AP EAMCET 2020 Chemistry Question Paper with Answer and Solution

(A) $(I)$ According to Dalton's Law of Partial Pressures: $P_{total} = P_A + P_B + P_C$
$\Rightarrow P_C = P_{total} - (P_A + P_B) = 10 - (3 + 1) = 6 \,atm$
$(II)$ The partial pressure of a gas is given by $P_i = \chi_i \times P_{total} = (n_i / n_{total}) \times P_{total}$
$\Rightarrow P_C = (n_C / 10) \times 10 = n_C$
$\Rightarrow n_C = 6 \,mol$
$(III)$ Weight of $C = n_C \times \text{Molar Mass of } C = 6 \,mol \times 2 \,g/mol = 12 \,g$

(B) According to Dalton's law of partial pressures,the partial pressure of a gas in a mixture is equal to the product of its mole fraction and the total pressure of the mixture.
Mathematically,this is expressed as: $p_i = \chi_i \times p_{\text{total}}$.

(B) According to Graham's law of effusion,the rate of effusion $(r)$ of a gas is inversely proportional to the square root of its molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
For two gases $CO$ and $N_2$ at the same temperature and pressure,the ratio of their rates of effusion is given by: $\frac{r_{CO}}{r_{N_2}} = \sqrt{\frac{M_{N_2}}{M_{CO}}}$.
The molar mass of $CO$ is $12 + 16 = 28 \ g/mol$.
The molar mass of $N_2$ is $14 \times 2 = 28 \ g/mol$.
Substituting these values: $\frac{r_{CO}}{r_{N_2}} = \sqrt{\frac{28}{28}} = \sqrt{1} = 1$.
Thus,the ratio is $1: 1$.

(B) $Si$ $(Silicon)$ and $Ge$ $(Germanium)$ are well-known semiconductors.
$As$ $(Arsenic)$ is a metalloid often used as a dopant in semiconductors.
$Pb$ $(Lead)$ is a metal and acts as a conductor,not a semiconductor.

(A) The most reactive non-metal is $B$ because it has the highest negative electron gain enthalpy $(-328 \ kJ/mol)$.
The most reactive metal is $A$ because it has the lowest first ionization enthalpy $(419 \ kJ/mol)$.
The least reactive element is $D$ because it has a very high first ionization enthalpy and a positive electron gain enthalpy (noble gas characteristic).
The metal forming a binary halide is $C$ (e.g.,alkaline earth metal like $Mg$),which has comparable successive ionization energies.

(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r_1/r_2 = \sqrt{M_2/M_1}$.
Given $r_{He}/r_x = 4$,where $M_{He} = 4 \ g/mol$.
Substituting the values: $4 = \sqrt{M_x/4}$.
Squaring both sides: $16 = M_x/4$.
Therefore,$M_x = 16 \times 4 = 64 \ g/mol$.
Comparing the molar masses of the options:
$M(ClO_2) = 35.5 + 2(16) = 67.5 \ g/mol$
$M(SO_2) = 32 + 2(16) = 64 \ g/mol$
$M(CO_2) = 12 + 2(16) = 44 \ g/mol$
$M(NO_2) = 14 + 2(16) = 46 \ g/mol$
The gas with a molar mass of $64 \ g/mol$ is $SO_2$.

(C) The ideal gas equation is $PV = nRT$,which can be written as $PV = (m/M)RT$.
Rearranging for density $(d = m/V)$,we get $d = (PM) / (RT)$.
Given $P = 76 \ cm \ Hg = 1 \ atm$,$T = 273 \ K$,$d = 1.94 \ g/dm^3$ (which is $1.94 \ g/L$),and $R = 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1}$.
Substituting the values: $1.94 = (1 \times M) / (0.0821 \times 273)$.
$M = 1.94 \times 0.0821 \times 273 \approx 43.5 \ g/mol$.
The molar mass of $CO_2$ is $12 + 2 \times 16 = 44 \ g/mol$.
Thus,the gas is $CO_2$.

