NEET 2014 Chemistry Question Paper with Answer and Solution

(B) : Aestivation is the arrangement of accessory floral organs (sepals or petals) in relation to one another in a floral bud. It may be of open,valvate,twisted,or imbricate type.
In imbricate aestivation,there is an irregular overlapping of petals or sepals by one another without any specific direction.
It has subtypes such as quincuncial,ascending imbricate,and descending imbricate (also known as vexillary).
Examples include $Cassia$ and $Gulmohur$.

(B) The correct answer is $B$.
In non-reducing sugars,a free aldehyde or ketonic group is absent.
Sucrose is a non-reducing sugar formed by the condensation of one molecule each of glucose and fructose with the release of a water molecule.
$A$ glycosidic bond is established between carbon atom $1$ of glucose and carbon atom $2$ of fructose,which masks the functional groups required for reduction.

(A) $Geitonogamy$ is the transfer of pollen grains from the anther to the stigma of another flower of the same plant.
Since the flowers are borne on the same plant,they are genetically identical.
Therefore,genetically,$Geitonogamy$ is a type of self-pollination ($Autogamy$ equivalent).
However,because it requires a pollinating agent to transfer the pollen,it is ecologically considered a type of cross-pollination.

(C) : The organs which have the same fundamental structure but are different in functions are called homologous organs.
These organs follow the same basic plan of organisation during their development.
However,in the adult condition,these organs are modified to perform different functions as an adaptation to different environments.
The forelimbs of cat,lizard,whale,and bat have the same basic structural plan.
In each case,the forelimb consists of humerus,radius-ulna,carpals,metacarpals,and digits.
The skeletal parts of the forelimbs of all these vertebrates are similar in structure and arrangement.
But the forelimbs of these animals have different shapes and functions.
In lizard,they are used for walking; in cat,for running; in whale,for swimming; and in bat,for flying.

(C) According to Avogadro's hypothesis,the volume of a gas $(V)$ is directly proportional to the number of moles $(n)$ at constant temperature and pressure.
Therefore,the ratio of the volumes of gases is equal to the ratio of their moles.
Let the mass of each gas be $m$.
The molar masses are: $M_{H_2} = 2 \ g/mol$,$M_{O_2} = 32 \ g/mol$,and $M_{CH_4} = 16 \ g/mol$.
The ratio of volumes is:
$V_{H_2} : V_{O_2} : V_{CH_4} = n_{H_2} : n_{O_2} : n_{CH_4} = \frac{m}{2} : \frac{m}{32} : \frac{m}{16}$.
Multiplying by $32/m$,we get:
$16 : 1 : 2$.

(A) The reaction is: $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$
At $S.T.P.$,$22.4 \ L$ of any gas corresponds to $1 \ mol$.
Initial moles of $H_2 = \frac{22.4 \ L}{22.4 \ L/mol} = 1 \ mol$.
Initial moles of $Cl_2 = \frac{11.2 \ L}{22.4 \ L/mol} = 0.5 \ mol$.
According to the stoichiometry,$1 \ mol$ of $H_2$ reacts with $1 \ mol$ of $Cl_2$.
Since $Cl_2$ is present in a smaller amount $(0.5 \ mol < 1 \ mol)$,it is the limiting reagent.
The amount of $HCl$ formed is determined by the limiting reagent $(Cl_2)$:
$1 \ mol$ of $Cl_2$ produces $2 \ mol$ of $HCl$.
Therefore,$0.5 \ mol$ of $Cl_2$ produces $2 \times 0.5 = 1.0 \ mol$ of $HCl$.

(A) The balanced chemical reaction is: $2Mg + O_2 \rightarrow 2MgO$.
Moles of $Mg = \frac{1.0 \ g}{24 \ g/mol} = 0.04167 \ mol$.
Moles of $O_2 = \frac{0.56 \ g}{32 \ g/mol} = 0.0175 \ mol$.
According to the stoichiometry,$2 \ mol$ of $Mg$ requires $1 \ mol$ of $O_2$.
Therefore,$0.04167 \ mol$ of $Mg$ would require $\frac{0.04167}{2} = 0.020835 \ mol$ of $O_2$.
Since we have only $0.0175 \ mol$ of $O_2$,$O_2$ is the limiting reagent.
$Mg$ consumed $= 2 \times \text{moles of } O_2 = 2 \times 0.0175 = 0.035 \ mol$.
$Mg$ left $= 0.04167 - 0.035 = 0.00667 \ mol$.
Mass of $Mg$ left $= 0.00667 \ mol \times 24 \ g/mol = 0.16 \ g$.

(A) The quantum numbers $n = 3$,$l = 1$,and $m_l = 0$ uniquely define a single orbital.
$n = 3$ indicates the third shell.
$l = 1$ indicates a $p$-subshell.
$m_l = 0$ specifies a particular orbital within the $p$-subshell (typically $3p_z$).
Since all three quantum numbers are fixed,they correspond to exactly $1$ orbital.

