TS EAMCET 2016 Chemistry Question Paper with Answer and Solution

(D) Alkaline $KMnO_4$ oxidizes any alkyl group directly attached to a benzene ring to a carboxylic acid group $(-COOH)$,provided the alkyl group has at least one $\alpha$-hydrogen atom.
In both ethylbenzene and propylbenzene,the $\alpha$-carbon has hydrogen atoms,so both are oxidized to benzoic acid.
Therefore,both $Y$ and $Z$ are benzoic acid.

(A) The structure of the compound is $CH_3-C(=O)-CH_2-C \equiv N$.
Let us label the carbons as follows:
$(i) CH_3-$,$(ii) -C(=O)-$,$(iii) -CH_2-$,$(iv) -CN$.
$1$. The carbon in the methyl group $(CH_3)$ is bonded to four atoms (three $H$ and one $C$),so it is $sp^3$ hybridised.
$2$. The carbonyl carbon $(C=O)$ is bonded to three atoms (one $C$ of $CH_3$,one $O$,and one $C$ of $CH_2$),so it is $sp^2$ hybridised.
$3$. The methylene carbon $(CH_2)$ is bonded to four atoms (two $H$,one $C$ of $C=O$,and one $C$ of $CN$),so it is $sp^3$ hybridised.
$4$. The cyano carbon $(CN)$ is bonded to two atoms (one $C$ of $CH_2$ and one $N$),so it is $sp$ hybridised.
Thus,the hybridisation of carbons $(i), (ii), (iii),$ and $(iv)$ is $sp^3, sp^2, sp^3, sp$ respectively.

(A) According to $VSEPR$ theory:
$1$. $PCl_3$: Central atom $P$ has $5$ valence electrons. It forms $3$ bonds with $Cl$ and has $1$ lone pair. Total electron pairs = $4$ ($3$ bond pairs + $1$ lone pair),resulting in a Trigonal pyramidal geometry $(A-III)$.
$2$. $BF_3$: Central atom $B$ has $3$ valence electrons. It forms $3$ bonds with $F$ and has $0$ lone pairs. Total electron pairs = $3$,resulting in a Trigonal planar geometry $(B-V)$.
$3$. $ClF_3$: Central atom $Cl$ has $7$ valence electrons. It forms $3$ bonds with $F$ and has $2$ lone pairs. Total electron pairs = $5$,resulting in a $T$-shape geometry $(C-II)$.
$4$. $XeF_4$: Central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ and has $2$ lone pairs. Total electron pairs = $6$,resulting in a Square planar geometry $(D-I)$.
Therefore,the correct match is $A-III, B-V, C-II, D-I$.

(D) According to Fajan's rule,the covalent character of an ionic bond increases with the increase in the size of the anion.
Since the cation $K^{+}$ is common in all the given compounds,the covalent character depends on the size of the halide ions $(F^{-}, Cl^{-}, I^{-})$.
The size order of the anions is $F^{-} < Cl^{-} < I^{-}$.
Therefore,the order of covalent character is $KF < KCl < KI$.

(B) The correct option is $B$.
$(A)$ The equilibrium constant $(K_c)$ is temperature-dependent.
$(B)$ The value of $K_c$ is a constant at a given temperature and is independent of the initial concentrations of reactants and products.
$(C)$ At equilibrium,the rate of the forward reaction is equal to the rate of the backward reaction.
$(D)$ For the reaction $Ni_{(s)} + 4CO_{(g)} \rightleftharpoons Ni(CO)_{4(g)}$,the equilibrium constant expression is $K_c = \frac{[Ni(CO)_4]}{[CO]^4}$. Since the expression in the option omitted the $K_c$ definition or was potentially ambiguous,$B$ is the most fundamentally correct statement regarding the nature of $K_c$.

(A) The atomic number of the element $Uus$ is $117$.
Ununseptium $(Uus)$ is the second heaviest known element and penultimate element of the $7^{th}$ period of the periodic table.
Electronic configuration of $Uus = [Rn] 5f^{14} 6d^{10} 7s^2 7p^5$.

