$(D)$ For $1 \ g$ of $O_2$: $\frac{1}{32} \times N_A \times 2 = \frac{1}{16} \times N_A \approx 0.0625 \times N_A$.
Comparing the values,$1 \ g$ of $Li$ has the maximum number of atoms.

(C) The correct matches are as follows:
$(a)$ $Bacillus$ $thuringiensis$ produces $Cry$ proteins, which are used as bio-insecticides.
$(b)$ $Thermus$ $aquaticus$ is the source of $Taq$ $DNA$ polymerase, which is heat-stable and used in $PCR$.
$(c)$ $Agrobacterium$ $tumefaciens$ is a natural genetic engineer used as a cloning vector to transfer genes into plants.
$(d)$ $Salmonella$ $typhimurium$ was used by Cohen and Boyer for the construction of the first $rDNA$ molecule.
Therefore, the correct sequence is $(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)$.

(B) The $Hilum$ is the point of attachment where the body of the ovule fuses with the funicle. It represents the junction between the ovule and the stalk (funicle).

(A) The menstrual cycle is regulated by various hormones.
During the follicular phase,estrogen levels rise,which triggers a surge in the secretion of Luteinizing Hormone $(LH)$ from the anterior pituitary gland.
This phenomenon is known as the $LH$ surge.
This peak level of $LH$ induces the rupture of the mature Graafian follicle,leading to the release of the secondary oocyte,a process called ovulation.
Therefore,a high concentration of $LH$ (not low) is required for ovulation,and high estrogen levels are the primary trigger for this $LH$ surge.

(A) In the $DNA$ double helix structure, nitrogenous bases are paired according to Chargaff's rules.
Adenine $(A)$ always pairs with thymine $(T)$ via two hydrogen bonds $(A=T)$.
Conversely, guanine $(G)$ pairs with cytosine $(C)$ via three hydrogen bonds $(G \equiv C)$.
Therefore, the statement that adenine pairs with thymine through two $H$-bonds is correct.

(D) The correct matches are as follows:
$(a)$ Floating ribs: These are the $11^{th}$ and $12^{th}$ pairs of ribs that do not connect with the sternum,hence they are called floating ribs. So,$(a) - (iv)$.
$(b)$ Acromion process: This is a flattened,expanded process of the scapula that articulates with the clavicle. So,$(b) - (iii)$.
$(c)$ Scapula: The scapula is a large triangular flat bone situated in the dorsal part of the thorax between the $2^{nd}$ and $7^{th}$ ribs. So,$(c) - (i)$.
$(d)$ Glenoid cavity: This is a depression in the scapula that articulates with the head of the humerus to form the shoulder joint. So,$(d) - (ii)$.
Therefore,the correct sequence is $(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)$.

(B) Gibberellins are plant growth regulators that are known to increase the length of the stem in sugarcane plants.
Sugarcane stores sugar as carbohydrates in their stems.
By spraying gibberellins on the sugarcane crop,the internodal length increases,which leads to an increase in the overall length of the stem.
This increase in stem length results in an increase in the total yield of the sugarcane crop by as much as $20$ tonnes per acre.

(C) Amino acids are classified based on the nature of their $R$-group.
Basic amino acids contain an extra amino group in their side chain,making them basic in nature.
Examples of basic amino acids include Lysine,Arginine,and Histidine.
Tyrosine is an aromatic amino acid.
Glutamic acid is an acidic amino acid.
Valine is a neutral,non-polar,aliphatic amino acid.
Therefore,Lysine is the correct answer.

(A) ग्लूकोकोर्टिकोइड्स (जैसे कोर्टिसोल) यकृत में ग्लूकोनियोजेनेसिस (गैर-कार्बोहाइड्रेट स्रोतों से ग्लूकोज का निर्माण) को उत्तेजित करते हैं,जिससे रक्त में ग्लूकोज का स्तर बढ़ जाता है। इसलिए,विकल्प $A$ सही है। ग्लूकागन एक हाइपरग्लाइसेमिक हार्मोन है,हाइपोग्लाइसेमिक नहीं। इंसुलिन एक हाइपोग्लाइसेमिक हार्मोन है,हाइपरग्लाइसेमिक नहीं।

(A) In monocotyledonous plants,the primary root is short-lived and is replaced by a large number of roots. These roots originate from the base of the stem and constitute the fibrous root system. Therefore,the correct answer is $A$.

(B) The correct matches are as follows:
$1$. $Clostridium \ butylicum$ produces $Butyric \ acid$ $(a-ii)$.
$2$. $Trichoderma \ polysporum$ produces $Cyclosporin-A$,which is an immunosuppressive agent $(b-i)$.
$3$. $Monascus \ purpureus$ produces $Statins$,which act as blood cholesterol-lowering agents $(c-iv)$.
$4$. $Aspergillus \ niger$ is used for the commercial production of $Citric \ acid$ $(d-iii)$.
Therefore,the correct sequence is $(a-ii, b-i, c-iv, d-iii)$.

