MHT CET 2025 Chemistry Question Paper with Answer and Solution

(D) The atomic ratio of $Xe : F$ is given as $0.4 : 2.4$.
Dividing both by $0.4$,we get the molar ratio as $1 : 6$.
Thus,the empirical formula of the compound is $XeF_6$.
In $XeF_6$,let the oxidation number of $Xe$ be $x$.
Since the oxidation number of $F$ is $-1$,we have:
$x + 6(-1) = 0$
$x - 6 = 0$
$x = +6$.
Therefore,the oxidation number of $Xe$ is $+6$.

(C) In the given reaction: $Ag_2O + H_2O + 2e^- \rightarrow 2Ag + 2OH^-$
$1$. The oxidation state of $Ag$ in $Ag_2O$ is $+1$,and it changes to $0$ in $Ag$ metal.
$2$. Since the oxidation number of $Ag$ decreases from $+1$ to $0$,it undergoes reduction.
$3$. Therefore,$Ag$ is reduced.
$4$. The oxidation states of $H$ $(+1)$ and $O$ $(-2)$ remain unchanged in $H_2O$ and $OH^-$,so neither oxidation nor reduction occurs for these elements.

(C) The chemical formula for calcium phosphate is $Ca_3(PO_4)_2$.
Let the oxidation number of phosphorus be $x$.
The oxidation number of calcium $(Ca)$ is $+2$ and oxygen $(O)$ is $-2$.
For the neutral compound $Ca_3(PO_4)_2$,the sum of oxidation numbers must be zero:
$3(+2) + 2[x + 4(-2)] = 0$
$6 + 2[x - 8] = 0$
$6 + 2x - 16 = 0$
$2x - 10 = 0$
$2x = 10$
$x = +5$
Therefore,the oxidation number of phosphorus is $+5$.

(C) Let the oxidation number of sulphur $(S)$ be $x$.
In $SO_3$,the oxidation number of oxygen $(O)$ is $-2$.
The sum of oxidation numbers of all atoms in a neutral molecule is $0$.
Therefore,$x + 3(-2) = 0$.
$x - 6 = 0$.
$x = +6$.
Thus,the oxidation number of sulphur in $SO_3$ is $+6$.

(A) To find the value of $x$,we need to balance the charge on both sides of the equation.
In $BiO_3^{-}$,the oxidation state of $Bi$ is $x + 3(-2) = -1$,so $x = +5$.
In $Bi^{3+}$,the oxidation state of $Bi$ is $+3$.
The change in oxidation state is $+5 - (+3) = 2$.
However,we must also balance the total charge.
Total charge on the left side: $(-1) + 6(+1) + x(-1) = 5 - x$.
Total charge on the right side: $(+3) + 3(0) = +3$.
Equating the charges: $5 - x = 3$,which gives $x = 2$.

(C) The chemical formula for methanal is $HCHO$.
Let the oxidation number of carbon be $x$.
The oxidation number of hydrogen is $+1$ and that of oxygen is $-2$.
The sum of oxidation numbers in a neutral molecule is $0$.
So,$1(+1) + x + 1(-2) + 1(+1) = 0$.
$1 + x - 2 + 1 = 0$.
$x = 0$.
Therefore,the oxidation number of carbon in methanal is $0$.

(C) To find the oxidation state of carbon in each compound,we use the rule that the sum of oxidation states of all atoms in a neutral molecule is $0$. Let the oxidation state of carbon be $x$.
$1$. In $HCHO$: $2(+1) + x + (-2) = 0 \implies x = 0$.
$2$. In $CH_3OH$: $x + 3(+1) + (-2) + (+1) = 0 \implies x + 2 = 0 \implies x = -2$.
$3$. In $CHCl_3$: $x + 1(+1) + 3(-1) = 0 \implies x - 2 = 0 \implies x = +2$.
$4$. In $C_{12}H_{22}O_{11}$ (Sucrose): $12x + 22(+1) + 11(-2) = 0 \implies 12x + 22 - 22 = 0 \implies 12x = 0 \implies x = 0$.
Comparing the values $(0, -2, +2, 0)$,the maximum oxidation state is $+2$ in $CHCl_3$.

(A) Lithium carbonate $(Li_2CO_3)$ is thermally unstable compared to other alkali metal carbonates. Upon heating,it decomposes to form lithium oxide $(Li_2O)$ and carbon dioxide $(CO_2)$.
The balanced chemical equation is:
$Li_2CO_3 \xrightarrow{\Delta} Li_2O + CO_2$

(C) Alkali metals react with oxygen to form different types of oxides depending on their size and ionization energy.
$Li$ forms only the monoxide $(Li_2O)$.
$Na$ forms the peroxide $(Na_2O_2)$.
$K$,$Rb$,and $Cs$ form superoxides $(MO_2)$ when reacted with excess air or oxygen.
Therefore,$K$ is the correct answer.

(C) The alkali metals are the elements of Group $1$ of the periodic table,which include Lithium $(Li)$,Sodium $(Na)$,Potassium $(K)$,Rubidium $(Rb)$,Caesium $(Cs)$,and Francium $(Fr)$.
Beryllium $(Be)$ belongs to Group $2$ of the periodic table,which are known as alkaline earth metals.
Therefore,Beryllium is not an alkali metal.

(C) Among the given options,$Na$ (Sodium) is an alkali metal. Alkali metals react with oxygen to form oxides,peroxides,or superoxides depending on their size. $Na$ reacts with excess oxygen to form sodium peroxide $(Na_2O_2)$.

(D) When magnesium burns in air,it reacts with both oxygen and nitrogen present in the atmosphere.
The chemical reactions are as follows:
$2Mg(s) + O_2(g) \rightarrow 2MgO(s)$
$3Mg(s) + N_2(g) \rightarrow Mg_3N_2(s)$
Therefore,the products formed are a mixture of magnesium oxide $(MgO)$ and magnesium nitride $(Mg_3N_2)$.

(B) When alkaline earth metals $(M)$ are dissolved in liquid ammonia,they form ammoniated metal cations and solvated electrons,represented by the reaction: $M + (x+y)NH_3 \rightarrow [M(NH_3)_x]^{2+} + 2[e(NH_3)_y]^-$.
These solvated electrons are responsible for the characteristic deep blue colour of the solution.
Therefore,the correct option is $B$.

(C) Quicklime $(CaO)$ is prepared by the thermal decomposition of limestone $(CaCO_3)$.
The chemical reaction is as follows:
$CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g)$
Therefore,calcium carbonate is the correct compound used for the preparation of quicklime.

