MHT CET 2026 Chemistry Question Paper with Answer and Solution

(B) Step $1$: Calculate the mass of solutes. Assuming density is $1 \text{ g/mL}$, mass of $50 \text{ mL}$ solution is $50 \text{ g}$.
Mass of $AgNO_3 = 16.9\% \text{ of } 50 \text{ g} = 0.169 \times 50 = 8.45 \text{ g}$.
Mass of $KCl = 7.45\% \text{ of } 50 \text{ g} = 0.0745 \times 50 = 3.725 \text{ g}$.
Step $2$: Calculate moles of reactants.
Moles of $AgNO_3 = \frac{8.45 \text{ g}}{169 \text{ g/mol}} = 0.05 \text{ mol}$.
Moles of $KCl = \frac{3.725 \text{ g}}{74.5 \text{ g/mol}} = 0.05 \text{ mol}$.
Step $3$: Reaction equation: $AgNO_3 + KCl \rightarrow AgCl(s) + KNO_3$.
Since stoichiometric ratio is $1:1$ and moles are equal, both are limiting reagents.
Moles of $AgCl$ formed $= 0.05 \text{ mol}$.
Step $4$: Mass of $AgCl = 0.05 \text{ mol} \times 143.3 \text{ g/mol} = 7.165 \text{ g} \approx 7 \text{ g}$.

(C) The balanced chemical equation for the synthesis of ammonia is: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$.
From the equation, $3 \text{ moles}$ of $H_2$ produce $2 \text{ moles}$ of $NH_3$.
Molar mass of $NH_3 = 14 + 3 \times 1 = 17 \text{ g/mol}$.
Moles of $NH_3$ produced $= \frac{3.4 \text{ g}}{17 \text{ g/mol}} = 0.2 \text{ mol}$.
Since $2 \text{ moles}$ of $NH_3$ require $3 \text{ moles}$ of $H_2$, $0.2 \text{ moles}$ of $NH_3$ require $\frac{3}{2} \times 0.2 = 0.3 \text{ moles}$ of $H_2$.
Number of $H_2$ molecules $= 0.3 \times 6.022 \times 10^{23} = 1.8066 \times 10^{23}$.
Note: The provided options appear incorrect based on the calculation. The closest value is $1.8066 \times 10^{23}$.

(D) The formula for percentage atom economy is:
$\text{Atom Economy} = \left( \frac{\text{Formula weight of desired product}}{\text{Sum of formula weights of all reactants}} \right) \times 100\%$
Given:
Formula weight of desired product = $65 \text{ u}$
Sum of formula weights of all reactants = $130 \text{ u}$
Calculation:
$\text{Atom Economy} = \left( \frac{65}{130} \right) \times 100\%$
$\text{Atom Economy} = 0.5 \times 100\% = 50\%$
Therefore, the correct option is $D$.

(A) Step $1$: Calculate the mass in $amu$ (or $u$).
The molar mass of $O_2$ is $32 \text{ g/mol}$. The mass of one molecule of $O_2$ is $32 \text{ u}$.
Mass of $100$ molecules $= 100 \times 32 \text{ u} = 3200 \text{ u}$.
Step $2$: Calculate the mass in grams.
Mass of one molecule of $O_2 = \frac{32 \text{ g}}{6.022 \times 10^{23}} \approx 5.314 \times 10^{-23} \text{ g}$.
Mass of $100$ molecules $= 100 \times 5.314 \times 10^{-23} \text{ g} = 5.314 \times 10^{-21} \text{ g}$.

(B) Step $1$: Calculate the molar mass of $CO_2$. Molar mass of $CO_2 = 12 + (2 \times 16) = 44 \text{ g/mol}$.
Step $2$: Calculate the number of moles of $CO_2$. $n = \frac{\text{mass}}{\text{molar mass}} = \frac{99 \text{ g}}{44 \text{ g/mol}} = 2.25 \text{ mol}$.
Step $3$: Calculate the volume at $STP$. At $STP$, $1 \text{ mole}$ of any gas occupies $22.4 \text{ L}$.
Volume $= n \times 22.4 \text{ L/mol} = 2.25 \times 22.4 \text{ L} = 50.4 \text{ L}$.

(C) The number of molecules is proportional to the number of moles $(n)$.
$(A)$ $n = \frac{0.4 \text{ g}}{2 \text{ g/mol}} = 0.2 \text{ mol}$.
$(B)$ $n = \frac{2 \text{ g}}{32 \text{ g/mol}} = 0.0625 \text{ mol}$.
$(C)$ At $STP$, $1 \text{ mol}$ occupies $22.4 \text{ dm}^3$. So, $n = \frac{8 \text{ dm}^3}{22.4 \text{ dm}^3/\text{mol}} \approx 0.357 \text{ mol}$.
$(D)$ $n = \frac{2 \text{ dm}^3}{22.4 \text{ dm}^3/\text{mol}} \approx 0.089 \text{ mol}$.
Comparing the values, $0.357 \text{ mol}$ is the highest. Thus, option $(C)$ contains the maximum number of molecules.

(A) Step $1$: Calculate the number of moles of $H_2$ in $1 \text{ g}$. Molar mass of $H_2 = 2 \text{ g/mol}$. Moles of $H_2 = \frac{1 \text{ g}}{2 \text{ g/mol}} = 0.5 \text{ mol}$.
Step $2$: Calculate the number of moles for each option:
$(A)$ $14 \text{ g of } N_2$: Molar mass of $N_2 = 28 \text{ g/mol}$. Moles $= \frac{14}{28} = 0.5 \text{ mol}$.
$(B)$ $18 \text{ g of } H_2O$: Molar mass of $H_2O = 18 \text{ g/mol}$. Moles $= \frac{18}{18} = 1 \text{ mol}$.
$(C)$ $16 \text{ g of } CO$: Molar mass of $CO = 28 \text{ g/mol}$. Moles $= \frac{16}{28} \approx 0.57 \text{ mol}$.
$(D)$ $28 \text{ g of } N_2$: Molar mass of $N_2 = 28 \text{ g/mol}$. Moles $= \frac{28}{28} = 1 \text{ mol}$.
Step $3$: Since the number of molecules is directly proportional to the number of moles, $14 \text{ g of } N_2$ contains the same number of molecules as $1 \text{ g of } H_2$.

(A) Step $1$: Identify the given values: Molarity $(M) = 0.1 \text{ mol/L}$, Volume $(V) = 100 \text{ ml} = 0.1 \text{ L}$, Molar mass $(MW) = 106 \text{ g/mol}$.
Step $2$: Use the formula for Molarity: $M = \frac{\text{mass}}{MW \times V(\text{in L})}$.
Step $3$: Rearrange to find mass: $\text{mass} = M \times MW \times V$.
Step $4$: Substitute the values: $\text{mass} = 0.1 \text{ mol/L} \times 106 \text{ g/mol} \times 0.1 \text{ L} = 1.06 \text{ g}$.

(A) Step $1$: Determine the molar mass of sulfur in the compound $Al_2(SO_4)_3$. There are $3$ atoms of sulfur in one formula unit. Total mass of sulfur = $3 \times 32 \text{ g/mol} = 96 \text{ g/mol}$.
Step $2$: Determine the molar mass of aluminium in the compound. There are $2$ atoms of aluminium. Total mass of aluminium = $2 \times 27 \text{ g/mol} = 54 \text{ g/mol}$.
Step $3$: Use the ratio of masses to find the amount of aluminium. The mass ratio of $Al$ to $S$ is $54 : 96$.
Step $4$: Calculate the mass of $Al$ for $4 \text{ g}$ of sulfur: $\text{Mass of } Al = (54 / 96) \times 4 \text{ g} = (9 / 16) \times 4 \text{ g} = 9 / 4 \text{ g} = 2.25 \text{ g}$.