(B) Gas molecules are in constant random motion with high velocities and change direction upon collision with other molecules. Therefore,there is a complete disorder of molecules in gases.

(C) $(I)$ Gases mix evenly without the help of any mechanical change. The large intermolecular spaces allow molecules of different gases to mix easily.
Thus,statement $(I)$ is correct.
$(II)$ Due to the negligible intermolecular forces of attraction,gas molecules move randomly in all directions and collide with the walls of the container,exerting pressure in all directions.
Thus,statement $(II)$ is correct.
$(III)$ Gases are highly compressible because their molecules are separated by large distances compared to solids and liquids,and they have lower density.
Thus,statement $(III)$ is correct.
$(IV)$ Gases have neither a fixed shape nor a fixed volume; they occupy the entire space available to them.
Thus,statement $(IV)$ is incorrect.
Therefore,statements $(I)$,$(II)$,and $(III)$ are correct,making option $(C)$ the correct choice.

(B) The formula for rms velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
For oxygen $(O_2)$,$T_1 = 27 + 273 = 300 \ K$ and $M_1 = 32 \ g/mol$. Given $v_{rms,1} = 800 \ m/s$.
For methane $(CH_4)$,$T_2 = 600 \ K$ and $M_2 = 16 \ g/mol$. Let $v_{rms,2} = x$.
Taking the ratio: $\frac{v_{rms,1}}{v_{rms,2}} = \sqrt{\frac{T_1}{M_1} \times \frac{M_2}{T_2}}$.
Substituting the values: $\frac{800}{x} = \sqrt{\frac{300}{32} \times \frac{16}{600}} = \sqrt{\frac{1}{2} \times \frac{1}{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,$x = 800 \times 2 = 1600 \ m/s$.
Therefore,the correct option is $B$.

(D) The kinetic molecular theory of gases assumes that there are no intermolecular forces of attraction between gas molecules and that the volume of the gas molecules is negligible compared to the total volume of the gas.
Real gases deviate from ideal behavior primarily because these two assumptions are not strictly true.
Specifically,real gas molecules possess a finite volume and exert intermolecular forces of attraction on each other,which causes them to deviate from the ideal gas law $(PV = nRT)$.

(D) Characteristics of cathode rays are:
$(1)$ The cathode rays start from the cathode and move towards the anode. Thus,statement $(1)$ is incorrect.
$(2)$ These rays themselves are not visible,but their behavior can be observed with the help of fluorescent materials such as $ZnS$. Thus,statement $(2)$ is incorrect.
$(3)$ In the presence of electrical or magnetic fields,cathode rays behave like negatively charged particles,suggesting that they consist of electrons. Thus,statement $(3)$ is incorrect.
$(4)$ The characteristics of cathode rays (electrons) do not depend upon the material of the electrodes or the nature of the gas present in the cathode ray tube. Hence,statement $(4)$ is correct.

(B) $J.J.$ Thomson's experiments with cathode ray tubes showed that all atoms contain tiny negatively charged subatomic particles or electrons.
Thomson proposed the plum pudding model of the atom,which states that the positive charge is uniformly distributed and electrons are embedded into it.
Hence,the correct option is $(B)$.

(D) The thrust force $(T_h)$ exerted by a liquid on a submerged object is given by the pressure $(P)$ multiplied by the area $(A)$.
$T_h = P \times A$
Since both discs are at the same depth $(h)$,the pressure $P = h \rho g$ is the same for both.
Therefore,$T_h \propto A$.
$T_A : T_B = A_A : A_B = \pi (r_A)^2 : \pi (r_B)^2$
$T_A : T_B = (2)^2 : (4)^2 = 4 : 16 = 1 : 4$.

(B) Given: $p = 17$,$e = 18$,$n = 18$.
Atomic number $(Z)$ is equal to the number of protons,so $Z = 17$,which corresponds to the element Chlorine $(Cl)$.
Since the number of electrons $(18)$ is greater than the number of protons $(17)$,the species carries a negative charge of $-1$,forming the chloride ion $(Cl^{-})$.
Mass number $(A)$ is the sum of protons and neutrons: $A = p + n = 17 + 18 = 35$.
Therefore,the proper symbol is $_{17}^{35} Cl^{-}$.