(D) The correct matches are as follows:
$(a)$ Placenta produces hormones like Human Chorionic Gonadotropin $(hCG)$, thus $(a)-(ii)$.
$(b)$ Zona pellucida is the non-cellular layer surrounding the ovum, thus $(b)-(iii)$.
$(c)$ Bulbourethral glands secrete a fluid that helps in the lubrication of the penis, thus $(c)-(iv)$.
$(d)$ Leydig cells (interstitial cells) are responsible for the synthesis and secretion of testicular hormones called androgens, thus $(d)-(i)$.
Therefore, the correct sequence is $(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)$.

(C) The correct matching is as follows:
$(a)$ Eosinophils: These cells release histaminase enzyme which helps in controlling allergic reactions. Thus,$(a)-(iii)$.
$(b)$ Basophils: These cells release granules containing histamine,serotonin,and heparin,which are involved in inflammatory reactions. Thus,$(b)-(iv)$.
$(c)$ Neutrophils: These are the most abundant phagocytic cells in the blood that destroy foreign organisms. Thus,$(c)-(ii)$.
$(d)$ Lymphocytes: These cells are responsible for the immune response of the body. Thus,$(d)-(i)$.
Therefore,the correct sequence is $(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)$.

(A) Robert May,a renowned ecologist,placed the global species diversity estimate at about $7$ million. While more than $1.5$ million species have been recorded to date,many species remain to be discovered and described,particularly in the tropics.

(A) $CH_3COONa$ is a salt of a strong base $(NaOH)$ and a weak acid $(CH_3COOH)$.
Therefore,its aqueous solution undergoes anionic hydrolysis to produce $OH^-$ ions,making the solution basic in nature.

(B) The enthalpy change of a reaction is given by the difference between the sum of bond energies of reactants and the sum of bond energies of products:
$\Delta H = \sum(BE)_{reactants} - \sum(BE)_{products}$
For the reaction $H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$,the equation is:
$\Delta H = [BE(H-H) + BE(Br-Br)] - [2 \times BE(H-Br)]$
Substituting the given values:
$-109 = [435 + 192] - 2 \times BE(H-Br)$
$-109 = 627 - 2 \times BE(H-Br)$
$2 \times BE(H-Br) = 627 + 109$
$2 \times BE(H-Br) = 736$
$BE(H-Br) = 368 \; kJ \ mol^{-1}$

(B) For a salt of the type $AB$,the dissociation is represented as: $AB(s) \rightleftharpoons A^+(aq) + B^-(aq)$.
The solubility product expression is $K_{sp} = [A^+][B^-] = s \times s = s^2$,where $s$ is the molar solubility.
Given $K_{sp} = 4 \times 10^{-8}$.
Therefore,$s^2 = 4 \times 10^{-8}$.
Taking the square root on both sides: $s = \sqrt{4 \times 10^{-8}} = 2 \times 10^{-4} \ mol/L$.

(B) The potential energy of two $H$ atoms at an infinite distance is represented by $a$.
The potential energy of the $H_2$ molecule at the equilibrium bond length is represented by $b$.
The bond energy is defined as the energy required to break the bond,which is the difference between the energy of the separated atoms and the energy of the bonded molecule.
Therefore,the bond energy is $(a-b)$.

(B) Steam distillation is a separation technique used for substances that are steam volatile and immiscible with water.
In this process,the compound vaporizes at a temperature lower than its boiling point,allowing for purification without decomposition.

(A) The purpose of adding gypsum $(CaSO_{4} \cdot 2H_{2}O)$ to cement is to retard or slow down the initial setting process of the cement.
This allows the cement to remain in a plastic state for a longer duration,ensuring it can be properly placed and hardened without forming cracks.

(D) Hydrolysis is a reaction where water molecules break down a compound without changing the oxidation states of the elements involved.
In the reaction $2F_{2(g)} + 2H_{2}O_{(l)} \rightarrow 4HF_{(aq)} + O_{2(g)}$,fluorine $(F_2)$ acts as a strong oxidizing agent.
It oxidizes oxygen in water from $-2$ to $0$ (in $O_2$) and itself gets reduced from $0$ to $-1$ (in $HF$).
Since this is a redox reaction and not a simple hydrolysis,option $D$ is the correct answer.