(B) At $STP$ (Standard Temperature and Pressure),the molar volume of an ideal gas is $22.4 \ dm^3 \ mol^{-1}$.
Given the amount of substance $n = 0.5 \ mol$.
The volume $V$ is calculated as $V = n \times \text{molar volume}$.
$V = 0.5 \ mol \times 22.4 \ dm^3 \ mol^{-1} = 11.2 \ dm^3$.
Therefore,the correct option is $B$.

(C) The balanced chemical equation for the thermal decomposition of potassium chlorate is:
$2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$
From the stoichiometry,$2 \ mol$ of $KClO_3$ produces $3 \ mol$ of $O_2$.
At $STP$,$1 \ mol$ of any gas occupies $22.4 \ dm^3$.
Therefore,$3 \ mol$ of $O_2$ occupies $3 \times 22.4 \ dm^3 = 67.2 \ dm^3$.
Mass of $2 \ mol$ of $KClO_3 = 2 \times 122.5 \ g = 245 \ g$.
Since $67.2 \ dm^3$ of $O_2$ is produced by $245 \ g$ of $KClO_3$,
$5.6 \ dm^3$ of $O_2$ will be produced by:
$\frac{245 \ g}{67.2 \ dm^3} \times 5.6 \ dm^3 = 20.416 \ g \approx 20.40 \ g$.

(C) The molar mass of water $(H_2O)$ is calculated as:
$M = (2 \times 1.008 \ g/mol) + (1 \times 16.00 \ g/mol) = 18.016 \ g/mol \approx 18 \ g/mol$.
To find the mass,we use the formula:
$\text{Mass} = \text{moles} \times \text{molar mass}$.
$\text{Mass} = 0.25 \ mol \times 18 \ g/mol = 4.5 \ g$.
Therefore,the correct option is $C$.

(D) At $STP$ (Standard Temperature and Pressure),the molar volume of an ideal gas is $22.4 \ L \ mol^{-1}$.
Number of moles $(n)$ is calculated using the formula: $n = \frac{\text{Given Volume}}{\text{Molar Volume at STP}}$.
Substituting the given values: $n = \frac{0.448 \ L}{22.4 \ L \ mol^{-1}}$.
$n = 0.02 \ mol$.

(B) $1$. Calculate the molar mass of $CaCl_2$: $M = 40 + 2 \times 35.5 = 111 \ g/mol$.
$2$. Calculate the number of moles of $CaCl_2$: $n = \frac{222 \ g}{111 \ g/mol} = 2 \ mol$.
$3$. Since $1 \ mol$ of $CaCl_2$ contains $1 \ mol$ of $Ca^{2+}$ ions,$2 \ mol$ of $CaCl_2$ contains $2 \ mol$ of $Ca^{2+}$ ions.
$4$. The number of $Ca^{2+}$ ions is $2 \times N_A$.

(C) At $STP$,$1 \ mole$ of an ideal gas occupies $22.4 \ dm^3$ volume.
Given volume of gas = $4.48 \ dm^3$.
Number of moles $(n)$ = $\frac{\text{Given Volume}}{\text{Molar Volume at STP}} = \frac{4.48 \ dm^3}{22.4 \ dm^3/mol} = 0.2 \ mol$.
We know that $n = \frac{\text{Mass}}{\text{Molar Mass (M)}}$.
$0.2 \ mol = \frac{5.6 \ g}{M}$.
$M = \frac{5.6 \ g}{0.2 \ mol} = 28 \ g/mol$.
The molar mass of $N_2$ is $2 \times 14 = 28 \ g/mol$.
Therefore,the gas is $N_2$.

(D) The molar mass of anhydrous calcium chloride $(CaCl_2)$ is $40 + 2 \times 35.5 = 111 \ g/mol$.
Number of moles of $CaCl_2 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{222 \ g}{111 \ g/mol} = 2 \ mol$.
Each mole of $CaCl_2$ dissociates to give $2 \ mol$ of $Cl^{-}$ ions.
Total moles of $Cl^{-}$ ions $= 2 \times 2 \ mol = 4 \ mol$.
Number of $Cl^{-}$ ions $= \text{moles} \times N_A = 4 \ N_A$.

(D) At $STP$,the molar volume of an ideal gas is $22400 \ mL \ mol^{-1}$.
Number of moles of water vapour in $1 \ mL$ is $n = \frac{1 \ mL}{22400 \ mL \ mol^{-1}} = 4.464 \times 10^{-5} \ mol$.
The number of molecules is given by $N = n \times N_A$,where $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
$N = 4.464 \times 10^{-5} \times 6.022 \times 10^{23} \approx 2.69 \times 10^{19}$ molecules.
Thus,the correct option is $D$.

(D) The general formula for alkanes is $C_nH_{2n+2}$.
Decane has $n = 10$,so its formula is $C_{10}H_{22}$.
Undecane has $n = 11$,so its formula is $C_{11}H_{24}$.
The difference between the two is one $CH_2$ group.
The molar mass of a $CH_2$ group is $(1 \times 12.01) + (2 \times 1.008) \approx 14.02 \ g \ mol^{-1}$.
Rounding to the nearest whole number,the difference is $14 \ g \ mol^{-1}$.

(C) Let the mass of $O_2$ be $1x$ and the mass of $CH_4$ be $4x$.
Number of moles of $O_2$ $(n_{O_2})$ = $\frac{\text{mass}}{\text{molar mass}} = \frac{1x}{32}$.
Number of moles of $CH_4$ $(n_{CH_4})$ = $\frac{\text{mass}}{\text{molar mass}} = \frac{4x}{16} = \frac{x}{4}$.
The ratio of molecules is equal to the ratio of moles.
Ratio = $n_{O_2} : n_{CH_4} = \frac{x}{32} : \frac{x}{4} = \frac{1}{32} : \frac{1}{4} = 1 : 8$.

(C) At $STP$ (Standard Temperature and Pressure),the molar volume of an ideal gas is $22.7 \ L \ mol^{-1}$ (using the current $IUPAC$ standard of $1 \ bar$ pressure).
Given volume $V = 1 \ m^3 = 1000 \ L$.
The number of moles $n$ is calculated as:
$n = \frac{V}{V_m} = \frac{1000 \ L}{22.7 \ L \ mol^{-1}} \approx 44.05 \ mol$.
If using the older $STP$ definition ($1 \ atm$ pressure),the molar volume is $22.4 \ L \ mol^{-1}$.
$n = \frac{1000 \ L}{22.4 \ L \ mol^{-1}} \approx 44.64 \ mol$.
Rounding to the nearest value provided in the options,the correct answer is $44.6$.