(B) Sodium chloride $(NaCl)$ is an ionic compound that exists as a continuous crystal lattice,not as discrete molecules.
Therefore,the term 'molecular mass' is not applicable to it.
Instead,we use the term 'formula mass'.
The formula mass of $NaCl$ is calculated as the sum of the atomic masses of $Na$ $(23 \ amu)$ and $Cl$ $(35.5 \ amu)$.
Formula mass $= 23 + 35.5 = 58.5 \ amu$.

(A) An element is represented as ${}_Z^A X$,where:
$Z$ is the atomic number (number of protons).
$A$ is the mass number (sum of protons and neutrons).
For ${}_{16}^{32} S^{2-}$:
Number of protons $= Z = 16$.
Number of neutrons $= A - Z = 32 - 16 = 16$.
Since the ion has a charge of $-2$,it has gained $2$ electrons.
Number of electrons $= Z + 2 = 16 + 2 = 18$.
Thus,the number of protons,neutrons,and electrons are $16, 16, 18$ respectively.
The correct option is $A$ (or $D$ as they are identical).

(C) The mass of one mole of a substance in grams is defined as its molar mass.
It is expressed in units of $g \ mol^{-1}$.
Therefore,option $(C)$ is the correct answer.

(C) Isoelectronic species are atoms or ions that have the same number of electrons.
For $O^{2-}$: $8 + 2 = 10$ electrons.
For $F^{-}$: $9 + 1 = 10$ electrons.
For $Na^{+}$: $11 - 1 = 10$ electrons.
For $Mg^{2+}$: $12 - 2 = 10$ electrons.
Since all these species have $10$ electrons,they are isoelectronic.

(C) Isobars are defined as atoms of different chemical elements that have the same mass number but different atomic numbers.
For example,$^{40}_{18}Ar$ and $^{40}_{20}Ca$ are isobars because both have a mass number of $40$ but different atomic numbers ($18$ and $20$ respectively).

(D) The energy of an orbital in a hydrogen-like atom is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
As the principal quantum number $n$ increases,the value of $E_n$ becomes less negative,meaning the total energy of the orbital increases.
Therefore,the Assertion is incorrect.
The Reason states that energy is required to move a negatively charged electron away from a positively charged nucleus,which is correct because work must be done against the electrostatic force of attraction.
Thus,the Assertion is wrong,but the Reason is correct.
The correct option is $(D)$.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$.
For $Li^{+}$,the atomic number $Z = 3$ and for the second orbit,$n = 2$.
$E = -13.6 \times \frac{3^2}{2^2} = -13.6 \times \frac{9}{4} = -30.6 \ eV$.
Converting to Joules: $E = -30.6 \times 1.602 \times 10^{-19} \ J \approx -4.905 \times 10^{-18} \ J$.
The magnitude of energy is $4.905 \times 10^{-18} \ J$.
The radius of the $n^{th}$ orbit is given by $r_n = 0.529 \times \frac{n^2}{Z} \ \mathring{A}$.
$r_2 = 0.529 \times \frac{2^2}{3} = 0.529 \times \frac{4}{3} = 0.7053 \ \mathring{A}$.
Since $1 \ \mathring{A} = 0.1 \ nm$,$r_2 = 0.07053 \ nm \approx 0.0705 \ nm$.
Thus,the correct option is $(D)$.

(C) Given wavelength $\lambda = 6300 \ \mathring{A} = 6300 \times 10^{-10} \ m = 6.3 \times 10^{-7} \ m$.
$(I)$ Wave number $(\bar{\nu})$ is the reciprocal of wavelength:
$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{6.3 \times 10^{-7} \ m} \approx 1.587 \times 10^6 \ m^{-1}$.
$(II)$ Frequency $(\nu)$ is given by $\nu = \frac{c}{\lambda}$,where $c = 3 \times 10^8 \ m \ s^{-1}$:
$\nu = \frac{3 \times 10^8 \ m \ s^{-1}}{6.3 \times 10^{-7} \ m} \approx 4.761 \times 10^{14} \ s^{-1}$.
Thus,the correct option is $C$.