(B) $(1) \ HAuCl_4 \Rightarrow (+1) + x + 4(-1) = 0, x = +3$
$(2) \ Cu_2\underline{O} \Rightarrow 2(+1) + x = 0, x = -2$
$(3) \ \underline{Cl}O_3^{-} \Rightarrow x + 3(-2) = -1, x = +5$
$(4) \ K_2\underline{Cr_2O_7} \Rightarrow 2(+1) + 2x + 7(-2) = 0, x = +6$
The oxidation number of oxygen in $Cu_2O$ is $-2$,not $-1$. Therefore,option $B$ is incorrect.

(B) According to Boyle's Law,for a fixed amount of gas at a constant temperature,the product of pressure and volume is constant: $P_{1}V_{1} = P_{2}V_{2}$.
Given:
$P_{1} = 1 \; bar$
$V_{1} = 600 \; dm^{3}$
$V_{2} = 150 \; dm^{3}$
Substituting the values into the equation:
$(1 \; bar) \times (600 \; dm^{3}) = P_{2} \times (150 \; dm^{3})$
$P_{2} = \frac{600 \; bar \cdot dm^{3}}{150 \; dm^{3}}$
$P_{2} = 4 \; bar$.

(B) For any orbital,the number of angular nodes is given by the azimuthal quantum number $l$.
For an $s$-orbital,$l = 0$,so the number of angular nodes is $0$.
The number of radial nodes is given by the formula $n - l - 1$.
For the $3s$ orbital,the principal quantum number $n = 3$ and $l = 0$.
Therefore,the number of radial nodes $= 3 - 0 - 1 = 2$.
Thus,the $3s$ orbital has $0$ angular nodes and $2$ radial nodes.

(A) To determine the geometry of the molecules,we look at their hybridization and the number of lone pairs on the central atom:
$1$. $N_2O$ (Nitrous oxide): The structure is $N \equiv N-O$ or $N=N=O$. The central nitrogen atom is $sp$ hybridized and has no lone pairs,resulting in a linear geometry.
$2$. $NO_2$ (Nitrogen dioxide): The nitrogen atom has one unpaired electron and is $sp^2$ hybridized,leading to a bent (angular) geometry.
$3$. $HOCl$ (Hypochlorous acid): The oxygen atom is $sp^3$ hybridized with two lone pairs,resulting in a bent geometry.
$4$. $O_3$ (Ozone): The central oxygen atom is $sp^2$ hybridized with one lone pair,resulting in a bent geometry.
Therefore,$N_2O$ is the only linear molecule among the given options.

(B) $1$. $NH_3$: The central atom $N$ has $3$ bond pairs and $1$ lone pair,resulting in a trigonal pyramidal shape.
$2$. $PCl_5$: The central atom $P$ has $5$ bond pairs and $0$ lone pairs,resulting in a trigonal bipyramidal geometry.
$3$. $SF_6$: The central atom $S$ has $6$ bond pairs and $0$ lone pairs,resulting in an octahedral geometry.
$4$. $BeCl_2$: The central atom $Be$ has $2$ bond pairs and $0$ lone pairs,resulting in a linear geometry.
Therefore,the pair $PCl_5$ - Trigonal planar is incorrectly matched,as it should be trigonal bipyramidal.

(C) Acid rain reacts with the marble $(CaCO_{3})$ of the Taj Mahal,causing damage to it.
As a result,the monument is being slowly disfigured,and the marble is becoming discoloured and lustreless.
Therefore,the statement that it causes no damage to monuments is false.

(D) The reaction of methane with $Br_2$ in the presence of light $(hv)$ proceeds via a free radical mechanism,which is a type of substitution reaction.
Step $1$: Initiation: $Br-Br \xrightarrow{hv} 2Br^{\bullet}$
Step $2$: Propagation: $CH_3-H \xrightarrow{hv} CH_3^{\bullet} + H^{\bullet}$ and $CH_3^{\bullet} + Br^{\bullet} \rightarrow CH_3-Br$.
Therefore,option $D$ is the correct answer.

(B) For a reaction to be spontaneous,the change in Gibbs free energy,$\Delta_{r} G$,must be negative $(\Delta_{r} G < 0)$.
The relationship is given by the equation: $\Delta_{r} G = \Delta_{r} H - T \Delta_{r} S$.
Substituting the condition for spontaneity: $\Delta_{r} H - T \Delta_{r} S < 0$.
Rearranging the inequality: $\Delta_{r} H < T \Delta_{r} S$ or $\Delta_{r} S > \frac{\Delta_{r} H}{T}$.
Given $\Delta_{r} H = 30 \; kJ \; mol^{-1} = 30,000 \; J \; mol^{-1}$ and $T = 450 \; K$.
$\Delta_{r} S > \frac{30,000 \; J \; mol^{-1}}{450 \; K} = 66.67 \; J \; K^{-1} \; mol^{-1}$.
Therefore,the reaction will be spontaneous if $\Delta_{r} S > 66.67 \; J \; K^{-1} \; mol^{-1}$.

