TS EAMCET 2023 Chemistry Question Paper with Answer and Solution

(A) The correct matches are as follows:
$(A)$ $Na_2O_2$ is used as an oxidising agent $(III)$.
$(B)$ $D_2O$ (heavy water) is used as a moderator in nuclear reactors $(IV)$.
$(C)$ $Cs$ (Cesium) is used in photoelectric cells due to its low ionisation energy $(I)$.
$(D)$ $Mg(OH)_2$ (milk of magnesia) is used as an antacid $(II)$.
Therefore,the correct matching is $A-III, B-IV, C-I, D-II$.

(B) $I$. Beryllium oxide $(BeO)$ is amphoteric in nature,reacting with both acids and bases. This is a correct statement.
$II$. Beryllium does not react directly with hydrogen to form $BeH_2$. This is an incorrect statement.
$III$. $BeH_2$ is prepared by the reaction of $BeCl_2$ with $LiAlH_4$: $2BeCl_2 + LiAlH_4 \rightarrow 2BeH_2 + LiCl + AlCl_3$. This is a correct statement.
$IV$. Beryllium sulphate $(BeSO_4)$ is highly soluble in water due to the high hydration enthalpy of the small $Be^{2+}$ ion. It is not the least soluble. This is an incorrect statement.
Therefore,statements $I$ and $III$ are correct.

(D) Alkali metals $(Group \ 1)$ have only one valence electron per atom,which leads to weaker metallic bonding compared to alkaline earth metals $(Group \ 2)$,which have two valence electrons.
As a result,alkali metals are softer,have lower boiling points,and larger atomic/ionic radii than their corresponding alkaline earth metal counterparts.
Furthermore,due to their larger atomic size and lower effective nuclear charge,alkali metals have lower ionization enthalpy values compared to alkaline earth metals.

(C) The correct statement analysis is as follows:
$1$. The molecular formula of calgon is $Na_2[Na_4(PO_3)_6]$,which is correct.
$2$. Beryllium halides are covalent in nature due to the small size and high polarizing power of $Be^{2+}$ ions,making them soluble in organic solvents,which is correct.
$3$. Among alkali metals,Lithium $(Li)$ has the highest reducing property in aqueous solution due to its high hydration energy. Sodium $(Na)$ does not have the least reducing property; the trend is $Li > K > Rb > Cs > Na$. Thus,the statement is incorrect.
$4$. White metal (also known as Babbitt metal) is an alloy typically composed of tin,copper,and antimony,not Lithium. This statement is also incorrect.
However,in the context of standard competitive chemistry questions,the most widely recognized incorrect statement regarding the specific properties of alkali metals is that sodium has the least reducing property,as it contradicts the standard electrochemical series behavior in aqueous media.

(B) $KOH \rightarrow K^{+} + OH^{-}$
$Cd(OH)_2 \rightleftharpoons Cd^{2+} + 2OH^{-}$
Solubility product $K_{sp} = [Cd^{2+}][OH^{-}]^2$
Given $[OH^{-}] = 0.1 \ M$ due to common ion effect.
$K_{sp} = [S] \times [0.1]^2$
$2.5 \times 10^{-14} = [S] \times 0.01$
$[S] = \frac{2.5 \times 10^{-14}}{10^{-2}} = 2.5 \times 10^{-12}$
$[S] = 25 \times 10^{-13} = x \times 10^{-y}$
Thus,$x = 25$ and $y = 13$.

(B) The balanced chemical equation for the combustion of glucose is:
$C_6H_{12}O_6 + 6O_2 \longrightarrow 6CO_2 + 6H_2O$
The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \ g/mol$.
From the stoichiometry,$1 \ mol$ of glucose $(180 \ g)$ produces $6 \ mol$ of $CO_2$ $(6 \times 44 = 264 \ g)$ and $6 \ mol$ of $H_2O$ $(6 \times 18 = 108 \ g)$.
For $10 \ g$ of glucose:
Mass of $CO_2 = \frac{264 \ g}{180 \ g} \times 10 \ g = 14.66 \ g$.
Mass of $H_2O = \frac{108 \ g}{180 \ g} \times 10 \ g = 6.0 \ g$.

(C) $1$. Calculate the number of atoms in $6 \,g$ of $H_2O$:
$n(H_2O) = \frac{6 \,g}{18 \,g/mol} = \frac{1}{3} \,mol$.
Since each $H_2O$ molecule has $3$ atoms ($2$ $H$ and $1$ $O$),total atoms $= 3 \times \frac{1}{3} \,N_A = N_A$.
$2$. Evaluate options:
$A$: $n(He) = \frac{0.4}{4} = 0.1 \,mol$. Atoms $= 0.1 \,N_A$.
$B$: $n(CO_2) = \frac{22}{44} = 0.5 \,mol$. Atoms $= 3 \times 0.5 \,N_A = 1.5 \,N_A$.
$C$: $n(H_2) = \frac{1}{2} = 0.5 \,mol$. Atoms $= 2 \times 0.5 \,N_A = N_A$.
$D$: $n(CO) = \frac{12}{28} \approx 0.43 \,mol$. Atoms $= 2 \times 0.43 \,N_A = 0.86 \,N_A$.
Thus,$1 \,g$ of $H_2$ has the same number of atoms.

(D) The sum is $12.0 + 19.034 + 2.0143 = 33.0483$.
According to the rules for addition in significant figures,the result should be reported to the same number of decimal places as the measurement with the fewest decimal places.
Here,$12.0$ has $1$ decimal place,$19.034$ has $3$ decimal places,and $2.0143$ has $4$ decimal places.
The least number of decimal places is $1$.
Rounding $33.0483$ to $1$ decimal place gives $33.0$.
The number $33.0$ has $3$ significant figures.

