WBJEE 2017 Chemistry Question Paper with Answer and Solution

(B) The reaction between $aniline$ and $acetic$ $anhydride$ is:
$C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH$
Here,$1$ mole of $aniline$ reacts with $1$ mole of $acetic$ $anhydride$ to form $1$ mole of $acetanilide$ $(C_6H_5NHCOCH_3)$.
Given: $2$ moles of $aniline$ and $1$ mole of $acetic$ $anhydride$.
Since $acetic$ $anhydride$ is the limiting reagent,$1$ mole of $acetanilide$ will be formed.
The molar mass of $acetanilide$ $(C_8H_9NO)$ is $135 \ g/mol$.
Therefore,the yield is $1 \ mol \times 135 \ g/mol = 135 \ g$.

(D) The given reactions are:
$(1) \ 2 \ H_2O_{(g)} \rightleftharpoons 2 \ H_{2(g)} + O_{2(g)}, \ K_1 = 6.4 \times 10^{-8}$
$(2) \ 2 \ CO_{2(g)} \rightleftharpoons 2 \ CO_{(g)} + O_{2(g)}, \ K_2 = 1.6 \times 10^{-6}$
We need the equilibrium constant $(K)$ for the reaction: $H_{2(g)} + CO_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$
Divide reaction $(1)$ by $2$ and reverse it: $H_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons H_2O_{(g)}$,$K_3 = \frac{1}{\sqrt{K_1}}$
Divide reaction $(2)$ by $2$: $CO_{2(g)} \rightleftharpoons CO_{(g)} + \frac{1}{2} O_{2(g)}$,$K_4 = \sqrt{K_2}$
Adding these two reactions gives the target reaction,so $K = K_3 \times K_4 = \frac{\sqrt{K_2}}{\sqrt{K_1}}$
$K = \sqrt{\frac{1.6 \times 10^{-6}}{6.4 \times 10^{-8}}} = \sqrt{\frac{16 \times 10^{-7}}{6.4 \times 10^{-8}}} = \sqrt{\frac{160}{6.4}} = \sqrt{25} = 5$

(A) The electronegativity of an atom depends on the percentage of $s$-character in its hybrid orbitals. Higher $s$-character leads to higher electronegativity.
In $Me_{3}N$,the nitrogen atom is $sp^{3}$-hybridized ($25\% \ s$-character).
In $C_{5}H_{5}N$ (pyridine),the nitrogen atom is $sp^{2}$-hybridized ($33.3\% \ s$-character).
In $MeCN$ (acetonitrile),the nitrogen atom is $sp$-hybridized ($50\% \ s$-character).
Therefore,the order of electronegativity of the nitrogen atom is $MeCN > C_{5}H_{5}N > Me_{3}N$.

(A) The electron affinity generally increases across a period from left to right due to an increase in effective nuclear charge and a decrease in atomic size.
Among the given elements ($C$,$N$,$O$,$F$),$F$ has the highest effective nuclear charge and the smallest atomic size.
Although $Cl$ has the highest electron affinity in the entire periodic table,among the provided options,$F$ has the highest $1^{st}$ electron affinity.

(A) The structure is $Ph-CH_{2}-CH_{2}-COOH$.
To name this compound according to $IUPAC$ rules,we identify the longest carbon chain containing the carboxylic acid group as the parent chain.
The carbon atom of the $-COOH$ group is assigned position $1$.
The chain consists of $3$ carbons,so the parent alkane is propane.
The carboxylic acid suffix is $-oic$ acid.
There is a phenyl group attached to the $3^{rd}$ carbon atom.
Therefore,the correct name is $3-$phenylpropanoic acid.

(D) The isomerisation of $1-$butyne $(CH_3-CH_2-C \equiv CH)$ to $2-$butyne $(CH_3-C \equiv C-CH_3)$ is a base-catalyzed reaction.
When $1-$butyne is treated with ethanolic $KOH$ at high temperatures,it undergoes base-catalyzed isomerisation to form the more stable internal alkyne,$2-$butyne.
The reaction is as follows:
$CH_3-CH_2-C \equiv CH \xrightarrow{\text{ethanolic } KOH, \Delta} CH_3-C \equiv C-CH_3$

(A) The reaction of hexadeuteriobenzene $(C_6D_6)$ with $Br_2$ in the presence of $Fe$ is an electrophilic aromatic substitution reaction.
Since $1 \ mole$ of $Br_2$ is used,one deuterium atom is replaced by a bromine atom to form bromopentadeuteriobenzene $(C_6D_5Br)$.
The addition of $H_2O$ is a standard workup procedure to remove inorganic salts or quench the reaction.
Thus,the major product is bromopentadeuteriobenzene.