(D) Sucrose is a disaccharide composed of $\alpha-D$-glucose and $\beta-D$-fructose units linked by a glycosidic linkage.
Upon hydrolysis in the presence of an acid or the enzyme invertase,sucrose breaks down into its constituent monosaccharides:
$\text{Sucrose} + H_2O \xrightarrow{H^+} \alpha-D\text{-Glucose} + \beta-D\text{-Fructose}$.

(C) The electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$.
For $Cr^{2+}$ ion,two electrons are removed,resulting in the configuration $[Ar] 3d^4$.
This means there are $n = 4$ unpaired electrons.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$
Substituting $n = 4$:
$\mu = \sqrt{4(4+2)} \ B.M. = \sqrt{4 \times 6} \ B.M. = \sqrt{24} \ B.M. \approx 4.90 \ B.M.$

(D) The $-O-O-$ linkage is known as a peroxy linkage.
Among the given options,$H_{2}S_{2}O_{8}$ (peroxodisulphuric acid) contains a peroxy linkage.
The structure of $H_{2}S_{2}O_{8}$ is $HO-SO_{2}-O-O-SO_{2}-OH$,which clearly shows the presence of the $-O-O-$ bond.
Therefore,the correct option is $D$.

(A) According to the spectrochemical series,the order of increasing field strength of the given ligands is $SCN^{-} < F^{-} < C_{2}O_{4}^{2-} < CN^{-}$.
This series is based on the experimental data of the crystal field splitting energy $(\Delta_o)$ produced by different ligands.

(B) The reduction reaction for calcium is: $Ca^{2+} + 2e^{-} \rightarrow Ca_{(s)}$
The valence factor ($n$-factor) for $Ca^{2+}$ is $2$.
According to Faraday's $1^{st}$ law of electrolysis,the number of Faradays required is equal to the number of gram equivalents of the product formed.
Number of moles of $Ca = \frac{\text{mass}}{\text{molar mass}} = \frac{20 \ g}{40 \ g \ mol^{-1}} = 0.5 \ mol$.
Number of gram equivalents = $\text{moles} \times n\text{-factor} = 0.5 \times 2 = 1 \ F$.

(D) The reaction of acetone $(CH_3COCH_3)$ with methylmagnesium chloride $(CH_3MgCl)$,which is a Grignard reagent,involves the nucleophilic attack of the methyl group on the carbonyl carbon of acetone.
This forms an intermediate magnesium alkoxide: $(CH_3)_3C-OMgCl$.
Subsequent hydrolysis of this intermediate with water $(H_2O)$ yields $2$-methylpropan-$2$-ol,commonly known as tert-butyl alcohol $((CH_3)_3COH)$.

(D) Cationic detergents are quaternary ammonium salts of amines with acetates,chlorides,or bromides as anions.
$Cetyltrimethyl \ ammonium \ bromide$ is a well-known cationic detergent,represented as $[CH_3(CH_2)_{15}N(CH_3)_3]^+ Br^-$.
It is used in hair conditioners.
Therefore,the correct option is $D$.

(A) In $CrO_{4}^{2-}$,the oxidation state of $Cr$ is $x + 4(-2) = -2$,so $x = +6$.
In $Cr_{2}O_{7}^{2-}$,the oxidation state of $Cr$ is $2x + 7(-2) = -2$,so $2x = +12$,$x = +6$.
Since the oxidation states of chromium in both ions are $+6$,the statement in option $A$ is incorrect.

(C) Urea $(NH_2CONH_2)$ reacts with water to form ammonium carbonate $(A)$: $NH_2CONH_2 + 2H_2O \rightarrow (NH_4)_2CO_3$ $(A)$.
Ammonium carbonate decomposes to form ammonia $(B)$: $(NH_4)_2CO_3 \rightarrow 2NH_3 + H_2O + CO_2$ $(B = NH_3)$.
When ammonia is passed through $Cu^{2+}_{(aq)}$ solution,it forms a deep blue complex,tetraamminocopper$(II)$ ion: $Cu^{2+}_{(aq)} + 4NH_3(aq) \rightarrow [Cu(NH_3)_4]^{2+}(aq)$ $(C)$.

(D) The formula for freezing point depression is $\Delta T_{f} = K_{f} \times m$.
Given: $K_{f} = 5.12 \ K \ kg \ mol^{-1}$ and $m = 0.078 \ m$.
Substituting the values: $\Delta T_{f} = 5.12 \times 0.078$.
$\Delta T_{f} = 0.39936 \ K$.
Rounding off to two decimal places,we get $\Delta T_{f} = 0.40 \ K$.