(B) Using the ideal gas equation,$PV = nRT$,for $1 \ mol$ of the mixture,the volume $V$ is given by $V = \frac{RT}{P}$.
Given $P = 760 \ torr = 1 \ atm$,$T = 300 \ K$,and $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,we have $V = \frac{0.0821 \times 300}{1} = 24.63 \ L$.
The total mass of $24.63 \ L$ of the mixture is $M = V \times d = 24.63 \ L \times 0.543 \ g \ L^{-1} = 13.37 \ g$.
Let $x$ be the mole fraction of $O_2$. Then the mole fraction of $He$ is $(1 - x)$.
The average molar mass of the mixture is $M_{avg} = x(32) + (1 - x)(4) = 13.37$.
Solving for $x$: $32x + 4 - 4x = 13.37$ $\Rightarrow 28x = 9.37$ $\Rightarrow x \approx 0.3346$.
The mass of $O_2$ in $1 \ mol$ of the mixture is $0.3346 \times 32 \approx 10.707 \ g$.
The mass percent of $O_2$ is $\frac{\text{mass of } O_2}{\text{total mass}} \times 100 = \frac{10.707}{13.37} \times 100 \approx 80.08 \% \approx 80 \%$.

(D) Using the ideal gas equation $PV = nRT = \frac{wRT}{M}$,we have $n = \frac{PV}{RT}$.
For gas $A$: $n_A = \frac{5V}{RT} = \frac{4}{M_A} \implies M_A = \frac{4RT}{5V}$.
When gas $B$ is added,the total pressure becomes $10 \ atm$. The partial pressure of gas $B$ is $P_B = P_{total} - P_A = 10 \ atm - 5 \ atm = 5 \ atm$.
For gas $B$: $n_B = \frac{5V}{RT} = \frac{16}{M_B} \implies M_B = \frac{16RT}{5V}$.
Taking the ratio: $\frac{M_A}{M_B} = \frac{4RT/5V}{16RT/5V} = \frac{4}{16} = \frac{1}{4}$.
Therefore,$4 M_A = M_B$.

(B) Using the ideal gas equation: $PV = nRT = \frac{m}{M} RT$
$\therefore m = \frac{PVM}{RT}$
Since both containers contain the same gas $(CO_2)$,the molar mass $(M)$ is constant.
Therefore,$\frac{m_A}{m_B} = \frac{P_A V_A}{P_B V_B} \times \frac{T_B}{T_A}$
Given: $P_A = 4P_B$,$V_A = 4V_B$,$T_A = 4T_B$,and $m_B = x \ g$.
Substituting these values: $\frac{m_A}{x} = \frac{(4P_B)(4V_B)}{P_B V_B} \times \frac{T_B}{4T_B}$
$\frac{m_A}{x} = 16 \times \frac{1}{4} = 4$
$m_A = 4x \ g$

(A) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of its molar mass $(M)$ and inversely proportional to the time taken $(t)$ for the same volume to diffuse:
$r \propto \frac{1}{\sqrt{M}}$ and $r \propto \frac{1}{t}$.
Therefore,$\frac{r_X}{r_{O_2}} = \sqrt{\frac{M_{O_2}}{M_X}} = \frac{t_{O_2}}{t_X}$.
Squaring both sides: $\frac{M_{O_2}}{M_X} = \frac{t_{O_2}^2}{t_X^2} \Rightarrow \frac{M_X}{t_X^2} = \frac{M_{O_2}}{t_{O_2}^2}$.
Given $M_{O_2} = 32 \ g/mol$ and $t_{O_2} = 20 \ s$:
$\frac{M_X}{t_X^2} = \frac{32}{20^2} = \frac{32}{400} = 0.08$.
Checking option $A$: For $H_2$,$M_X = 2$ and $t_X = 5 \ s$.
$\frac{2}{5^2} = \frac{2}{25} = 0.08$.
Since the ratio matches,the correct pair is $H_2$ and $5 \ s$.

(C) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molecular weight $M$ of the gas: $r \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of rates of diffusion of gases $X$ and $Y$ is given by: $\frac{r_x}{r_y} = \sqrt{\frac{M_y}{M_x}}$.
Given $M_x = 36$ and $M_y = 64$,we have: $\frac{r_x}{r_y} = \sqrt{\frac{64}{36}}$.
Simplifying the expression: $\frac{r_x}{r_y} = \frac{8}{6} = \frac{4}{3}$.

(B) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $\frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}}$.
Given that $r_A = \sqrt{5} \ r_B$,we have $\frac{r_A}{r_B} = \sqrt{5}$.
Substituting the values: $\sqrt{5} = \sqrt{\frac{M_B}{x}}$.
Squaring both sides: $5 = \frac{M_B}{x}$.
Therefore,$M_B = 5 \ x \ g \ mol^{-1}$.

(D) The radius of an orbit in a hydrogen-like species is given by the formula: $r_n = a_0 \times \frac{n^2}{Z}$, where $a_0$ is the Bohr radius, $n$ is the principal quantum number, and $Z$ is the atomic number.
For the $3^{rd}$ orbit of $H$ atom $(n=3, Z=1)$: $R = a_0 \times \frac{3^2}{1} = 9 a_0$.
For the $2^{nd}$ orbit of $He^{+}$ ion $(n=2, Z=2)$: $r_2 = a_0 \times \frac{2^2}{2} = 2 a_0$.
From the first equation, $a_0 = \frac{R}{9}$.
Substituting this into the second equation: $r_2 = 2 \times (\frac{R}{9}) = \frac{2}{9} R$.