(A) The strength of oxo-acids increases with the increase in the oxidation number of the central atom.
For $HClO_{3}$,the oxidation number of $Cl$ is $+5$.
For $HClO_{2}$,the oxidation number of $Cl$ is $+3$.
For $HOCl$,the oxidation number of $Cl$ is $+1$.
For $HOBr$,the oxidation number of $Br$ is $+1$.
Since $HClO_{3}$ has the highest oxidation number of the central atom,it is the strongest Bronsted acid among the given compounds.

(D) When $NaCN$ is dissolved in de-ionised water,it undergoes hydrolysis as follows:
$CN^{-} + H_2O \rightleftharpoons HCN + OH^{-}$
Since $NaCN$ is a salt of a strong base $(NaOH)$ and a weak acid $(HCN)$,the $CN^{-}$ ion undergoes anionic hydrolysis.
This reaction produces $OH^{-}$ ions in the solution,making it basic.
Therefore,the resulting solution will have a $pH > 7$.

(C) To determine the shape of $XeF_{5}^{-}$,we calculate the number of electron pairs around the central $Xe$ atom:
Number of electron pairs $= \frac{1}{2} [V + M - C + A] = \frac{1}{2} [8 + 5 - 0 + 1] = 7$.
Here,$V = 8$ (valence electrons of $Xe$),$M = 5$ (monovalent atoms),and $A = 1$ (anionic charge).
With $7$ electron pairs,the electron geometry is pentagonal bipyramidal ($sp^{3}d^{3}$ hybridization).
Out of these $7$ pairs,$5$ are bond pairs and $2$ are lone pairs.
The two lone pairs occupy the axial positions to minimize repulsion,leaving the $5$ $Xe-F$ bonds in a single equatorial plane.
Therefore,the molecular shape of $XeF_{5}^{-}$ is pentagonal planar.

(B) Lime juice contains citric acid,which acts as a source of $H^{+}$ ions. When $H^{+}$ is added to a mixture containing iodide $(I^{-})$ and iodate $(IO_{3}^{-})$ ions,a comproportionation reaction occurs to produce iodine $(I_{2})$,which appears violet in solution. The balanced chemical equation is: $5I^{-} + IO_{3}^{-} + 6H^{+} \longrightarrow 3I_{2} + 3H_{2}O$.

(D) The number of milli-equivalents of $NaOH$ is given by $N \times V = 0.1 \times 20 = 2 \ mEq$.
Since $1 \ mEq = 10^{-3} \ Eq$,the number of equivalents is $2 \times 10^{-3} = 0.002 \ Eq$.
At the point of neutralisation,the number of equivalents of acid equals the number of equivalents of base.
Therefore,$0.002 = \frac{\text{mass of acid}}{\text{equivalent weight}} = \frac{0.126}{E}$.
Solving for $E$,we get $E = \frac{0.126}{0.002} = 63$.

(A) According to molecular orbital theory,the electronic configuration for $B_{2}$ ($10$ electrons) is: $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \pi 2p_{x}^{1} = \pi 2p_{y}^{1}$.
Since it contains two unpaired electrons,$B_{2}$ is paramagnetic.
The electronic configuration for $C_{2}$ ($12$ electrons) is: $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \pi 2p_{x}^{2} = \pi 2p_{y}^{2}$.
Since all electrons are paired,$C_{2}$ is diamagnetic.

(C) Given weight ratio of $CH_4$ and $SO_2$ is $1:2$.
Let the weight of $CH_4 = w$ and weight of $SO_2 = 2w$.
Molar mass of $CH_4 (M_1) = 16 \ g/mol$.
Molar mass of $SO_2 (M_2) = 64 \ g/mol$.
Number of moles of $CH_4 (n_1) = \frac{w}{16}$.
Number of moles of $SO_2 (n_2) = \frac{2w}{64} = \frac{w}{32}$.
The ratio of the number of molecules is equal to the ratio of the number of moles.
Ratio $= \frac{n_{SO_2}}{n_{CH_4}} = \frac{w/32}{w/16} = \frac{16}{32} = \frac{1}{2}$.
Thus,the ratio is $1:2$.