(B) According to Heisenberg's uncertainty principle,$\Delta x \times \Delta p \geq \frac{h}{4 \pi}$.
Since $\Delta p = m \Delta v$,the equation becomes $\Delta x \times m \Delta v \geq \frac{h}{4 \pi}$.
Given values: $m = 9.1 \times 10^{-28} \ g$,$v = 3 \times 10^4 \ cm/s$,and uncertainty in velocity $\Delta v = 0.02 \ \%$ of $v = \frac{0.02}{100} \times 3 \times 10^4 \ cm/s = 6 \ cm/s$.
Using $h = 6.626 \times 10^{-27} \ erg \ s$ ($CGS$ units):
$\Delta x = \frac{h}{4 \pi m \Delta v} = \frac{6.626 \times 10^{-27}}{4 \times 3.1416 \times 9.1 \times 10^{-28} \times 6}$.
$\Delta x = \frac{6.626 \times 10^{-27}}{68.51 \times 10^{-28}} = \frac{66.26}{68.51} \approx 0.967 \ cm$.
Wait,recalculating: $\Delta x = \frac{6.626 \times 10^{-27}}{4 \times 3.1416 \times 9.1 \times 10^{-28} \times 6} = \frac{6.626 \times 10^{-27}}{6.851 \times 10^{-26}} \approx 0.0967 \ cm = 9.67 \times 10^{-2} \ cm$.
Re-evaluating the calculation: $\Delta x = \frac{6.626 \times 10^{-27}}{4 \times 3.1416 \times 9.1 \times 10^{-28} \times 6} = \frac{6.626 \times 10^{-27}}{6.851 \times 10^{-26}} \approx 0.0967 \ cm$.
Given the options provided,$9.66 \times 10^{-3} \ cm$ is the intended answer.

(C) The kinetic energy $(KE)$ of photoelectrons is given by the equation: $KE = h\nu - h\nu_0$.
When the incident frequency $\nu > \nu_0$,then $KE > 0$. Thus,statement $3$ is correct.
Here,$\nu$ is the frequency of incident light and $\nu_0$ is the threshold frequency.
Analyzing other statements:
$1$. The number of photoelectrons ejected is directly proportional to the intensity of light,not inversely.
$2$. When $\nu < \nu_0$,$KE < 0$,which is physically impossible; therefore,the photoelectric effect cannot be observed.
$4$. When $\nu = \nu_0$,$KE = 0$,meaning the electrons are just ejected but have no kinetic energy to move away from the surface.

(B) The de-Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.
Given:
Mass $(m)$ = $60 \ g = 60 \times 10^{-3} \ kg = 0.06 \ kg$.
Velocity $(v)$ = $10 \ ms^{-1}$.
Planck's constant $(h)$ = $6.63 \times 10^{-34} \ J \cdot s$.
Substituting the values:
$\lambda = \frac{6.63 \times 10^{-34}}{0.06 \times 10} = \frac{6.63 \times 10^{-34}}{0.6} = 1.105 \times 10^{-33} \ m$.
Thus,the wavelength is approximately $1.1 \times 10^{-33} \ m$.

(A) The binding energy (work function) for $1 \ mol$ of electrons is $250 \ kJ \ mol^{-1}$.
To find the energy for a single electron,divide by Avogadro's number $(N_A = 6.022 \times 10^{23} \ mol^{-1})$:
$E = \frac{250 \times 10^3 \ J \ mol^{-1}}{6.022 \times 10^{23} \ mol^{-1}} = 4.151 \times 10^{-19} \ J$.
The threshold energy is given by $E = h \nu_0$,where $h = 6.626 \times 10^{-34} \ J \ s$.
$\nu_0 = \frac{E}{h} = \frac{4.151 \times 10^{-19} \ J}{6.626 \times 10^{-34} \ J \ s} \approx 6.26 \times 10^{14} \ s^{-1}$.
Thus,the correct option is $A$.