(B) The basic character of an oxide increases with an increase in the metallic character of the element.
As we move down the group,the metallic character increases.
Among the given elements ($B$,$Al$,$Ga$,$Tl$),$Tl$ $(Thallium)$ is at the bottom of the group and thus possesses the highest metallic character.
Therefore,$Tl$ forms a basic oxide.
$B$ $(Boron)$ forms an acidic oxide,while $Al$ $(Aluminium)$ and $Ga$ $(Gallium)$ form amphoteric oxides.

(C) In an acidic medium,the permanganate ion $(MnO_4^{-})$ acts as a strong oxidizing agent and reacts with hydrogen peroxide $(H_2O_2)$.
The balanced chemical equation for this reaction is: $2MnO_4^{-} + 5H_2O_2 + 6H^{+} \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O$.
In $MnO_4^{-}$,the oxidation state of $Mn$ is calculated as: $x + 4(-2) = -1$,which gives $x = +7$.
In the product $Mn^{2+}$,the oxidation state of $Mn$ is $+2$.
Therefore,the oxidation state of $Mn$ changes from $+7$ to $+2$.

(B) By the triangle inequality,for any complex numbers $z_1$ and $z_2$,we have $|z_1| + |z_2| \geq |z_1 + z_2|$.
More specifically,for the expression $|z-1| + |z-5|$,we can use the property $|a| + |b| \geq |a - b|$.
Let $a = z-1$ and $b = z-5$.
Then $|z-1| + |z-5| \geq |(z-1) - (z-5)|$.
$|z-1| + |z-5| \geq |z - 1 - z + 5|$.
$|z-1| + |z-5| \geq |4|$.
$|z-1| + |z-5| \geq 4$.
Thus,the minimum value is $4$.

(D) First,find the prime factorization of $7!$:
$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 2^4 \times 3^2 \times 5^1 \times 7^1$.
The number of divisors of a number $N = p_1^{a} p_2^{b} p_3^{c} p_4^{d}$ is given by $(a+1)(b+1)(c+1)(d+1)$.
Here,$a=4, b=2, c=1, d=1$.
Therefore,the number of divisors is $(4+1)(2+1)(1+1)(1+1) = 5 \times 3 \times 2 \times 2 = 60$.

(D) In environmental chemistry,the medium or component that is adversely affected by a pollutant is known as a $receptor$ or $target$. $Receptors$ can be biotic (living organisms) or abiotic (non-living components) that suffer negative impacts due to the presence of pollutants.

(D) The equations of the circles are:
$x^2+y^2-2 \lambda x-2 y-7=0$ $(i)$
$3(x^2+y^2)-8 x+29 y=0 \Rightarrow x^2+y^2-\frac{8}{3} x+\frac{29}{3} y=0$ $(ii)$
Comparing these with the general form $x^2+y^2+2gx+2fy+c=0$:
For circle $(i)$: $g_1 = -\lambda, f_1 = -1, c_1 = -7$
For circle $(ii)$: $g_2 = -\frac{4}{3}, f_2 = \frac{29}{6}, c_2 = 0$
Since the circles are orthogonal,the condition is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Substituting the values:
$2(-\lambda)(-\frac{4}{3}) + 2(-1)(\frac{29}{6}) = -7 + 0$
$\frac{8}{3}\lambda - \frac{29}{3} = -7$
$\frac{8}{3}\lambda = -7 + \frac{29}{3} = \frac{-21+29}{3} = \frac{8}{3}$
$\lambda = 1$

(A) The reaction sequence is as follows:
$1$. The reaction of a vicinal dibromide with $Zn$ dust (dehalogenation) leads to the formation of an alkene.
$2$. $X = (CH_3)_2CBrCH_2Br$ reacts with $Zn$ to form $Y = (CH_3)_2C=CH_2$ (isobutylene).
$3$. Ozonolysis of isobutylene $(CH_3)_2C=CH_2$ in the presence of $Zn/H_2O$ yields acetone $(CH_3)_2CO$ and formaldehyde $(CH_2O)$.