(C) The radius of the $n^{\text{th}}$ orbit of a hydrogen-like species is given by the formula: $r_n = 0.529 \times \frac{n^2}{Z} \mathring{A}$.
For the first orbit of the hydrogen atom $(n=1, Z=1)$,the radius is $r_1 = 0.529 \mathring{A}$.
For the radius of the $n^{\text{th}}$ orbit of species $X$ to be equal to the first Bohr orbit of hydrogen,we set $r_n = r_1$,which implies $\frac{n^2}{Z} = 1$,or $n^2 = Z$.
Testing the options:
For $Be^{3+}$,the atomic number $Z = 4$. Substituting $Z=4$ into $n^2 = Z$,we get $n^2 = 4$,which gives $n = 2$.
Thus,for $n = 2$ and $X = Be^{3+}$,the condition is satisfied.

(C) The energy of an electron in a hydrogen atom is given by the formula: $E_n = \frac{-13.6 \ eV}{n^2}$.
For $n=1$,$E_1 = \frac{-13.6}{1^2} = -13.6 \ eV$.
For $n=2$,$E_2 = \frac{-13.6}{2^2} = -3.4 \ eV$.
For $n=3$,$E_3 = \frac{-13.6}{3^2} = -1.51 \ eV$.
The ratio $E_1 : E_2 : E_3$ is $\frac{-13.6}{1} : \frac{-13.6}{4} : \frac{-13.6}{9}$.
Dividing by $-13.6$,we get $1 : \frac{1}{4} : \frac{1}{9}$.
Multiplying by $36$ to clear the denominators,we get $36 : 9 : 4$.

(B) The energy of an orbit in a hydrogen-like species is given by the formula $E_n = -E_0 \times \frac{Z^2}{n^2}$,where $E_0$ is the ground state energy of hydrogen $(2.18 \times 10^{-18} \ J)$.
For the second orbit of hydrogen $(n=2, Z=1)$: $E_2 = -2.18 \times 10^{-18} \times \frac{1^2}{2^2} = -5.45 \times 10^{-19} \ J$.
For the first orbit of $Li^{2+}$ ion $(n=1, Z=3)$:
$E_1 = -2.18 \times 10^{-18} \times \frac{3^2}{1^2} \ J$
$E_1 = -2.18 \times 10^{-18} \times 9 \ J$
$E_1 = -1.962 \times 10^{-17} \ J$.

(B) The energy of the $n^{th}$ Bohr orbit is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
For the hydrogen atom $(Z=1)$,the energy difference between orbits $n_1$ and $n_2$ is $\Delta E = 13.6 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \text{ eV}$.
Difference in energy between the first $(n=1)$ and second $(n=2)$ orbits:
$\Delta E_{1-2} = 13.6 \times (\frac{1}{1^2} - \frac{1}{2^2}) = 13.6 \times (1 - \frac{1}{4}) = 13.6 \times \frac{3}{4} \text{ eV}$.
Difference in energy between the second $(n=2)$ and third $(n=3)$ orbits:
$\Delta E_{2-3} = 13.6 \times (\frac{1}{2^2} - \frac{1}{3^2}) = 13.6 \times (\frac{1}{4} - \frac{1}{9}) = 13.6 \times \frac{5}{36} \text{ eV}$.
The required ratio is $\frac{\Delta E_{1-2}}{\Delta E_{2-3}} = \frac{13.6 \times (3/4)}{13.6 \times (5/36)} = \frac{3}{4} \times \frac{36}{5} = \frac{27}{5}$.

(A) The Rydberg formula is given by $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the second line of the Balmer series of the $H$-atom $(Z=1, n_1=2, n_2=4)$:
$\frac{1}{\lambda} = R(1)^2 \left[ \frac{1}{2^2} - \frac{1}{4^2} \right] = R \left[ \frac{1}{4} - \frac{1}{16} \right] = \frac{3R}{16}$ $\Rightarrow R = \frac{16}{3\lambda}$.
For the first line of the Lyman series of the $He^{+}$ ion $(Z=2, n_1=1, n_2=2)$:
$\frac{1}{\lambda_2} = R(2)^2 \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = 4R \left[ 1 - \frac{1}{4} \right] = 4R \left( \frac{3}{4} \right) = 3R$.
Substituting $R = \frac{16}{3\lambda}$ into the expression for $\lambda_2$:
$\frac{1}{\lambda_2} = 3 \left( \frac{16}{3\lambda} \right) = \frac{16}{\lambda}$ $\Rightarrow \lambda_2 = \frac{\lambda}{16}$.

(B) Given threshold frequency,$\nu_0 = 1.0 \times 10^{15} \ s^{-1}$.
According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E._{max})$ is given by $K.E._{max} = h(\nu - \nu_0)$,where $\nu$ is the frequency of incident radiation.
For $\nu_1 = 1.5 \times 10^{15} \ s^{-1}$:
$K.E._1 = h(1.5 \times 10^{15} - 1.0 \times 10^{15}) = h(0.5 \times 10^{15})$.
For $\nu_2 = 2.0 \times 10^{15} \ s^{-1}$:
$K.E._2 = h(2.0 \times 10^{15} - 1.0 \times 10^{15}) = h(1.0 \times 10^{15})$.
The ratio of maximum kinetic energies is:
$\frac{K.E._1}{K.E._2} = \frac{h(0.5 \times 10^{15})}{h(1.0 \times 10^{15})} = \frac{0.5}{1.0} = 1:2$.

(C) The angular momentum $(L)$ of an electron in a hydrogen atom is given by Bohr's postulate:
$L = \frac{nh}{2\pi}$
For the ground state,the principal quantum number $n = 1$.
Substituting the values:
$L = \frac{1 \times 6.625 \times 10^{-34}}{2 \times 3.14159} \ J \ s$
$L \approx \frac{6.625 \times 10^{-34}}{6.283} \ J \ s$
$L \approx 1.0545 \times 10^{-34} \ J \ s$
To match the options,we express this as:
$1.0545 \times 10^{-34} = 1055 \times 10^{-37} \ J \ s$
Therefore,the correct option is $C$.

