JEE Main 2014 Chemistry Question Paper with Answer and Solution

(B) Freons and chlorofluorocarbons reach the stratosphere without being destroyed in the troposphere due to their low reactivity.
In the stratosphere,the $Cl$ and $Br$ atoms are liberated from the parent compound by the action of ultraviolet light.
For example:
$CF_2Cl_2 + h\nu \to CF_2Cl^{\bullet} + Cl^{\bullet}$
$CFCl_3 + h\nu \to CFCl_2^{\bullet} + Cl^{\bullet}$
The $Cl^{\bullet}$ atoms destroy ozone molecules through catalytic cycles.
The problem is more serious at the poles because ice crystals in polar stratospheric clouds provide a surface for these reactions to occur more efficiently.

(D) Given,percentage of $H = 12.5\%$.
Therefore,percentage of $N = 100 - 12.5 = 87.5\%$.

Element	Percentage	Atomic ratio	Simple ratio
$H$	$12.5\%$	$\frac{12.5}{1} = 12.5$	$\frac{12.5}{6.25} = 2$
$N$	$87.5\%$	$\frac{87.5}{14} = 6.25$	$\frac{6.25}{6.25} = 1$

Empirical formula $= NH_2$.
Empirical formula mass $= 14 + (2 \times 1) = 16 \ g/mol$.
Molecular mass $= 2 \times \text{Vapour density} = 2 \times 16 = 32 \ g/mol$.
$n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{32}{16} = 2$.
Molecular formula $= n \times (\text{Empirical formula}) = 2 \times (NH_2) = N_2H_4$.

(C) The given equilibrium is $AgCl_{(s)} + 2NH_{3(aq)} \rightleftharpoons [Ag(NH_3)_2]^+_{(aq)} + Cl^-_{(aq)}$.
When $HNO_3$ is added,it reacts with $NH_3$ to form $NH_4^+$ ions: $NH_{3(aq)} + H^+_{(aq)} \to NH_{4(aq)}^+$.
This decreases the concentration of $NH_3$,which shifts the equilibrium to the left according to Le Chatelier's principle.
Consequently,the complex ion $[Ag(NH_3)_2]^+$ dissociates,releasing $Ag^+$ ions,which then react with $Cl^-$ ions to precipitate $AgCl_{(s)}$.
The overall reaction is: $[Ag(NH_3)_2]^+_{(aq)} + Cl^-_{(aq)} + 2H^+_{(aq)} \to AgCl_{(s)} + 2NH_{4(aq)}^+$.
Thus,adding $HNO_3$ causes the white precipitate of $AgCl$ to reappear.

(B) In the case of $(SiH_3)_3N$,the nitrogen atom is $sp^2$ hybridized because the lone pair of electrons on the nitrogen atom is donated into the empty $d$-orbital of the silicon atom,forming a $d\pi - p\pi$ back-bonding.
This results in a trigonal planar geometry for $(SiH_3)_3N$.
In contrast,$(CH_3)_3N$,$P(CH_3)_3$,and $P(SiH_3)_3$ all have a lone pair on the central atom and exhibit a pyramidal shape due to $sp^3$ hybridization.

(C) The reduction of an internal alkyne to a trans-alkene is achieved using dissolving metal reduction,typically with sodium or lithium in liquid ammonia $(NH_3(l))$.
$Ph-C \equiv C-Ph \xrightarrow{Li / NH_3(l)} \text{trans-}Ph-CH=CH-Ph$
Catalytic hydrogenation with $H_2 / \text{Lindlar catalyst}$ yields the cis-alkene,while complete catalytic hydrogenation yields the alkane. $LiAlH_4$ does not reduce alkynes to alkenes.

(C) The compound $A$ reacts with nucleophiles like $H_2 O$,$NH_3$,and $CH_3 COOH$ to form a carboxylic acid,an amide,and an acid anhydride respectively. This reactivity is characteristic of a ketene.
Given the molecular formula $C_5 H_8 O$,the structure $CH_3-CH_2-C(CH_3)=C=O$ (ethylmethylketene) fits the description.
$1$. Reaction with $H_2 O$: $CH_3-CH_2-C(CH_3)=C=O + H_2 O \rightarrow CH_3-CH_2-CH(CH_3)-COOH$
$2$. Reaction with $NH_3$: $CH_3-CH_2-C(CH_3)=C=O + NH_3 \rightarrow CH_3-CH_2-CH(CH_3)-CONH_2$
$3$. Reaction with $CH_3 COOH$: $CH_3-CH_2-C(CH_3)=C=O + CH_3 COOH \rightarrow CH_3-CH_2-CH(CH_3)-COOCOCH_3$

(B) The structure of allene is $CH_2=C=CH_2$.
The terminal carbon atoms are bonded to two hydrogen atoms and one carbon atom via a double bond,resulting in $sp^2$ hybridization.
The central carbon atom is bonded to two carbon atoms via two double bonds,resulting in $sp$ hybridization.
Therefore,the hybridization types present are $sp^2$ and $sp$.

(C) The electrolysis of an aqueous solution of dipotassium succinate is an example of the Kolbe electrolysis reaction.
At the anode,the succinate ion undergoes oxidation to form ethene and carbon dioxide:
$CH_2COO^{-} - CH_2COO^{-} \xrightarrow{-2e^-} CH_2=CH_2 + 2CO_2$
At the cathode,water is reduced to produce hydrogen gas:
$2H_2O + 2e^- \to H_2 + 2OH^-$
Therefore,the gas liberated at the anode is ethene $(CH_2=CH_2)$.

(D) The total energy $(E)$ of an electron in a hydrogen atom is the sum of its kinetic energy $(K.E.)$ and potential energy $(P.E.)$.
For an electron in an orbit of radius $r$,the electrostatic potential energy is given by $P.E. = -\frac{e^2}{r}$.
The kinetic energy is given by $K.E. = \frac{1}{2} \frac{e^2}{r}$.
Therefore,the total energy is $E = K.E. + P.E. = \frac{e^2}{2r} - \frac{e^2}{r} = -\frac{e^2}{2r}$.