(D) The reaction sequence is the industrial preparation of benzaldehyde from toluene.
$1$. Toluene reacts with chlorine in the presence of light $(Cl_2/h\nu)$ to undergo side-chain chlorination.
$2$. The reaction proceeds through the formation of benzyl chloride,then benzal chloride $(C_6H_5CHCl_2)$,and finally benzotrichloride.
$3$. To obtain benzaldehyde,the reaction is controlled to produce benzal chloride $(C_6H_5CHCl_2)$ as the intermediate $X$.
$4$. Benzal chloride $(C_6H_5CHCl_2)$ on hydrolysis with water at $373 \ K$ yields benzaldehyde $(C_6H_5CHO)$.
Therefore,compound $X$ is benzal chloride $(C_6H_5CHCl_2)$.

(A) is the correct statement. Pig iron contains impurities like $C, S, Si, P$ etc. and is relatively brittle,but it can be cast into a variety of shapes.
$B$ is incorrect because wrought iron is the purest form of iron with very low carbon content $(< 0.5 \%)$.
$C$ is incorrect because the blistered appearance of copper is due to the evolution of $SO_2$ gas,not $CO_2$.
$D$ is incorrect because Nickel is refined by the Mond process,whereas the Van Arkel method is used for Zirconium $(Zr)$ and Titanium $(Ti)$.

(B) Natural rubber is a natural polymer known as $cis-1,4-polyisoprene$.
It is obtained from the latex of rubber trees.
The other options,such as poly(Butadiene-acrylonitrile) (Buna-$N$),poly(Butadiene-styrene) (Buna-$S$),and polybutadiene,are synthetic polymers.

(A) The reaction between two different carbonyl compounds (benzaldehyde and acetophenone),where at least one of them has an $\alpha$-hydrogen atom (acetophenone has three $\alpha$-hydrogens),in the presence of a dilute base like $NaOH$ is called $Cross$ $Aldol$ $condensation$.
Benzaldehyde $(C_6H_5CHO)$ does not have an $\alpha$-hydrogen,while acetophenone $(C_6H_5COCH_3)$ has $\alpha$-hydrogens.
This specific reaction leads to the formation of a $\beta$-hydroxy ketone,which upon dehydration gives an $\alpha,\beta$-unsaturated ketone.

(B) positive deviation from Raoult's law occurs when the solute-solvent intermolecular interactions are weaker than the solute-solute and solvent-solvent interactions.
In the mixture of $Ethanol + Acetone$,the hydrogen bonding between ethanol molecules is disrupted by the addition of acetone,leading to weaker intermolecular forces in the solution compared to pure ethanol.
Therefore,this mixture exhibits a positive deviation from Raoult's law.

(D) For a first order reaction,the rate equation is given by: $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$
Given: $k = 4.606 \times 10^{-3} \ s^{-1}$,$[A]_0 = 2.0 \ g$,$[A]_t = 0.2 \ g$
Substituting the values: $4.606 \times 10^{-3} = \frac{2.303}{t} \log_{10} \frac{2.0}{0.2}$
$4.606 \times 10^{-3} = \frac{2.303}{t} \log_{10} (10)$
Since $\log_{10} (10) = 1$,we have: $4.606 \times 10^{-3} = \frac{2.303}{t}$
$t = \frac{2.303}{4.606 \times 10^{-3}} = \frac{1}{2} \times 10^3 = 500 \ s$

(B) The carbylamine test is a characteristic reaction for primary amines (both aliphatic and aromatic).
When a primary amine is heated with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$,it forms an isocyanide (carbylamine),which has a foul,offensive smell.
Secondary and tertiary amines do not contain the necessary $-NH_2$ group to form an isocyanide and therefore do not give this test.
Among the given options,aniline $(C_6H_5NH_2)$ is a primary amine and will give the carbylamine test.

(A) Anisole $(C_6H_5OCH_3)$ on heating with $HI$ undergoes cleavage of the $C-O$ bond between the methyl group and the oxygen atom to form $Phenol$ $(C_6H_5OH)$ and $Methyl \ iodide$ $(CH_3I)$.
This reaction proceeds via an $S_N2$ mechanism where the iodide ion attacks the less sterically hindered methyl carbon.

(B) The reaction of $2-$bromopentane with a base (like $NaOEt$) to form pent$-2-$ene is a $\beta-$elimination reaction.
In this process,a hydrogen atom is removed from the $\beta-$carbon and a halogen atom is removed from the $\alpha-$carbon,which is known as dehydrohalogenation.
Since the major product formed is the more substituted alkene (pent$-2-$ene),the reaction follows Zaitsev rule.
Dehydration refers to the removal of water,which is not the case here.
Therefore,statements $a$,$b$,and $c$ are correct.