(B) Let the mass of '$x$' atoms of the element be $M_e$ and the mass of '$x$' atoms of carbon be $M_c$.
Given ratio: $\frac{M_e}{M_c} = \frac{9}{1}$.
We know that the mass of '$x$' atoms is proportional to their molar mass $(M)$:
$\frac{M_e}{M_c} = \frac{M_{\text{element}}}{M_{\text{carbon}}}$.
Substitute the given values: $\frac{M_{\text{element}}}{12 \text{ g/mol}} = \frac{9}{1}$.
$M_{\text{element}} = 9 \times 12 \text{ g/mol} = 108 \text{ g/mol}$.

(A) $1$. At $STP$, $1 \text{ mole}$ of any gas occupies a volume of $22.4 \text{ dm}^3$.
$2$. The number of moles $(n)$ in $2.24 \text{ dm}^3$ of $CO_2$ is calculated as: $n = \frac{\text{Given Volume}}{\text{Molar Volume}} = \frac{2.24 \text{ dm}^3}{22.4 \text{ dm}^3/\text{mol}} = 0.1 \text{ mol}$.
$3$. The number of molecules is given by: $\text{Number of molecules} = n \times N_A$, where $N_A$ is Avogadro's number $(6.022 \times 10^{23} \text{ molecules/mol})$.
$4$. Therefore, $\text{Number of molecules} = 0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}$ molecules.

(B) Step $1$: Write the balanced chemical equation for the thermal decomposition of potassium chlorate:
$2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$
Step $2$: From the stoichiometry, $2 \text{ moles}$ of $KClO_3$ produce $3 \text{ moles}$ of $O_2$.
Step $3$: At $STP$, $1 \text{ mole}$ of any gas occupies $22.4 \text{ L}$. Therefore, $11.2 \text{ L}$ of $O_2$ corresponds to $n = \frac{11.2 \text{ L}}{22.4 \text{ L/mol}} = 0.5 \text{ moles}$ of $O_2$.
Step $4$: Using the mole ratio, $3 \text{ moles}$ of $O_2$ are produced by $2 \text{ moles}$ of $KClO_3$. So, $0.5 \text{ moles}$ of $O_2$ are produced by $\frac{2}{3} \times 0.5 = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3} \text{ moles}$ of $KClO_3$.

(D) The balanced chemical equation for the thermal decomposition of potassium chlorate is:
$2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$
From the stoichiometry of the reaction, $2 \text{ moles}$ of $KClO_3$ produce $3 \text{ moles}$ of $O_2$.
At $STP$, $1 \text{ mole}$ of any ideal gas occupies $22.4 \text{ L}$.
Therefore, $22.4 \text{ L}$ of $O_2$ corresponds to $1 \text{ mole}$ of $O_2$.
Using the mole ratio: $\frac{2 \text{ moles } KClO_3}{3 \text{ moles } O_2} = \frac{x \text{ moles } KClO_3}{1 \text{ mole } O_2}$.
Solving for $x$: $x = \frac{2}{3} \text{ moles}$ of $KClO_3$.

(A) Step $1$: Calculate the molar mass of dinitrogen $(N_2)$. Molar mass of $N = 14 \text{ g/mol}$, so molar mass of $N_2 = 2 \times 14 = 28 \text{ g/mol}$.
Step $2$: Calculate the number of moles $(n)$ in $70 \text{ g}$ of $N_2$. $n = \frac{\text{mass}}{\text{molar mass}} = \frac{70 \text{ g}}{28 \text{ g/mol}} = 2.5 \text{ mol}$.
Step $3$: Calculate the number of molecules using Avogadro's number $(N_A = 6.022 \times 10^{23} \text{ molecules/mol})$. Number of molecules = $n \times N_A = 2.5 \times 6.022 \times 10^{23} = 15.055 \times 10^{23} = 1.5055 \times 10^{24}$.

(D) $1$. The molar mass of water $(H_2O)$ is calculated as: $(2 \times 1.008) + 16.00 = 18.016 \text{ g/mol}$, approximately $18 \text{ g/mol}$.
$2$. The formula to calculate mass is: $\text{Mass} = \text{Number of moles} \times \text{Molar mass}$.
$3$. Substituting the given values: $\text{Mass} = 0.25 \text{ mol} \times 18 \text{ g/mol} = 4.5 \text{ g}$.

(C) The average atomic mass is calculated using the formula: $\text{Average atomic mass} = \sum (\text{abundance} \times \text{mass number})$.
$\text{Average atomic mass} = (0.75 \times 35) + (0.25 \times 37)$.
$\text{Average atomic mass} = 26.25 + 9.25$.
$\text{Average atomic mass} = 35.5 \text{ u}$.

(B) The average atomic mass is calculated by the weighted average of the isotopes:
$\text{Average atomic mass} = (\text{Mass of isotope 1} \times \text{Abundance 1}) + (\text{Mass of isotope 2} \times \text{Abundance 2})$
$\text{Average atomic mass} = (35 \times 0.75) + (37 \times 0.25)$
$\text{Average atomic mass} = 26.25 + 9.25 = 35.5 \text{ u}$

(C) $1$. Calculate the molar mass of ammonia $(NH_3)$: $M = 14 + (3 \times 1) = 17 \text{ g/mol}$.
$2$. Calculate the number of moles $(n)$: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{2.125 \text{ g}}{17 \text{ g/mol}} = 0.125 \text{ mol}$.
$3$. Calculate the number of molecules: $\text{Number of molecules} = n \times N_A = 0.125 \times 6.022 \times 10^{23} = 7.5275 \times 10^{22} \approx 7.53 \times 10^{22}$.
Given the options, the closest value is $7.51 \times 10^{22}$.

(A) The formula for atom economy is defined as:
$\text{Atom Economy} = \left( \frac{\text{Formula weight of desired product}}{\text{Sum of formula weights of all reactants}} \right) \times 100\%$
Given:
$\text{Atom Economy} = 50\% = 0.5$
$\text{Sum of formula weights of all reactants} = 274 \text{ u}$
Let the formula weight of the desired product be $x$.
$0.5 = \frac{x}{274 \text{ u}}$
$x = 0.5 \times 274 \text{ u} = 137 \text{ u}$
Thus, the formula weight of the desired product is $137 \text{ u}$.

(C) The electronic configuration of argon $(Z = 18)$ is $1s^2 2s^2 2p^6 3s^2 3p^6$.
Total electrons in $s$ orbitals = $2 (1s) + 2 (2s) + 2 (3s) = 6$.
Total electrons in $p$ orbitals = $6 (2p) + 6 (3p) = 12$.
Therefore, the total number of electrons in $s$ and $p$ orbitals are $6$ and $12$ respectively.

(B) Step $1$: Determine the valency of element $A$. The configuration $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$ shows $2$ electrons in the outermost shell. Thus, $A$ loses $2$ electrons to form $A^{2+}$ ion.
Step $2$: Determine the valency of element $B$. The configuration $1s^2 2s^2 2p^6 3s^2 3p^5$ shows $7$ electrons in the outermost shell. Thus, $B$ gains $1$ electron to form $B^-$ ion.
Step $3$: Determine the formula of the ionic compound. To balance the charges, one $A^{2+}$ ion reacts with two $B^-$ ions. The formula is $AB_2$.