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.
Given:
Mass $m = 6.63 \ g = 6.63 \times 10^{-3} \ kg$
Velocity $v = 100 \ ms^{-1}$
Planck's constant $h = 6.63 \times 10^{-34} \ J \cdot s$
Substituting the values:
$\lambda = \frac{6.63 \times 10^{-34} \ J \cdot s}{(6.63 \times 10^{-3} \ kg) \times (100 \ ms^{-1})}$
$\lambda = \frac{6.63 \times 10^{-34}}{6.63 \times 10^{-1}} \ m$
$\lambda = 10^{-33} \ m$

(D) When an inert gas is added to a system at equilibrium at constant volume,the total pressure of the system increases.
However,the partial pressure of each reactant and product remains the same because the concentration (moles/volume) of each species does not change.
According to Le Chatelier's principle,since the concentrations of the reacting species remain unchanged,there is no effect on the state of equilibrium.

(B) The balanced chemical equation is: $BaCl_2 + H_2SO_4 \to BaSO_4 + 2HCl$
First,calculate the mass of reactants:
Mass of $BaCl_2 = 20.8 \ g$ (assuming $100 \ mL$ solution contains $20.8 \ g$ solute).
Moles of $BaCl_2 = \frac{20.8 \ g}{208 \ g/mol} = 0.1 \ mol$
Mass of $H_2SO_4 = 9.8 \ g$ (assuming $50 \ mL$ solution contains $9.8 \ g$ solute).
Moles of $H_2SO_4 = \frac{9.8 \ g}{98 \ g/mol} = 0.1 \ mol$
Since the stoichiometry is $1:1$,both reactants are consumed completely.
Therefore,$0.1 \ mol$ of $BaSO_4$ is formed.
Mass of $BaSO_4 = 0.1 \ mol \times 233 \ g/mol = 23.3 \ g$

(B) The entropy change for the reaction is calculated using the formula:
$\Delta S^o = \sum S^o(\text{products}) - \sum S^o(\text{reactants})$
For the reaction $CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2H_2O_{(l)}$:
$\Delta S^o = [S^o(CO_2) + 2 \times S^o(H_2O)] - [S^o(CH_4) + 2 \times S^o(O_2)]$
Substituting the given values:
$\Delta S^o = [213.6 + 2 \times 69.9] - [186.2 + 2 \times 205.2]$
$\Delta S^o = [213.6 + 139.8] - [186.2 + 410.4]$
$\Delta S^o = 353.4 - 596.6$
$\Delta S^o = -242.8 \ J \ K^{-1} \ mol^{-1}$

(B) The dissociation of hydrazoic acid $(HN_3)$ is given by the equation:
$HN_3 \rightleftharpoons N_3^{-} + H^{+}$
By definition,a conjugate base is formed when an acid loses a proton $(H^{+})$.
Therefore,the conjugate base of hydrazoic acid $(HN_3)$ is the azide ion,$N_3^{-}$.

(C) To calculate the average enthalpy of a $C-H$ bond in methane,we use the Born-Haber cycle approach based on the reaction: $C_{(graphite)} + 2H_{2(g)} \to CH_{4(g)}$.
To find the bond dissociation energy of $C-H$,we need to convert the reactants into gaseous atoms:
$(i)$ The enthalpy of sublimation of carbon: $C_{(graphite)} \to C_{(g)}$
$(ii)$ The dissociation energy of the hydrogen molecule: $H_{2(g)} \to 2H_{(g)}$
By knowing these values,we can calculate the total energy required to atomize the reactants and then relate it to the formation enthalpy to find the average $C-H$ bond energy.

(C) The energy of a photon is given by the equation $E = h \nu$.
Given:
Planck's constant,$h = 6.63 \times 10^{-34} \ J \cdot s$
Frequency,$\nu = 2.47 \times 10^{15} \ Hz$
Substituting the values:
$E = (6.63 \times 10^{-34} \ J \cdot s) \times (2.47 \times 10^{15} \ s^{-1})$
$E = 16.3761 \times 10^{-19} \ J$
$E = 1.63761 \times 10^{-18} \ J \approx 1.640 \times 10^{-18} \ J$.

(B) When copper is exposed to moist air containing $CO_2$,it reacts to form a green coating on its surface.
This green layer consists of basic copper carbonate,which has the chemical formula $CuCO_3 \cdot Cu(OH)_2$.

(D) Elements in the same group of the Periodic table exhibit similar chemical properties because they possess the same number of valence electrons.
These valence electrons determine the chemical reactivity and bonding behavior of the atoms.
While the atomic number determines the position of an element,the specific configuration of valence electrons is the direct cause of the similarity in chemical properties within a group.

(A) The ionic radii of species depend on the number of shells and the effective nuclear charge.
For species in the same period,the ionic radius decreases as the atomic number increases (e.g.,$N^{3-} > O^{2-}$).
For species in the same group,the ionic radius increases as we move down the group due to the addition of a new shell (e.g.,$P^{3-} > N^{3-}$ and $S^{2-} > O^{2-}$).
Comparing the given species:
$1$. $O^{2-}$ and $N^{3-}$ are in the second period,so $N^{3-} > O^{2-}$.
$2$. $S^{2-}$ and $P^{3-}$ are in the third period,so $P^{3-} > S^{2-}$.
$3$. Since $S^{2-}$ and $P^{3-}$ have more shells than $O^{2-}$ and $N^{3-}$,they are larger.
Combining these,the order of increasing ionic radii is $O^{2-} < N^{3-} < S^{2-} < P^{3-}$.

(B) Global warming is primarily caused by the increase in greenhouse gases in the atmosphere. The most significant contributor is $CO_2$ (carbon dioxide),followed by methane $(CH_4)$. These gases trap heat in the Earth's atmosphere,leading to a rise in global temperatures.

(A) $H_2O_2$ acts as a reducing agent when it reacts with strong oxidising agents. In an acidic medium,$H_2O_2$ reduces $MnO_4^-$ to $Mn^{2+}$.
The balanced chemical equation is:
$2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \to K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$
In this reaction,the oxidation state of $Mn$ changes from $+7$ to $+2$,and $H_2O_2$ is oxidized to $O_2$.