(D) The energy $E$ of a photon is related to its wavelength $\lambda$ by the equation: $E = \frac{hc}{\lambda}$.
Given values are: $h = 6.63 \times 10^{-34} \ J \ s$,$c = 3 \times 10^8 \ m \ s^{-1}$,and $\lambda = 45 \ nm = 45 \times 10^{-9} \ m$.
Substituting these values into the equation:
$E = \frac{(6.63 \times 10^{-34} \ J \ s) \times (3 \times 10^8 \ m \ s^{-1})}{45 \times 10^{-9} \ m}$
$E = \frac{19.89 \times 10^{-26}}{45 \times 10^{-9}} \ J$
$E = 0.442 \times 10^{-17} \ J = 4.42 \times 10^{-18} \ J$.

(B) Isoelectronic species contain the same number of electrons.
$Be^{2+}$ has an atomic number of $4$,so $Be^{2+}$ contains $4 - 2 = 2$ electrons.
Among the given options:
$H^{+}$ contains $1 - 1 = 0$ electrons.
$Li^{+}$ has an atomic number of $3$,so $Li^{+}$ contains $3 - 1 = 2$ electrons.
$Na^{+}$ has an atomic number of $11$,so $Na^{+}$ contains $11 - 1 = 10$ electrons.
$Mg^{2+}$ has an atomic number of $12$,so $Mg^{2+}$ contains $12 - 2 = 10$ electrons.
Since $Li^{+}$ has $2$ electrons,it is isoelectronic with $Be^{2+}$.

(D) Cations lose electrons and are smaller in size than the parent atom,whereas anions gain electrons and are larger in size than the parent atom.
For the hydrogen species,the order is $H^{-} > H > H^{+}$.
For isoelectronic species,the ionic radius decreases as the atomic number (nuclear charge) increases.
Comparing $O^{2-}$,$F^{-}$,and $Na^{+}$,all have $10$ electrons. Their atomic numbers are $8$,$9$,and $11$ respectively.
Therefore,the correct order of ionic radii is $O^{2-} > F^{-} > Na^{+}$.
Since none of the provided options $A$,$B$,or $C$ correctly represent this order,the correct answer is $D$.

(C) $CO_2$ and $CH_4$ have zero dipole moment as these are symmetrical in nature.
Between $NH_3$ and $NF_3$,$NH_3$ has a greater dipole moment than $NF_3$. In both $NH_3$ and $NF_3$,the nitrogen atom possesses one lone pair of electrons.
In $NH_3$,the net $N-H$ bond dipole is in the same direction as the dipole of the lone pair,leading to their addition. However,in $NF_3$,the net bond dipole of the three $N-F$ bonds is in the opposite direction to that of the lone pair dipole,leading to partial cancellation. Thus,$NH_3$ has the maximum dipole moment.

(B) Species with $sp^2$ hybridization exhibit a plane triangular shape.
In $NO_3^-$,the central $N$ atom is $sp^2$ hybridized and has no lone pair of electrons,resulting in a trigonal planar (plane triangular) geometry.
$N_3^-$ (azide ion),$NO_2^-$ (nitrite ion),and $CO_2$ (carbon dioxide) are $sp$ hybridized and possess a linear shape.

(B) The relation between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.
Given $\Delta U = 2.1 \ kcal$,$\Delta n_g = 2 - 0 = 2$ (since $X_2O_4$ is liquid),$R = 2 \times 10^{-3} \ kcal \ K^{-1} \ mol^{-1}$,and $T = 300 \ K$.
$\Delta H = 2.1 + (2 \times 2 \times 10^{-3} \times 300) = 2.1 + 1.2 = 3.3 \ kcal$.
Now,using the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$.
Given $\Delta S = 20 \ cal \ K^{-1} = 20 \times 10^{-3} \ kcal \ K^{-1}$.
$\Delta G = 3.3 - (300 \times 20 \times 10^{-3}) = 3.3 - 6.0 = -2.7 \ kcal$.

(C) $Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$.
When dissolved in water,it undergoes anionic hydrolysis to produce $OH^-$ ions,making the solution basic with a $pH > 7$.
$KCl$ and $NaCl$ are salts of strong acids and strong bases,resulting in a neutral solution $(pH \approx 7)$.
$CuSO_4$ is a salt of a weak base $(Cu(OH)_2)$ and a strong acid $(H_2SO_4)$,resulting in an acidic solution $(pH < 7)$.

(B) The relationship between standard Gibbs energy change and the equilibrium constant is given by $\Delta G^{o} = -RT \ln K_{sp}$ or $\Delta G^{o} = -2.303 \ RT \log K_{sp}$.
Given $\Delta G^{o} = +63.3 \ kJ \ mol^{-1} = 63.3 \times 10^{3} \ J \ mol^{-1}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,and $T = 25 + 273 = 298 \ K$.
Substituting these values into the equation:
$63.3 \times 10^{3} = -2.303 \times 8.314 \times 298 \times \log K_{sp}$
$63.3 \times 10^{3} = -5705.84 \times \log K_{sp}$
$\log K_{sp} = -\frac{63300}{5705.84} \approx -11.094$
$K_{sp} = 10^{-11.094} = 8.05 \times 10^{-12} \approx 8.0 \times 10^{-12}$.