(C) Power $(P) = 100 \ W = 100 \ J \cdot s^{-1}$.
Number of photons emitted per second $(n) = 2.0 \times 10^{20} \ s^{-1}$.
Energy of one photon $(E) = \frac{hc}{\lambda}$.
Total power emitted is given by $P = n \times E = n \times \frac{hc}{\lambda}$.
Substituting the values: $100 = \frac{2.0 \times 10^{20} \times 6.63 \times 10^{-34} \times 3 \times 10^8}{\lambda}$.
$\lambda = \frac{2.0 \times 6.63 \times 3 \times 10^{20-34+8}}{100} \ m$.
$\lambda = \frac{39.78 \times 10^{-6}}{100} \ m = 39.78 \times 10^{-8} \ m$.
Converting to $\mathring{A}$: $\lambda = 39.78 \times 10^{-8} \times 10^{10} \ \mathring{A} = 3978 \ \mathring{A}$.
Thus,$x = 3978$.

(A) Statement $I$ is incorrect because the electron spin quantum number describes the orientation of the spin of the electron,not the nucleus.
Statement $II$ is correct because orbitals with the same $n$ and $l$ values (degenerate orbitals) have the same energy in the absence of an external magnetic field.
Statement $III$ is incorrect because the energy of a photon $(E = h\nu = \frac{hc}{\lambda})$ is inversely proportional to the wavelength $(\lambda)$ and directly proportional to the wave number $(\bar{\nu} = \frac{1}{\lambda})$.
Statement $IV$ is correct because the Lyman series corresponds to electronic transitions to the $n=1$ energy level,which emits radiation in the ultra-violet region.
Therefore,statements $II$ and $IV$ are correct.

(C) The rules for quantum numbers are:
$1$. The principal quantum number '$n$' can be any positive integer $(1, 2, 3, \dots)$.
$2$. The azimuthal quantum number '$l$' can have values from $0$ to $(n-1)$.
$3$. The magnetic quantum number '$m$' can have values from $-l$ to $+l$ including $0$.
$4$. The spin quantum number '$s$' can be $\pm\frac{1}{2}$.
In option $C$,$n=3$,so the maximum value of '$l$' is $(n-1) = 2$.
Since the given value is $l=3$,which is not possible for $n=3$,this set is impossible.

(C) The atomic number $Z = 24$ corresponds to Chromium $(Cr)$.
The electronic configuration of $Cr$ is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$.
For $(n+\ell) = 3$: The orbitals are $2p$ $(n=2, \ell=1)$ and $3s$ $(n=3, \ell=0)$. Total electrons $= 6 + 2 = 8$.
For $(n+\ell) = 4$: The orbitals are $3p$ $(n=3, \ell=1)$ and $4s$ $(n=4, \ell=0)$. Total electrons $= 6 + 1 = 7$.
For $(n+\ell) = 5$: The orbital is $3d$ $(n=3, \ell=2)$. Total electrons $= 5$.

(C) The first ionisation enthalpy of oxygen is less than that of nitrogen.
This is due to the extra stability of half-filled electronic configurations.
Nitrogen $(1s^2 2s^2 2p^3)$ has a half-filled $p$-orbital,which is more stable,and hence its ionisation energy is greater than that of oxygen $(1s^2 2s^2 2p^4)$.
Therefore,the assertion $(A)$ is true,but the reason $(R)$ is false because half-filled or completely filled orbitals are more stable,not less stable.

(A) Intensive properties are those properties of a system that do not depend on the quantity or size of matter present in the system.
Analyzing the given list:
$I$. Molar volume: Intensive (volume per mole).
$II$. Mass: Extensive.
$III$. Internal energy: Extensive.
$IV$. Volume: Extensive.
$V$. Enthalpy: Extensive.
$VI$. Temperature: Intensive.
$VII$. Density: Intensive (mass per unit volume).
Therefore,the intensive properties are $I, VI,$ and $VII$.

(C) The formula for reversible isothermal work is: $W = -2.303 nRT \log \left(\frac{V_2}{V_1}\right)$
Given: $n = 2.5 \ mol$,$V_1 = 2 \ dm^3$,$V_2 = 20 \ dm^3$,$W = -16.5 \ kJ = -16500 \ J$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$
Substituting the values: $-16500 = -2.303 \times 2.5 \times 8.314 \times T \times \log \left(\frac{20}{2}\right)$
$-16500 = -2.303 \times 2.5 \times 8.314 \times T \times \log(10)$
Since $\log(10) = 1$,we have: $-16500 = -2.303 \times 2.5 \times 8.314 \times T$
$-16500 = -47.87 \times T$
$T = \frac{16500}{47.87} \approx 344.68 \ K$
Rounding to the nearest value,$T \approx 345 \ K$.

(D) For a reversible isothermal expansion,the work done is given by the formula: $W = -nRT \ln \frac{P_1}{P_2} = -2.303 nRT \log \frac{P_1}{P_2}$.
Given: $n = 2 \ mol$,$P_1 = 10 \ atm$,$P_2 = 1 \ atm$,$R = 8.3 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values: $W = -2.303 \times 2 \times 8.3 \times T \times \log \frac{10}{1}$.
Since $\log 10 = 1$,we get $W = -2.303 \times 2 \times 8.3 \times T \times 1 = -38.2298 \times T \ J$.
To convert to $kJ$,divide by $1000$: $W = -38.2298 \times 10^{-3} \times T \ kJ \approx -3.82 \times 10^{-2} \times T \ kJ$.