(C) The $(n + l)$ rule states that the energy of an orbital depends on the sum of the principal quantum number $(n)$ and the azimuthal quantum number $(l)$.
$(1)$ For $3s$: $n = 3, l = 0 \implies n + l = 3 + 0 = 3$. For $3p$: $n = 3, l = 1 \implies n + l = 3 + 1 = 4$.
$(2)$ For $2p$: $n = 2, l = 1 \implies n + l = 2 + 1 = 3$. For $4s$: $n = 4, l = 0 \implies n + l = 4 + 0 = 4$.
$(3)$ For $3d$: $n = 3, l = 2 \implies n + l = 3 + 2 = 5$. For $4p$: $n = 4, l = 1 \implies n + l = 4 + 1 = 5$.
$(4)$ For $2p$: $n = 2, l = 1 \implies n + l = 3$. For $3p$: $n = 3, l = 1 \implies n + l = 4$.
Since $3d$ and $4p$ both have $(n + l) = 5$, they have similar values.

(B) $1$. The atomic number of Silicon $(Si)$ is $14$. Its electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^2$. In the $3p$ subshell, there are $2$ unpaired electrons.
$2$. The atomic number of Chromium $(Cr)$ is $24$. Its electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$. In the $4s$ subshell, there is $1$ unpaired electron, and in the $3d$ subshell, there are $5$ unpaired electrons. Total unpaired electrons = $1 + 5 = 6$.
$3$. Thus, the number of unpaired electrons in $Si$ and $Cr$ are $2$ and $6$ respectively.

(B) The number of neutrons $n$ is calculated as $n = A - Z$, where $A$ is the mass number and $Z$ is the atomic number.
$(A)$ $^{24}_{12}Mg$: $n = 24 - 12 = 12$ (Matches $II$).
$(B)$ $^{27}_{13}Al$: $n = 27 - 13 = 14$ (Matches $IV$).
$(C)$ $^{9}_{4}Be$: $n = 9 - 4 = 5$ (Matches $I$).
$(D)$ $^{12}_{6}C$: $n = 12 - 6 = 6$ (Matches $III$).
Therefore, the correct matching is $(A)$-$(II)$, $(B)$-$(IV)$, $(C)$-$(I)$, $(D)$-$(III)$.

(C) $1$. Isobars are atoms of different chemical elements that have the same mass number but different atomic numbers.
$2$. In the pair $^{40}K_{19}$ and $^{40}Ca_{20}$, both atoms have a mass number of $A = 40$ but different atomic numbers ($Z = 19$ for Potassium and $Z = 20$ for Calcium).
$3$. Therefore, they are isobars.

(D) $1$. As we move from left to right across a period, the nuclear charge increases by one unit for each successive element.
$2$. The additional electrons are added to the same principal energy shell.
$3$. The increased nuclear charge exerts a stronger pull on the electrons, drawing them closer to the nucleus.
$4$. Consequently, the effective nuclear charge increases, which results in a decrease in the atomic size (atomic radius) across the period.

(C) $1$. Atomic radius decreases across a period from left to right due to an increase in effective nuclear charge $(Z_{\text{eff}})$.
$2$. As we move from $Group \ 1$ to $Group \ 17$, the atomic radius decreases.
$3$. $Group \ 18$ elements (noble gases) have larger atomic radii than $Group \ 17$ elements because their radii are defined as van der Waals radii, which are significantly larger than covalent radii.
$4$. Therefore, $Group \ 17$ elements (halogens) possess the smallest covalent atomic radii in their respective periods.

(C) $1$. Electronegativity is the tendency of an atom to attract a shared pair of electrons towards itself.
$2$. In the periodic table, electronegativity generally increases across a period from left to right and decreases down a group.
$3$. Among the given elements, $F$, $O$, $N$, and $Cl$, $Fluorine$ $(F)$ is located at the top right of the periodic table (excluding noble gases).
$4$. Therefore, $Fluorine$ has the highest electronegativity value of $4.0$ on the $Pauling$ scale.

(A) Step $1$: Electron gain enthalpy is the energy change when an electron is added to a neutral gaseous atom.
Step $2$: Noble gases like $Xenon$ $(Xe)$ have a stable $ns^2 np^6$ electronic configuration.
Step $3$: Adding an electron to a noble gas requires energy to overcome the stability of the filled shell, resulting in a positive electron gain enthalpy.
Step $4$: Halogens like $Bromine$, $Iodine$, and $Chlorine$ have high electron affinity and release energy upon electron gain, resulting in negative electron gain enthalpy.
Step $5$: Therefore, $Xenon$ is the correct answer.

(D) Step $1$: Identify the group and period requirements. Group $2$ elements are alkaline earth metals. The sixth period corresponds to the shell $n = 6$.
Step $2$: Analyze the options. $Rubidium$ $(Rb)$ is in group $1$, period $5$. $Strontium$ $(Sr)$ is in group $2$, period $5$. $Caesium$ $(Cs)$ is in group $1$, period $6$. $Barium$ $(Ba)$ is in group $2$, period $6$.
Step $3$: Conclusion. $Barium$ $(Ba)$ is the element that belongs to group $2$ and period $6$.

(B) $1$. The dipole moments of the given molecules are approximately:
$BeF_2$ (linear): $0 \text{ D}$
$CHCl_3$: $1.04 \text{ D}$
$NH_3$: $1.47 \text{ D}$
$H_2O$: $1.85 \text{ D}$
$2$. Comparing these values, $H_2O$ has the highest dipole moment due to the high electronegativity difference between $O$ and $H$ and its bent molecular geometry where the bond dipoles do not cancel out.

(D) $1$. $BF_3$ is a planar molecule with a symmetric structure, so its net dipole moment is $0 \text{ D}$.
$2$. Both $NH_3$ and $NF_3$ are pyramidal. In $NH_3$, the dipole moments of the $N-H$ bonds and the lone pair on $N$ are in the same direction, reinforcing each other. In $NF_3$, the dipole moments of the $N-F$ bonds are in the opposite direction to the lone pair, partially cancelling each other out.
$3$. Therefore, the order of dipole moments is $NH_3 (1.46 \text{ D}) > NF_3 (0.24 \text{ D}) > BF_3 (0 \text{ D})$.
$4$. The correct order is $NH_3 > NF_3 > BF_3$.

(B) According to Fajan's rule, the covalent character of an ionic bond increases with an increase in the oxidation state of the metal cation.
$1$. The oxidation states of the metal ions in the given compounds are: $SnCl_2$ $(Sn^{2+})$, $PbCl_4$ $(Pb^{4+})$, $SbCl_3$ $(Sb^{3+})$, and $PbCl_2$ $(Pb^{2+})$.
$2$. $A$ higher positive charge on the cation results in greater polarizing power, which leads to more covalent character.
$3$. Since $Pb^{4+}$ has the highest oxidation state among the given options, $PbCl_4$ exhibits the most covalent character.

(B) Step $1$: An ionic bond is formed by the complete transfer of electrons from a metal atom to a non-metal atom.
Step $2$: For the metal to lose electrons easily, it must have a low ionization enthalpy.
Step $3$: For the non-metal to accept electrons easily and release energy, it must have a high negative value of electron gain enthalpy (high electron affinity).
Step $4$: Therefore, the combination of low ionization enthalpy of the metal and high negative electron gain enthalpy of the non-metal favors ionic bond formation.