(C) The molar mass of sodium hydroxide $(NaOH)$ is calculated as:
$M = 23.0 + 16.0 + 1.0 = 40.0 \ g/mol$.
Number of moles $(n)$ = $\frac{\text{mass}}{\text{molar mass}} = \frac{0.160 \ g}{40.0 \ g/mol} = 0.004 \ mol$.
Since $1 \ mol = 1000 \ mmol$,the number of millimoles is:
$0.004 \ mol \times 1000 \ mmol/mol = 4.0 \ mmol$.

(A) $1$. Molar mass of methane $(CH_4)$ = $12 + (4 \times 1) = 16 \ g/mol$.
$2$. Number of moles of $CH_4$ = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{1.6 \ g}{16 \ g/mol} = 0.1 \ mol$.
$3$. Number of molecules of $CH_4$ = $0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}$ molecules.
$4$. Each molecule of $CH_4$ contains $6$ (from $C$) + $4 \times 1$ (from $H$) = $10$ electrons.
$5$. Total number of electrons = $6.022 \times 10^{22} \times 10 = 6.022 \times 10^{23}$ electrons.

(C) The molar mass of dinitrogen $(N_2)$ is $2 \times 14 = 28 \ g/mol$.
Number of moles $(n)$ = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{56 \ g}{28 \ g/mol} = 2 \ mol$.
At $STP$,$1 \ mol$ of any gas occupies $22.4 \ L$.
Therefore,the volume of $2 \ mol$ of $N_2$ = $2 \times 22.4 \ L = 44.8 \ L$.

(B) $1$. Calculate the molar mass of methane $(CH_4)$: $12 + 4 \times 1 = 16 \ g/mol$.
$2$. Calculate the number of moles in $3.2 \ g$ of $CH_4$: $n = \frac{3.2 \ g}{16 \ g/mol} = 0.2 \ mol$.
$3$. Calculate the number of molecules: $0.2 \ mol \times 6.022 \times 10^{23} \text{ molecules/mol} = 1.2044 \times 10^{23} \text{ molecules}$.
$4$. Determine the number of electrons in one molecule of $CH_4$: Carbon has $6$ electrons and each Hydrogen has $1$ electron,so $6 + 4(1) = 10$ electrons per molecule.
$5$. Calculate the total number of electrons: $1.2044 \times 10^{23} \text{ molecules} \times 10 \text{ electrons/molecule} = 1.2044 \times 10^{24} \text{ electrons}$.

(A) Step $1$: Calculate the number of moles of urea.
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{5.4 \ g}{60 \ g \ mol^{-1}} = 0.09 \ mol$.
Step $2$: Calculate the number of molecules.
$\text{Number of molecules} = n \times N_A = 0.09 \ mol \times 6.022 \times 10^{23} \ mol^{-1} = 5.4198 \times 10^{22} \approx 5.42 \times 10^{22}$ molecules.

(B) The formula to calculate mass is: $\text{Mass} = \text{Number of moles} \times \text{Molar mass}$.
Given: $\text{Number of moles} = 0.25 \ mol$,$\text{Molar mass} = 56 \ g \ mol^{-1}$.
$\text{Mass} = 0.25 \ mol \times 56 \ g \ mol^{-1} = 14 \ g$.
To convert the mass into kilograms: $14 \ g = 14 \times 10^{-3} \ kg = 1.4 \times 10^{-2} \ kg$.
Therefore,the correct option is $B$.

(C) The chemical formula of urea is $NH_2CONH_2$ or $CH_4N_2O$.
The molar mass of urea is $(12 + 4 \times 1 + 2 \times 14 + 16) = 60 \ g/mol$.
The number of moles of urea in $5.4 \ g$ is $n = \frac{5.4}{60} = 0.09 \ mol$.
Each molecule of urea contains $4$ hydrogen atoms.
Therefore,the number of hydrogen atoms = $n \times 4 \times N_A$,where $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Number of hydrogen atoms = $0.09 \times 4 \times 6.022 \times 10^{23} = 0.36 \times 6.022 \times 10^{23} = 2.16792 \times 10^{23} \approx 2.168 \times 10^{23}$.

(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:
$\text{Sum of formula weights of all reactants} = 274 \ u$
$\text{Atom Economy} = 50 \%$
Substituting the values into the formula:
$50 = \left( \frac{\text{Formula weight of desired product}}{274 \ u} \right) \times 100$
$0.5 = \frac{\text{Formula weight of desired product}}{274 \ u}$
$\text{Formula weight of desired product} = 0.5 \times 274 \ u = 137 \ u$
Therefore,the correct option is $A$.

(B) The vapour density of a gas is defined as the ratio of the mass of a certain volume of the gas to the mass of an equal volume of hydrogen gas at the same temperature and pressure.
It is calculated using the formula: $\text{Vapour Density} = \frac{\text{Molar Mass of Gas}}{2}$.
The molar mass of $O_2$ gas is $32 \ g/mol$.
Therefore,$\text{Vapour Density} = \frac{32}{2} = 16$.

(B) The molar mass $(M)$ of a gas is related to its vapour density $(VD)$ by the formula: $M = 2 \times VD$.
Given $VD = 16$,so $M = 2 \times 16 = 32 \ g/mol$.
The number of moles $(n)$ is calculated as: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{8 \ g}{32 \ g/mol} = 0.25 \ mol$.
At $STP$,$1 \ mol$ of an ideal gas occupies $22.4 \ dm^3$.
Therefore,the volume occupied by $0.25 \ mol$ of gas is: $V = 0.25 \ mol \times 22.4 \ dm^3/mol = 5.6 \ dm^3$.

(A) According to the ideal gas equation,$PV = nRT$,where $P$ is pressure,$V$ is volume,$n$ is the number of moles,$R$ is the universal gas constant,and $T$ is temperature.
Since $V$,$n$,and $T$ are constant for all four gases,the pressure $P$ is given by $P = \frac{nRT}{V}$.
Since $n$,$R$,$T$,and $V$ are identical for all gases,the pressure exerted by each gas will be the same.
However,in real gas behavior,the pressure exerted depends on the compressibility factor and intermolecular forces. For ideal gases,the pressure is equal. If the question implies real gases,the pressure exerted by a gas is influenced by the van der Waals equation: $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
For gases with smaller molecular size and weaker intermolecular forces (like $H_2$),the deviation from ideal behavior is minimal,and they exert pressure closest to the ideal value. Given the options and the standard context of such problems,if all are treated as ideal,the pressure is the same. If the question implies which gas behaves most ideally or exerts the highest pressure due to lower attractive forces,$H_2$ is the correct choice.