(A) According to collision theory,the collision frequency $(Z)$ is directly proportional to the number of reactant molecules per unit volume.
Therefore,increasing the concentration of reactants increases the number of collisions per unit time,which is known as the collision frequency.
Activation energy,heat of reaction,and threshold energy are characteristic properties of a reaction and do not change with concentration.

(A) An amino acid is classified as basic if it contains more number of $-NH_2$ groups than $-COOH$ groups.
Lysine has the structure $H_2N-(CH_2)_4-CH(NH_2)-COOH$.
It contains two $-NH_2$ groups and one $-COOH$ group.
Therefore,it is a basic amino acid.

(D) The Zeta potential is the potential difference between the fixed layer and the diffused layer of ions surrounding a colloidal particle.
Greater the magnitude of the Zeta potential,the more stable the colloidal system will be,as it indicates stronger electrostatic repulsion between particles.

(B) For a body-centered cubic $(bcc)$ structure, the relationship between the atomic radius $(r)$ and the edge length $(a)$ is given by the formula: $\sqrt{3} a = 4 r$.
Given that the edge length $a = 288 \ pm$, we can calculate the radius $r$ as follows:
$r = \frac{\sqrt{3}}{4} \times a$
$r = \frac{\sqrt{3}}{4} \times 288 \ pm$.

(C) During the electrolysis of dilute $H_2SO_4$,the following reactions occur:
At the cathode: $2H^+_{(aq)} + 2e^- \rightarrow H_{2(g)}$
At the anode: $2H_2O_{(l)} \rightarrow O_{2(g)} + 4H^+_{(aq)} + 4e^-$
Since the oxidation potential of water is higher than that of the sulphate ion $(SO_4^{2-})$,water is oxidized at the anode to release oxygen gas.

(B) According to Raoult's law for non-volatile solutes,the relative lowering of vapour pressure is given by:
$\frac{P_{A}^{0} - P_{s}}{P_{A}^{0}} = \frac{n_{B}}{n_{A} + n_{B}} \approx \frac{n_{B}}{n_{A}}$ (for dilute solutions).
Given:
$w_{B} = 8 \; g$,$W_{A} = 114 \; g$,$M_{A} = 114 \; g \; mol^{-1}$.
Since the vapour pressure is reduced to $80 \%$,$P_{s} = 0.8 P_{A}^{0}$.
Therefore,$\frac{P_{A}^{0} - 0.8 P_{A}^{0}}{P_{A}^{0}} = \frac{0.2 P_{A}^{0}}{P_{A}^{0}} = 0.2$.
Using the formula $\frac{\Delta P}{P_{A}^{0}} = \frac{w_{B} \times M_{A}}{m_{B} \times W_{A}}$:
$0.2 = \frac{8 \times 114}{m_{B} \times 114}$.
$0.2 = \frac{8}{m_{B}}$.
$m_{B} = \frac{8}{0.2} = 40 \; g \; mol^{-1}$.

(B) The given reaction is the Rosenmund reduction.
In this reaction,an acid chloride $(RCOCl)$ is hydrogenated to an aldehyde $(RCHO)$ using $H_2$ gas in the presence of a palladium catalyst supported on barium sulfate $(Pd/BaSO_4)$.
The reaction is:
$C_6H_5COCl + H_2 \xrightarrow{Pd/BaSO_4} C_6H_5CHO + HCl$
Here,compound $(A)$ is benzoyl chloride $(C_6H_5COCl)$.

(C) Lanthanoids exhibit a primary oxidation state of $+3$. However, they can also show $+2$ or $+4$ oxidation states in certain solutions or solid compounds.
Therefore, the statement that they reveal only $+3$ oxidation state is incorrect.

(B) For a zero order reaction,the half-life $(t_{1/2})$ is given by the formula:
$t_{1/2} = \frac{[A]_0}{2k}$
Where $[A]_0$ is the initial concentration and $k$ is the rate constant.
Rearranging for $k$:
$k = \frac{[A]_0}{2 \times t_{1/2}}$
Given $[A]_0 = 0.02 \; M$ and $t_{1/2} = 100 \; s$:
$k = \frac{0.02}{2 \times 100} = \frac{0.02}{200} = 1.0 \times 10^{-4} \; mol \; L^{-1} s^{-1}$

(C) According to Valence Bond Theory:
$1. 4, sp^3$ hybridisation results in a $tetrahedral$ geometry $(a-iii)$.
$2. 4, dsp^2$ hybridisation results in a $square planar$ geometry $(b-iv)$.
$3. 5, sp^3d$ hybridisation results in a $trigonal bipyramidal$ geometry $(c-i)$.
$4. 6, d^2sp^3$ hybridisation results in an $octahedral$ geometry $(d-ii)$.
Therefore,the correct matching is $a-iii, b-iv, c-i, d-ii$.