(C) The atomic number of fluorine $(F)$ is $9$. Therefore, an $F_2$ molecule has $9 \times 2 = 18$ electrons.
The molecular orbital configuration for $F_2$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^2, \pi^* 2p_y^2$.
Bonding electrons are those in orbitals without an asterisk $(*)$. Total bonding electrons = $2 (\sigma 1s) + 2 (\sigma 2s) + 2 (\sigma 2p_z) + 2 (\pi 2p_x) + 2 (\pi 2p_y) = 10$.
Antibonding electrons are those in orbitals with an asterisk $(*)$. Total antibonding electrons = $2 (\sigma^* 1s) + 2 (\sigma^* 2s) + 2 (\pi^* 2p_x) + 2 (\pi^* 2p_y) = 8$.
Thus, bonding electrons are $10$ and antibonding electrons are $8$.

(C) According to Molecular Orbital Theory $(MOT)$, the electronic configuration of $O_2$ ($16$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
Since $O_2$ has two unpaired electrons in its antibonding $\pi^*$ orbitals, it is paramagnetic.
$Li_2$, $N_2$, and $F_2$ have all paired electrons and are diamagnetic.

(A) According to Molecular Orbital Theory $(MOT)$, the electronic configuration of $O_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
Since $O_2$ contains two unpaired electrons in the $\pi^*$ antibonding molecular orbitals, it exhibits paramagnetism.
$O_3$, $N_2$, and $F_2$ are diamagnetic because they do not contain any unpaired electrons.

(A) Step $1$: Bond length increases as the size of the bonded atoms increases because the distance between the nuclei increases. Thus, it is directly proportional to the size of the atom.
Step $2$: Bond length decreases as the bond multiplicity (number of bonds between two atoms) increases, because higher multiplicity leads to stronger attraction and shorter distance. Thus, it is inversely proportional to the bond multiplicity.
Step $3$: Therefore, $(I)$ is direct and $(II)$ is inverse.

(D) Step $1$: Calculate the number of valence electrons in $IF_5$: $7 + 5 \times 7 = 42$ electrons.
Step $2$: Determine the hybridization: The central atom $I$ is bonded to $5$ $F$ atoms and has $1$ lone pair, giving a steric number of $6$ ($sp^3d^2$ hybridization).
Step $3$: Predict the geometry: With $5$ bond pairs and $1$ lone pair, the molecular geometry is square pyramidal.

(C) $1$. The structure of benzylic alcohol is $C_6H_5CH_2OH$.
$2$. In this molecule, the $-OH$ group is attached to a methylene group $(-CH_2-)$.
$3$. The carbon atom in the $-CH_2-$ group is bonded to two hydrogen atoms, one phenyl ring, and one oxygen atom.
$4$. Since this carbon atom forms four sigma bonds and has no lone pairs, its steric number is $4$.
$5$. Therefore, the hybridization of this carbon atom is $sp^3$.

(D) $1$. Identify the hybridization of $C_2$: It is attached to one double bond and two single bonds, so it has $3$ sigma bonds and $0$ lone pairs, making it $sp^2$ hybridized.
$2$. Identify the hybridization of $C_3$: It is attached to four single bonds, so it has $4$ sigma bonds and $0$ lone pairs, making it $sp^3$ hybridized.
$3$. Therefore, the $C_2 - C_3$ bond is formed by the overlap of an $sp^2$ orbital and an $sp^3$ orbital, which is of the $sp^2 - sp^3$ type.

(C) The central atom $Br$ has $7$ valence electrons.
It forms $3$ covalent bonds with $3$ $F$ atoms, utilizing $3$ electrons.
Remaining electrons = $7 - 3 = 4$ electrons.
These $4$ electrons form $2$ lone pairs.
Thus, there are $2$ lone pairs and $3$ bond pairs.

(C) $1$. Both $NH_3$ and $NF_3$ have a pyramidal geometry with a lone pair on the $N$ atom.
$2$. In $NH_3$, the dipole moments of the three $N-H$ bonds are directed towards the more electronegative $N$ atom, and the dipole moment of the lone pair is also in the same direction. Thus, they add up to a large resultant dipole moment $(1.46 \ D)$.
$3$. In $NF_3$, the $F$ atoms are more electronegative than $N$. The dipole moments of the three $N-F$ bonds are directed away from the $N$ atom, which opposes the direction of the lone pair's dipole moment.
$4$. Consequently, the resultant dipole moment of $NF_3$ $(0.24 \ D)$ is much smaller than that of $NH_3$.
$5$. Therefore, the statement that the orbital dipole due to the lone pair is in the same direction as the resultant dipole moment of the three bonds is incorrect for $NF_3$.

(D) The structure of the given molecule is $HO-CH_2-CH_2-CH(CH_3)_2$.
In this molecule, all carbon atoms are bonded to four other atoms via single bonds.
Therefore, all $5$ carbon atoms are $sp^3$ hybridized.
The correct option is $D$.

(B) According to the $VSEPR$ (Valence Shell Electron Pair Repulsion) theory, the magnitude of repulsion between electron pairs follows the order: $Lp-Lp > Lp-Bp > Bp-Bp$.
This is because a lone pair $(Lp)$ occupies more space around the central atom than a bond pair $(Bp)$, leading to greater repulsion.

(B) $1$. Chalcogens (Group $16$ elements) have $6$ electrons in their valence shell.
$2$. In their hydrides $(H_2X)$, the central chalcogen atom forms $2$ covalent bonds with hydrogen atoms.
$3$. Out of $6$ valence electrons, $2$ are used in bonding, leaving $4$ electrons as non-bonding electrons.
$4$. These $4$ non-bonding electrons form $2$ lone pairs.
$5$. Therefore, there are $2$ bond pairs and $2$ lone pairs.

(C) $1$. Determine the hybridization and geometry of each molecule:
$2$. $SO_2$: $sp^2$ hybridized, bent geometry.
$3$. $BCl_3$: $sp^2$ hybridized, trigonal planar geometry.
$4$. $CO_2$: $sp$ hybridized, linear geometry with a bond angle of $180^\circ$.
$5$. $NH_3$: $sp^3$ hybridized, trigonal pyramidal geometry.
$6$. Therefore, $CO_2$ is the linear molecule.

(C) Step $1$: The electronic configuration of Nitrogen $(Z=7)$ is $1s^2 2s^2 2p^3$.
Step $2$: To complete its octet, each Nitrogen atom shares three electrons, forming a triple bond.
Step $3$: The triple bond consists of one $\sigma$ bond (formed by head-on $p_z-p_z$ overlap) and two $\pi$ bonds (formed by lateral $p_x-p_x$ and $p_y-p_y$ overlap).
Step $4$: Thus, the $N_2$ molecule contains a triple bond involving $p-p$ orbital overlapping.

(D) $1$. $A$ $\sigma$ bond is formed by head-on overlapping, while a $\pi$ bond is formed by lateral (sideways) overlapping.
$2$. Lateral overlapping is less effective than head-on overlapping, making the $\pi$ bond weaker than the $\sigma$ bond.
$3$. $A$ $\pi$ bond can only form between atoms that are already bonded by a $\sigma$ bond.
$4$. The formation of a $\pi$ bond increases the bond order, which results in a decrease in the bond length, not an increase. Therefore, statement $(d)$ is incorrect.

(D) The formal charge is calculated using the formula: $\text{Formal Charge} = V - L - \frac{1}{2}B$, where $V$ is the number of valence electrons, $L$ is the number of lone pair electrons, and $B$ is the number of bonding electrons.
For the sulphur atom in $H_2SO_4$:
$1$. Valence electrons $(V)$ = $6$.
$2$. Lone pair electrons $(L)$ = $0$ (all valence electrons are involved in bonding).
$3$. Bonding electrons $(B)$ = $12$ (sulphur forms $6$ bonds: $2$ double bonds with oxygen atoms and $2$ single bonds with hydroxyl groups).
$4$. $\text{Formal Charge} = 6 - 0 - \frac{1}{2}(12) = 6 - 6 = 0$.