(C) Gay-Lussac's law states that the pressure of a given mass of a gas is directly proportional to its absolute temperature,provided the volume remains constant.
Mathematically,this is expressed as $P \propto T$ or $\frac{P}{T} = \text{constant}$ at constant volume $(V)$ and fixed mass $(n)$.

(B) $1$. Calculate the number of moles for each gas:
$n(N_2) = \frac{28 \ g}{28 \ g/mol} = 1 \ mol$
$n(He) = \frac{8 \ g}{4 \ g/mol} = 2 \ mol$
$n(Ne) = \frac{40 \ g}{20 \ g/mol} = 2 \ mol$
$2$. Calculate the total number of moles:
$n_{total} = 1 + 2 + 2 = 5 \ mol$
$3$. Calculate the mole fraction of $N_2$:
$x(N_2) = \frac{n(N_2)}{n_{total}} = \frac{1}{5} = 0.2$
$4$. Calculate the partial pressure of $N_2$ using Dalton's Law:
$P(N_2) = x(N_2) \times P_{total} = 0.2 \times 20 \ bar = 4 \ bar$

(C) The rate of diffusion is defined as the amount of gas (in moles) that diffuses per unit time.
Mathematically,$\text{Rate} = \frac{\text{Amount of substance}}{\text{Time}}$.
The $SI$ unit for the amount of substance is $mol$ and the $SI$ unit for time is $s$.
Therefore,the $SI$ unit for the rate of diffusion is $mol \cdot s^{-1}$.

(A) According to Dalton's Law of partial pressure,the partial pressure of a gas is directly proportional to its mole fraction $(P_i = X_i P_{total})$.
Since the temperature and volume are constant,the mole fraction $X_i$ is proportional to the number of moles $n_i$.
Let the mass of each gas be $m$.
The number of moles $n_i$ is given by $n_i = \frac{m}{M_i}$,where $M_i$ is the molar mass.
For $H_2$: $n = \frac{m}{2}$
For $He$: $n = \frac{m}{4}$
For $CO_2$: $n = \frac{m}{44}$
For $Ne$: $n = \frac{m}{20}$
Comparing the number of moles,the gas with the smallest molar mass will have the highest number of moles.
Since $H_2$ has the smallest molar mass $(2 \ g/mol)$,it will have the maximum number of moles and therefore exert the maximum partial pressure.

(C) According to Dalton's Law of Partial Pressure,the partial pressure of a gas is directly proportional to its mole fraction $(x_i)$ in the mixture,i.e.,$P_i = x_i \times P_{total}$.
Since $x_i = \frac{n_i}{n_{total}}$,the partial pressure is directly proportional to the number of moles $(n_i)$ of the gas.
Given that the masses $(m)$ of all gases are equal,the number of moles is calculated as $n = \frac{m}{M}$,where $M$ is the molar mass.
Thus,$n \propto \frac{1}{M}$.
The molar masses are: $M(H_2) = 2 \ g/mol$,$M(He) = 4 \ g/mol$,$M(Ne) = 20 \ g/mol$,and $M(CO_2) = 44 \ g/mol$.
Since $CO_2$ has the highest molar mass $(44 \ g/mol)$,it will have the minimum number of moles $(n = \frac{m}{44})$ for a fixed mass $m$.
Therefore,$CO_2$ exerts the minimum partial pressure.

(D) According to Dalton's Law of Partial Pressures,the partial pressure of a gas is directly proportional to its number of moles in a mixture at constant temperature and volume $(P_i = n_i \times \frac{RT}{V})$.
For the partial pressure of $Ne$ to be equal to the partial pressure of $He$,their number of moles must be equal $(n_{Ne} = n_{He})$.
First,calculate the moles of $He$ $(n_{He})$:
$n_{He} = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \ g}{4 \ g/mol} = 1 \ mol$.
Since $n_{Ne} = n_{He}$,the required moles of $Ne$ is $1 \ mol$.
Now,calculate the mass of $Ne$:
$\text{Mass of } Ne = n_{Ne} \times \text{molar mass of } Ne = 1 \ mol \times 20 \ g/mol = 20 \ g$.

(C) Let the mass of both helium $(He)$ and oxygen $(O_2)$ be $m \ g$.
The molar mass of $He$ is $4 \ g/mol$ and the molar mass of $O_2$ is $32 \ g/mol$.
The number of moles of $He$ $(n_{He})$ = $\frac{m}{4}$.
The number of moles of $O_2$ $(n_{O_2})$ = $\frac{m}{32}$.
The total number of moles $(n_{total})$ = $\frac{m}{4} + \frac{m}{32} = \frac{8m + m}{32} = \frac{9m}{32}$.
According to Dalton's law of partial pressures,the fraction of total pressure exerted by a gas is equal to its mole fraction.
Mole fraction of $He$ $(x_{He})$ = $\frac{n_{He}}{n_{total}} = \frac{m/4}{9m/32} = \frac{m}{4} \times \frac{32}{9m} = \frac{8}{9}$.

(D) The partial pressure $(P_i)$ of a gas in a mixture is directly proportional to its number of moles $(n_i)$ according to Dalton's Law of Partial Pressures ($P_i = x_i P_{total}$,where $x_i = \frac{n_i}{n_{total}}$).
Number of moles $(n)$ is calculated as $n = \frac{\text{mass}}{\text{molar mass}}$.
For $H_2$: $n = \frac{5 \text{ g}}{2 \text{ g/mol}} = 2.5 \text{ mol}$.
For $He$: $n = \frac{8 \text{ g}}{4 \text{ g/mol}} = 2.0 \text{ mol}$.
For $CO_2$: $n = \frac{50 \text{ g}}{44 \text{ g/mol}} \approx 1.136 \text{ mol}$.
For $Ne$: $n = \frac{20 \text{ g}}{20 \text{ g/mol}} = 1.0 \text{ mol}$.
Since $Ne$ has the minimum number of moles $(1.0 \text{ mol})$,it exerts the minimum partial pressure.