(B) The neutralization reaction is $HCl + NaOH \longrightarrow NaCl + H_2O$.
Number of milliequivalents of $HCl = x \times y$.
Number of milliequivalents of $NaOH = y \times x$.
Since the milliequivalents are equal,the solution is neutral and the number of milliequivalents of $NaCl$ formed is $xy$.
Total volume of solution = $x \ mL$ $(HCl)$ + $y \ mL$ $(NaOH)$ + $(x+y) \ mL$ (distilled water) = $2(x+y) \ mL$.
$\text{Normality} = \frac{\text{Total milliequivalents of salt}}{\text{Total volume in } mL} = \frac{xy}{2(x+y)} \ N$.

(A) For an ideal gas,the ideal gas equation is $pV = \frac{w}{M}RT$,where $w$ is the mass,$M$ is the molecular weight,$R$ is the gas constant,and $T$ is the temperature.
Since $w$,$R$,and $T$ are constant,let $wRT = K$. Then $pV = \frac{K}{M}$,or $V = \frac{K}{Mp}$.
Taking the logarithm on both sides: $\log V = \log \left( \frac{K}{M} \right) - \log p$.
This is in the form of $y = mx + c$,where $y = \log V$,$x = \log p$,$m = -1$,and the intercept $c = \log \left( \frac{K}{M} \right)$.
From the graph,the intercept for $M_{2}$ is higher than the intercept for $M_{1}$.
Therefore,$\log \left( \frac{K}{M_{2}} \right) > \log \left( \frac{K}{M_{1}} \right)$.
This implies $\frac{K}{M_{2}} > \frac{K}{M_{1}}$,which means $M_{1} > M_{2}$.

(A) The energy of a photon is given by $E = h \nu = \frac{hc}{\lambda} = hc \cdot \bar{\nu}$,where $\bar{\nu} = \frac{1}{\lambda}$.
For option $(A)$: $\lambda = 300 \ nm = 300 \times 10^{-9} \ m$. $E = \frac{(6.626 \times 10^{-34} \ J \cdot s)(3 \times 10^{8} \ m/s)}{300 \times 10^{-9} \ m} = 6.626 \times 10^{-19} \ J$.
For option $(B)$: $\nu = 3 \times 10^{4} \ s^{-1}$. $E = h \nu = (6.626 \times 10^{-34} \ J \cdot s)(3 \times 10^{4} \ s^{-1}) = 1.9878 \times 10^{-29} \ J$.
For option $(C)$: $\bar{\nu} = 30 \ cm^{-1} = 3000 \ m^{-1}$. $E = hc \cdot \bar{\nu} = (6.626 \times 10^{-34} \ J \cdot s)(3 \times 10^{8} \ m/s)(3000 \ m^{-1}) = 5.9634 \times 10^{-22} \ J$.
For option $(D)$: $E = 6.626 \times 10^{-27} \ J$.
Comparing the values,the energy in option $(A)$ is the highest.

(A) According to the $(n+l)$ rule:
$1$. The orbital with a lower $(n+l)$ value has lower energy.
$2$. If the $(n+l)$ values are the same,the orbital with the lower $n$ value has lower energy.
Calculating $(n+l)$ values:
$(i)$ $n+l = 4+1 = 5$
$(ii)$ $n+l = 4+0 = 4$
$(iii)$ $n+l = 3+2 = 5$
$(iv)$ $n+l = 3+1 = 4$
Comparing the values:
For $(iv)$ and $(ii)$,both have $(n+l) = 4$. Since $(iv)$ has $n=3$ and $(ii)$ has $n=4$,$(iv) < (ii)$.
For $(iii)$ and $(i)$,both have $(n+l) = 5$. Since $(iii)$ has $n=3$ and $(i)$ has $n=4$,$(iii) < (i)$.
Combining these,the order of increasing energy is $(iv) < (ii) < (iii) < (i)$.

(B) The electronic configuration of $Cr$ $(Z=24)$ is $[Ar]_{18} 4s^1 3d^5$.
Electrons $1$ to $18$ are filled in the $Ar$ core.
The $19^{\text{th}}$ electron is the first electron to enter the $4s$ orbital.
For the $4s$ orbital: principal quantum number $n=4$,azimuthal quantum number $l=0$,magnetic quantum number $m_l=0$,and spin quantum number $m_s=+\frac{1}{2}$.
Thus,the set of quantum numbers is $(4, 0, 0, +\frac{1}{2})$.