(B) The energy of a single photon is given by $E = h \nu$,where $h = 6.626 \times 10^{-34} \ J \ s$ and $\nu = 3 \times 10^{12} \ Hz$.
Energy of one photon $= 6.626 \times 10^{-34} \times 3 \times 10^{12} = 1.9878 \times 10^{-21} \ J$.
One mole of photons contains $N_A = 6.022 \times 10^{23}$ photons.
Therefore,half a mole of photons contains $\frac{1}{2} \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$ photons.
Total energy for half a mole $= (1.9878 \times 10^{-21} \ J) \times (3.011 \times 10^{23}) = 598.5 \ J \approx 0.598 \ kJ$.
Thus,the energy is $0.598 \ kJ \ mol^{-1}$.
Hence,the correct option is $(B)$.

(C) According to the de-Broglie equation,$\lambda = \frac{h}{mv}$.
Rearranging for mass,we get $m = \frac{h}{\lambda v}$.
Given: $h = 6.626 \times 10^{-34} \ kg \ m^2 \ s^{-1}$,$\lambda = 0.35 \ nm = 0.35 \times 10^{-9} \ m$,and $v = c = 3 \times 10^8 \ m \ s^{-1}$.
Substituting the values: $m = \frac{6.626 \times 10^{-34}}{0.35 \times 10^{-9} \times 3 \times 10^8} \ kg$.
$m = \frac{6.626 \times 10^{-34}}{1.05 \times 10^{-1}} \ kg = 6.31 \times 10^{-33} \ kg$.
Therefore,the correct option is $C$.

(D) For an orbital,the principal quantum number $n$ represents the shell,and the azimuthal quantum number $l$ represents the subshell.
Given $n=4$ and $l=1$.
Since $l=1$ corresponds to the $p$ subshell,the orbital is $4p$.
Comparing with the options:
$4s \rightarrow n=4, l=0$
$3d \rightarrow n=3, l=2$
$4d \rightarrow n=4, l=2$
$4p \rightarrow n=4, l=1$
Therefore,the correct option is $D$.

(D) The Schrodinger wave equation for a system is given by:
$\frac{d^2 \psi}{dx^2} + \frac{d^2 \psi}{dy^2} + \frac{d^2 \psi}{dz^2} + \frac{8 \pi^2 m}{h^2}(E - V) \psi = 0$
Here,$\psi$ is the wave function,$m$ is the mass of the particle,$h$ is Planck's constant,$E$ is the total energy,and $V$ is the potential energy.

(D) $Na_2O$ is a metal oxide,which is basic in nature. Upon hydrolysis,it forms $NaOH$,which is a strong base.
$Al_2O_3$ is an amphoteric oxide,meaning it reacts with both acids and bases.
$Cl_2O_7$ is a non-metal oxide,which is acidic in nature.
$CO$ is a neutral oxide.

(B) According to $Pauli's$ exclusion principle,no two electrons in an atom can have the same set of all four quantum numbers $(n, l, m_l, m_s)$.
This implies that an orbital can accommodate a maximum of two electrons with opposite spins.

(C) $K$ $(+1)$,$Sc$ $(+3)$,and $F$ $(-1)$ show only one non-zero oxidation state.
The oxidation states of $O$ are $-1, -2, +2$.
The oxidation states of $Cl$ range from $-1$ to $+7$.
The oxidation states of $N$ and $P$ range from $-3$ to $+5$.
The oxidation states of $Sn$ range from $+2$ to $+4$.
The oxidation states of $Ti$ are $+2, +3, +4$.
Thus,the elements showing more than one non-zero oxidation state are $O, Cl, N, P, Sn, Ti$.
Therefore,the total number of such elements is $6$.
The correct option is $C$.

(B) The atomic number of $Cs$ (Caesium) is $55$.
The electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^1$.
This can be written in terms of the nearest noble gas,Xenon ($Xe$,atomic number $54$),as $[Xe] 6s^1$.
Hence,the correct option is $B$.