(D) According to Le Chatelier's principle,if a change is applied to a system at equilibrium,the system will shift in a direction that counteracts the change.
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} + \text{Heat}$
$(a)$ Increasing the concentration of $NH_{3(g)}$ shifts the equilibrium in the backward direction to consume the added product.
$(b)$ Decreasing the pressure shifts the equilibrium toward the side with more moles of gas (the reactant side,$4 \text{ moles}$ vs $2 \text{ moles}$),which is the backward direction.
$(c)$ Decreasing the concentration of $N_{2(g)}$ and $H_{2(g)}$ shifts the equilibrium in the backward direction to produce more reactants.
$(d)$ Increasing pressure shifts the equilibrium toward the side with fewer moles of gas (the product side,$2 \text{ moles}$ vs $4 \text{ moles}$),which is the forward direction. Since the reaction is exothermic,decreasing the temperature also favors the forward direction.

(A) The relationship between equilibrium constants at two different temperatures is given by the van't Hoff equation: $\log \frac{K'_p}{K_p} = -\frac{\Delta H}{2.303 R} \left[ \frac{1}{T_2} - \frac{1}{T_1} \right]$.
For an exothermic reaction,the enthalpy change $\Delta H$ is negative $(\Delta H < 0)$.
Given that $T_2 > T_1$,the term $\left( \frac{1}{T_2} - \frac{1}{T_1} \right)$ is negative.
Substituting these into the equation: $\log \frac{K'_p}{K_p} = -\frac{(-ve)}{2.303 R} \times (-ve) = -ve$.
Since $\log \frac{K'_p}{K_p} < 0$,it implies $\frac{K'_p}{K_p} < 1$,which means $K'_p < K_p$ or $K_p > K'_p$.

(C) Two compounds can exist together if they do not react with each other via a redox reaction.
$FeCl_3$ and $SnCl_2$ react as $2FeCl_3 + SnCl_2 \rightarrow 2FeCl_2 + SnCl_4$.
$HgCl_2$ and $SnCl_2$ react as $2HgCl_2 + SnCl_2 \rightarrow Hg_2Cl_2 + SnCl_4$.
$FeCl_3$ and $KI$ react as $2FeCl_3 + 2KI \rightarrow 2FeCl_2 + 2KCl + I_2$.
$FeCl_2$ and $SnCl_2$ do not react with each other because both are in their lower oxidation states and cannot act as oxidizing agents for each other.

(C) In reaction $I$: The oxidation state of oxygen in $H_2O_2$ is $-1$. It changes to $0$ in $O_2$. Since the oxidation state increases,$H_2O_2$ acts as a reducing agent.
In reaction $II$: The oxidation state of oxygen in $H_2O_2$ is $-1$. It changes to $0$ in $O_2$. Since the oxidation state increases,$H_2O_2$ acts as a reducing agent.
Therefore,$H_2O_2$ acts as a reducing agent in both reactions.

(C) When $H_2O_2$ is added to an acidified solution of a dichromate $Cr_2O_7^{2-}$,a deep blue coloured complex,chromic peroxide $CrO_5$ (or $CrO(O_2)_2$) is formed.
$Cr_2O_7^{2-} + 2H^{+} + 4H_2O_2 \rightarrow 2CrO(O_2)_2 + 5H_2O$
This deep blue coloured complex has a butterfly structure with two peroxide linkages $(-O-O-)$ and one double-bonded oxygen.
Let the oxidation state of $Cr$ be $x$.
In $CrO_5$,there is one oxygen atom with oxidation state $-2$ and four oxygen atoms in two peroxide linkages with oxidation state $-1$ each.
Thus,$x + 1(-2) + 4(-1) = 0$.
$x - 2 - 4 = 0$.
$x = +6$.

(A) The reaction between ammonia and sulfuric acid is: $2NH_3 + H_2SO_4 \longrightarrow (NH_4)_2SO_4$.
Number of millimoles of $H_2SO_4$ used = $M \times V(mL) = 1 \ M \times 10 \ mL = 10 \ mmol$.
Since $1 \ mol$ of $H_2SO_4$ reacts with $2 \ mol$ of $NH_3$,the number of millimoles of $NH_3$ produced = $2 \times 10 \ mmol = 20 \ mmol = 20 \times 10^{-3} \ mol$.
Mass of nitrogen = $20 \times 10^{-3} \ mol \times 14 \ g/mol = 0.28 \ g$.
Percentage of nitrogen = $\frac{\text{Mass of Nitrogen}}{\text{Mass of Sample}} \times 100 = \frac{0.28}{0.75} \times 100 = 37.33 \ \%$.

(C) The $-CH_3$ group is an $o, p$-directing group. In $m$-xylene ($1,3$-dimethylbenzene),the positions available for electrophilic substitution are $2, 4, 5,$ and $6$.
$1$. The position between the two $-CH_3$ groups (position $2$) is sterically hindered,so no significant substitution occurs there.
$2$. The $4$ and $6$ positions are equivalent and are ortho to one $-CH_3$ group and para to the other,making them the most favorable sites for substitution.
$3$. The $5$ position is meta to both $-CH_3$ groups and is less activated.
Therefore,the major products formed are $4$-bromo-$1,3$-dimethylbenzene and $6$-bromo-$1,3$-dimethylbenzene (which are equivalent due to symmetry) or substitution at the $4$ and $6$ positions relative to the methyl groups.