(C) The compressibility factor $(Z)$ is defined as the ratio of the molar volume of a real gas $(V_m)$ to the molar volume of an ideal gas $(V_{ideal})$ at the same temperature and pressure.
$Z = \frac{V_m}{V_{ideal}}$
At $STP$,the molar volume of an ideal gas $(V_{ideal})$ is $22.40 \ dm^3 \ mol^{-1}$.
Given $Z = 1.05$,we can calculate the molar volume of the real gas $(V_m)$ as:
$V_m = Z \times V_{ideal}$
$V_m = 1.05 \times 22.40 \ dm^3 \ mol^{-1}$
$V_m = 23.52 \ dm^3 \ mol^{-1}$

(D) Isotopes are defined as atoms of the same element having the same atomic number $(Z)$ but different mass numbers $(A)$.
Since they have the same atomic number,they have the same number of protons and electrons,which leads to similar chemical properties.
They occupy the same position in the periodic table because the periodic table is arranged based on atomic number.
However,isotopes have different numbers of neutrons because their mass numbers $(A = Z + N)$ differ while their atomic numbers $(Z)$ remain the same.
Therefore,the statement that they have an equal number of neutrons is false.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -2.18 \times 10^{-18} \times \frac{Z^2}{n^2} \ J$.
For $He^{+}$,the atomic number $Z = 2$.
The orbit number $n = 3$.
Substituting these values into the formula:
$E_3 = -2.18 \times 10^{-18} \times \frac{2^2}{3^2} \ J$
$E_3 = -2.18 \times 10^{-18} \times \frac{4}{9} \ J$
$E_3 = -2.18 \times 10^{-18} \times 0.4444 \ J$
$E_3 \approx -0.9688 \times 10^{-18} \ J$
$E_3 \approx -9.69 \times 10^{-19} \ J$.

(C) hydrogen-like species is an atom or ion that contains only one electron.
To determine this,we check the number of electrons in each species:
$A) \ Li^{2+}$: Lithium has $3$ electrons. $Li^{2+}$ has $3 - 2 = 1$ electron. It is hydrogen-like.
$B) \ Be^{3+}$: Beryllium has $4$ electrons. $Be^{3+}$ has $4 - 3 = 1$ electron. It is hydrogen-like.
$C) \ Li^{+}$: Lithium has $3$ electrons. $Li^{+}$ has $3 - 1 = 2$ electrons. It is $NOT$ hydrogen-like.
$D) \ He^{+}$: Helium has $2$ electrons. $He^{+}$ has $2 - 1 = 1$ electron. It is hydrogen-like.
Therefore,$Li^{+}$ is the correct answer.

(B) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Lyman series,$n_1 = 1$.
The longest wavelength corresponds to the smallest energy transition,which occurs from $n_2 = 2$ to $n_1 = 1$.
Substituting the values: $\frac{1}{\lambda} = 109677 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \ cm^{-1}$.
$\frac{1}{\lambda} = 109677 \left( 1 - \frac{1}{4} \right) = 109677 \times \frac{3}{4} \ cm^{-1}$.
$\frac{1}{\lambda} = 82257.75 \ cm^{-1}$.
$\lambda = \frac{1}{82257.75} \ cm \approx 1.216 \times 10^{-5} \ cm$.

(A) The wavenumber $(\bar{\nu})$ for a transition in a hydrogen atom is given by the Rydberg formula: $\bar{\nu} = R_{H} \times Z^2 \times \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Here,$R_{H} = 109677 \ cm^{-1}$,$Z = 1$ (for hydrogen atom),$n_1 = 2$,and $n_2 = 5$.
Substituting the values: $\bar{\nu} = 109677 \times 1^2 \times \left( \frac{1}{2^2} - \frac{1}{5^2} \right)$.
$\bar{\nu} = 109677 \times \left( \frac{1}{4} - \frac{1}{25} \right)$.
$\bar{\nu} = 109677 \times \left( \frac{25 - 4}{100} \right) = 109677 \times \frac{21}{100}$.
$\bar{\nu} = 109677 \times 0.21 = 23032.17 \ cm^{-1}$.
Rounding to the nearest integer,we get $23032 \ cm^{-1}$.

(A) The hexacyanoferrate$(II)$ ion is $[Fe(CN)_6]^{4-}$.
$1$. The central metal ion is $Fe^{2+}$.
$2$. The atomic number $(Z)$ of $Fe$ is $26$.
$3$. The number of electrons in $Fe^{2+}$ is $26 - 2 = 24$.
$4$. Each $CN^-$ ligand donates $2$ electrons. Since there are $6$ ligands,the total electrons donated by ligands = $6 \times 2 = 12$.
$5$. $EAN = (\text{Number of electrons in metal ion}) + (\text{Number of electrons donated by ligands})$.
$6$. $EAN = 24 + 12 = 36$.

(A) The $EAN$ (Effective Atomic Number) is calculated using the formula: $EAN = Z - \text{oxidation state} + 2 \times \text{coordination number}$.
For $[Co(NH_3)_6]^{3+}$:
$1$. The atomic number $(Z)$ of $Co$ is $27$.
$2$. The oxidation state of $Co$ is calculated as $x + 6(0) = +3$,so $x = +3$.
$3$. The coordination number is $6$ (since $NH_3$ is a monodentate ligand).
$4$. $EAN = 27 - 3 + 2(6) = 24 + 12 = 36$.

(A) Anionic ligands are those that carry a negative charge.
$A$. Isothiocyanato $(NCS^-)$ is an anionic ligand.
$B$. Ammine $(NH_3)$ is a neutral ligand.
$C$. Aqua $(H_2O)$ is a neutral ligand.
$D$. Ethylenediamine $(en)$ is a neutral ligand.
Therefore,the correct option is $A$.

(C) The $EAN$ (Effective Atomic Number) is calculated using the formula: $EAN = Z - ON + 2 \times CN$,where $Z$ is the atomic number,$ON$ is the oxidation number,and $CN$ is the coordination number.
For $[Zn(NH_3)_4]^{2+}$:
$Z$ of $Zn = 30$.
$ON$ of $Zn$: $x + 4(0) = +2 \implies x = +2$.
$CN$ of $Zn = 4$.
$EAN = 30 - 2 + 2(4) = 28 + 8 = 36$.

(C) According to the spectrochemical series,ligands are arranged in increasing order of their field strength:
$I^{-} < Br^{-} < S^{2-} < SCN^{-} < Cl^{-} < F^{-} < OH^{-} < H_2O < NCS^{-} < EDTA^{4-} < NH_3 < en < CN^{-} < CO$.
Ligands like $F^{-}$,$Cl^{-}$,$Br^{-}$,and $I^{-}$ are considered weak field ligands because they cause small crystal field splitting $(\Delta_o)$.
Among the given options,$F^{-}$ is a weak field ligand,while $EDTA^{4-}$,$NH_3$,and $CO$ are strong field ligands.