(B) The enthalpy change of the reaction is calculated using the formula: $\Delta H_{r}^{\circ} = \sum \Delta H_{f}^{\circ}(\text{products}) - \sum \Delta H_{f}^{\circ}(\text{reactants})$.
For the reaction $N_2O_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow 2 NO_{(g)}$,the expression is: $\Delta H_{r}^{\circ} = [2 \times \Delta H_{f}^{\circ}(NO)] - [\Delta H_{f}^{\circ}(N_2O) + \frac{1}{2} \times \Delta H_{f}^{\circ}(O_2)]$.
Given that $\Delta H_{f}^{\circ}(O_2) = 0 \ kJ \ mol^{-1}$ (standard state of an element),$\Delta H_{f}^{\circ}(N_2O) = 82.0 \ kJ \ mol^{-1}$,and $\Delta H_{f}^{\circ}(NO) = 90.0 \ kJ \ mol^{-1}$.
Substituting the values: $\Delta H_{r}^{\circ} = [2 \times 90.0] - [82.0 + 0] = 180.0 - 82.0 = +98.0 \ kJ$.

(A) The combustion reaction of $CO_{(g)}$ is given by:
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}$
For the standard enthalpy of combustion,we use the formula:
$\Delta H_{comb}^{\circ} = \sum \Delta H_{f, products}^{\circ} - \sum \Delta H_{f, reactants}^{\circ}$
Given that $\Delta H_{f}^{\circ}$ for $O_{2(g)} = 0 \ kJ \ mol^{-1}$,$\Delta H_{f}^{\circ}$ for $CO_{(g)} = -110 \ kJ \ mol^{-1}$,and $\Delta H_{f}^{\circ}$ for $CO_{2(g)} = -393 \ kJ \ mol^{-1}$.
Substituting these values:
$\Delta H_{comb}^{\circ} = \Delta H_{f}^{\circ}(CO_{2}) - [\Delta H_{f}^{\circ}(CO) + \frac{1}{2} \Delta H_{f}^{\circ}(O_{2})]$
$\Delta H_{comb}^{\circ} = -393 - [-110 + 0]$
$\Delta H_{comb}^{\circ} = -393 + 110 = -283 \ kJ \ mol^{-1}$.

(A) Given: $\Delta U = 2.1 \ kCal = 2100 \ cal$,$\Delta S = 20 \ cal \ K^{-1}$,$T = 25 + 273 = 298 \ K$,$R = 2 \ cal \ K^{-1} \ mol^{-1}$,$\Delta n_g = 2 - 0 = 2$.
$\Delta H = \Delta U + \Delta n_g RT = 2100 + (2 \times 2 \times 298) = 2100 + 1192 = 3292 \ cal$.
$\Delta G = \Delta H - T \Delta S = 3292 - (298 \times 20) = 3292 - 5960 = -2668 \ cal$.
$\Delta G = -2.668 \ kCal \approx -2.67 \ kCal$.

(B, C) Statement $(I)$ is incorrect because both $Au$ and $Pt$ dissolve in aqua regia to form soluble compounds $HAuCl_4$ and $H_2PtCl_6$.
Statement $(II)$ is correct because in perchloric acid $(HClO_4)$,chlorine exhibits an oxidation state of $+7$.
Statement $(III)$ is correct because $HCl$ has the lowest boiling point among hydrogen halides due to the absence of hydrogen bonding and lower molecular mass compared to $HBr$ and $HI$.
Statement $(IV)$ is incorrect because the stability of halogen oxides decreases in the order $I > Cl > Br$.

(B) Deacon's method involves the preparation of $Cl_2$ from $HCl$ gas by oxidation at about $723 \ K$ in the presence of a catalyst like $CuCl_2$.
The chemical reaction is:
$4 HCl + O_2 \xrightarrow[723 \ K]{CuCl_2} 2 H_2 O + 2 Cl_2$

(A) $Cl_2$ reacts with $CO$ to form phosgene $(COCl_2)$: $CO + Cl_2 \rightarrow COCl_2$.
$I_2O_5$ is a strong oxidizing agent used for the quantitative estimation of carbon monoxide $(CO)$: $I_2O_5 + 5CO \rightarrow I_2 + 5CO_2$.
$O_3$ (Ozone) acts as a powerful disinfectant and bleaching agent due to its ability to release nascent oxygen.
Therefore,all three pairs are correctly matched.

(D) The thermal stability of $OF_2$ at room temperature is due to the small difference between the electronegativities of $O$ and $F$.
The order of stability of the oxides of halogens is $I > Cl > Br$.
$I_2O_5$ is a very good oxidizing agent and is used for the estimation of $CO$.
$ClO_2$ is a well-known bleaching agent.
Thus,statements $A$,$C$,and $D$ are correct.

(C) Interhalogen compounds are generally more reactive than halogens (except $F_2$) because the $X-X'$ bond is weaker than the $X-X$ bond.
They are diamagnetic in nature as all electrons are paired.
During the hydrolysis of interhalogen compounds,the less electronegative halogen forms an oxyacid,and the more electronegative halogen forms a hydrohalic acid.
For $ICl$,the reaction is: $ICl + H_2O \rightarrow HCl + HOI$.
Therefore,the statement that the products are $HI + HOCl$ is incorrect.

(A) Elastomers are polymers that have weak intermolecular forces of attraction,allowing them to be stretched.
These include: Natural rubber,vulcanized rubber,Buna-$N$,and Neoprene.
Therefore,there are $4$ elastomers in the given list.