(D) The $3d$ transition series corresponds to elements where the $3d$ orbital is being filled, which are elements with atomic numbers $21$ to $30$.
Electronic configuration of $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$.
Electronic configuration of $Cu$ $(Z=29)$ is $[Ar] 3d^{10} 4s^1$.
Both elements have an $ns^1$ (specifically $4s^1$) configuration and belong to the $3d$ transition series.
Therefore, the correct answer is $24$ and $29$.

(B) $1$. The atomic number of $Cu$ is $29$.
$2$. According to the Aufbau principle, the expected configuration is $[Ar] 3d^9 4s^2$.
$3$. However, a completely filled $d$-subshell $(d^{10})$ provides extra stability due to symmetry and exchange energy.
$4$. Therefore, one electron from the $4s$ orbital shifts to the $3d$ orbital, resulting in the observed configuration: $[Ar] 3d^{10} 4s^1$.

(D) $1$. The reducing property of hydrides of Group $16$ elements depends on the bond dissociation enthalpy of the $E-H$ bond.
$2$. As we move down the group from $O$ to $Te$, the atomic size increases, which leads to an increase in the bond length and a decrease in the bond dissociation enthalpy.
$3$. $A$ weaker $E-H$ bond is more easily broken, making the hydride a stronger reducing agent.
$4$. Since the $H-Te$ bond is the weakest among the given hydrides, $H_2Te$ has the highest reducing property.

(D) $1$. Acidic strength of hydrides of group $16$ elements depends on the bond dissociation enthalpy of the $H-E$ bond.
$2$. As we move down the group from $O$ to $Te$, the atomic size increases, which leads to an increase in the bond length and a decrease in the $H-E$ bond dissociation enthalpy.
$3$. $A$ weaker $H-E$ bond is easier to break, making it easier to release $H^+$ ions.
$4$. Therefore, the acidic strength increases in the order: $H_2O < H_2S < H_2Se < H_2Te$.
$5$. Thus, $H_2Te$ is the most acidic compound.

(A) $1$. The acidity of hydrides of group $16$ elements increases down the group.
$2$. As we move down the group from $S$ to $Te$, the atomic size increases.
$3$. The increase in atomic size leads to a decrease in the bond dissociation enthalpy of the $H-E$ bond (where $E = S, Se, Te$).
$4$. $A$ weaker $H-E$ bond makes it easier for the molecule to release $H^+$ ions in an aqueous solution.
$5$. Therefore, the acidity increases in the order: $H_2S < H_2Se < H_2Te$.

(A) Step $1$: $Astatine$ $(At)$ is a chemical element with atomic number $85$.
Step $2$: It belongs to the halogen family, which is located in Group $17$ of the modern periodic table.
Step $3$: Therefore, the correct group is $17$.

(D) $1$. Solubility of a gas in water depends on the nature of the gas and its interaction with water.
$2$. $NH_3$ and $HCl$ are polar gases that react with water or form strong hydrogen bonds, making them highly soluble.
$3$. $CO_2$ is slightly polar and reacts to some extent with water to form carbonic acid $(H_2CO_3)$, making it moderately soluble.
$4$. $N_2$ is a non-polar, diatomic molecule with very weak van der Waals forces, resulting in very low solubility in water.
$5$. Therefore, $N_2$ is the least soluble gas among the given options.

(D) $1$. Solubility of a gas in water depends on the nature of the gas and its interaction with water.
$2$. $HCl$, $NH_3$, and $CO_2$ are polar or capable of reacting with water (e.g., $NH_3$ forms $NH_4OH$, $CO_2$ forms $H_2CO_3$, $HCl$ ionizes).
$3$. $O_2$ is a non-polar diatomic molecule with very weak van der Waals forces, making it the least soluble in water compared to the others.
$4$. Therefore, $O_2$ is the least soluble gas.

(D) $1$. Boiling point depends on the molecular mass and the magnitude of van der Waals forces (dipole-dipole and London dispersion forces).
$2$. As the number of bromine atoms increases in the haloalkane, the molecular mass increases significantly.
$3$. Higher molecular mass leads to stronger van der Waals forces, resulting in a higher boiling point.
$4$. Comparing the compounds: $CHBr_3$ (tribromomethane) has the highest molecular mass, followed by $CH_2Br_2$ (dibromomethane), then $CH_3Br$ (bromomethane), and finally $CH_3Cl$ (chloromethane) which has the lowest molecular mass.
$5$. Therefore, the decreasing order is $CHBr_3 > CH_2Br_2 > CH_3Br > CH_3Cl$.

(A) $1$. The physical state of an ether depends on its molecular mass and intermolecular forces.
$2$. Dimethyl ether $(CH_3OCH_3)$ has the lowest molecular mass among the given options.
$3$. Due to its low molecular mass, it has a very low boiling point of approximately $-24 \ ^\circ C$.
$4$. Since room temperature is typically around $25 \ ^\circ C$, dimethyl ether exists as a gas at room temperature, while the others are liquids.

(D) $1$. Calculate the steric number for $XeOF_4$: $Steric \ number = \frac{1}{2} (V + M - C + A) = \frac{1}{2} (8 + 4 + 0 - 0) = 6$.
$2$. $A$ steric number of $6$ corresponds to $sp^3d^2$ hybridization.
$3$. In $XeOF_4$, the central atom $Xe$ is bonded to $4$ fluorine atoms and $1$ oxygen atom (double bond), leaving $1$ lone pair.
$4$. Due to the presence of $5$ bond pairs and $1$ lone pair, the geometry is octahedral, but the shape is square pyramidal.

(D) $1$. Quaternary ammonium salts have the general formula $[R_4N]^+X^-$.
$2$. The four $C-N$ bonds are covalent bonds.
$3$. One of these $C-N$ bonds is formed by the donation of a lone pair from the nitrogen atom to the fourth alkyl group, which is a coordinate (dative) bond.
$4$. The electrostatic attraction between the quaternary ammonium cation $[R_4N]^+$ and the anion $X^-$ constitutes an ionic bond.
$5$. Therefore, all three types of bonds are present.

(C) $1$. In the titration of $NaCl$ with $AgNO_3$ using fluorescein as an adsorption indicator, the fluorescein ions are adsorbed on the surface of the $AgCl$ precipitate at the equivalence point.
$2$. Before the equivalence point, the precipitate is negatively charged due to the adsorption of $Cl^-$ ions.
$3$. At the equivalence point, the surface of the $AgCl$ precipitate becomes positively charged due to the excess of $Ag^+$ ions.
$4$. The negatively charged fluorescein ions are then adsorbed onto the positively charged $AgCl$ surface, forming a reddish-pink complex.
$5$. Therefore, the colour change observed at the endpoint is from yellowish-green (or pale yellow) to reddish-pink.

(B) In a dry cell, the cathode reaction involves the reduction of manganese dioxide $(MnO_2)$ to manganese oxyhydroxide $(MnO(OH))$.
In $MnO_2$, the oxidation state of $Mn$ is $+4$.
In $MnO(OH)$, the oxidation state of $Mn$ is $+3$.
Therefore, the oxidation number of $Mn$ decreases by $1$ $(+4 \rightarrow +3)$.