(B) The Victor-Meyer's test is used to distinguish between primary $(1^o)$,secondary $(2^o)$,and tertiary $(3^o)$ alcohols.
$(i)$ Primary $(1^o)$ alcohols react to form nitrolic acid,which gives a blood red colour with $KOH$.
(ii) Secondary $(2^o)$ alcohols react to form pseudonitrol,which gives a blue colour with $KOH$.
(iii) Tertiary $(3^o)$ alcohols do not react with $HNO_2$ and remain colourless in the presence of $KOH$.
Thus,the colours are Red,Blue,and Colourless respectively.

(A) Anti-Markownikoff addition (peroxide effect) is observed only with $HBr$.
In the mechanism,the addition of the halogen radical to the alkene is exothermic for all hydrogen halides.
However,the step involving the abstraction of a hydrogen atom from the hydrogen halide by the alkyl radical is crucial.
For $HBr$,this step is exothermic,making the overall reaction favorable.
For $HCl$,the $H-Cl$ bond is very strong,making this step endothermic.
For $HI$,the $I-I$ bond formation is exothermic,but the addition of the iodine radical to the alkene is endothermic,making the overall process unfavorable.
Therefore,$HCl$ and $HI$ do not undergo anti-Markownikoff's addition.

(D) The bromination of alkanes follows a free radical mechanism.
Bromine is highly selective and preferentially replaces the hydrogen atom on the carbon that forms the most stable radical.
In $2-$methylbutane $(CH_3-CH(CH_3)-CH_2-CH_3)$,the $C2$ position is a tertiary $(3^\circ)$ carbon.
The $3^\circ$ radical is significantly more stable than $2^\circ$ or $1^\circ$ radicals.
Thus,the major product is $2-$bromo$-2-$methylbutane.

(D) In the molecule $HCN$,the structure is $H-C \equiv N$.
There is one $\sigma$ bond between $H$ and $C$.
There is one $\sigma$ bond and two $\pi$ bonds between $C$ and $N$.
Total bonds in $HCN$: $2\,\sigma$ bonds and $2\,\pi$ bonds.
Therefore,the correct option is $D$.

(C) Optical isomerism requires the presence of an asymmetric (chiral) carbon atom,which is a carbon atom bonded to four different groups.
$1$. Lactic acid $(CH_3CH(OH)COOH)$ has a chiral carbon.
$2$. Tartaric acid $(HOOC-CH(OH)-CH(OH)-COOH)$ has chiral carbons.
$3$. $\alpha -$ amino acids $(R-CH(NH_2)-COOH)$ have a chiral carbon (except glycine).
$4$. Maleic acid $(HOOC-CH=CH-COOH)$ is a geometric isomer (cis-isomer) and does not contain any chiral carbon atom.
Therefore,maleic acid does not exhibit optical isomerism.

(B) Given: Mass $m = 160\, g = 0.16\, kg$,initial velocity $v = 10\, m/s$,angle $\theta = 60^\circ$,and acceleration due to gravity $g = 10\, m/s^2$.
At the highest point,the vertical component of velocity is zero,and the horizontal component is $v_x = v \cos \theta = 10 \cos 60^\circ = 10 \times 0.5 = 5\, m/s$.
The height of the highest point is $H = \frac{v^2 \sin^2 \theta}{2g} = \frac{10^2 \times (\sin 60^\circ)^2}{2 \times 10} = \frac{100 \times 0.75}{20} = 3.75\, m$.
The angular momentum $L$ at the highest point with respect to the point of projection is given by $L = m \times v_x \times H$.
Substituting the values: $L = 0.16 \times 5 \times 3.75 = 0.8 \times 3.75 = 3.0\, kg\, m^2/s$.

(D) The energy required to ionize one atom of $Na$ is given by $E = \frac{IE}{N_A}$.
$E = \frac{495.5 \times 10^3 \ J \ mol^{-1}}{6.022 \times 10^{23} \ mol^{-1}} = 8.228 \times 10^{-19} \ J$.
Using the relation $E = h\nu$,the frequency $\nu$ is $\nu = \frac{E}{h}$.
$\nu = \frac{8.228 \times 10^{-19} \ J}{6.626 \times 10^{-34} \ J \ s} = 1.2417 \times 10^{15} \ s^{-1}$.
Thus,the lowest frequency is approximately $1.24 \times 10^{15} \ s^{-1}$.

(B) The molecular orbital configuration of the molecules is determined by the total number of electrons.
For $NO$:
Total number of electrons $= 7(N) + 8(O) = 15$.
The molecular orbital configuration is $KK(\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1$.
Due to the presence of one unpaired electron in the $\pi^* 2p_x$ orbital,$NO$ is paramagnetic.
In contrast,$N_2$ ($14$ electrons),$CO$ ($14$ electrons),and $O_3$ ($24$ electrons) have all electrons paired,making them diamagnetic.

(C) The dissociation reaction is: $[Zr_3(PO_4)_4] \rightleftharpoons 3Zr^{4+} + 4PO_4^{3-}$
If $S$ is the molar solubility,the concentration of $Zr^{4+}$ is $3S$ and $PO_4^{3-}$ is $4S$.
The solubility product expression is: $K_{sp} = [Zr^{4+}]^3 [PO_4^{3-}]^4$
Substituting the values: $K_{sp} = (3S)^3 (4S)^4$
$K_{sp} = (27S^3) \times (256S^4)$
$K_{sp} = 6912S^7$
Therefore,$S = (K_{sp} / 6912)^{1/7}$

(B) The decomposition reaction is $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$.
The equilibrium constant expression is $K_p = (P_{NH_3})^2 \times (P_{CO_2})$.
From the stoichiometry,$NH_3$ and $CO_2$ are produced in a $2:1$ molar ratio. If $P$ is the total pressure at equilibrium,then $P_{NH_3} = \frac{2P}{3}$ and $P_{CO_2} = \frac{P}{3}$.
Substituting these into the $K_p$ expression: $K_p = (\frac{2P}{3})^2 \times (\frac{P}{3}) = \frac{4P^3}{27}$.
Given $K_p = 2.9 \times 10^{-5}$,we have $2.9 \times 10^{-5} = \frac{4P^3}{27}$.
Solving for $P$: $P^3 = \frac{2.9 \times 10^{-5} \times 27}{4} = 19.575 \times 10^{-5} = 195.75 \times 10^{-6}$.
$P = (195.75)^{1/3} \times 10^{-2} \ atm \approx 5.82 \times 10^{-2} \ atm$.