(C) Copper is a transition metal $(a-ii)$.
Fluorine is a non-metal $(b-i)$.
Silicon is a metalloid $(c-iv)$.
Cerium is a lanthanoid $(d-iii)$.
Therefore,the correct match is $a-ii, b-i, c-iv, d-iii$.

(C) In the collision theory of chemical kinetics,the number of collisions per unit volume per unit time between reactant molecules $A$ and $B$ is defined as the collision frequency,denoted by $Z_{AB}$.

(B) Boron is purified by Zone refining $(iv)$.
$(b)$ Tin $(Sn)$ is purified by Liquation $(iii)$.
$(c)$ Zirconium $(Zr)$ is purified by Van Arkel method $(i)$.
$(d)$ Nickel $(Ni)$ is purified by Mond's process $(ii)$.
Therefore,the correct match is $(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)$.

(B) The acidity of substituted phenols depends on the electronic effects of the substituent groups attached to the benzene ring.
Electron-withdrawing groups $(EWG)$ increase acidity by stabilizing the phenoxide ion through $-I$ and $-R$ effects.
Electron-donating groups $(EDG)$ decrease acidity by destabilizing the phenoxide ion through $+I$ and $+R$ effects.
In the given options:
$(A)$ $-CH_3$ is an $EDG$ ($+I$ and hyperconjugation).
$(B)$ $-NO_2$ is a strong $EWG$ ($-I$ and $-R$ effects).
$(C)$ $-OCH_3$ is an $EDG$ ($+R$ effect dominates over $-I$).
$(D)$ $-C_2H_5$ is an $EDG$ ($+I$ effect).
Since $-NO_2$ is the only strong electron-withdrawing group among the choices,$p$-nitrophenol is the strongest acid.

(C) Osteomalacia is a disease caused by the deficiency of Vitamin $D$.
In this condition,the bone density decreases,and the bones become soft and weak.

(B) $AgBr$ is a unique compound that exhibits both Frenkel and Schottky defects.
In $AgBr$,the $Ag^+$ ion is small enough to occupy interstitial sites (Frenkel defect),while the crystal lattice also allows for the formation of vacancy pairs (Schottky defect) due to the ionic nature of the compound.

(B) Chloramphenicol is a well-known antibiotic.
It is classified as a bacteriostatic antibiotic,meaning it inhibits the growth of bacteria rather than killing them directly.
It is a broad-spectrum antibiotic,which means it is effective against a wide range of microorganisms,including both gram-positive and gram-negative bacteria.
Therefore,the statement that it inhibits the growth of only gram-positive bacteria is incorrect.

(D) The $S_{N}1$ reaction mechanism proceeds via the formation of a carbocation intermediate. The stability of the carbocation determines the reactivity towards $S_{N}1$.
$(A)$ $C_6H_5CH_2Cl$ forms a resonance-stabilized benzyl carbocation $(C_6H_5CH_2^+)$.
$(B)$ $CH_2=CHCH_2Cl$ forms a resonance-stabilized allyl carbocation $(CH_2=CHCH_2^+)$.
$(C)$ $(CH_3)_3CCl$ forms a stable tertiary carbocation $((CH_3)_3C^+)$.
$(D)$ $C_6H_5CH_2CH_2Cl$ is a primary alkyl halide. The formation of a primary carbocation $(C_6H_5CH_2CH_2^+)$ is highly unfavorable due to its low stability. Therefore,it does not undergo $S_{N}1$ reaction.

(B) The reaction of an amide with bromine $(Br_2)$ in the presence of an aqueous or ethanolic solution of a strong base like sodium hydroxide $(NaOH)$ is known as the Hofmann bromamide degradation reaction.
In this reaction,the amide is converted into a primary amine containing one carbon atom less than the original amide.
The reaction for propanamide $(CH_3CH_2CONH_2)$ is:
$CH_3CH_2CONH_2 + Br_2 + 4NaOH \rightarrow CH_3CH_2NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
As shown in the reaction,propanamide ($3$ carbons) yields ethylamine ($2$ carbons).

(D) The charge on colloidal particles is determined by the nature of the dispersed phase and the dispersion medium. Based on standard chemical classifications:
$1$. Hydrated $Al_{2}O_{3}$ is a positively charged sol.
$2$. $TiO_{2}$ is a positively charged sol.
$3$. Haemoglobin is a positively charged sol.
$4$. Starch is a negatively charged sol.
Therefore,the correct option is $D$.

(B) Bakelite is a thermosetting polymer formed by the condensation reaction of phenol and formaldehyde. It possesses a highly cross-linked structure,which makes it hard and infusible.