(D) Let the oxidation state of $Xe$ be $x$.
The oxidation state of $O$ is $-2$ and $F$ is $-1$.
Sum of oxidation states in a neutral molecule is $0$.
$x + (-2) + 4(-1) = 0$
$x - 2 - 4 = 0$
$x - 6 = 0$
$x = +6$
Therefore, the oxidation state of $Xe$ is $+6$.

(D) Step $1$: $\Delta G^{\circ}$ (Standard Gibbs energy) depends on the amount of substance, so it is an extensive property. This statement is correct.
Step $2$: $E^{\circ}_{\text{cell}}$ (Standard cell potential) is independent of the amount of substance, so it is an intensive property. This statement is correct.
Step $3$: For a spontaneous process at constant temperature and pressure, $\Delta G < 0$ (negative). This statement is correct.
Step $4$: Electrical work done by a cell is given by $W_{\text{elec}} = -\Delta G$. Therefore, electrical work is equal to the negative of $\Delta G$, not $\Delta G$ itself. This statement is incorrect.

(B) Step $1$: Entropy $(S)$ is a measure of the randomness or disorder of a system. $A$ negative $\Delta S$ implies a decrease in randomness.
Step $2$: Dissolving sugar in water increases disorder as the solid lattice breaks down into solution $(\Delta S > 0)$.
Step $3$: Adsorption of a gas on a solid surface restricts the movement of gas molecules, leading to a decrease in disorder $(\Delta S < 0)$.
Step $4$: The reaction $2NH_3(g) \rightarrow N_2(g) + 3H_2(g)$ increases the number of moles of gas from $2$ to $4$, increasing disorder $(\Delta S > 0)$.
Step $5$: Sublimation of dry ice $(CO_2(s) \rightarrow CO_2(g))$ involves a transition from solid to gas, which significantly increases disorder $(\Delta S > 0)$.
Conclusion: Option $B$ is the correct answer.

(C) Step $1$: Enantiomers are non-superimposable mirror images of each other. Therefore, the statement in option $(C)$ is false.
Step $2$: $S_N1$ reactions proceed via a carbocation intermediate, leading to racemization ($1$:$1$ mixture of enantiomers).
Step $3$: $A$ racemic mixture contains equal amounts of two enantiomers, resulting in zero net optical rotation.
Step $4$: $S_N2$ reactions involve a backside attack, resulting in the Walden inversion of configuration.

(B) compound is optically active if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
$(1)$ $3-\text{Bromohexane}$: $CH_3CH_2CH(Br)CH_2CH_2CH_3$. The $C3$ atom is bonded to $-H, -Br, -CH_2CH_3, -CH_2CH_2CH_3$. It is chiral.
$(2)$ $2-\text{Bromo}-2-\text{methylbutane}$: $CH_3C(Br)(CH_3)CH_2CH_3$. The $C2$ atom is bonded to two identical $-CH_3$ groups. It is achiral.
$(3)$ $2-\text{Bromopentane}$: $CH_3CH(Br)CH_2CH_2CH_3$. The $C2$ atom is bonded to $-H, -Br, -CH_3, -CH_2CH_2CH_3$. It is chiral.
$(4)$ $2-\text{Bromo}-3-\text{methylbutane}$: $CH_3CH(Br)CH(CH_3)_2$. The $C2$ atom is bonded to $-H, -Br, -CH_3, -CH(CH_3)_2$. It is chiral.
Therefore, $2-\text{Bromo}-2-\text{methylbutane}$ is not optically active.

(C) $1$. Acidic strength of phenols depends on the stability of the phenoxide ion formed after the loss of a proton.
$2$. Electron-withdrawing groups $(-EWG)$ increase acidic strength by stabilizing the phenoxide ion, while electron-donating groups $(-EDG)$ decrease it.
$3$. The substituents at the $p$-position are:
- $p$-nitrophenol $(III)$: $-NO_2$ is a strong $-I$ and $-M$ group (strongest acid).
- $p$-chlorophenol $(I)$: $-Cl$ is $-I$ (electron-withdrawing) but $+M$ (electron-donating). The $-I$ effect dominates, making it more acidic than phenol.
- $p$-Cresol $(II)$: $-CH_3$ is $+I$ and hyperconjugation (weakly electron-donating).
- $p$-methoxyphenol $(IV)$: $-OCH_3$ is $-I$ but strong $+M$ (strongly electron-donating).
$4$. The order of electron-donating/withdrawing ability is: $-NO_2$ $(-M)$ > $-Cl$ $(-I > +M)$ > $-CH_3$ $(+I)$ > $-OCH_3$ $(+M)$.
$5$. Thus, the decreasing order of acidic strength is: $III > I > II > IV$.

(B) An aromatic alcohol is a compound in which the hydroxyl group $(-OH)$ is attached to a carbon atom that is part of a side chain attached to an aromatic ring, not directly to the aromatic ring itself.
$(I)$ Phenol: The $-OH$ group is directly attached to the benzene ring. This is a phenol, not an alcohol.
$(II)$ Benzyl alcohol $(C_6H_5CH_2OH)$: The $-OH$ group is attached to a carbon atom of the side chain. This is an aromatic alcohol.
$(III)$ $m$-Ethylbenzyl alcohol: The $-OH$ group is attached to a carbon atom of the side chain. This is an aromatic alcohol.
$(IV)$ $m$-Ethylphenol: The $-OH$ group is directly attached to the benzene ring. This is a phenol.
Therefore, $(II)$ and $(III)$ are aromatic alcohols.

(B) The chemical formula of mustard gas is $(ClCH_2CH_2)_2S$.
It consists of carbon $(C)$, hydrogen $(H)$, sulfur $(S)$, and chlorine $(Cl)$ atoms.
Nitrogen $(N)$ is not present in mustard gas.

(C) $1$. Acrolein is $CH_2=CH-CHO$ (one $-CHO$ group).
$2$. o-Tolualdehyde is $C_6H_4(CH_3)CHO$ (one $-CHO$ group).
$3$. Phthaldehyde (also known as $o$-phthalaldehyde) has the structure $C_6H_4(CHO)_2$, which contains two $-CHO$ groups attached to the benzene ring at ortho positions.
$4$. Butyraldehyde is $CH_3CH_2CH_2CHO$ (one $-CHO$ group).
$5$. Therefore, Phthaldehyde is the correct answer.

(A) $1$. $A$ benzylic alcohol is one where the $-OH$ group is attached to a carbon atom that is bonded to a benzene ring.
$2$. $A$ primary benzylic alcohol has the structure $Ar-CH_2OH$, where the carbon atom attached to the $-OH$ group is bonded to only one other carbon atom (the benzene ring).
$3$. In $C_6H_5CH_2OH$ (benzyl alcohol), the $-OH$ group is attached to a $CH_2$ group which is directly bonded to the benzene ring $(C_6H_5)$. Thus, it is a primary benzylic alcohol.
$4$. $C_6H_5 - CH(CH_3)OH$ is a secondary benzylic alcohol.
$5$. The other options are not benzylic alcohols as they lack a benzene ring.

(A) $1$. The structure of Methanal is $H-CHO$. Here, the $-CHO$ group is attached to a hydrogen atom, not an $sp^3$ hybridized carbon atom.
$2$. The structure of Ethanal is $CH_3-CHO$. Here, the $-CHO$ group is attached to a $CH_3$ group, where the carbon is $sp^3$ hybridized.
$3$. The structure of $2-$Methylpropanal is $(CH_3)_2CH-CHO$. Here, the $-CHO$ group is attached to a $CH$ group, where the carbon is $sp^3$ hybridized.
$4$. The structure of Propanal is $CH_3-CH_2-CHO$. Here, the $-CHO$ group is attached to a $CH_2$ group, where the carbon is $sp^3$ hybridized.
$5$. Therefore, Methanal is the only aldehyde where the $-CHO$ group is not attached to an $sp^3$ hybridized carbon atom.