(B) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$ of the gas: $r \propto \frac{1}{\sqrt{M}}$.
The rate of diffusion is defined as $r = \frac{V}{t}$,where $V$ is the volume and $t$ is the time.
Thus,$\frac{r_{SO_2}}{r_{O_2}} = \frac{V_{SO_2} / t_{SO_2}}{V_{O_2} / t_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{SO_2}}}$.
Given: $V_{SO_2} = 20 \ dm^3$,$t_{SO_2} = 60 \ s$,$t_{O_2} = 30 \ s$.
Molar mass of $SO_2 = 32 + (2 \times 16) = 64 \ g/mol$.
Molar mass of $O_2 = 2 \times 16 = 32 \ g/mol$.
Substituting the values: $\frac{20 / 60}{V_{O_2} / 30} = \sqrt{\frac{32}{64}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
$\frac{1/3}{V_{O_2}/30} = \frac{1}{\sqrt{2}} \implies \frac{10}{V_{O_2}} = \frac{1}{1.414}$.
$V_{O_2} = 10 \times 1.414 = 14.14 \ dm^3 \approx 14.1 \ dm^3$.

(C) In the given reaction: $\overset{+4}{H_2SO_3}_{(aq)} + Sn^{4+}_{(aq)} + H_2O_{(l)} \to Sn^{2+}_{(aq)} + \overset{+6}{HSO_4^-}_{(aq)} + 3H^{+}_{(aq)}$.
$1$. The oxidation state of $S$ in $H_2SO_3$ increases from $+4$ to $+6$,which means $H_2SO_3$ undergoes oxidation.
$2$. The oxidation state of $Sn$ decreases from $+4$ to $+2$,which means $Sn^{4+}$ undergoes reduction.
$3$. Since $H_2SO_3$ undergoes oxidation,it acts as the reducing agent.
Therefore,the correct statement is that $H_2SO_3$ is the reducing agent because it undergoes oxidation.

(A) All the given ions $(O_2^{2-}, F^-, Li^+, B^{3+})$ are isoelectronic with the neon configuration $(1s^2 2s^2 2p^6)$,having $10$ electrons.
For isoelectronic species,the ionic radius increases as the nuclear charge $(Z)$ decreases.
The atomic numbers are: $O (Z=8)$,$F (Z=9)$,$Li (Z=3)$,$B (Z=5)$.
Since $O_2^{2-}$ has the lowest nuclear charge $(Z=8)$,it exerts the least attraction on the electrons,resulting in the largest ionic radius.

(C) The given unbalanced reaction is $Cr_2O_7^{2-} + Fe^{2+} + C_2O_4^{2-} \to Cr^{3+} + Fe^{3+} + CO_2$.
Step $1$: Oxidation half-reactions:
$Fe^{2+} \to Fe^{3+} + e^-$ $(i)$
$C_2O_4^{2-} \to 2CO_2 + 2e^-$ $(ii)$
Adding $(i)$ and $(ii)$: $Fe^{2+} + C_2O_4^{2-} \to Fe^{3+} + 2CO_2 + 3e^-$.
Step $2$: Reduction half-reaction:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$ $(iii)$.
Step $3$: To balance the electrons,multiply the oxidation half-reaction by $2$ and the reduction half-reaction by $1$:
$2(Fe^{2+} + C_2O_4^{2-} \to Fe^{3+} + 2CO_2 + 3e^-) \implies 2Fe^{2+} + 2C_2O_4^{2-} \to 2Fe^{3+} + 4CO_2 + 6e^-$.
Step $4$: Adding the balanced half-reactions:
$Cr_2O_7^{2-} + 2Fe^{2+} + 2C_2O_4^{2-} + 14H^+ \to 2Cr^{3+} + 2Fe^{3+} + 4CO_2 + 7H_2O$.
The total number of electrons involved in the balanced redox reaction is $6$.

(B) According to Fajan's rules,the ionic character of a compound depends on the polarizing power of the cation.
Smaller and highly charged cations have greater polarizing power,leading to higher covalent character and lower ionic character.
Larger and less charged cations have lower polarizing power,leading to higher ionic character.
Comparing the cations: $Rb^+$ (largest size,$+1$ charge),$Li^+$ ($+1$ charge),$Mg^{2+}$ ($+2$ charge),and $Be^{2+}$ (smallest size,$+2$ charge).
$RbCl$ has the largest cation with the lowest charge,making it the most ionic.
$BeCl_2$ has the smallest cation with a high charge,making it the most covalent (least ionic).
Therefore,the compound with the greatest ionic character is $RbCl$ and the one with the least ionic character is $BeCl_2$.

(C) Feldspars are the most abundant aluminosilicate minerals on the Earth's surface.
In these structures,silicon and aluminium atoms occupy the centers of interlinked tetrahedra of $SiO_4^{4-}$ and $AlO_4^{5-}$.
These tetrahedra connect at each corner to other tetrahedra,forming an intricate,three-dimensional,negatively charged framework.
Cations like $Na^+$,$K^+$,or $Ca^{2+}$ occupy the voids within this structure.

(A) To find the oxidation state of the central atom in $CrO_2Cl_2$:
Let the oxidation state of $Cr$ be $x$.
Oxygen $(O)$ has an oxidation state of $-2$ and Chlorine $(Cl)$ has an oxidation state of $-1$.
Setting the sum of oxidation states to zero: $x + 2(-2) + 2(-1) = 0$.
$x - 4 - 2 = 0$.
$x - 6 = 0$.
$x = +6$.
Thus,$CrO_2Cl_2$ contains $Cr$ in the $+6$ oxidation state.