(A) Statement $(I)$ is correct: Starch is a polymer of $\alpha-D(+)$-glucose.
Statement $(II)$ is incorrect: Amylose is the water-soluble component of starch.
Statement $(III)$ is incorrect: Amylose is a long unbranched chain polymer of $\alpha-D(+)$-glucose.
Statement $(IV)$ is correct: Cellulose is a straight chain polymer of $\beta-D(+)$-glucose units.
Therefore,the correct statements are $(I)$ and $(IV)$.

(B) The correct matching is as follows:
$(A)$ Fibre: $(IV)$ Dacron
$(B)$ Elastomer: $(III)$ Neoprene
$(C)$ Thermosetting polymer: $(I)$ Bakelite
$(D)$ Thermoplastic polymer: $(II)$ Polystyrene
Therefore,the correct sequence is $A-IV, B-III, C-I, D-II$.

(B) Bakelite is a condensation polymer of phenol and formaldehyde. Thus,$A \rightarrow III$.
Natural rubber is an addition polymer of $2$-methyl-$1,3$-butadiene (isoprene). Thus,$B \rightarrow I$.
Glyptal is a condensation polymer of phthalic acid and ethylene glycol. Thus,$C \rightarrow IV$.
Nylon $2$-Nylon $6$ is a condensation polymer of glycine and aminocaproic acid. Thus,$D \rightarrow II$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(A) Glyptal is a condensation polymer of ethylene glycol and phthalic acid.
Bakelite is a thermosetting polymer that is used in making electrical switches and other electrical appliances.
Nylon-$2$-nylon-$6$ is a biodegradable polymer.
Thus,all the statements are correct.

(A) The reaction sequence shows the conversion of $p$-nitrotoluene to terephthalic acid $(R)$.
$R$ is terephthalic acid $(C_6H_4(COOH)_2)$.
Terephthalic acid and ethylene glycol $(HOCH_2CH_2OH)$ are the monomers used for the synthesis of the polymer Dacron (also known as Terylene or Polyethylene terephthalate).

(C) Ziegler-Natta catalyst is a mixture of $TiCl_4$ and $Al(C_2H_5)_3$.
It is specifically used for the polymerization of ethene to produce $HDPE$ (High Density Polythene) under low pressure conditions.

(D) The reducing power of oxides of group $16$ elements decreases down the group.
This is because the stability of the $+4$ oxidation state increases down the group due to the inert pair effect.
$SO_2$ acts as a reducing agent because $S$ can easily be oxidized from $+4$ to $+6$ oxidation state.
In contrast,$TeO_2$ is more stable in the $+4$ state and does not easily oxidize to $+6$.
Therefore,$SO_2$ is the strongest reducing agent among the given options.

$(A)$ For a crystal system, the parameters $a = b \neq c$ and $\alpha = \beta = \gamma = 90^{\circ}$ correspond to the tetragonal crystal system.
In this case, $a = b = 200 \text{ pm}$ and $c = 300 \text{ pm}$, which satisfies the condition $a = b \neq c$.
Therefore, the correct crystal system is tetragonal.

(C) The density $(d)$ of a unit cell is given by the formula: $d = \frac{Z \times M}{N_A \times a^3}$
For $fcc$ structure,the number of atoms per unit cell $(Z)$ is $4$.
The edge length $(a)$ is $x \ \mathring{A} = x \times 10^{-8} \ cm$.
Given $M = 63.5 \ g \ mol^{-1}$ and $N_A = 6.0 \times 10^{23} \ mol^{-1}$.
Substituting these values:
$d = \frac{4 \times 63.5}{(6.0 \times 10^{23}) \times (x \times 10^{-8})^3} \ g \ cm^{-3}$
$d = \frac{254}{6.0 \times 10^{23} \times x^3 \times 10^{-24}} \ g \ cm^{-3}$
$d = \frac{254}{0.6 \times x^3} \ g \ cm^{-3} = \frac{423.33}{x^3} \ g \ cm^{-3}$
Thus,the approximate density is $\frac{423}{x^3} \ g \ cm^{-3}$.

(B) Number of $X$ atoms per unit cell $= 1 \times 1 = 1$.
Number of $Y$ atoms per unit cell $= (8 - 1) \times \frac{1}{8} = \frac{7}{8}$.
Ratio of $X : Y = 1 : \frac{7}{8} = 8 : 7$.
Therefore,the empirical formula is $X_8 Y_7$.

(C) Let the number of oxygen atoms in the $ccp$ lattice be $1$.
Since the number of octahedral voids is equal to the number of atoms,the number of octahedral voids is $1$.
The number of tetrahedral voids is twice the number of atoms,so the number of tetrahedral voids is $2$.
Atoms of $A$ occupy $\frac{1}{8}$ of the tetrahedral voids,so the number of $A$ atoms $= \frac{1}{8} \times 2 = \frac{1}{4}$.
Atoms of $B$ occupy half of the octahedral voids,so the number of $B$ atoms $= \frac{1}{2} \times 1 = \frac{1}{2}$.
The ratio of $A:B:O$ is $\frac{1}{4} : \frac{1}{2} : 1$.
Multiplying the ratio by $4$,we get $1 : 2 : 4$.
Thus,the molecular formula of the compound is $AB_2O_4$.

(A) Contribution of $W$ atoms at the corners $= 8 \times \frac{1}{8} = 1$.
Contribution of $O$ atoms at the edges $= 12 \times \frac{1}{4} = 3$.
Contribution of $Na$ atoms at the body center $= 1 \times 1 = 1$.
Thus,the ratio of atoms $Na : W : O$ is $1 : 1 : 3$.
Therefore,the formula of the compound is $NaWO_3$.