(C) According to the spectrochemical series,ligands are arranged in increasing order of their field strength:
$I^{-} < Br^{-} < S^{2-} < SCN^{-} < Cl^{-} < F^{-} < OH^{-} < C_2O_4^{2-} < H_2O < NCS^{-} < NH_3 < en < NO_2^{-} < CN^{-} < CO$.
Ligands like $Br^{-}$ are considered weak field ligands because they cause small crystal field splitting $(\Delta_o)$.
In contrast,$NH_3$,$CN^{-}$,and $CO$ are strong field ligands.

(C) In the coordination complex $K_4[Fe(CN)_6]$,the ligand is the cyanide ion,$CN^-$.
Each $CN^-$ ligand acts as a monodentate ligand,meaning it donates one electron pair through a single donor atom (the carbon atom).
There are $6$ $CN^-$ ligands coordinated to the central metal ion $Fe^{2+}$.
Therefore,in one mole of the complex,there are $6$ moles of $CN^-$ ligands.
Since each $CN^-$ ligand has one donor atom (carbon),the total number of moles of donor atoms is $6 \times 1 = 6$ moles.

(B) The $EAN$ (Effective Atomic Number) is calculated using the formula: $EAN = Z - \text{oxidation state} + 2 \times \text{coordination number}$.
For $[Fe(CN)_6]^{3-}$:
$1$. The atomic number $(Z)$ of $Fe$ is $26$.
$2$. Let the oxidation state of $Fe$ be $x$. Then $x + 6(-1) = -3$,which gives $x = +3$.
$3$. The coordination number of $Fe$ is $6$.
$4$. $EAN = 26 - 3 + 2(6) = 23 + 12 = 35$.

(D) The $EAN$ (Effective Atomic Number) rule states that the total number of electrons around the central metal ion in a complex should be equal to the atomic number of the next noble gas.
$1$. For $[Co(NH_3)_6]^{3+}$: $Co$ $(Z=27)$. $EAN = 27 - 3 + (6 \times 2) = 36$ (Obeys $EAN$ rule).
$2$. For $[Cr(CO)_6]$: $Cr$ $(Z=24)$. $EAN = 24 - 0 + (6 \times 2) = 36$ (Obeys $EAN$ rule).
$3$. For $[Zn(NH_3)_4]^{2+}$: $Zn$ $(Z=30)$. $EAN = 30 - 2 + (4 \times 2) = 36$ (Obeys $EAN$ rule).
$4$. For $[Cu(NH_3)_4]^{2+}$: $Cu$ $(Z=29)$. $EAN = 29 - 2 + (4 \times 2) = 35$ (Does $NOT$ obey $EAN$ rule).
Therefore,the correct option is $D$.

(A) complex salt containing both complex cations and complex anions is known as a coordination isomer or a salt where both parts are coordination entities.
In the compound $[Pt(NH_3)_4][PtCl_6]$,the cation is $[Pt(NH_3)_4]^{2+}$ and the anion is $[PtCl_6]^{2-}$.
Therefore,this compound contains both complex cations and complex anions.

(A) complex anion is a negatively charged coordination entity.
In $Na_3[Co(NO_2)_6]$,the coordination entity is $[Co(NO_2)_6]^{3-}$,which is a complex anion.
In $Na_3[Co(NO_2)_6]$ (Sodium hexanitrocobaltate$(III)$),the complex part is the anion.
In $K_4[Fe(CN)_6]$,the complex part is also an anion.
Option $A$ represents a compound with a complex anion.

(C) neutral complex is one that carries no net charge.
$1$. $[Ni(CN)_4]^{2-}$ is an anionic complex with a charge of $-2$.
$2$. $Na_3[AlF_6]$ is an ionic compound consisting of $Na^+$ ions and $[AlF_6]^{3-}$ complex ions.
$3$. $[Co(NH_3)_3(NO_2)_3]$ has no net charge because the oxidation state of $Co$ is $+3$,and it is balanced by three neutral $NH_3$ ligands $(0 \times 3 = 0)$ and three anionic $NO_2^-$ ligands $(-1 \times 3 = -3)$. Thus,$+3 + 0 - 3 = 0$.
$4$. $[Cu(NH_3)_4]^{2+}$ is a cationic complex with a charge of $+2$.
Therefore,the neutral complex is $[Co(NH_3)_3(NO_2)_3]$.

(C) An anionic ligand is a negatively charged ligand.
In the complex $[Ni(CN)_4]^{2-}$,the ligand is cyanide $(CN^-)$,which is an anionic ligand.
In the other options:
- Tetraamminecopper$(II)$ ion: $[Cu(NH_3)_4]^{2+}$ contains neutral $NH_3$ ligands.
- Pentaammineaquacobalt$(III)$ iodide: $[Co(NH_3)_5(H_2O)]I_3$ contains neutral $NH_3$ and $H_2O$ ligands.
- Pentacarbonyliron$(0)$: $[Fe(CO)_5]$ contains neutral $CO$ ligands.
Therefore,the correct option is $C$.

(B) The general formula $MA_2BC$ represents a coordination complex with a central metal atom $M$,two identical ligands $A$,and two different ligands $B$ and $C$.
In the complex $[Pt(NH_3)(H_2O)Cl_2]$,the central metal is $Pt$,$A = Cl^-$,$B = NH_3$,and $C = H_2O$.
This complex exhibits geometric isomerism (cis and trans forms) and is a classic example of the $MA_2BC$ type.
Therefore,the correct option is $B$.

(A) Solvate isomerism (also known as hydrate isomerism when water is the solvent) occurs when the solvent molecule acts as a ligand in one isomer and as a lattice molecule (outside the coordination sphere) in another.
In the pair $[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$,the first complex has $6$ water molecules as ligands,while the second has $5$ water molecules as ligands and $1$ water molecule as a lattice molecule.
Therefore,this pair exhibits solvate isomerism.

(B) In the complex $[Ni(CN)_4]^{2-}$,the central metal ion is $Ni^{2+}$.
The atomic number of $Ni$ is $28$,so the electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
$CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
This results in one empty $3d$ orbital,one $4s$ orbital,and two $4p$ orbitals being available for hybridisation.
Thus,the hybridisation involved is $dsp^2$,which corresponds to a square planar geometry.

(D) The field strength of ligands is determined by the spectrochemical series.
According to the spectrochemical series,the order of field strength for the given ligands is: $OH^{-} < C_2O_4^{2-} < H_2O < CO$.
$CO$ is a strong field ligand that acts as a $\pi$-acid ligand,capable of forming $\pi$-backbonding with the metal center,which results in the highest crystal field splitting energy $(\Delta_o)$.
Therefore,$CO$ has the highest field strength among the given options.