(C) The reaction of propyne $(CH_3-C \equiv CH)$ with $2$ equivalents of $HBr$ proceeds via Markovnikov's addition.
In the first step,$HBr$ adds to the triple bond to form $CH_3-CBr=CH_2$ ($2$-bromopropene).
In the second step,another molecule of $HBr$ adds to the double bond following Markovnikov's rule,where the hydrogen atom attaches to the carbon with more hydrogens,resulting in the geminal dihalide $CH_3-CBr_2-CH_3$ ($2$,$2$-dibromopropane).

(C) $NH_4Cl$ is a salt formed from a weak base $(NH_4OH)$ and a strong acid $(HCl)$.
In water,it undergoes hydrolysis: $NH_4^+ + H_2O \rightleftharpoons NH_4OH + H^+$.
Due to the production of $H^+$ ions,the solution becomes acidic.
Therefore,the $pH$ of an aqueous solution of $NH_4Cl$ is $< 7$.

(C) The gas produced by the passage of air over hot coke is called producer gas.
The reaction proceeds as:
$2 C (s) + (O_2 + 3.76 N_2) (g) \xrightarrow{\Delta} 2 CO (g) + 3.76 N_2 (g)$
This mixture of $CO$ and $N_2$ is known as producer gas.

(D) Beryllium $(Be)$ has the smallest atomic size in group $2$. Due to its high ionization enthalpy,its outermost electrons are tightly held and do not get excited to higher energy levels by the energy provided in a flame test. Therefore,$Be$ does not impart any characteristic color to the flame.

(B) In the given reaction in acidic medium,$KMnO_4$ acts as an oxidizing agent.
$KMnO_4$ gains $5$ electrons per molecule to reduce $Mn^{7+}$ to $Mn^{2+}$.
Therefore,the $n$-factor for $KMnO_4$ in acidic medium is $5$.
Equivalent weight = $\frac{\text{Molecular weight}}{n\text{-factor}}$
Equivalent weight = $\frac{158}{5} = 31.6 \ g/\text{equivalent}$.

(A) The balanced chemical equation is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.
First,calculate the moles of reactants:
$n_{N_2} = \frac{50.0 \ g}{28.0 \ g/mol} \approx 1.786 \ mol$.
$n_{H_2} = \frac{10.0 \ g}{2.016 \ g/mol} \approx 4.96 \ mol$.
According to the stoichiometry,$1 \ mol$ of $N_2$ requires $3 \ mol$ of $H_2$.
For $1.786 \ mol$ of $N_2$,we need $1.786 \times 3 = 5.358 \ mol$ of $H_2$.
Since we only have $4.96 \ mol$ of $H_2$,$H_2$ is the limiting reagent.
The moles of $NH_3$ produced depend on the limiting reagent $(H_2)$:
$n_{NH_3} = n_{H_2} \times \frac{2 \ mol \ NH_3}{3 \ mol \ H_2} = 4.96 \times \frac{2}{3} \approx 3.31 \ mol$.
Using the exact mass calculation: $n_{NH_3} = \frac{10 \ g}{6 \ g} \times 2 = \frac{20}{6} = 3.33 \ mol$.

(D) According to the Maxwell-Boltzmann distribution,as the temperature of a gas increases,the most probable speed $(u_{mp})$ increases,and the peak of the distribution curve shifts to the right and flattens.
In the given figure,the peak for $T_2$ is at the highest speed,followed by $T_3$,and then $T_1$.
Therefore,the most probable speed follows the order $u_{mp}(T_2) > u_{mp}(T_3) > u_{mp}(T_1)$.
Since $u_{mp} = \sqrt{\frac{2RT}{M}}$,it implies that $u_{mp} \propto \sqrt{T}$.
Thus,the correct order of temperature is $T_2 > T_3 > T_1$.

(A) The average kinetic energy of an ideal gas is given by the formula $KE = \frac{3}{2} RT$.
Since the kinetic energy depends only on the temperature $T$ and the gas constant $R$,it is independent of the nature or molar mass of the gas.
Therefore,at the same temperature $T$,the kinetic energy of $CH_4$ and $O_2$ will be equal.
Thus,$KE_{(CH_4)} = KE_{(O_2)} = X$.