(B) In $CO_2$,the carbon atom is $sp$ hybridized.
Among the given options,Ethyne $(HC \equiv CH)$ also has $sp$ hybridized carbon atoms.
Ethane $(CH_3-CH_3)$ has $sp^3$ hybridization,Ethene $(CH_2=CH_2)$ has $sp^2$ hybridization,and Ethanol $(CH_3-CH_2-OH)$ has $sp^3$ hybridized carbon atoms.

(D) Photochemical smog is formed by the reaction of nitrogen oxides $(NO_x)$,volatile organic compounds $(VOCs)$,and sunlight.
Common components of photochemical smog include ozone $(O_3)$,acrolein $(CH_2=CH-CHO)$,formaldehyde,and peroxyacetyl nitrate $(PAN)$.
Chlorofluorocarbons $(CFCs)$ are primarily responsible for the depletion of the ozone layer in the stratosphere and are not components of photochemical smog.

(B) The reaction of aqueous $KMnO_4$ with $H_2O_2$ in acidic medium is $3H_2SO_4 + 2KMnO_4 + 5H_2O_2 \rightarrow 5O_2 + 2MnSO_4 + 8H_2O + K_2SO_4$.
In the above reaction,$KMnO_4$ oxidises $H_2O_2$ to $O_2$ and itself $[MnO_4^-]$ gets reduced to $Mn^{2+}$ ion as $MnSO_4$.
Hence,aqueous solution of $KMnO_4$ with $H_2O_2$ yields $Mn^{2+}$ and $O_2$ in acidic conditions.

(D) Tubectomy is a surgical sterilization procedure performed in females.
In this method,a small part of the fallopian tubes is either removed or tied up (ligated) to block the passage of the ovum.
This prevents the fertilization of the egg by the sperm,thereby acting as a permanent method of contraception.
Option $A$ refers to Oophorectomy.
Option $B$ refers to Vasectomy (performed in males).
Option $C$ refers to Hysterectomy.

(A) Anoxygenic photosynthesis is a process in which light energy is used to synthesize organic compounds from carbon dioxide,but water is not used as an electron donor,and therefore,oxygen is not produced as a byproduct.
This process is characteristic of purple and green photosynthetic bacteria,such as $Rhodospirillum$.
In contrast,$Spirogyra$,$Chlamydomonas$,and $Ulva$ are algae that perform oxygenic photosynthesis,similar to higher plants,where water is used as an electron donor and oxygen is released.

(A) In the $ABO$ blood grouping system,individuals with blood group $AB$ possess both $A$ and $B$ antigens on the surface of their red blood cells ($RBC$s).
Because their immune system recognizes both $A$ and $B$ antigens as 'self',they do not produce anti-$A$ or anti-$B$ antibodies in their blood plasma.
Consequently,they can receive blood from any $ABO$ blood group ($A, B, AB,$ or $O$) without triggering an immune reaction,making them universal recipients.

(A) The stimulation of a muscle fiber by a motor neuron occurs at the neuromuscular junction (also known as the motor end plate).
This is a chemical synapse between a motor neuron and a muscle fiber.
When an action potential reaches the axon terminal of the motor neuron,it triggers the release of the neurotransmitter acetylcholine into the synaptic cleft.
This neurotransmitter binds to receptors on the sarcolemma of the muscle fiber,initiating an action potential that leads to muscle contraction.

(A) Aldosterone is a steroid hormone secreted by the adrenal cortex.
It acts primarily on the distal convoluted tubules $(DCT)$ and collecting ducts of the nephron.
Its main function is to stimulate the reabsorption of sodium ions $(Na^+)$ and water from the renal filtrate back into the blood,while promoting the excretion of potassium ions $(K^+)$.
Therefore,an increase in aldosterone levels leads to an increase in sodium reabsorption.

(B) The correct match is $B$.
$1$. Melatonin is secreted by the pineal gland and plays a crucial role in regulating the $24$-hour (diurnal) rhythm of our body,such as the sleep-wake cycle and body temperature.
$2$. Oxytocin is secreted by the posterior pituitary,but its function is to stimulate contraction of smooth muscles of the uterus and milk ejection,not the growth of mammary glands.
$3$. Progesterone is secreted by the corpus luteum,but its primary function is to support pregnancy,whereas the stimulation of growth and activities of female secondary sex organs is the function of estrogen.
$4$. Atrial natriuretic factor $(ANF)$ is secreted by the atrial wall of the heart,not the ventricular wall,and it decreases blood pressure by causing vasodilation.

(C) The 'fight-or-flight' response is a physiological reaction that occurs in response to a perceived harmful event,attack,or threat to survival.
This response is primarily mediated by the sympathetic nervous system.
When the body perceives a threat,the sympathetic nerves stimulate the adrenal medulla.
The adrenal medulla then secretes catecholamines,specifically epinephrine (adrenaline) and norepinephrine (noradrenaline),into the bloodstream.
These hormones prepare the body for immediate physical action by increasing heart rate,dilating airways,and mobilizing energy reserves.