(A) According to Raoult's Law for an ideal solution:
$P_{total} = x_A P_A^{\circ} + x_B P_B^{\circ}$
Given:
$P_A^{\circ} = 70 \ torr$
$x_B = 0.2$
$x_A = 1 - 0.2 = 0.8$
$P_{total} = 84 \ torr$
Substituting the values into the equation:
$84 = (0.8 \times 70) + (0.2 \times P_B^{\circ})$
$84 = 56 + 0.2 P_B^{\circ}$
$0.2 P_B^{\circ} = 84 - 56$
$0.2 P_B^{\circ} = 28$
$P_B^{\circ} = \frac{28}{0.2} = 140 \ torr$

(D) Given: $P_A^0 = 50 \ mm \ Hg$,$P_B^0 = 32 \ mm \ Hg$.
Mole fraction in liquid phase: $X_A = \frac{1}{1+1} = 0.5$ and $X_B = \frac{1}{1+1} = 0.5$.
Partial pressures: $P_A = P_A^0 \times X_A = 50 \times 0.5 = 25 \ mm \ Hg$ and $P_B = P_B^0 \times X_B = 32 \times 0.5 = 16 \ mm \ Hg$.
Total pressure: $P_{total} = P_A + P_B = 25 + 16 = 41 \ mm \ Hg$.
Mole fraction of $A$ in vapour phase $(Y_A)$: $Y_A = \frac{P_A}{P_{total}} = \frac{25}{41} \approx 0.6097 \approx 0.61$.

(B) The osmotic pressure formula is given by $\pi = iCRT$.
Given: $\pi = 4.82 \ atm$,$C = 0.1 \ M$ (for $NaCl$,$N = M$),$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$,and $T = 300 \ K$.
Calculating the van't Hoff factor $(i)$:
$i = \frac{\pi}{CRT} = \frac{4.82}{0.1 \times 0.082 \times 300} = \frac{4.82}{2.46} \approx 1.959 \approx 1.96$.
For $NaCl$,the dissociation is $NaCl \rightarrow Na^+ + Cl^-$,so $n = 2$.
The degree of dissociation $(\alpha)$ is related to $i$ by $i = 1 + \alpha(n - 1)$.
$1.96 = 1 + \alpha(2 - 1) \Rightarrow \alpha = 0.96$.
Since $\alpha = x \times 10^{-2}$,we have $0.96 = x \times 10^{-2}$,which gives $x = 96$.

(A) The elevation in boiling point is given by $\Delta T_b = T_b - T_b^{\circ} = K_b \times m$,where $m$ is the molality of the solution.
$\Delta T_b = 373.202 \ K - 373.15 \ K = 0.052 \ K$.
Using the formula $\Delta T_b = K_b \times \frac{x \times 1000}{y \times M_{\text{urea}}}$,where $M_{\text{urea}} = 60 \ g \ mol^{-1}$.
$0.052 = 0.52 \times \frac{x \times 1000}{y \times 60}$.
$0.052 = \frac{x}{y} \times \frac{520}{60}$.
$\frac{x}{y} = \frac{0.052 \times 60}{520} = \frac{3.12}{520} = 0.006 = 6.0 \times 10^{-3}$.

(C) According to Raoult's law for a solution containing a non-volatile solute,the vapour pressure of the solution is given by $P = P^{\circ} \cdot x_{\text{solvent}}$.
Here,$P = 167 \ torr$ and $P^{\circ} = 268 \ torr$.
So,the mole fraction of benzene is $x_{\text{benzene}} = \frac{P}{P^{\circ}} = \frac{167}{268} \approx 0.6231$.
Since $x_{\text{benzene}} = \frac{n_{\text{benzene}}}{n_{\text{benzene}} + n_{\text{solute}}}$,we set $n_{\text{benzene}} = 1 \ mol$.
Then,$0.6231 = \frac{1}{1 + n_{\text{solute}}}$.
$1 + n_{\text{solute}} = \frac{1}{0.6231} \approx 1.6048$.
$n_{\text{solute}} = 1.6048 - 1 = 0.6048 \ mol \approx 0.605 \ mol$.

(A) The elevation in boiling point is given by $\Delta T_{b} = T_{b} - T_{b}^{\circ} = 100.17^{\circ} C - 100^{\circ} C = 0.17^{\circ} C$.
Using the formula $\Delta T_{b} = K_{b} \times m$,where $m$ is the molality:
$0.17 = 0.512 \times m \implies m = \frac{0.17}{0.512} \ mol \ kg^{-1}$.
The depression in freezing point is given by $\Delta T_{f} = K_{f} \times m$.
$\Delta T_{f} = 1.86 \times \frac{0.17}{0.512} \approx 0.6176^{\circ} C \approx 0.62^{\circ} C$.
The freezing point of the solution is $T_{f} = T_{f}^{\circ} - \Delta T_{f} = 0^{\circ} C - 0.62^{\circ} C = -0.62^{\circ} C$.

(B) The weight average molecular mass is calculated using the formula: $\bar{M}_w = \frac{\sum n_i M_i^2}{\sum n_i M_i}$
Given:
$n_1 = 3, M_1 = 1000$
$n_2 = 3, M_2 = 500$
$n_3 = 4, M_3 = 200$
Numerator: $(3 \times 1000^2) + (3 \times 500^2) + (4 \times 200^2) = 3,000,000 + 750,000 + 160,000 = 3,910,000$
Denominator: $(3 \times 1000) + (3 \times 500) + (4 \times 200) = 3000 + 1500 + 800 = 5300$
$\bar{M}_w = \frac{3,910,000}{5300} \approx 737.7$

(A) Let $x$ be the number of $M^{3+}$ ions and $y$ be the number of $M^{2+}$ ions.
Total number of metal ions is $x + y = 0.96$.
Therefore,$y = 0.96 - x$.
Since the oxide is electrically neutral,the total positive charge must equal the total negative charge ($-2$ for $O^{2-}$).
$3x + 2y = 2$.
Substituting $y = 0.96 - x$ into the equation:
$3x + 2(0.96 - x) = 2$.
$3x + 1.92 - 2x = 2$.
$x = 0.08$.
Thus,$y = 0.96 - 0.08 = 0.88$.
The fraction of $M^{3+} = \frac{0.08}{0.96} = 0.083$.
The fraction of $M^{2+} = \frac{0.88}{0.96} = 0.917$.