(B) The longest carbon chain contains $6$ carbon atoms,so the parent alkane is hexane.
There are two substituents: an ethyl group and a methyl group.
Numbering the chain from either end gives the locants $3$ and $4$.
According to $IUPAC$ rules,when two different substituents are present at equivalent positions,the substituent that comes first in alphabetical order (ethyl) is assigned the lower number.
Therefore,the correct name is $3-ethyl-4-methylhexane$.

(B) For the block to remain in equilibrium without slipping,the angle of inclination $\theta$ at the point of placement must satisfy the condition $\tan \theta \le \mu$.
The maximum height corresponds to the limiting case where $\tan \theta = \mu = 0.5$.
The slope of the surface at any point $(x, y)$ is given by $\frac{dy}{dx} = \tan \theta$.
Given the equation of the surface $y = \frac{x^3}{6}$,we find the slope:
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{x^3}{6} \right) = \frac{3x^2}{6} = \frac{x^2}{2}$.
Equating the slope to the coefficient of friction:
$\frac{x^2}{2} = 0.5$
$x^2 = 1$
$x = 1$.
Now,substitute $x = 1$ into the equation of the surface to find the maximum height $y$:
$y = \frac{(1)^3}{6} = \frac{1}{6} \, m$.

(B) According to Fajan's rule,the ionic character of a compound depends on the polarizability of the cation.
As the size of the cation increases,its polarizing power decreases,which leads to an increase in the ionic character of the bond.
Among the given cations ($Li^+$,$Rb^+$,$Be^{2+}$,$Mg^{2+}$),$Rb^+$ has the largest ionic radius,resulting in the highest ionic character for $RbCl$.
Conversely,$Be^{2+}$ has the smallest size and highest charge density,causing the greatest polarization of the chloride ion,which results in the least ionic (most covalent) character for $BeCl_2$.
Therefore,the compound with the greatest ionic character is $RbCl$ and the compound with the least ionic character is $BeCl_2$.

(A) The angular momentum of the bob about the point of suspension $O$ is given by $\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$.
As the bob moves in a horizontal circle,the position vector $\vec{r}$ and the velocity vector $\vec{v}$ change their directions continuously.
Specifically,the vector $\vec{L}$ is always perpendicular to the plane containing $\vec{r}$ and $\vec{v}$. As the bob revolves,this plane rotates about the vertical axis,causing the direction of $\vec{L}$ to change continuously.
However,the magnitude of the angular momentum $L = mvr \sin \theta$ remains constant because the speed $v$,the distance $r$ from the suspension point,and the angle $\theta$ between the string and the vertical axis remain constant during the uniform circular motion.
Therefore,the angular momentum changes in direction but not in magnitude.

(D) For the falling mass $m$,the equation of motion is: $mg - T = ma$ ... $(1)$
For the rotation of the hollow cylinder,the torque $\tau$ is given by $\tau = I\alpha$. Since the cylinder is hollow,its moment of inertia is $I = mR^2$.
The torque is provided by the tension $T$,so $T \cdot R = I\alpha = (mR^2) \cdot \frac{a}{R}$,where $a = R\alpha$ is the linear acceleration of the string.
This simplifies to $T = ma$ ... $(2)$
Substituting equation $(2)$ into equation $(1)$:
$mg - ma = ma$
$mg = 2ma$
$a = \frac{g}{2}$

(D) Since the particle starts from rest,we can represent its position as $x = A \cos(\omega t)$,where $A$ is the amplitude and $x$ is the displacement from the mean position. At $t=0$,$x=A$.
At $t=\tau$,the distance traveled is $a$,so the position is $x = A - a$.
At $t=2\tau$,the total distance traveled is $a + 2a = 3a$,so the position is $x = A - 3a$.
From the equation $x = A \cos(\omega t)$,we get:
$A - a = A \cos(\omega \tau) \implies \cos(\omega \tau) = \frac{A-a}{A} \quad ...(i)$
$A - 3a = A \cos(2\omega \tau) \implies \cos(2\omega \tau) = \frac{A-3a}{A} \quad ...(ii)$
Using the identity $\cos(2\theta) = 2\cos^2(\theta) - 1$:
$\frac{A-3a}{A} = 2\left(\frac{A-a}{A}\right)^2 - 1$
$A-3a = \frac{2(A^2 + a^2 - 2Aa)}{A} - A$
$A^2 - 3aA = 2A^2 + 2a^2 - 4Aa - A^2$
$A^2 - 3aA = A^2 + 2a^2 - 4Aa$
$a^2 = aA \implies A = 2a$ (Wait,re-evaluating: $A^2 - 3aA = A^2 + 2a^2 - 4Aa \implies aA = 2a^2 \implies A = 2a$ is incorrect,let's re-solve: $A^2 - 3aA = A^2 + 2a^2 - 4Aa \implies aA = 2a^2 \implies A = 2a$ is indeed correct).
Substitute $A=2a$ into $(i)$:
$\cos(\omega \tau) = \frac{2a-a}{2a} = \frac{1}{2}$
$\omega \tau = \frac{\pi}{3}$
$\frac{2\pi}{T} \tau = \frac{\pi}{3} \implies T = 6\tau$.

(A) Given: Number of turns $N = 1000$.
Face area $A = 4 \, cm^2 = 4 \times 10^{-4} \, m^2$.
Change in magnetic field $\Delta B = 10^{-2} \, Wb \, m^{-2}$.
Time interval $\Delta t = 0.01 \, s = 10^{-2} \, s$.
Since the axis of the coil is parallel to the magnetic field,the angle $\theta$ between the area vector and the magnetic field is $0^\circ$,so $\cos(0^\circ) = 1$.
The induced $e.m.f.$ is given by Faraday's law: $e = N \frac{\Delta \phi}{\Delta t} = N A \frac{\Delta B}{\Delta t}$.
Substituting the values:
$e = 1000 \times (4 \times 10^{-4} \, m^2) \times \frac{10^{-2} \, Wb \, m^{-2}}{10^{-2} \, s}$.
$e = 1000 \times 4 \times 10^{-4} \, V = 0.4 \, V$.
Converting to $mV$: $0.4 \, V = 0.4 \times 1000 \, mV = 400 \, mV$.