(A) According to the spectrochemical series,the order of field strength of ligands is: $I^{-} < Br^{-} < S^{2-} < SCN^{-} < Cl^{-} < N_{3} < F^{-} < OH^{-} < C_{2}O_{4}^{2-} < H_{2}O < NCS^{-} < EDTA^{4-} < NH_{3} < en < NO_{2}^{-} < CN^{-} < CO$.
Comparing the given options ($I^{-}$,$S^{2-}$,$en$,$CO$),the ligand $I^{-}$ is at the beginning of the spectrochemical series,indicating it has the lowest field strength.

(D) The central metal ion is $Co^{3+}$. The atomic number of $Co$ is $27$,so the electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$NH_3$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
In the presence of $NH_3$,the $6$ electrons in the $3d$ subshell pair up in the first three orbitals ($t_{2g}$ set).
Thus,all $3d$ electrons are paired,and there are no unpaired electrons in the $[Co(NH_3)_6]^{3+}$ complex.

(B) According to the spectrochemical series,the field strength of ligands is determined by their ability to cause crystal field splitting. The order of increasing field strength for the given ligands is:
$I^{-} < S^{2-} < Cl^{-} < OH^{-}$
Thus,the correct increasing order is $I^{-} < S^{2-} < Cl^{-} < OH^{-}$.

(D) The stability of coordination complexes formed by divalent metal ions with the same ligand generally follows the Irving-Williams series.
According to the Irving-Williams series,the stability of complexes increases with the decrease in ionic radius and increase in the charge-to-size ratio.
For the given ions,the order of stability is $Mn^{2+} < Cd^{2+} < Cu^{2+}$.
However,the question asks for the decreasing order of stability.
Comparing the given options,$Cu^{2+}$ is the most stable due to its smaller ionic radius and higher crystal field stabilization energy compared to $Mn^{2+}$ and $Cd^{2+}$.
Thus,the correct decreasing order is $Cu^{2+} > Cd^{2+} > Mn^{2+}$.

(A) The stability of complexes formed by transition metal ions with the same ligand is governed by the Irving-Williams series.
According to the Irving-Williams series,the stability of complexes for divalent metal ions of the $3d$ series follows the order: $Mn^{2+} < Fe^{2+} < Co^{2+} < Ni^{2+} < Cu^{2+} > Zn^{2+}$.
Among the given options,$Cu^{2+}$ has the highest stability constant for complex formation with a given ligand due to its smaller ionic radius and higher charge density compared to the other listed ions.
Therefore,$Cu^{2+}$ forms the most stable complex.

(B) The stability of complexes formed by transition metal ions with the same ligand generally follows the Irving-Williams series.
The order of stability for divalent metal ions is: $Mn^{2+} < Fe^{2+} < Co^{2+} < Ni^{2+} < Cu^{2+} > Zn^{2+}$.
Among the given options ($Co^{2+}$,$Fe^{2+}$,$Cd^{2+}$,$Cu^{2+}$),$Fe^{2+}$ has the lowest stability constant because it appears earliest in the series among the transition metals listed.
$Cd^{2+}$ is a $d^{10}$ ion and generally forms less stable complexes compared to the $3d$ transition series ions due to its larger size and lower charge density.

(A) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$1$. For $V^{3+}$ $([Ar] 3d^2)$: $n = 2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.
$2$. For $Cr^{3+}$ $([Ar] 3d^3)$: $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
$3$. For $Mn^{2+}$ $([Ar] 3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
$4$. For $Fe^{2+}$ $([Ar] 3d^6)$: $n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
Comparing the values,$V^{3+}$ has the lowest number of unpaired electrons $(n=2)$,resulting in the lowest spin-only magnetic moment.

(D) transition element is defined as an element which has an incompletely filled $d$-orbital in its ground state or in any one of its oxidation states.
$Ni$ $(3d^8 4s^2)$,$Fe$ $(3d^6 4s^2)$,and $Ag$ ($4d^{10} 5s^1$ in ground state,but $Ag^{2+}$ has $4d^9$) are considered transition elements.
$Hg$ $([Xe] 4f^{14} 5d^{10} 6s^2)$ has a completely filled $d$-orbital in its ground state $(5d^{10})$ and also in its common oxidation state ($Hg^{2+}$ is $5d^{10}$).
Therefore,$Hg$ is not regarded as a transition element.

(C) The element $Co$ (Cobalt) has an atomic number of $27$,which places it in the $3d$ transition series (first transition series).
The element $Mo$ (Molybdenum) has an atomic number of $42$,which places it in the $4d$ transition series (second transition series).
Therefore,the correct sequence is $3d$ and $4d$.

(B) The electronic configuration of the given elements are:
$Dy$ $(Z=66)$: $[Xe] 4f^{10} 6s^2$. The last electron enters the $4f$ orbital.
$Ag$ $(Z=47)$: $[Kr] 4d^{10} 5s^1$. The last electron enters the $4d$ orbital,which is $(n-1)d$ where $n=5$.
$Pu$ $(Z=94)$: $[Rn] 5f^6 7s^2$. The last electron enters the $5f$ orbital.
$Pa$ $(Z=91)$: $[Rn] 5f^2 6d^1 7s^2$. The last electron enters the $5f$ orbital.
Therefore,$Ag$ is the element where the last electron is placed in the $(n-1)d$ orbital.

(C) The atomic number of $Ti$ (Titanium) is $22$.
The electronic configuration of $Ti$ is $[Ar] 3d^2 4s^2$.
In the $+3$ oxidation state,$Ti$ loses three electrons (two from $4s$ and one from $3d$).
The electronic configuration of $Ti^{3+}$ is $[Ar] 3d^1$.
Since there is only one electron in the $3d$ orbital,the number of unpaired electrons is $1$.

(A) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
$1$. For $Cu^{2+}$ $(Z=29)$: Electronic configuration is $[Ar] 3d^9$. Number of unpaired electrons $(n)$ = $1$.
$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ BM}$.
$2$. For $Cr^{3+}$ $(Z=24)$: Electronic configuration is $[Ar] 3d^3$. Number of unpaired electrons $(n)$ = $3$.
$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM}$.
$3$. For $Co^{2+}$ $(Z=27)$: Electronic configuration is $[Ar] 3d^7$. Number of unpaired electrons $(n)$ = $3$.
$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM}$.
$4$. For $Fe^{2+}$ $(Z=26)$: Electronic configuration is $[Ar] 3d^6$. Number of unpaired electrons $(n)$ = $4$.
$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM}$.
Comparing the values,$Cu^{2+}$ has the least magnetic moment.