(D) $1$. $A$ phenol is an organic compound in which a hydroxyl group $(-OH)$ is directly attached to a carbon atom of an aromatic ring.
$2$. In options $(A)$, $(B)$, and $(C)$, the $-OH$ group is directly bonded to the benzene or naphthalene ring, making them phenols.
$3$. In option $(D)$, the $-OH$ group is attached to a side-chain carbon $(-CH_2-)$ which is then attached to the benzene ring. This compound is benzyl alcohol, not a phenol.
$4$. Therefore, the correct answer is $(D)$.

(C) An allylic halide is a compound in which the halogen atom is bonded to an $sp^3$ hybridized carbon atom adjacent to a carbon-carbon double bond $(C=C)$.
In option $(C)$, $CH_2 = CH - CH_2 - X$, the halogen atom $X$ is attached to a carbon atom that is adjacent to the $C=C$ double bond, making it an allylic halide.
Option $(A)$ is a primary alkyl halide.
Option $(B)$ is a benzylic halide.
Option $(D)$ is a vinylic halide.

(D) benzylic alcohol is one where the $-OH$ group is attached to a carbon atom that is bonded to a benzene ring.
$(1)$ $A$ primary benzylic alcohol has the $-OH$ group on a carbon attached to one other carbon (besides the benzene ring).
$(2)$ $A$ secondary benzylic alcohol has the $-OH$ group on a carbon attached to two other carbons.
$(3)$ $A$ tertiary benzylic alcohol has the $-OH$ group on a carbon attached to three other carbons (one of which is the benzene ring).
In option $D$, the carbon atom bearing the $-OH$ group is directly attached to the benzene ring and two methyl groups. Thus, it is a tertiary benzylic alcohol.

(A) $1$. $A$ haloalkyne is an organic compound containing a halogen atom bonded to a carbon atom that is part of a carbon-carbon triple bond $(C \equiv C)$.
$2$. In option $A$, $CH_3-C \equiv C-Cl$, the chlorine atom is directly attached to a carbon involved in a triple bond.
$3$. Option $B$ is a haloalkene, option $C$ is a haloalkane, and option $D$ is a haloarene.

(C) $1$. Schottky defect is a type of point defect in ionic crystals where an equal number of cations and anions are missing from their lattice sites.
$2$. Because equal numbers of cations and anions are missing, the stoichiometry remains the same.
$3$. Since the charges of the missing cations and anions balance each other, the electrical neutrality is maintained.
$4$. Because atoms/ions are missing from the lattice, the mass of the crystal decreases while the volume remains constant, leading to a decrease in density.
$5$. Therefore, the statement that the density remains unchanged is incorrect.

(A) Step $1$: $Schottky$ defect is a type of vacancy defect in ionic solids.
Step $2$: In this defect, an equal number of cations and anions are missing from their lattice sites to maintain electrical neutrality.
Step $3$: Due to the missing ions, the mass of the crystal decreases while the volume remains constant, leading to a decrease in the density of the substance.
Step $4$: It is typically found in ionic compounds where the cation and anion sizes are similar. Therefore, option $A$ is correct.

(B) Steel is an alloy of iron $(Fe)$ and carbon $(C)$. In the crystal lattice of iron, carbon atoms are much smaller than iron atoms. These small carbon atoms occupy the interstitial sites (the empty spaces between the lattice points) of the iron crystal lattice. Therefore, steel exhibits an interstitial impurity defect.

(B) In a close-packed three-dimensional structure, a tetrahedral hole (or void) is formed by the arrangement of $4$ spheres.
These $4$ spheres are located at the corners of a regular tetrahedron.
Therefore, the total number of spheres surrounding a tetrahedral void is $4$.

(C) $1$. Identify the number of formula units per unit cell $(Z)$: The anions are at the $8$ corners, contributing $8 \times \frac{1}{8} = 1$ anion. The cation is at the body center, contributing $1$ cation. Thus, $Z = 1$.
$2$. Use the density formula: $d = \frac{Z \times M}{N_A \times V}$, where $V$ is the volume of the unit cell.
$3$. Rearrange to solve for $V$: $V = \frac{Z \times M}{d \times N_A}$.
$4$. Substitute the values: $V = \frac{1 \times 168.6 \text{ g mol}^{-1}}{4 \text{ g cm}^{-3} \times 6.022 \times 10^{23} \text{ mol}^{-1}}$.
$5$. Calculate: $V = \frac{168.6}{24.088 \times 10^{23}} \approx 7.0 \times 10^{-23} \text{ cm}^3$.

(A) For an $FCC$ (face-centered cubic) unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $4r = a\sqrt{2}$.
Given, edge length $a = 392 \text{ pm}$.
Substituting the value: $r = \frac{a\sqrt{2}}{4} = \frac{a}{2\sqrt{2}}$.
$r = \frac{392}{2 \times 1.414} = \frac{392}{2.828} \approx 138.6 \text{ pm}$.

(D) In an $fcc$ crystal lattice, each particle is in contact with $12$ other particles.
Specifically, each atom has $4$ neighbours in its own layer, $4$ neighbours in the layer above, and $4$ neighbours in the layer below.
Therefore, the coordination number is $12$.

(B) Step $1$: Identify the given parameters. For a $bcc$ unit cell, the number of atoms per unit cell $Z = 2$. Edge length $a = 400 \text{ pm} = 400 \times 10^{-10} \text{ cm} = 4 \times 10^{-8} \text{ cm}$. Density $d = 3.5 \text{ g cm}^{-3}$. Avogadro's number $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$.
Step $2$: Use the density formula for a unit cell: $d = \frac{Z \times M}{a^3 \times N_A}$.
Step $3$: Rearrange the formula to solve for molar mass $M$: $M = \frac{d \times a^3 \times N_A}{Z}$.
Step $4$: Substitute the values: $M = \frac{3.5 \times (4 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{2}$.
Step $5$: Calculate $a^3 = 64 \times 10^{-24} \text{ cm}^3$. $M = \frac{3.5 \times 64 \times 10^{-24} \times 6.022 \times 10^{23}}{2} = \frac{3.5 \times 64 \times 0.6022}{2} = 3.5 \times 32 \times 0.6022 = 67.4464 \text{ g mol}^{-1}$.
Thus, the molar mass is approximately $67.4 \text{ g mol}^{-1}$.

(B) $1$. Number of corners in a unit cell = $8$. Each corner atom contributes $1/8$ to the unit cell.
$2$. Number of $A$ atoms = $7 \times (1/8) = 7/8$.
$3$. Number of face centers in a unit cell = $6$. Each face center atom contributes $1/2$ to the unit cell.
$4$. Number of $B$ atoms = $6 \times (1/2) = 3$.
$5$. Ratio of $A:B = 7/8 : 3 = 7 : 24$.
$6$. Therefore, the simplest formula is $A_7B_{24}$.

(A) Step $1$: Calculate the number of atoms of $A$ per unit cell. Since $A$ atoms are at the $8$ corners, each contributes $1/8$ to the unit cell. Number of $A = 8 \times (1/8) = 1$.
Step $2$: Calculate the number of atoms of $B$ per unit cell. Since $B$ atom is at the body center, it contributes $1$ to the unit cell. Number of $B = 1$.
Step $3$: Determine the ratio of $A$ to $B$. The ratio is $1:1$.
Step $4$: The formula of the compound is $AB$.