(B) The critical angle $\theta_c$ is given by the relation $\sin \theta_c = \frac{1}{\mu}$.
Since the refractive index $\mu$ depends on the wavelength $\lambda$ (Cauchy's equation: $\mu \approx A + \frac{B}{\lambda^2}$),$\mu$ decreases as wavelength increases.
Frequency $f$ is inversely proportional to wavelength $(f = \frac{c}{\lambda})$,so $\mu$ decreases as frequency decreases.
For green light,$\sin \theta_c = \frac{1}{\mu_{green}}$.
Light with a frequency less than green light has a longer wavelength,resulting in a smaller refractive index $\mu$ and a larger critical angle $\theta_c$.
Since the incident angle is fixed at $\theta_c$ (for green light),these colors (lower frequency) will have an incident angle less than their respective critical angles,allowing them to refract into the air.
Conversely,light with a higher frequency has a larger $\mu$ and a smaller critical angle,causing it to undergo total internal reflection.

(C) The horizontal position is given by $x = u_x t$. Given $x = 40 \text{ m}$ at $t = 2 \text{ s}$,we have $40 = u_x \times 2$,which gives $u_x = 20 \text{ m/s}$.
The vertical position is given by $y = u_y t - \frac{1}{2} g t^2$. Given $y = 50 \text{ m}$ at $t = 2 \text{ s}$ and $g = 10 \text{ m/s}^2$,we have $50 = u_y(2) - \frac{1}{2}(10)(2)^2$.
$50 = 2u_y - 20 \Rightarrow 2u_y = 70 \Rightarrow u_y = 35 \text{ m/s}$.
The angle of projection $\theta$ is given by $\tan \theta = \frac{u_y}{u_x}$.
$\tan \theta = \frac{35}{20} = \frac{7}{4}$.
Therefore,$\theta = \tan^{-1} \frac{7}{4}$.

(D) The condition for simple harmonic motion $(S.H.M.)$ is that the restoring force $F$ must be directly proportional to the displacement $x$ and directed towards the equilibrium position.
Mathematically,this is expressed as $F = -kx$,where $k$ is a positive force constant.
Comparing this with the given options,the expression $-bx$ (where $b$ is a positive constant) represents the force in $S.H.M.$
Therefore,the correct option is $D$.

(B) The monoclinic crystal system is one of the $7$ primitive crystal systems.
In a monoclinic unit cell,the side lengths are unequal,represented as $a \neq b \neq c$.
The angles are defined such that two angles are $90^{\circ}$ and one is not,specifically $\alpha = \gamma = 90^{\circ}$ and $\beta \neq 90^{\circ}$.

(D) The hydrolysis of xenon fluorides ($XeF_4$ and $XeF_6$) yields various xenon-oxo compounds.
$XeF_6 + H_2O \rightarrow XeOF_4 + 2HF$
$XeF_6 + 2H_2O \rightarrow XeO_2F_2 + 4HF$
$XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$
$XeO_4$ (Xenon tetroxide) is an unstable compound and cannot be prepared by the direct hydrolysis of xenon fluorides. It is typically prepared by the reaction of barium perxenate $(Ba_2XeO_6)$ with concentrated sulfuric acid $(H_2SO_4)$.

(D) The transition elements exhibit variable oxidation states due to the participation of $(n-1)d$ and $ns$ electrons in bonding.
Among the given elements,$Mn \ (Z=25)$ has the electronic configuration $[Ar] \ 3d^5 \ 4s^2$.
It exhibits the largest number of oxidation states ranging from $+2$ to $+7$ $(+2, +3, +4, +6, +7)$ because it has the maximum number of unpaired electrons available for bonding.

(D) The magnitude of crystal field splitting energy $(\Delta _0)$ depends on the strength of the ligand,which is described by the spectrochemical series.
The spectrochemical series for the given ligands is: $F^{-} < NH_3 < CN^{-} < CO$.
$CO$ is a strong $\pi$-acid ligand and sits at the end of the spectrochemical series,causing the largest splitting of $d$-orbitals.
Therefore,$CO$ causes the highest $\Delta _0$.

(C) The absorption of visible light and the resulting coloured nature of transition metal complexes are due to the $d-d$ electronic transition,which requires the presence of at least one unpaired $d$-electron.
Let us analyze the electronic configuration of the metal ions in the given complexes:
$1. [Sc(H_2O)_6]^{3+}: Sc^{3+} (Z=21) = [Ar] 3d^0$. No unpaired electrons.
$2. [Ti(NH_3)_6]^{4+}: Ti^{4+} (Z=22) = [Ar] 3d^0$. No unpaired electrons.
$3. [V(NH_3)_6]^{3+}: V^{3+} (Z=23) = [Ar] 3d^2$. Two unpaired electrons.
$4. [Zn(NH_3)_6]^{2+}: Zn^{2+} (Z=30) = [Ar] 3d^{10}$. No unpaired electrons.
Since $[V(NH_3)_6]^{3+}$ contains unpaired $d$-electrons,it can undergo $d-d$ transitions and is the most likely to absorb visible light.

(A) Oxymercuration-demercuration is a reaction that follows Markovnikov's rule for the addition of water across a double bond without rearrangement.
For allylbenzene $(C_6H_5-CH_2-CH=CH_2)$,the electrophilic mercuric species attacks the terminal carbon $(CH_2)$ to form a more stable carbocation-like intermediate,followed by the attack of water at the more substituted carbon $(CH)$.
This results in the formation of $1-$phenylpropan$-2-$ol as the major product.
The reaction is: $C_6H_5-CH_2-CH=CH_2 \xrightarrow[(ii) NaBH_4]{(i) Hg(OAc)_2} C_6H_5-CH_2-CH(OH)-CH_3$.

(C) The conversion of benzene diazonium chloride to chlorobenzene using $CuCl/HCl$ is a classic example of the Sandmeyer reaction.
The chemical equation is: $C_6H_5N_2Cl \xrightarrow{CuCl/HCl} C_6H_5Cl + N_2$
Therefore,the correct option is $C$.