(A) . $R.H. Whittaker$'s classification was based on five main criteria: complexity of cell structure,complexity of body organization,mode of nutrition,phylogenetic relationships,and mode of reproduction.
While cell structure complexity distinguishes prokaryotes from eukaryotes,the 'presence or absence of a well-defined nucleus' is a characteristic feature of cell structure rather than a separate primary criterion in his system.

(A) Viruses are non-cellular organisms that are characterized by having an inert crystalline structure outside the living cell.
Structurally,a virus consists of a genetic material (either $DNA$ or $RNA$) which is enclosed within a protective protein coat called a capsid.
Viruses never contain both $DNA$ and $RNA$ simultaneously.
They lack a cellular structure,meaning they do not have a nucleus,cytoplasm,or organelles.
Therefore,the correct description is that they have genetic material enclosed in a protein coat.

(D) Isogamy is a type of sexual reproduction where the fusing gametes are morphologically similar.
In $Spirogyra$,sexual reproduction occurs through conjugation,where the gametes are non-flagellated (non-motile) and isogamous.
$Sargassum$ shows oogamy.
$Ectocarpus$ shows isogamy or anisogamy with flagellated gametes.
$Ulothrix$ shows isogamy with flagellated gametes.
Therefore,the correct option is $D$.

(D) The phylum $Cnidaria$ (also known as $Coelenterata$) includes organisms that are primarily marine,but it also contains some species that inhabit fresh water. $A$ classic example of a freshwater cnidarian is $Hydra$.
In contrast,$Ctenophora$,$Cephalochordata$,and the majority of $Mammals$ (excluding aquatic mammals which are marine) do not fit the description of having significant representation in both freshwater and marine environments in the same way $Cnidaria$ does. Therefore,$Cnidaria$ is the correct taxon.

(B) $Planaria$ belongs to the phylum $Platyhelminthes$.
It exhibits a remarkable capacity for $Regeneration$.
When the body of $Planaria$ is cut into pieces,each fragment can develop into a complete individual.
This process is a form of asexual reproduction where the lost body parts are regrown through the proliferation of specialized cells called neoblasts.

(B) $Torpedo$ is a marine cartilaginous fish (Chondrichthyes) that possesses specialized electric organs on either side of its head.
These organs are modified muscle tissues capable of generating a powerful electric discharge,which the fish uses for defense and to stun its prey.
$Pristis$ is the sawfish,$Trygon$ is the stingray,and $Scoliodon$ is the dogfish (shark).

(C) To distinguish between an old dicot stem and an old dicot root,the position of the $Protoxylem$ is the most reliable anatomical feature.
In dicot stems,the $Protoxylem$ is situated towards the center (endarch condition).
In dicot roots,the $Protoxylem$ is situated towards the periphery (exarch condition).
Even in older specimens where secondary growth has occurred,the primary xylem remains at its original position,allowing for clear identification.

(B) Tracheary elements include tracheids and vessels.
Tracheids are elongated,tube-like cells with tapering ends.
Unlike vessels,which have perforated end walls allowing for the continuous flow of water,tracheids are imperforate,meaning they lack open pores or perforations at their ends.
Both tracheids and vessels are lignified and lack a nucleus at maturity,but the lack of perforations is the distinguishing feature of tracheids.

(D) Sphagnum,a moss,provides peat that has long been used as fuel. It is also used as packing material for trans-shipment of living material because of its high water-holding capacity. Therefore,it is responsible for the formation of peat.

(A) The balanced chemical equation for the reaction is: $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}$
Calculate the initial moles of reactants at $S.T.P.$:
$n_{H_2} = \frac{22.4 \ L}{22.4 \ L/mol} = 1 \ mol$
$n_{Cl_2} = \frac{11.2 \ L}{22.4 \ L/mol} = 0.5 \ mol$
According to the stoichiometry,$1 \ mol$ of $H_2$ reacts with $1 \ mol$ of $Cl_2$. Since we have $0.5 \ mol$ of $Cl_2$,it acts as the limiting reagent.
Therefore,$0.5 \ mol$ of $Cl_2$ will react with $0.5 \ mol$ of $H_2$ to produce $2 \times 0.5 = 1 \ mol$ of $HCl_{(g)}$.

(B) The $CFSE$ is calculated using the formula: $CFSE = [(-0.4 \times n_{t_{2g}}) + (0.6 \times n_{e_g})] \Delta_0$.
For a complex to have zero $CFSE$,the configuration must be such that the stabilization energy of electrons in $t_{2g}$ orbitals is exactly cancelled by the destabilization energy of electrons in $e_g$ orbitals.
Let's analyze the electronic configurations of the metal ions in the given complexes (all are high-spin complexes with $H_2O$ as a weak field ligand):
$A) [Mn(H_2O)_6]^{3+}: Mn^{3+} = 3d^4$. Configuration: $t_{2g}^3 e_g^1$. $CFSE = [(-0.4 \times 3) + (0.6 \times 1)] \Delta_0 = [-1.2 + 0.6] \Delta_0 = -0.6 \Delta_0$.
$B) [Fe(H_2O)_6]^{3+}: Fe^{3+} = 3d^5$. Configuration: $t_{2g}^3 e_g^2$. $CFSE = [(-0.4 \times 3) + (0.6 \times 2)] \Delta_0 = [-1.2 + 1.2] \Delta_0 = 0$.
$C) [Co(H_2O)_6]^{2+}: Co^{2+} = 3d^7$. Configuration: $t_{2g}^5 e_g^2$. $CFSE = [(-0.4 \times 5) + (0.6 \times 2)] \Delta_0 = [-2.0 + 1.2] \Delta_0 = -0.8 \Delta_0$.
$D) [Co(H_2O)_6]^{3+}: Co^{3+} = 3d^6$. Configuration: $t_{2g}^4 e_g^2$. $CFSE = [(-0.4 \times 4) + (0.6 \times 2)] \Delta_0 = [-1.6 + 1.2] \Delta_0 = -0.4 \Delta_0$.
Thus,$[Fe(H_2O)_6]^{3+}$ has zero $CFSE$.