(D) Isoelectronic species are atoms or ions that have the same number of electrons.
For $(I)$: $O^{2-} (8+2=10)$,$F^{-} (9+1=10)$,$Na^{+} (11-1=10)$,$Mg^{2+} (12-2=10)$. All have $10$ electrons,so $(I)$ is isoelectronic.
For $(II)$: $Na^{+} (10)$,$Mg^{+} (11)$,$Al^{3+} (10)$,$F^{-} (10)$. Since $Mg^{+}$ has $11$ electrons,$(II)$ is not isoelectronic.
For $(III)$: $N^{3-} (7+3=10)$,$O^{2-} (8+2=10)$,$F^{-} (9+1=10)$,$Ne (10)$. All have $10$ electrons,so $(III)$ is isoelectronic.
Therefore,$(I)$ and $(III)$ are isoelectronic species.

(A) The given electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1$.
Summing the electrons: $2+2+6+2+6+10+1 = 29$.
The element with atomic number $29$ is Copper $(Cu)$.
This configuration represents the stable ground state of $Cu$ where the $3d$ subshell is completely filled.

(A) The extra stability of half-filled and fully filled electronic configurations can be explained in terms of symmetry and exchange energy.
All the orbitals of the same subshell that are either completely filled or half-filled have a more symmetrical distribution of electrons.
Consequently,the shielding of electrons is relatively small,and the electrons are more strongly attracted by the nucleus.
Furthermore,the number of possible exchanges is maximum in these configurations,which leads to higher exchange energy and,therefore,greater stability of the atom.
Thus,both $(A)$ and $(R)$ are correct,and $(R)$ is the correct explanation of $(A)$.

(D) The standard molar enthalpy of formation $(\Delta_f H^{\circ})$ is defined as the enthalpy change when $1 \ mol$ of a substance is formed from its constituent elements in their standard states.
For the reaction: $N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$,the enthalpy change is $\Delta_r H^{\circ} = -92 \ kJ$ for the formation of $2 \ mol$ of $NH_{3(g)}$.
To find the enthalpy of formation for $1 \ mol$ of $NH_{3(g)}$,we divide the reaction by $2$:
$\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \longrightarrow NH_{3(g)}$
$\Delta_f H^{\circ} = \frac{\Delta_r H^{\circ}}{2} = \frac{-92 \ kJ}{2} = -46 \ kJ \ mol^{-1}$.

(B) The synthesis of Aspirin (acetylsalicylic acid) involves the acetylation of salicylic acid.
Salicylic acid $(X)$ reacts with acetic anhydride $(Y)$ in the presence of an acid catalyst $(H^+)$ to form acetylsalicylic acid (Aspirin) and acetic acid $(CH_3COOH)$.
The reaction is:
$\text{Salicylic acid} + (CH_3CO)_2O \xrightarrow{H^+} \text{Aspirin} + CH_3COOH$.
Thus,$X$ is salicylic acid and $Y$ is acetic anhydride $(CH_3CO)_2O$.

(C) The reaction of nitriles $(R-CN)$ with stannous chloride $(SnCl_2)$ in the presence of hydrochloric acid $(HCl)$,followed by hydrolysis $(H_3O^+)$,yields an aldehyde $(R-CHO)$. This specific reduction method is known as the Stephen's reaction.
The mechanism involves the formation of an imine hydrochloride intermediate,which is subsequently hydrolyzed to the corresponding aldehyde:
$R-C \equiv N + SnCl_2 + 2HCl \rightarrow R-CH=NH \cdot HCl$
$R-CH=NH \cdot HCl + H_2O \rightarrow R-CHO + NH_4Cl$

(D) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom for protonation.
In aromatic amines like $C_6H_5NH_2$,$C_6H_5NHCH_3$,and $C_6H_5N(CH_3)_2$,the lone pair of electrons on the nitrogen atom is involved in resonance with the benzene ring,which decreases its availability for protonation.
In aliphatic amines like $CH_3NH_2$,there is no such resonance,and the electron-donating inductive effect of the alkyl group increases the electron density on the nitrogen atom,making it more available for protonation.
Therefore,$CH_3NH_2$ is a much stronger base than the given aromatic amines.

(B) Insulin is a peptide hormone secreted by the pancreas.
It regulates blood glucose metabolism by signaling the liver,muscles,and fat cells to take up excess glucose from the blood,thereby maintaining the level within the normal limit.