(A) An ion has zero magnetic moment if it has no unpaired electrons (diamagnetic).
$V^{2+} = [Ar] 3d^3$ ($3$ unpaired electrons)
$Zn^{2+} = [Ar] 3d^{10}$ ($0$ unpaired electrons)
$Cu^{2+} = [Ar] 3d^9$ ($1$ unpaired electron)
$Fe^{2+} = [Ar] 3d^6$ ($4$ unpaired electrons)
$Fe^{3+} = [Ar] 3d^5$ ($5$ unpaired electrons)
$Ti^{3+} = [Ar] 3d^1$ ($1$ unpaired electron)
$Sc^{3+} = [Ar] 3d^0$ ($0$ unpaired electrons)
$Ti^{4+} = [Ar] 3d^0$ ($0$ unpaired electrons)
$Ni^{3+} = [Ar] 3d^7$ ($3$ unpaired electrons)
$Co^{3+} = [Ar] 3d^6$ ($4$ unpaired electrons)
$Cu^{+} = [Ar] 3d^{10}$ ($0$ unpaired electrons)
The ions with zero magnetic moment are $Zn^{2+}$,$Sc^{3+}$,$Ti^{4+}$,and $Cu^+$.
There are $4$ such ions.

(A) $[CoF_6]^{3-}: Co$ is in $+3$ oxidation state $(3d^6)$. $F^-$ is a weak-field ligand,so electrons remain unpaired $(n=4)$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \ B.M.$
$[Co(C_2O_4)_3]^{3-}: Co$ is in $+3$ oxidation state $(3d^6)$. $C_2O_4^{2-}$ is a strong-field ligand,so electrons pair up $(n=0)$. $\mu = 0 \ B.M.$
$[FeF_6]^{3-}: Fe$ is in $+3$ oxidation state $(3d^5)$. $F^-$ is a weak-field ligand,so electrons remain unpaired $(n=5)$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \ B.M.$
$[Mn(CN)_6]^{3-}: Mn$ is in $+3$ oxidation state $(3d^4)$. $CN^-$ is a strong-field ligand,causing pairing $(n=2)$. $\mu = \sqrt{2(2+2)} = \sqrt{8} \ B.M.$
Thus,the correct match is $A-II, B-I, C-IV, D-III$.

(B) Physical adsorption (physisorption) is an exothermic process,so it is favoured at low temperatures according to Le Chatelier's principle.
High surface area of the adsorbent provides more active sites for the accumulation of adsorbate molecules.
High pressure increases the concentration of gas molecules on the surface of the adsorbent,thereby increasing the extent of adsorption.
Therefore,factors $(I)$,$(II)$,and $(V)$ favour physical adsorption.

(B) Adsorption is an exothermic process. According to Le Chatelier's principle,for physical adsorption,the extent of adsorption $(x / m)$ decreases with an increase in temperature at a constant pressure.
From the given graph,at a constant pressure,the extent of adsorption follows the order: $T_2 > T_3 > T_4$.
Since the extent of adsorption is inversely proportional to temperature for physisorption,the correct order of temperatures is $T_4 > T_3 > T_2$.

(B) Sodium lauryl sulphate $(CH_3(CH_2)_{11}SO_4^-Na^+)$ is an anionic surfactant.
It consists of a long hydrophobic hydrocarbon chain and a hydrophilic ionic head.
In water,these molecules aggregate to form ionic micelles above the Critical Micelle Concentration $(CMC)$.

(A) Bredig's Arc (Electrical disintegration) method is used to prepare colloidal sols of metals such as gold,silver,platinum,etc.
In this method,an electric arc is struck between electrodes of the metal immersed in the dispersion medium.
The intense heat produced vaporizes the metal,which then condenses to form particles of the colloidal size.
Metal sols typically acquire a negative charge due to the preferential adsorption of anions from the dispersion medium.

(D) The Tyndall effect is the scattering of light by colloidal particles.
Statement $(I)$ is correct: It is a standard method to distinguish between a true solution (which does not show the effect) and a colloidal solution (which does).
Statement $(II)$ is correct: The scattering of light depends on the difference in refractive indices between the dispersed phase and the dispersion medium.
Statement $(III)$ is incorrect: The Tyndall effect is observed when the diameter of the dispersed particles is not much smaller than the wavelength of the light used; if particles are too small,they do not scatter light effectively.
Therefore,statements $(I)$ and $(II)$ are correct.

(B) . Negatively charged sol: When $FeCl_3$ is added to excess $NaOH$ solution,it forms a negatively charged $Fe(OH)_3$ sol due to the adsorption of $OH^-$ ions. Thus,$A-III$.
$B$. Milk: Milk is a well-known example of an emulsion (liquid in liquid). Thus,$B-I$.
$C$. Gold number: It is a measure of the protective power of a lyophilic colloid. Thus,$C-IV$.
$D$. Colloidal antimony: It is used in the treatment of the disease kala-azar. Thus,$D-II$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(A) The Ellingham diagram is a graphical representation of the variation of the change in standard Gibbs free energy $(\Delta G^{\circ})$ with temperature $(T)$.
Thus,$X = \Delta G^{\circ}$ and $Y = T$.