(A) The acidity of hydrides of group $16$ elements increases down the group as the bond dissociation enthalpy decreases.
As we move from $S$ to $Se$ to $Te$,the size of the central atom increases,which leads to a weaker $H-E$ bond.
Therefore,the ease of release of $H^+$ ions increases,making the acidity order: $H_2S < H_2Se < H_2Te$.

(C) The dipole moment of $CO_2$ is $0 \, D$ because it is a linear molecule where the bond dipoles cancel each other.
The dipole moment of $CH_4$ is $0 \, D$ because it is a symmetrical tetrahedral molecule where the bond dipoles cancel each other.
In $NH_3$,the bond dipoles of the three $N-H$ bonds and the lone pair on $N$ are in the same direction,resulting in a net dipole moment of $1.46 \, D$.
In $NF_3$,the bond dipoles of the three $N-F$ bonds are in the opposite direction to the lone pair dipole,which significantly reduces the net dipole moment to $0.24 \, D$.
Therefore,$NH_3$ has the maximum dipole moment among the given molecules.

(A) $CO_2$ is transported from the respiratory tissues to the lungs by the blood in $3$ ways:
$(i)$ In dissolved state or as a physical solution: $A$ very small amount is physically dissolved in plasma ($7\%$ i.e.,$0.3\, ml$ of $CO_2$ per $100\, ml$ of blood).
$(ii)$ Bicarbonate ions: About $70\%$ (i.e.,$2.5\, ml$ per $100\, ml$ of blood) of $CO_2$ diffuses into plasma and then into $RBCs$,where it combines with $H_2O$ in the presence of the enzyme carbonic anhydrase to form carbonic acid,which then dissociates into hydrogen ions and bicarbonate ions.
$(iii)$ Carbaminohaemoglobin: $23\%$ (i.e.,$1\, ml$ of $CO_2$ per $100\, ml$ of blood) combines with haemoglobin to form an unstable compound called carbaminohaemoglobin.

(B) The spontaneity of a process is determined by the Gibbs free energy equation: $\Delta G = \Delta H - T \Delta S$.
For a process to be spontaneous,$\Delta G$ must be negative.
During the adsorption of a gas on a solid surface,the gas molecules become restricted,leading to a decrease in randomness,so $\Delta S < 0$ (negative).
Substituting this into the equation: $\Delta G = \Delta H - T(\text{negative}) = \Delta H + T \Delta S$.
For $\Delta G$ to be negative,$\Delta H$ must be sufficiently negative to overcome the positive $T \Delta S$ term.
Therefore,$\Delta H$ must be highly negative.

(A) The reaction of $CH_3CH_2CH=CH_2$ with $HBr$ in the presence of $H_2O_2$ (peroxide effect or Kharasch effect) follows anti-Markovnikov addition to give $CH_3CH_2CH_2CH_2Br$ $(Y)$.
Then,$CH_3CH_2CH_2CH_2Br$ reacts with sodium ethoxide $(C_2H_5ONa)$ via the Williamson ether synthesis mechanism to form $CH_3(CH_2)_3 - O - CH_2CH_3$ $(Z)$.

(D) In a body-centered cubic $(bcc)$ unit cell,the body-centered atom is located at the center of the cube.
The distance from the center of the cube to any corner is equal to half of the body diagonal.
The length of the body diagonal of a cube with side length $a$ is $\sqrt{3} a$.
Therefore,the distance between the body-centered atom and one corner atom is $\frac{1}{2} \times \sqrt{3} a = \frac{\sqrt{3}}{2} a$.

(C) The depression in freezing point $(\Delta T_f)$ is given by the formula $\Delta T_f = i \times K_f \times m$,where $i$ is the van't Hoff factor,$K_f$ is the cryoscopic constant,and $m$ is the molality.
Since $K_f$ and $m$ are constant for all solutions,$\Delta T_f \propto i$.
We calculate the van't Hoff factor $(i)$ for each solute:
$1. KCl \rightarrow K^+ + Cl^-$,so $i = 2$.
$2. C_6H_{12}O_6$ (glucose) is a non-electrolyte,so $i = 1$.
$3. Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$,so $i = 5$.
$4. K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$,so $i = 3$.
Since $Al_2(SO_4)_3$ has the largest van't Hoff factor $(i = 5)$,it will exhibit the largest freezing point depression.