(D) The dimension $[ML^{2} T^{-2}]$ represents energy or work.
$\because$ Kinetic energy $(E_k) = \frac{1}{2} mv^{2} = [M][LT^{-1}]^{2} = [ML^{2} T^{-2}]$.
$\therefore$ Kinetic energy has the dimension of $[ML^{2} T^{-2}]$.

(C) Given $f(x) = \int_{-1}^{x} |t| \, dt$.
For $x \geq 0$,we can split the integral at $t=0$:
$f(x) = \int_{-1}^{0} |t| \, dt + \int_{0}^{x} |t| \, dt$.
Since $|t| = -t$ for $t < 0$ and $|t| = t$ for $t \geq 0$,we have:
$f(x) = \int_{-1}^{0} (-t) \, dt + \int_{0}^{x} t \, dt$.
Evaluating the integrals:
$f(x) = \left[ -\frac{t^2}{2} \right]_{-1}^{0} + \left[ \frac{t^2}{2} \right]_{0}^{x}$.
$f(x) = \left( 0 - (-\frac{(-1)^2}{2}) \right) + \left( \frac{x^2}{2} - 0 \right)$.
$f(x) = \frac{1}{2} + \frac{x^2}{2} = \frac{1}{2}(1+x^2)$.

(C) lactol exists in equilibrium with its open-chain hydroxy-aldehyde form.
In the presence of $NaBH_4$,the aldehyde group is reduced to a primary alcohol.
The lactol $S$ ($2$-hydroxytetrahydropyran) opens to form $5-$hydroxypentanal,which is then reduced by $NaBH_4$ to pentane$-1,5-$diol.

(D) The reaction is a Kolbe-Schmitt reaction.
$1$. $4$-Methoxyphenol reacts with $NaOH$ to form sodium $4$-methoxyphenoxide.
$2$. The phenoxide ion is highly reactive towards electrophilic aromatic substitution.
$3$. The $-O^-$ group is a strong ortho/para directing group. Since the para position is already occupied by the $-OMe$ group,the electrophile $(CO_2)$ attacks the ortho position relative to the $-O^-$ group.
$4$. Subsequent acidification with $H_3O^+$ yields the final product,$2$-hydroxy-$5$-methoxybenzoic acid.

(D) The reduction of benzenediazonium chloride $(C_{6}H_{5}N_{2}^{+}Cl^{-})$ to phenylhydrazine $(C_{6}H_{5}NHNH_{2})$ is a standard chemical transformation.
It can be achieved using $SnCl_{2} / HCl$ (stannous chloride in hydrochloric acid) or $Na_{2}SO_{3}$ (sodium sulfite).
Therefore,both reagents are effective for this reduction.

(C) The reaction described is the $Carbylamine$ reaction,which is a characteristic test for primary amines ($R-NH_2$ or $Ar-NH_2$).
When a primary amine is heated with chloroform $(CHCl_3)$ and ethanolic potassium hydroxide $(KOH)$,it forms an isocyanide (carbylamine),which is characterized by a highly offensive or nauseating smell.
Among the given options,$PhNH_2$ (aniline) is a primary amine.
The reaction is: $PhNH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} PhNC + 3KCl + 3H_2O$.
Thus,$PhNH_2$ produces the nauseating smell.

(B) The reaction of an alkyl nitrile with a Grignard reagent $(PhMgBr)$ followed by acidic hydrolysis $(H_3O^+)$ proceeds as follows:
$1$. The nucleophilic phenyl group $(Ph^-)$ from $PhMgBr$ attacks the electrophilic carbon of the nitrile group $(-CN)$ to form an imine salt intermediate.
$2$. Upon acidic hydrolysis $(H_3O^+)$,the imine salt is converted into an imine,which is unstable and further hydrolyzes to form a ketone.
$3$. The reaction sequence is: $R-CN + PhMgBr$ $\rightarrow R-C(Ph)=N-MgBr$ $\xrightarrow{H_3O^+} R-C(Ph)=NH$ $\xrightarrow{H_3O^+} R-C(=O)Ph$.
$4$. In this case,$R$ is an isopropyl group,so the final product is $3$-methyl-$1$-phenylbutan-$1$-one.