(C) For the given reaction: $5Br^-{_{\text{(aq)}}} + 6H^+{_{\text{(aq)}}} + BrO_3^-{_{\text{(aq)}}} \rightarrow 3Br_{2\text{(aq)}} + 3H_2O_{\text{(l)}}$
The rate of reaction expression is:
$Rate = -\frac{1}{5} \frac{\Delta[Br^{-}]}{\Delta t} = -\frac{\Delta[BrO_3^{-}]}{\Delta t} = +\frac{1}{3} \frac{\Delta[Br_2]}{\Delta t}$
Given that $-\frac{\Delta[BrO_3^{-}]}{\Delta t} = 0.01 \ mol \ L^{-1} \ min^{-1}$,we equate the terms:
$-\frac{\Delta[BrO_3^{-}]}{\Delta t} = \frac{1}{3} \frac{\Delta[Br_2]}{\Delta t}$
Substituting the given value:
$0.01 = \frac{1}{3} \frac{\Delta[Br_2]}{\Delta t}$
Therefore,$\frac{\Delta[Br_2]}{\Delta t} = 3 \times 0.01 = 0.03 \ mol \ L^{-1} \ min^{-1}$.

(A) Chloroxylenol is an example of an antiseptic.
Dettol is a mixture of chloroxylenol and $\alpha$-terpineol.
Chloroxylenol is responsible for the antiseptic and disinfectant properties of Dettol.

(C) Geometrical isomerism is observed in complexes where the relative positions of ligands around the central metal atom can change.
$1$. Octahedral complexes of the type $[MX_2L_4]$ exhibit geometrical isomerism (cis and trans forms).
$2$. Square planar complexes of the type $[MX_2L_2]$ exhibit geometrical isomerism (cis and trans forms).
$3$. Tetrahedral complexes of the type $[MABXL]$ do not exhibit geometrical isomerism because all four positions in a tetrahedron are adjacent to each other,making the relative positions of ligands identical.
$4$. Octahedral complexes of the type $[MX_2(L^{-}L)_2]$ exhibit geometrical isomerism due to the arrangement of $X$ ligands.
Therefore,the correct answer is the tetrahedral complex.

(C) Copper matte is a mixture of $Cu_2S$ and $FeS$.
During the smelting of roasted copper ore (copper pyrites,$CuFeS_2$),the ore is heated with silica $(SiO_2)$ in a blast furnace.
The iron oxide $(FeO)$ formed reacts with silica to form slag $(FeSiO_3)$,while the remaining $Cu_2S$ and $FeS$ form the molten copper matte.
$2CuFeS_2 + O_2 \longrightarrow Cu_2S + 2FeS + SO_2$

(B) The variation of molar conductivity $(\Lambda_m)$ with concentration $(C)$ for strong electrolytes like $KCl$ is given by the Kohlrausch equation: $\Lambda_m = \Lambda_m^{\circ} - A \sqrt{C}$.
Here,$\Lambda_m^{\circ}$ is the molar conductivity at infinite dilution and $A$ is a constant.
This equation follows the linear form $y = mx + c$,where $y = \Lambda_m$,$x = \sqrt{C}$,$m = -A$ (slope),and $c = \Lambda_m^{\circ}$ (intercept).
Since the slope is negative $(-A)$,the plot of $\Lambda_m$ versus $\sqrt{C}$ is a straight line with a negative slope,indicating that $\Lambda_m$ decreases as $\sqrt{C}$ increases.

(A) The correct answer is $H_3CCH_2CH_2CH_2Cl$ (n-butyl chloride).
For isomeric alkyl halides,the boiling point decreases with an increase in branching.
Branching of the carbon chain makes the molecule more compact,which reduces the surface area.
As the surface area decreases,the magnitude of intermolecular van der Waals forces of attraction also decreases.
Therefore,straight-chain isomers have higher boiling points compared to their branched-chain counterparts with the same molecular formula.
Among the given options,$H_3CCH_2CH_2CH_2Cl$ is a straight-chain isomer,while the others are branched,making it the one with the highest boiling point.