(C) The density of a unit cell is given by the formula: $d = \frac{Z \times M}{a^3 \times N_A}$
Where $d$ is density, $Z$ is the number of atoms per unit cell, $M$ is molar mass, $a^3$ is the volume of the unit cell, and $N_A$ is Avogadro's number.
Given: $d = 2.7 \text{ g cm}^{-3}$, $M = 27 \text{ g mol}^{-1}$, and $a^3 \times N_A = 40 \text{ cm}^3 \text{ mol}^{-1}$.
Rearranging the formula to solve for $Z$: $Z = \frac{d \times a^3 \times N_A}{M}$
Substituting the values: $Z = \frac{2.7 \times 40}{27}$
$Z = \frac{108}{27} = 4$
Thus, the number of atoms present in the unit cell is $4$.

(A) $1$. The volume of the unit cell $V_{cell} = a^3 = (3.36 \times 10^{-8} \text{ cm})^3 = 37.933 \times 10^{-24} \text{ cm}^3 = 3.7933 \times 10^{-23} \text{ cm}^3$.
$2$. For a simple cubic unit cell, the packing efficiency is $\frac{\pi}{6} \approx 0.524$.
$3$. The occupied volume $V_{occ} = 0.524 \times V_{cell} = 0.524 \times 3.7933 \times 10^{-23} \text{ cm}^3 \approx 1.987 \times 10^{-23} \text{ cm}^3$.
$4$. The void volume $V_{void} = V_{cell} - V_{occ} = V_{cell} \times (1 - 0.524) = 3.7933 \times 10^{-23} \times 0.476 \approx 1.8056 \times 10^{-23} \text{ cm}^3$.
$5$. The closest value is $1.807 \times 10^{-23} \text{ cm}^3$.

(A) $1$. Let the number of atoms of element $Y$ in the $FCC$ lattice be $n = 4$.
$2$. The number of tetrahedral voids in an $FCC$ lattice is $2n = 2 \times 4 = 8$.
$3$. Element $X$ occupies one-third of the tetrahedral voids, so the number of atoms of $X$ is $\frac{1}{3} \times 8 = \frac{8}{3}$.
$4$. The ratio of atoms $X:Y$ is $\frac{8}{3} : 4 = 8 : 12 = 2 : 3$.
$5$. Therefore, the formula of the compound is $X_2Y_3$.

(D) Step $1$: The mass of a single unit cell is given by the product of density $(\rho)$ and volume of the unit cell $(V_{uc})$.
Mass of one unit cell = $\rho \times V_{uc} = 2.5 \times 10^{-22} \text{ g}$.
Step $2$: The number of unit cells in a given mass $(m)$ is calculated using the formula: $\text{Number of unit cells} = \frac{\text{Total mass}}{\text{Mass of one unit cell}}$.
Step $3$: Substitute the given values: $\text{Number of unit cells} = \frac{1 \text{ g}}{2.5 \times 10^{-22} \text{ g}} = \frac{1}{2.5} \times 10^{22} = 0.4 \times 10^{22} = 4.0 \times 10^{21}$.
Thus, the correct option is $D$.

(C) For an $fcc$ unit cell, the number of atoms per unit cell $Z = 4$.
Given: Density $d = 8.92 \text{ g cm}^{-3}$, edge length $a = 3.608 \times 10^{-8} \text{ cm}$, $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$.
The formula for density is $d = \frac{Z \times M}{a^3 \times N_A}$.
Rearranging for molar mass $M$: $M = \frac{d \times a^3 \times N_A}{Z}$.
$M = \frac{8.92 \times (3.608 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{4}$.
$M = \frac{8.92 \times 46.97 \times 10^{-24} \times 6.022 \times 10^{23}}{4}$.
$M = \frac{252.36}{4} \approx 63.09 \text{ g/mol}$.
Thus, the atomic mass is approximately $63 \text{ g/mol}$.

(B) $1$. Let the number of anions $(B)$ in the $hcp$ lattice be $N$.
$2$. The number of tetrahedral voids in an $hcp$ lattice is $2N$.
$3$. Cations $(A)$ occupy $1/2$ of the tetrahedral voids, so the number of cations = $\frac{1}{2} \times 2N = N$.
$4$. The ratio of cations $(A)$ to anions $(B)$ is $N : N = 1 : 1$.
$5$. Therefore, the formula of the ionic compound is $AB$.

(A) $1$. The packing efficiency of a simple cubic unit cell is $52.4\%$, meaning $52.4\%$ of the total volume is occupied by atoms.
$2$. The void volume (empty space) is $100\% - 52.4\% = 47.6\%$ of the total volume.
$3$. Let the total volume be $V$. Given that $47.6\% \text{ of } V = 2.0 \times 10^{-23} \text{ cm}^3$.
$4$. $0.476 \times V = 2.0 \times 10^{-23} \text{ cm}^3$.
$5$. $V = \frac{2.0 \times 10^{-23}}{0.476} \approx 4.2 \times 10^{-23} \text{ cm}^3$.

(A) For a $bcc$ unit cell, the number of atoms per unit cell, $Z = 2$.
The formula for density $d$ is given by: $d = \frac{Z \times M}{a^3 \times N_A}$.
Given: $M = 225 \text{ g mol}^{-1}$, $Z = 2$, and $a^3 \times N_A = 75 \text{ cm}^3 \text{ mol}^{-1}$.
Substituting the values: $d = \frac{2 \times 225}{75}$.
$d = \frac{450}{75} = 6.0 \text{ g cm}^{-3}$.
Thus, the density is $6.0 \text{ g cm}^{-3}$.

(C) In a simple cubic unit cell, each atom is at the corner of the cube.
Each corner atom is shared by $8$ adjacent unit cells.
In the crystal lattice, each atom is in direct contact with $6$ nearest neighbors (one along each axis: $x, -x, y, -y, z, -z$).
Therefore, the coordination number of a particle in a simple cubic structure is $6$.

(B) For an $fcc$ (face-centered cubic) unit cell, the relationship between the edge length '$a$' and the atomic radius '$r$' is given by: $4r = a\sqrt{2}$.
Given: $a = 360 \text{ pm}$.
Substituting the value: $r = \frac{a\sqrt{2}}{4} = \frac{a}{2\sqrt{2}}$.
$r = \frac{360}{2 \times 1.414} = \frac{360}{2.828} \approx 127.29 \text{ pm}$.
Rounding to the nearest option, the radius is $127.26 \text{ pm}$.

(B) $1$. For an $fcc$ unit cell, the number of particles per unit cell $(Z)$ is $4$.
$2$. The volume of one particle (assuming spherical shape) is $V_p = \frac{4}{3} \pi r^3$.
$3$. For $fcc$, the relation between edge length $(a)$ and radius $(r)$ is $a = 2\sqrt{2}r$, so $r = \frac{a}{2\sqrt{2}}$.
$4$. Volume of $4$ particles = $4 \times \frac{4}{3} \pi (\frac{a}{2\sqrt{2}})^3 = \frac{16}{3} \pi \frac{a^3}{16\sqrt{2}} = \frac{\pi a^3}{3\sqrt{2}}$.
$5$. Given $a = 200 \text{ pm} = 2 \times 10^{-8} \text{ cm}$.
$6$. Volume = $\frac{3.14159 \times (2 \times 10^{-8})^3}{3 \times 1.414} = \frac{3.14159 \times 8 \times 10^{-24}}{4.242} \approx 5.92 \times 10^{-24} \text{ cm}^3$.