(C) The oxidation half-reaction is:
$MnO_4^{2-} \longrightarrow MnO_4^- + e^-$
From the stoichiometry of the reaction,$1 \ mol$ of $MnO_4^{2-}$ requires $1 \ mol$ of electrons for oxidation to $MnO_4^-$.
Therefore,$0.1 \ mol$ of $MnO_4^{2-}$ requires $0.1 \ mol$ of electrons.
The quantity of electricity $Q$ is given by $Q = n \times F$,where $n$ is the number of moles of electrons and $F$ is Faraday's constant $(96500 \ C \ mol^{-1})$.
$Q = 0.1 \ mol \times 96500 \ C \ mol^{-1} = 9650 \ C$.

(D) According to Faraday's second law of electrolysis,the number of equivalents of substances displaced are equal.
Equivalent weight of $Ag = \frac{108}{1} = 108 \ g/eq$.
Equivalent volume of $O_2$ at $STP = \frac{22400 \ mL}{4} = 5600 \ mL/eq$.
Number of equivalents of $O_2 = \frac{5600 \ mL}{5600 \ mL/eq} = 1 \ eq$.
Since the number of equivalents is equal,the equivalents of $Ag$ displaced = $1 \ eq$.
Weight of $Ag = \text{Equivalents} \times \text{Equivalent weight} = 1 \times 108 = 108 \ g$.

(B) The $Tyndall \ effect$ is independent of the charge on the colloidal particles.
It is an optical property that depends on the size of the colloidal particles and the wavelength of light used.
The $Tyndall \ effect$,also known as $Tyndall \ scattering$,is the scattering of light by particles in a colloid or a fine suspension.
In contrast,$electro-osmosis$,$coagulation$,and $electrophoresis$ are electrical properties that directly depend on the charge present on the colloidal particles.

(D) The $Tyndall \ effect$ is an optical phenomenon that occurs due to the scattering of light by colloidal particles.
It depends on the size of the particles and the refractive index difference between the dispersed phase and the dispersion medium,but it is independent of the charge on the particles.
In contrast,$Coagulation$,$Electrophoresis$,and $Electro-osmosis$ are all properties that depend on the electrical charge of the colloidal particles.

(A) The acidic strength of hydrides increases as the size of the central atom increases,which weakens the $M-H$ bond.
Since the atomic size increases from $S$ to $Se$ to $Te$,the bond dissociation enthalpy decreases,leading to an increase in acidic strength.
The order is $H_2S < H_2Se < H_2Te$.
Acidic nature $\propto \frac{1}{\text{Bond dissociation enthalpy}}$.

(B) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 2.83 \ BM$,we have:
$2.83 = \sqrt{n(n+2)}$
$(2.83)^2 = n(n+2)$
$8.00 \approx n^2 + 2n$
$n^2 + 2n - 8 = 0$
$(n+4)(n-2) = 0$
Since $n$ cannot be negative,$n = 2$.
Now,let us check the number of unpaired electrons for each ion:
$1. Ti^{3+} (Z=22): [Ar] 3d^1 \rightarrow n = 1$
$2. Ni^{2+} (Z=28): [Ar] 3d^8 \rightarrow n = 2$
$3. Cr^{3+} (Z=24): [Ar] 3d^3 \rightarrow n = 3$
$4. Mn^{2+} (Z=25): [Ar] 3d^5 \rightarrow n = 5$
Since $Ni^{2+}$ has $2$ unpaired electrons,it corresponds to a magnetic moment of $2.83 \ BM$.

(A) Lanthanoid contraction is the regular decrease in atomic and ionic radii of lanthanides.
This is due to the imperfect shielding (or poor screening effect) of $f$-orbitals.
Due to their diffused shape,$f$-orbitals are unable to effectively counterbalance the effect of the increased nuclear charge.
Hence,the net result is a contraction in the size of lanthanoids.

(B) $H_2O$ is a weak field ligand,hence $\Delta_o < $ pairing energy.
$CFSE = (-0.4x + 0.6y) \Delta_o$
where $x$ and $y$ are the number of electrons occupying $t_{2g}$ and $e_g$ orbitals respectively.
For $[Fe(H_2O)_6]^{3+}$ complex ion:
$Fe^{3+}$ is a $3d^5$ system. In a weak field,the configuration is $t_{2g}^3 e_g^2$.
$CFSE = (-0.4 \times 3) + (0.6 \times 2) = -1.2 + 1.2 = 0 \Delta_o$.

(B) Cis-platin is a well-known anticancer agent.
The chemical formula of cis-platin is $cis-[PtCl_2(NH_3)_2]$.
In this complex,the $cis$ isomer is specifically effective in inhibiting the growth of cancer cells,making it a vital antitumour agent.

(D) Racemisation occurs in $S_N1$ reactions where a chiral carbocation intermediate is formed,allowing the nucleophile to attack from both sides.
Compound $(iv)$,$CH_3CH(Cl)C_2H_5$,is a secondary alkyl halide that can form a stable carbocation and possesses a chiral center,leading to a racemic mixture upon hydrolysis.
Compounds $(i)$,$(ii)$,and $(iii)$ are primary alkyl halides. Primary halides typically undergo $S_N2$ reactions,which result in Walden inversion rather than racemisation. Therefore,only $(iv)$ undergoes racemisation.