(D) $Zn$ reacts with very dilute nitric acid to form 'laughing gas' $(N_2O)$.
$4 Zn + 10 HNO_3 (\text{very dilute}) \longrightarrow 4 Zn(NO_3)_2 + N_2O + 5 H_2O$

(A) $XeF_4$ is obtained by heating a mixture of xenon and fluorine in the molar ratio of $1: 5$ at $873 \ K$ and $7 \ bar$ pressure in an enclosed nickel vessel for a few hours.
The reaction proceeds as:
$Xe_{(g)} + 2 \ F_{2(g)} \xrightarrow{873 \ K, 7 \ bar} XeF_4$
The extra fluorine is added to ensure the complete conversion of xenon to $XeF_4$.

(D) The polydispersity index $(PDI)$ of a polymer is defined as the ratio of the weight average molecular mass $(\bar{M}_w)$ to the number average molecular mass $(\bar{M}_n)$.
$PDI = \frac{\bar{M}_w}{\bar{M}_n}$
It serves as a measure of the breadth or distribution of molecular weights in a polymer sample.

(A) The packing efficiency is the percentage of total space occupied by the constituent particles in a crystal lattice.
For simple cubic $(sc)$ lattice,the packing efficiency is $52.4 \%$.
For body-centred cubic $(bcc)$ lattice,the packing efficiency is $68 \%$.
For cubic close packing $(ccp)$ or face-centred cubic $(fcc)$ lattice,the packing efficiency is $74 \%$.
Therefore,the order of packing efficiency is $sc < bcc < ccp$.

(B) The formula for boiling point elevation is $\Delta T_b = K_b \times m$.
Given,$K_b = 0.52 \ K \ kg \ mol^{-1}$ and molality $m = 0.1 \ m$.
Calculating the elevation in boiling point: $\Delta T_b = 0.52 \times 0.1 = 0.052 \ K$ (or $^{\circ}C$).
The boiling point of the solution is $T_b = T_b^{\circ} + \Delta T_b$.
Since the boiling point of pure water $T_b^{\circ} = 100^{\circ}C$,we have $T_b = 100 + 0.052 = 100.052^{\circ}C$.

(C) The van't Hoff factor $(i)$ is defined as the ratio of the observed colligative property to the calculated (theoretical) colligative property.
$i = \frac{\Delta T_f \text{ (observed)}}{\Delta T_f \text{ (calculated)}}$
Given:
$\Delta T_f \text{ (observed)} = 0.025 \ K$
$i = 2.0$
Substituting the values:
$2.0 = \frac{0.025}{\Delta T_f \text{ (calculated)}}$
$\Delta T_f \text{ (calculated)} = \frac{0.025}{2.0} = 0.0125 \ K$

(B) The magnetic moment $(\mu)$ is related to the number of unpaired electrons '$n$' by the formula: $\mu_{eff} = \sqrt{n(n+2)} \ BM$.
Given $\mu_{eff} = \sqrt{24} \ BM$,we equate $\sqrt{n(n+2)} = \sqrt{24}$,which gives $n(n+2) = 24$.
Solving for $n$,we get $n^2 + 2n - 24 = 0$,which factors to $(n+6)(n-4) = 0$. Since $n$ must be positive,$n = 4$.
Now,let us check the number of unpaired electrons for each ion:
$Mn^{2+}$ $(Z=25)$: $[Ar] 3d^5$,$n = 5$.
$Fe^{2+}$ $(Z=26)$: $[Ar] 3d^6$,$n = 4$.
$Fe^{3+}$ $(Z=26)$: $[Ar] 3d^5$,$n = 5$.
$Co^{2+}$ $(Z=27)$: $[Ar] 3d^7$,$n = 3$.
Thus,$Fe^{2+}$ has $4$ unpaired electrons and corresponds to a magnetic moment of $\sqrt{24} \ BM$.

(A) An emulsion is a colloidal system in which both the dispersed phase and the dispersion medium are liquids.
Milk is a classic example of an emulsion where fat droplets are dispersed in water.
While butter and vanishing cream are also emulsions,milk is the most standard example cited in textbooks for a liquid-in-liquid colloidal system.