(B) Concentrated sulphuric acid $(H_2SO_4)$ acts as a strong dehydrating agent.
It removes water molecules from carbohydrates,such as sucrose $(C_{12}H_{22}O_{11})$,leaving behind carbon as a black residue.
The reaction is: $C_{12}H_{22}O_{11} \xrightarrow{H_2SO_4} 12C + 11H_2O$.
Therefore,this process is known as dehydration.

(A) The reactivity of aromatic compounds towards electrophilic aromatic substitution depends on the electron density of the benzene ring.
Groups that donate electrons to the ring (via resonance or induction) increase electron density and thus increase reactivity.
Groups that withdraw electrons from the ring decrease electron density and thus decrease reactivity.
$1$. $-OH$ group: Shows a strong $+R$ (resonance) effect,which significantly increases the electron density of the benzene ring,making it highly reactive.
$2$. Benzene: Has no substituent,serving as the reference.
$3$. $-Cl$ group: Shows a $-I$ (inductive) effect,which withdraws electron density,making it less reactive than benzene.
$4$. $-NO_2$ group: Shows a strong $-R$ (resonance) effect,which strongly withdraws electron density,making it the least reactive.
Therefore,the order of reactivity is: $\text{Phenol} > \text{Benzene} > \text{Chlorobenzene} > \text{Nitrobenzene}$.
Thus,phenol is the most reactive.

(C) The most common type of fuel cell is the hydrogen-oxygen fuel cell.
In this cell,hydrogen gas $(H_{2(g)})$ and oxygen gas $(O_{2(g)})$ are the reactants $(R)$.
The overall cell reaction is:
$2H_{2(g)} + O_{2(g)} \longrightarrow 2H_{2}O_{(l)}$
Thus,the product $(P)$ is water $(H_{2}O_{(l)})$.

(A) The molecular structures of the given $Xe$ compounds are determined by $VSEPR$ theory:
$1$. $XeF_2$: $Xe$ has $2$ bond pairs and $3$ lone pairs,resulting in a linear geometry.
$2$. $XeF_4$: $Xe$ has $4$ bond pairs and $2$ lone pairs,resulting in a square planar geometry.
$3$. $XeO_3$: $Xe$ has $3$ bond pairs and $1$ lone pair,resulting in a pyramidal geometry.
$4$. $XeOF_4$: $Xe$ has $5$ bond pairs and $1$ lone pair,resulting in a square pyramidal geometry.
Matching these with the columns:
$(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)$.

(A) The molecular formula of glucose is $C_6H_{12}O_6$.
It contains an aldehyde group $(-CHO)$ and six carbon atoms,making it an aldohexose.
It has five hydroxyl $(-OH)$ groups attached to different carbon atoms.
Due to the presence of the free aldehyde group,it acts as a reducing sugar.
Therefore,the statement that it is an aldopentose is incorrect.

(D) The standard reduction potentials $(E^{0})$ at $298 \; K$ are as follows:
$Au^{3+} + 3e^{-} \rightarrow Au_{(s)}, E^{0} = +1.40 \; V$
$Fe^{2+} + 2e^{-} \rightarrow Fe_{(s)}, E^{0} = -0.44 \; V$
$Mg^{2+} + 2e^{-} \rightarrow Mg_{(s)}, E^{0} = -2.36 \; V$
$K^{+} + 1e^{-} \rightarrow K_{(s)}, E^{0} = -2.93 \; V$
According to the electrochemical series,the species with the highest positive standard reduction potential occupies the top position. Therefore,$Au^{3+} + 3e^{-} \rightarrow Au_{(s)}$ is at the top.

(A) Phthalic acid $(C_6H_4(COOH)_2)$ on heating loses a water molecule to form phthalic anhydride. When heated strongly with ammonia,it forms phthalimide,which is an acid imide. Succinic acid also behaves similarly,but phthalic acid is the standard example for this reaction in the context of aromatic dicarboxylic acids.

(C) In the $3d$-series,$Manganese$ $(Mn)$ exhibits the maximum number of oxidation states,ranging from $+2$ to $+7$.
$Zinc$ $(Zn)$ has a completely filled $d$-orbital $(3d^{10})$ in its ground state and its common oxidation state $(+2)$,therefore it is not considered a transition element.
$Scandium$ $(Sc)$ exhibits only one oxidation state,which is $+3$.
$Copper$ $(Cu^{+})$ undergoes a disproportionation reaction in an aqueous solution: $2Cu^{+}_{(aq)} \longrightarrow Cu^{2+}_{(aq)} + Cu_{(s)}$.
Thus,the correct matching is: $a-iii, b-iv, c-i, d-ii$.

(D) Solutions that exhibit the same osmotic pressure at a given temperature are defined as isotonic solutions.