(B) The reaction of phenol $(C_6H_5OH)$ with sodium hydroxide $(NaOH)$ forms sodium phenoxide $(C_6H_5ONa)$.
Sodium phenoxide then undergoes a nucleophilic substitution reaction $(S_N2)$ with methyl iodide $(CH_3I)$ to produce anisole $(C_6H_5OCH_3)$.
This process is an example of Williamson's ether synthesis.

(C) The solubility of an organic acid in $NaHCO_3$ depends on its acidity. An acid will react with $NaHCO_3$ to form a salt and evolve $CO_2$ gas if it is stronger than carbonic acid $(H_2CO_3)$.
The general reaction is: $\text{Acid} + NaHCO_3 \longrightarrow \text{Salt} + H_2O + CO_2$.
$1$. $2,4,6-$Trinitrophenol (picric acid),benzoic acid,and benzenesulphonic acid are stronger acids than $H_2CO_3$ and thus dissolve in $NaHCO_3$ solution.
$2$. $o-$Nitrophenol is a weaker acid than $H_2CO_3$ due to intramolecular hydrogen bonding,which stabilizes the molecule and hinders the release of the proton. Therefore,it does not react with $NaHCO_3$ and is not soluble in it.

(D) The reactivity of carbonyl compounds towards nucleophilic addition reactions is primarily governed by the electrophilicity of the carbonyl carbon.
$1$. Electron-withdrawing groups ($-I, -M$ effects) increase the positive charge on the carbonyl carbon,thereby increasing its reactivity towards nucleophilic attack.
$2$. Electron-donating groups ($+I, +M$ effects) decrease the positive charge on the carbonyl carbon,thereby decreasing its reactivity.
$3$. Steric hindrance also plays a role; aldehydes are generally more reactive than ketones due to less steric hindrance.
Comparing the given compounds:
- $p$-Nitrobenzaldehyde has a strong electron-withdrawing $-NO_2$ group ($-I, -M$ effect),which significantly increases the electrophilicity of the carbonyl carbon.
- Benzaldehyde has no substituent.
- $p$-Methylbenzaldehyde has an electron-donating $-CH_3$ group ($+I$ effect).
- Acetophenone is a ketone and has an electron-donating $-CH_3$ group attached to the carbonyl carbon,making it the least reactive.
Thus,the order of reactivity is: $p$-Nitrobenzaldehyde > Benzaldehyde > $p$-Methylbenzaldehyde > Acetophenone.
Therefore,$p$-Nitrobenzaldehyde is the most reactive.

(A) The reaction shown is a coupling reaction between benzenediazonium chloride and aniline.
This reaction is an electrophilic aromatic substitution where the diazonium cation acts as the electrophile.
Since the $-NH_2$ group is a strong ortho/para-directing group,the electrophile attacks the para-position to the $-NH_2$ group,yielding $p$-aminoazobenzene,which is a yellow dye.

(B) The diazonium salt containing an aryl group directly linked to the nitrogen atom $(C_6H_5N_2^+X^-)$ is the most stable.
This stability is due to the resonance stabilization between the benzene ring and the diazonium group $(-N_2^+)$,which delocalizes the positive charge over the benzene ring.
Aliphatic diazonium salts are highly unstable and decompose readily to form carbocations and nitrogen gas.
The resonance structures of the benzene diazonium ion are shown below:
(The image provided shows the resonance structures where the positive charge is delocalized into the ortho and para positions of the benzene ring.)

(D) $D(+)$-glucose contains an aldehyde group at $C1$.
When it reacts with hydroxylamine $(NH_2OH)$,the carbonyl oxygen is replaced by $=NOH$ to form an oxime.
The configuration of the chiral centers ($C2$ to $C5$) remains the same as in $D$-glucose ($C2-OH$ right,$C3-OH$ left,$C4-OH$ right,$C5-OH$ right).
The structure is: $CH=NOH - C(H)(OH) - C(OH)(H) - C(H)(OH) - C(H)(OH) - CH_2OH$.

(C) Adrenaline is known as the emergency hormone.
It is released during stress conditions.
It stimulates glycogenolysis in the liver,which leads to the release of glucose into the bloodstream to provide energy for the 'fight or flight' response.

(D) Bakelite is a thermosetting polymer. These polymers are cross-linked or heavily branched molecules,which on heating undergo extensive cross-linking in moulds and become infusible. They cannot be reused.
Option $(A)$ is Neoprene (elastomer),$(B)$ is $PVC$ (thermoplastic),and $(C)$ is Nylon $6,6$ (fiber).

(C) Dacron,also known as terylene,is a condensation polymer formed by the polymerization of terephthalic acid $(HOOC-C_6H_4-COOH)$ and ethylene glycol $(HO-CH_2-CH_2-OH)$.
This reaction occurs by heating the mixture at $420-460 \ K$ in the presence of zinc acetate and antimony trioxide as a catalyst,resulting in the elimination of water molecules.

(C) Aspartame is the most successful and widely used artificial sweetener.
It is roughly $180$ times as sweet as cane sugar.
It is unstable at cooking temperatures,which limits its use to cold foods and soft drinks.