(A) Aminoglycosides are a class of compounds that act as potent $antibiotics$. They are effective against Gram-negative bacteria and are among the oldest classes of antibiotics used in clinical practice.

(D) Mutarotation is a property exhibited by reducing sugars that have a free hemiacetal or hemiketal group,allowing them to exist in equilibrium between their open-chain and cyclic forms.
Sucrose is a non-reducing sugar because its glycosidic linkage is formed between the $C1$ of glucose and the $C2$ of fructose,effectively blocking both anomeric carbons.
Since sucrose does not contain a free aldehydic or ketonic group,it cannot undergo ring-opening to exhibit mutarotation.

(D) The reaction between phthalic acid and resorcinol in the presence of concentrated $H_2SO_4$ is a condensation reaction.
Phthalic acid undergoes dehydration with two molecules of resorcinol to form a dye known as Fluorescein.
The reaction is: $\text{Phthalic acid} + 2 \times \text{Resorcinol} \xrightarrow{\text{Conc. } H_2SO_4} \text{Fluorescein} + 2H_2O$.

(B) The vapour pressure of a liquid increases with an increase in temperature.
According to the Clausius-Clapeyron equation,the relationship between vapour pressure $(P)$ and temperature $(T)$ is exponential,not linear.
Therefore,if we plot a graph between the vapour pressure and the temperature,we obtain a curve that rises faster as $T$ increases,indicating a non-linear increase.

(A) For a general reaction $aA + bB \to cC + dD$,the rate of reaction is expressed as:
Rate $= -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt} = k[A]^n[B]^m$
Given the reaction $3A + 2B \to C + D$,the rate expression is:
Rate $= -\frac{1}{3} \frac{d[A]}{dt} = -\frac{1}{2} \frac{d[B]}{dt} = \frac{d[C]}{dt} = \frac{d[D]}{dt} = k[A]^n[B]^m$
Comparing this with the given options,the correct representation is $-\frac{1}{3} \frac{d[A]}{dt} = \frac{d[C]}{dt} = k[A]^n[B]^m$.

(A) The dissociation of $Fe(NH_4)_2(SO_4)_2$ is given by: $Fe(NH_4)_2(SO_4)_2 \rightarrow Fe^{2+} + 2NH_4^+ + 2SO_4^{2-}$.
The total number of ions produced per formula unit is $1 + 2 + 2 = 5$. Thus,the expected van't Hoff factor $(i_{expected}) = 5$.
The normal osmotic pressure $(\pi_{nor})$ is calculated using the formula $\pi = CRT$:
$\pi_{nor} = 0.10 \times 0.082 \times 298 = 2.4436 \ atm$.
The observed osmotic pressure $(\pi_{ob})$ is given as $10.8 \ atm$.
The experimental van't Hoff factor $(i_{exp})$ is calculated as:
$i_{exp} = \frac{\pi_{ob}}{\pi_{nor}} = \frac{10.8}{2.4436} \approx 4.42$.
Therefore,the expected and experimental values are $5$ and $4.42$ respectively.

(C) In a cubic close packed $(ccp)$ or face-centered cubic $(fcc)$ structure,the number of atoms per unit cell is $Z = 4$.
There is one octahedral void at the body center of the cube and $12$ octahedral voids at the edges,each shared by $4$ unit cells.
Number of octahedral voids at the body center $= 1$.
Number of octahedral voids at the edges $= 12 \times \frac{1}{4} = 3$.
Total number of octahedral voids per unit cell $= 1 + 3 = 4$.
Since there are $4$ atoms per unit cell,the number of octahedral voids per atom $= \frac{4}{4} = 1$.

(D) An ideal solution is defined as a solution that obeys Raoult's law over the entire range of concentration.
For an ideal solution,the enthalpy of mixing,$\Delta H_{mixing}$,is equal to $0$.
Furthermore,the volume of mixing,$\Delta V_{mixing}$,is also equal to $0$.
This occurs because the intermolecular forces of attraction between the solute-solute $(A-A)$,solvent-solvent $(B-B)$,and solute-solvent $(A-B)$ molecules are identical in nature and magnitude.

(B) The two possible isomers for the given octahedral complex are $[M(NH_3)_5SO_4]Cl$ and $[M(NH_3)_5Cl]SO_4$.
These isomers produce different ions in solution.
Isomer $A$ gives a white precipitate with $AgNO_3$ due to the presence of free $Cl^-$ ions,indicating the structure $[M(NH_3)_5SO_4]Cl$.
Isomer $B$ gives a white precipitate with $BaCl_2$ due to the presence of free $SO_4^{2-}$ ions,indicating the structure $[M(NH_3)_5Cl]SO_4$.
Since the isomers differ in the ions they provide in solution,the type of isomerism exhibited is ionisation isomerism.

(B) The atomic number of Nickel $(Ni)$ is $28$. Its electronic configuration is $[Ar] 3d^8 4s^2$.
In the complex $[NiL_4]^{2-}$,let the oxidation state of $Ni$ be $x$. Since $L$ is a uninegative ligand,its charge is $-1$.
$x + 4(-1) = -2 \implies x = +2$.
So,$Ni^{2+}$ has the configuration $[Ar] 3d^8$.
For a complex to be diamagnetic,all electrons must be paired.
In a $d^8$ system,if the complex is diamagnetic,the ligand must be a strong field ligand that causes pairing of electrons in the $3d$ orbitals,resulting in $dsp^2$ hybridisation.
With $dsp^2$ hybridisation,the $3d$ orbitals are arranged such that there are no unpaired electrons.
Therefore,the hybridisation is $dsp^2$ and the number of unpaired electrons is $0$.

(C) $(a) \ NO^{+} = 7+8-1 = 14 \ e^-$. $O_2 = 16 \ e^-$. Thus,they are not isoelectronic. This statement is false.
$(b)$ Boron forms only covalent compounds due to its extremely high ionization energy. This statement is true.
$(c)$ In aqueous solution,$Ti^{3+}$ is more stable than $Ti^{+}$. This statement is false.
$(d) \ LiAlH_4$ is a well-known versatile reducing agent in organic synthesis. This statement is true.
Note: The question asks for the statement that is not true. Both $(a)$ and $(c)$ are technically false,but $(c)$ is a more standard textbook error regarding stability in aqueous media.