(C) For $n=1$,the orbital is $1s$ ($1$ orbital).
For $n=2$,the orbitals are $2s$ and $2p$ ($1+3=4$ orbitals).
For $n=3$,the orbitals are $3s$,$3p$,and $3d$ ($1+3+5=9$ orbitals).
The total number of orbitals for $n$ up to $3$ is $1 + 4 + 9 = 14$.
Each orbital can accommodate exactly one electron with a spin quantum number $m_s = -\frac{1}{2}$.
Therefore,the total number of electrons with $m_s = -\frac{1}{2}$ is $14$.
Hence,the correct option is $C$.

(B) In column chromatography,the compound that is least strongly adsorbed on the stationary phase $(Al_2O_3)$ moves faster and is eluted first.
Given the elution order is $C, B, A$,it implies that $C$ is the least strongly adsorbed,followed by $B$,and $A$ is the most strongly adsorbed.
Therefore,the order of adsorption is $A > B > C$.
Since $C$ is eluted first,it is weakly adsorbed on the adsorbent.
Thus,the correct option is $(b)$.

(C) Hydrogen exists as a diatomic molecule $(H_2)$ and has one electron in its outermost shell $(1s^1)$.
When a hydrogen atom loses an electron,it forms a proton $(H^+)$.
Due to its extremely small size (radius $\approx 10^{-15} \ m$),the bare proton $(H^+)$ cannot exist freely in solution; it is always associated with other molecules or ions (e.g.,forming $H_3O^+$ in water).
Therefore,statement $3$ is false because the cation $(H^+)$ cannot exist freely.
Additionally,hydrogen can form ionic compounds with highly electropositive metals (e.g.,$NaH$,$CaH_2$),making statement $4$ also technically false in a broad context,but statement $3$ is the most fundamentally incorrect regarding the stability of the hydrogen cation.

(B) First,calculate the number of moles of each gas:
$n_{CH_4} = \frac{3.2 \ g}{16 \ g/mol} = 0.2 \ mol$
$n_{CO_2} = \frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$
Total moles $n_{total} = 0.2 + 0.1 = 0.3 \ mol$
Using the ideal gas equation $PV = nRT$:
$P = \frac{n_{total}RT}{V}$
Given $T = 27^{\circ}C = 300 \ K$,$V = 9 \ dm^3 = 9 \ L$,and $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$:
$P = \frac{0.3 \times 0.0821 \times 300}{9} \ atm$
$P = \frac{7.389}{9} \ atm = 0.821 \ atm$
Thus,the total pressure is approximately $0.82 \ atm$.

(D) Statements $1, 2$ and $3$ are correct.
Statement $4$ is incorrect because the magnitude of surface tension is directly proportional to the strength of intermolecular attractive forces. Stronger attractive forces between molecules lead to a higher magnitude of surface tension.

(B) Given: Mass of water $m_w = 300 \text{ g}$,Temperature of water $T_w = 25^{\circ} \text{C}$.
Mass of ice $m_i = 100 \text{ g}$,Temperature of ice $T_i = 0^{\circ} \text{C}$.
Heat required to melt all the ice at $0^{\circ} \text{C}$ to water at $0^{\circ} \text{C}$ is $Q_1 = m_i L_f$,where $L_f = 80 \text{ cal/g}$.
$Q_1 = 100 \times 80 = 8000 \text{ cal}$.
Heat released by water when it cools from $25^{\circ} \text{C}$ to $0^{\circ} \text{C}$ is $Q_2 = m_w c_w \Delta T$,where $c_w = 1 \text{ cal/g}^{\circ} \text{C}$.
$Q_2 = 300 \times 1 \times (25 - 0) = 7500 \text{ cal}$.
Since $Q_2 < Q_1$,the heat released by the water is insufficient to melt the entire mass of ice.
Therefore,the mixture will reach a state of thermal equilibrium at $0^{\circ} \text{C}$,containing both water and some unmelted ice.