(A) $ADP$ stands for adenosine diphosphate,which contains $2$ phosphate groups.
$ATP$ stands for adenosine triphosphate,which contains $3$ phosphate groups.
Therefore,they differ in the number of phosphate units.

(C) $p-$nitrobenzoic acid $(Z)$ contains a $-NO_2$ group,which is a strong electron-withdrawing group. This stabilizes the carboxylate anion through both inductive and resonance effects,making it the most acidic.
Benzoic acid $(X)$ is stabilized by resonance of the carboxylate group with the benzene ring.
Peroxybenzoic acid $(Y)$ has the structure $C_6H_5CO_3H$. The $-O-O-$ bond is less stable and the resonance stabilization of the carboxylate anion is less effective compared to benzoic acid due to the presence of the extra oxygen atom,making it less acidic than benzoic acid.
Therefore,the correct order of acid strength is $Z > X > Y$.

(A, D) The conversion of propanoic acid $(CH_{3}CH_{2}COOH)$ to the cyclic acetal $(CH_{3}CH_{2}CH(OCH_{2}CH_{2}O))$ involves the following steps:
$1$. Conversion of carboxylic acid to acid chloride using $SOCl_{2}$ or $PCl_{5}$.
$2$. Reduction of acid chloride to aldehyde using $LiAlH_{4}$ (at low temperature) or Rosenmund reduction $(H_{2}/Pd-BaSO_{4})$.
$3$. Protection of the aldehyde group by forming a cyclic acetal using ethylene glycol in the presence of an acid catalyst.
Both options $(A)$ and $(D)$ describe valid pathways for this transformation,as $SnCl_{2}/HCl$ is also a standard reagent for the reduction of acid chlorides to aldehydes (Stephen reduction).

(A) In positron emission,the atomic number of the nucleus decreases by $1$ while the mass number remains constant.
The decay reaction for $F^{18}$ is:
${ }_{9}F^{18} \longrightarrow { }_{8}O^{18} + { }_{+1}e^{0}$
Since the $F^{18}$ atom is part of the $C_{6}H_{5}F^{18}$ molecule,the $F^{18}$ nucleus transforms into an $O^{18}$ nucleus.
Therefore,the resulting decay product is $C_{6}H_{5}O^{18}$.

(B) $1$. For $[NiCl_{4}]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^8)$. $Cl^-$ is a weak field ligand,so no pairing occurs. It has $2$ unpaired electrons.
$2$. For $Ni(CO)_{4}$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of $4s$ electrons into $3d$ orbitals,resulting in $0$ unpaired electrons.
$3$. For $[Cu(NH_{3})_{4}]^{2+}$: $Cu$ is in $+2$ oxidation state $(3d^9)$. It has $1$ unpaired electron.
Thus,the number of unpaired electrons are $2, 0, 1$ respectively. The correct option is $B$.

(D) The basicity of lanthanide ions depends on their ionic character.
As the atomic number increases from $La$ to $Lu$,the ionic radius decreases due to lanthanide contraction.
As the size of the cation decreases,the covalent character increases and the ionic character decreases.
Therefore,the basicity decreases as the ionic radius decreases.
The correct order of basicity is $La^{3+} > Ce^{3+} > Eu^{3+} > Lu^{3+}$.

(C) In the Hall-Heroult process,pure alumina $(Al_2O_3)$ has a very high melting point and is a poor conductor of electricity.
To overcome these issues,it is dissolved in molten cryolite $(Na_3AlF_6)$.
Fluorspar $(CaF_2)$ is added in small quantities to the mixture to lower the melting point of the electrolyte and to increase its electrical conductivity.
Therefore,both $(b)$ and $(c)$ are correct roles of fluorspar.

(D) The reaction sequence is as follows:
$1$. The starting material is a mono-methyl ester of a dicarboxylic acid,$HO_2C-(CH_2)_4-CO_2Me$.
$2$. Treatment with $Ag_2O$ converts the carboxylic acid group $(-COOH)$ into its silver salt $(-COO^-Ag^+)$.
$3$. Subsequent treatment with $Br_2$ in $CCl_4$ is the Borodine-Hunsdiecker reaction,which decarboxylates the silver salt and replaces the carboxyl group with a bromine atom.
$4$. The reaction is: $HO_2C-(CH_2)_4-CO_2Me$ $\xrightarrow{Ag_2O} AgO_2C-(CH_2)_4-CO_2Me$ $\xrightarrow{Br_2, CCl_4} Br-(CH_2)_4-CO_2Me$.
$5$. Thus,the product $R$ is $MeO_2C-(CH_2)_4-Br$.