(D) Malachite is a green-colored copper carbonate hydroxide mineral.
Its chemical formula is $Cu(OH)_2 \cdot CuCO_3$.

(D) When $1, 1, 1-$trichloropropane $(CH_3-CH_2-CCl_3)$ is treated with aqueous $KOH$,a nucleophilic substitution reaction occurs. The three chlorine atoms are replaced by three hydroxyl $(-OH)$ groups to form an unstable intermediate gem-triol. This intermediate immediately loses a water molecule to form propionic acid (propanoic acid).
Reaction:
$CH_3-CH_2-CCl_3 + 3KOH_{(aq)}$ $\rightarrow [CH_3-CH_2-C(OH)_3] + 3KCl$ $\rightarrow CH_3-CH_2-COOH + H_2O$

(D) Polymers which change irreversibly into hard and rigid material on heating are known as thermosetting polymers.
For example,$Bakelite$ is a thermosetting polymer.

(C) The stability of the phenoxide ion is increased by electron-withdrawing groups $(EWG)$ through the delocalization of the negative charge on the oxygen atom into the benzene ring.
$-CH_3$ and $-CH_2OH$ are electron-donating groups $(EDG)$ and decrease the stability of the phenoxide ion.
$-OCH_3$ acts as an electron-donating group via the resonance effect ($+R$ effect),which outweighs its inductive electron-withdrawing effect ($-I$ effect),thus destabilizing the phenoxide ion.
$-COCH_3$ is a strong electron-withdrawing group due to both the inductive effect $(-I)$ and the resonance effect $(-R)$.
Therefore,the $-COCH_3$ group is the most effective in stabilizing the phenoxide ion at the para position.

(B) When $Methyl \ amine$ $(CH_3NH_2)$ reacts with nitrous acid $(HNO_2)$,which is generated in situ by the reaction of $NaNO_2$ and $HCl$,it forms an unstable diazonium salt $(CH_3N_2^+Cl^-)$.
This salt immediately undergoes hydrolysis to produce $Methyl \ alcohol$ $(CH_3OH)$,nitrogen gas $(N_2)$,and water $(H_2O)$.
The overall reaction is: $CH_3NH_2 + NaNO_2 + HCl \rightarrow CH_3OH + N_2 + H_2O$.

(C) Sodium bicarbonate $(NaHCO_3)$ is a weak base. Only acids stronger than carbonic acid $(H_2CO_3)$ can react with it to evolve $CO_2$ gas and become soluble.
$2, 4, 6-$Trinitrophenol (picric acid),benzoic acid,and benzene sulphonic acid are stronger acids than carbonic acid and are soluble in $NaHCO_3$.
$o-$Nitrophenol is a weaker acid than carbonic acid. Additionally,it forms intramolecular hydrogen bonding between the $-OH$ group and the $-NO_2$ group,which makes the acidic proton less available for reaction with the base. Therefore,it is not soluble in sodium bicarbonate.

(D) The Williamson synthesis involves the reaction of an alkoxide ion $(RO^-)$ with an alkyl halide $(R'X)$.
This reaction proceeds via an $S_N2$ mechanism,where the alkoxide ion acts as a nucleophile and attacks the alkyl halide,displacing the halide ion.
Therefore,it is an example of a nucleophilic substitution reaction.

(D) The two polynucleotide chains of $DNA$ molecules are twisted around a common axis but run in opposite directions to form a right-handed helix.
These two chains are held together by specific hydrogen bonds between the complementary nitrogenous bases ($A$ pairs with $T$,and $G$ pairs with $C$).

(C) Butyric acid,also known as butanoic acid $(CH_3CH_2CH_2COOH)$,is found in milk and butter.
It is a product of anaerobic fermentation and is responsible for the characteristic unpleasant smell and acrid taste of rancid butter.

(C) $1$. The starting material is $p$-nitrotoluene.
$2$. Electrophilic aromatic substitution with $Br_2/FeBr_3$ introduces a bromine atom at the ortho position relative to the methyl group (since the $-CH_3$ group is ortho/para directing and the $-NO_2$ group is meta directing). This gives $B$ as $2$-bromo-$4$-nitrotoluene.
$3$. Reduction of the nitro group with $Sn/HCl$ converts the $-NO_2$ group to an $-NH_2$ group,yielding $C$ ($2$-bromo-$4$-aminotoluene).
$4$. Diazotization with $NaNO_2/HCl$ at $0-5 \ ^\circ C$ converts the $-NH_2$ group into a diazonium salt,$D$ ($2$-bromo-$4$-methylbenzenediazonium chloride).
$5$. The Sandmeyer reaction with $CuBr/HBr$ replaces the diazonium group with a bromine atom,resulting in $E$ ($2,4$-dibromotoluene).

(A) In the Victor-Meyer's test,the alcohols are converted into nitroalkanes,which are then treated with nitrous acid $(HNO_2)$ and finally made alkaline with $NaOH$.
For $1^o$ alcohols,the product is a nitrolic acid,which gives a blood-red colour in alkaline solution.
For $2^o$ alcohols,the product is a pseudonitrol,which gives a blue colour in alkaline solution.
For $3^o$ alcohols,the product does not react with nitrous acid and remains colourless in alkaline solution.
Therefore,the colours are red,blue,and colourless respectively.

(B) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 1.73 \ BM$,we have $1.73 = \sqrt{n(n+2)}$.
Squaring both sides,$3 = n(n+2)$,which gives $n^2 + 2n - 3 = 0$.
Solving for $n$,we get $(n+3)(n-1) = 0$,so $n = 1$ (since $n$ cannot be negative).
Vanadium $(V)$ has the atomic number $23$ and electronic configuration $[Ar] 3d^3 4s^2$.
For $n=1$,the vanadium ion must have one unpaired electron.
In $VCl_4$,vanadium is in the $+4$ oxidation state $(V^{4+})$.
The configuration of $V^{4+}$ is $[Ar] 3d^1$,which contains $1$ unpaired electron.
Thus,the formula is $VCl_4$.