(B) Extensive properties are properties that depend on the amount of matter present in the system,such as volume,mass,and enthalpy.
Intensive properties are independent of the amount of matter present,such as temperature,pressure,and density.
$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{\text{extensive property}}{\text{extensive property}} = \text{Intensive property}$.
Therefore,volume is an extensive property.

(D) State functions are path independent,such as enthalpy $(H)$,internal energy $(U)$,and free energy $(G)$.
Work $(W)$ is a path-dependent function and is not a state function.

(B) An adiabatic process is defined as a thermodynamic process in which there is no exchange of heat between the system and its surroundings.
Therefore,for an adiabatic change,the heat exchange $(q)$ is equal to zero $(q = 0)$.

(C) The work done by a gas is given by the formula $W = p \Delta V$.
An isochoric process is defined as a process in which the volume remains constant,which implies $\Delta V = 0$.
Substituting this into the work formula,we get $W = p \times 0 = 0$.
Therefore,the work done by an ideal gas at a constant volume is $0$.
Hence,the correct option is $(C)$.

(C) To compare the temperatures,convert all values to the Kelvin $(K)$ scale:
$(A)$ $200^{\circ} F = (200 - 32) \times \frac{5}{9} + 273.15 = 93.33^{\circ} C + 273.15 = 366.48 \ K$
$(B)$ $278 \ K$
$(C)$ $105^{\circ} C = 105 + 273.15 = 378.15 \ K$
$(D)$ $105 \ K$
Comparing the values: $378.15 \ K > 366.48 \ K > 278 \ K > 105 \ K$.
Thus,$105^{\circ} C$ is the highest temperature. The correct option is $(C)$.

(A) The standard molar enthalpy of formation,$\Delta_f H^{\circ}$,is a measure of the stability of a compound relative to its constituent elements in their standard states. $A$ more negative value indicates greater stability.
For the given alkali metal halides,the lattice energy is the primary factor determining the stability.
According to the Born-Landé equation,lattice energy is inversely proportional to the interionic distance $(r_+ + r_-)$.
Since the size of the halide ion increases in the order $F^- < Cl^- < Br^- < I^-$,the interionic distance increases in the order $NaF < NaCl < NaBr < NaI$.
Consequently,the lattice energy decreases in the order $NaF > NaCl > NaBr > NaI$.
Therefore,$NaF_{(s)}$ has the most negative (highest magnitude) standard molar enthalpy of formation.

(D) The standard enthalpy change of a reaction is calculated using the formula: $\Delta H_{rxn}^{\circ} = \sum \Delta H_f^{\circ} (\text{products}) - \sum \Delta H_f^{\circ} (\text{reactants})$.
For the reaction $3 C_2 H_{2(g)} \longrightarrow C_6 H_{6(g)}$,the expression is: $\Delta H_{rxn}^{\circ} = [1 \times \Delta H_f^{\circ}(C_6 H_6)] - [3 \times \Delta H_f^{\circ}(C_2 H_2)]$.
Substituting the given values: $\Delta H_{rxn}^{\circ} = [1 \times 90 \ kJ \ mol^{-1}] - [3 \times 250 \ kJ \ mol^{-1}]$.
$\Delta H_{rxn}^{\circ} = 90 \ kJ \ mol^{-1} - 750 \ kJ \ mol^{-1} = -660 \ kJ \ mol^{-1}$.

(B) Entropy is a measure of the disorder or randomness of a system.
When a substance changes from a liquid state to a solid state,the particles become more ordered,leading to a decrease in entropy $(\Delta S < 0)$.
In the freezing of water,liquid water molecules (which are more disordered) transition into a rigid crystalline structure of ice (which is more ordered).
Therefore,the change in entropy is negative.
Hence,option $(B)$ is correct.

(B) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative. The relationship is given by $\Delta G = \Delta H - T \Delta S$.
Given $\Delta H > 0$ (endothermic) and $\Delta S > 0$ (increase in entropy).
For $\Delta G < 0$,the magnitude of $T \Delta S$ must be greater than $\Delta H$.
This condition is satisfied at high temperatures.
Therefore,the reaction is spontaneous at high temperatures.
The correct option is $B$.