(A) $PbCl_{2}$ is sparingly soluble in cold water. Upon adding $HCl$,the excess $Cl^{-}$ ions react with $PbCl_{2}$ to form soluble complex anions.
The reaction is as follows:
$PbCl_{2(s)} + Cl^{-} \rightarrow [PbCl_{3}]^{-}_{(aq)}$
$PbCl_{2(s)} + 2Cl^{-} \rightarrow [PbCl_{4}]^{2-}_{(aq)}$
These complex ions are soluble in water,which increases the overall solubility of $PbCl_{2}$ in the presence of $HCl$.

(D) When $BaCl_{2}$ is added to an aqueous solution containing $CO_{3}^{2-}$,$SO_{3}^{2-}$,or $SO_{4}^{2-}$ ions,white precipitates are formed in all three cases due to the formation of $BaCO_{3}$,$BaSO_{3}$,and $BaSO_{4}$ respectively.
Therefore,the presence of a white precipitate does not uniquely identify any one of these anions,as all three anions react with $Ba^{2+}$ to produce a white precipitate.

(D) The reduction half-reaction is: $Cr_{2}O_{7}^{2-} + 14H^+ + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O$.
In this reaction,the oxidation state of $Cr$ changes from $+6$ to $+3$.
Since there are $2$ atoms of $Cr$ in $Cr_{2}O_{7}^{2-}$,the total change in oxidation state is $2 \times (6 - 3) = 6$.
Therefore,$6 \ mol$ of electrons are required to reduce $1 \ mol$ of $Cr_{2}O_{7}^{2-}$,which corresponds to $6 \ F$ of electricity.

(C) For $HCl$,the normality $(N)$ is equal to the molarity $(M)$ because the n-factor is $1$. Thus,$2 \ N = 2 \ M$ and $5 \ N = 5 \ M$.
Let $V_1$ be the volume of $2 \ M \ HCl$ used and $V_2$ be the volume of $5 \ M \ HCl$ used.
We are given $V_1 = 500 \ mL$ and we have a maximum of $500 \ mL$ of $5 \ M \ HCl$ available $(V_2 \leq 500 \ mL)$.
The mixture equation is $M_1 V_1 + M_2 V_2 = M_{final} (V_1 + V_2)$.
Substituting the values: $2(500) + 5(V_2) = 3(500 + V_2)$.
$1000 + 5V_2 = 1500 + 3V_2$.
$2V_2 = 500$,which gives $V_2 = 250 \ mL$.
Since $250 \ mL < 500 \ mL$,this is feasible.
The total volume of $3 \ M \ HCl$ prepared is $V_1 + V_2 = 500 \ mL + 250 \ mL = 750 \ mL$.

(B, C, D) When water is added to molten $NaCl$ during electrolysis,the following occurs:
$1$. The reaction $2Na + 2H_2O \longrightarrow 2NaOH + H_2$ takes place.
$2$. Hydrogen gas $(H_2)$ is evolved at the cathode because the discharge potential of $H^+$ ions is lower than that of $Na^+$ ions.
$3$. Caustic soda $(NaOH)$ is formed as a byproduct.
$4$. The reaction between sodium and water is highly exothermic,which can lead to the ignition of the evolved hydrogen gas,making a fire likely.
Therefore,options $B$,$C$,and $D$ are correct.

(D) In a body-centred cubic $(bcc)$ unit cell,the atoms touch each other along the body diagonal.
The length of the body diagonal is $\sqrt{3} a$.
Since the body diagonal consists of two radii of the corner atoms and the full diameter of the central atom,we have $\sqrt{3} a = 4r$.
Therefore,the atomic radius is $r = \frac{\sqrt{3}}{4} a$.

(C) For isotonic solutions,the effective concentration (osmolarity) must be the same: $i_1C_1 = i_2C_2$.
Assuming complete dissociation:
For $0.03 \ M \ NaCl$: $i = 2$,so $iC = 2 \times 0.03 = 0.06 \ M$.
For $0.02 \ M \ MgCl_2$: $i = 3$,so $iC = 3 \times 0.02 = 0.06 \ M$.
Since the $iC$ values are equal,the solutions are isotonic.