(D) Among the given transition metals,$Cr$ (Chromium),$V$ (Vanadium),and $Co$ (Cobalt) are hard metals with high melting points due to strong metallic bonding involving $d$-electrons.
$Cd$ (Cadmium) belongs to group $12$ (along with $Zn$ and $Hg$). These elements have a fully filled $d^{10}$ configuration,which results in weaker metallic bonding compared to other transition metals.
Therefore,$Cd$ is relatively soft compared to $Cr$,$V$,and $Co$.

(D) The colour of transition metal ions is primarily due to $d-d$ transitions.
For an ion to be coloured,it must have an incomplete $d$-subshell (i.e.,$d^1$ to $d^9$ configuration).
$1$. $Sc^{3+}$ $(3d^0)$: No $d$-electrons,hence colourless.
$2$. $Ti^{4+}$ $(3d^0)$: No $d$-electrons,hence colourless.
$3$. $Zn^{2+}$ $(3d^{10})$: Completely filled $d$-subshell,hence colourless.
$4$. $Cr^{3+}$ $(3d^3)$: Has three unpaired electrons in the $d$-subshell,allowing for $d-d$ transitions,hence it forms coloured compounds.
Therefore,the correct option is $D$.

(C) The basic strength of lanthanoid hydroxides,$Ln(OH)_3$,decreases as the ionic radius of the $Ln^{3+}$ ion decreases due to lanthanoid contraction.
As we move from $La$ to $Lu$ in the lanthanoid series,the ionic radius decreases.
Therefore,$La(OH)_3$ is the most basic,and $Lu(OH)_3$ is the least basic (weakest base).

(C) The atomic number of $Zn$ is $30$.
The electronic configuration of $Zn$ is $[Ar] 3d^{10} 4s^2$.
The electronic configuration of $Zn^{2+}$ is $[Ar] 3d^{10}$.
Since all $10$ electrons in the $3d$ subshell are paired,the number of unpaired electrons $(n)$ is $0$.
The formula for spin-only magnetic moment is $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n = 0$,we get $\mu = \sqrt{0(0+2)} = 0 \ BM$.

(D) The atomic number of $Mn$ is $25$. The electronic configuration of $Mn$ is $[Ar] 3d^5 4s^2$.
For $Mn^{2+}$,the configuration is $[Ar] 3d^5$.
This means there are $5$ unpaired electrons $(n = 5)$.
The spin only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n = 5$,we get $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.

(C) Trophies are typically made of bronze,which is an alloy of copper $(Cu)$ and tin $(Sn)$.
Therefore,the correct pair of elements is $Cu$ and $Sn$.

(A) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$1$. For $Mn^{2+}$ $(Z=25)$: Electronic configuration is $[Ar] 3d^5$. Number of unpaired electrons $(n)$ = $5$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
$2$. For $Ti^{3+}$ $(Z=22)$: Electronic configuration is $[Ar] 3d^1$. Number of unpaired electrons $(n)$ = $1$. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
$3$. For $Cu^{2+}$ $(Z=29)$: Electronic configuration is $[Ar] 3d^9$. Number of unpaired electrons $(n)$ = $1$. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
$4$. For $Ni^{2+}$ $(Z=28)$: Electronic configuration is $[Ar] 3d^8$. Number of unpaired electrons $(n)$ = $2$. $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.
Comparing the values,$Mn^{2+}$ has the highest number of unpaired electrons and thus the highest magnetic moment.

(B) The magnetic moment of an ion is determined by the number of unpaired electrons,calculated as $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$1$. For $Cr^{3+}$ $([Ar] 3d^3)$,$n = 3$,so it is paramagnetic.
$2$. For $Sc^{3+}$ $([Ar] 3d^0)$,$n = 0$,so it has no unpaired electrons and exhibits no magnetic moment (diamagnetic).
$3$. For $Cu^{2+}$ $([Ar] 3d^9)$,$n = 1$,so it is paramagnetic.
$4$. For $V^{3+}$ $([Ar] 3d^2)$,$n = 2$,so it is paramagnetic.
Therefore,$Sc^{3+}$ exhibits no magnetic moment.

(C) The atomic number of $Cu$ is $29$. The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.
For $Cu^{2+}$, the electronic configuration is $[Ar] 3d^9$.
This means there is $1$ unpaired electron in the $3d$ orbital $(n = 1)$.
The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n = 1$, we get $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.

(A) Metal ions form colourless compounds if they have a $d^0$ or $d^{10}$ electronic configuration (i.e.,no unpaired electrons).
$Zn^{2+}$ has a $3d^{10}$ configuration (fully filled,no unpaired electrons).
$Cu^{+}$ has a $3d^{10}$ configuration (fully filled,no unpaired electrons).
Since both $Zn^{2+}$ and $Cu^{+}$ have $d^{10}$ configurations,they form colourless compounds.
Therefore,the correct option is $A$.

(A) Inner transition elements are the elements in which the last electron enters the $f$-orbital. These include the lanthanoids $(58-71)$ and actinoids $(90-103)$.
$Cm$ ($Curium$,atomic number $96$) is an actinoid,which is a type of inner transition element.
$W$ $(Tungsten)$,$Mo$ $(Molybdenum)$,and $Ru$ $(Ruthenium)$ are transition elements belonging to the $d$-block.

(A) The lanthanoids are the elements with atomic numbers $57$ to $71$ ($La$ to $Lu$).
$Er$ (Erbium) has an atomic number of $68$,which falls within the range of the lanthanoid series.
$Am$ (Americium),$Np$ (Neptunium),and $Lr$ (Lawrencium) are actinoid elements.

(D) The electronic configuration of $Tb$ $(Z = 65)$ is $[Xe] 4f^9 6s^2$.
In the $+4$ oxidation state,$Tb^{4+}$ loses four electrons,resulting in the configuration $[Xe] 4f^7$.
This $f^7$ configuration is half-filled and therefore highly stable.
Thus,$Tb$ exhibits a $+4$ oxidation state with an $f^7$ configuration.

(C) The most common and stable oxidation state for lanthanoid elements $(Ln)$ is $+3$.
Therefore,when a lanthanoid reacts with the hydroxide ion $(OH^-)$,it forms a compound where the charge is balanced.
Since the hydroxide ion has a charge of $-1$,three hydroxide ions are required to balance the $+3$ charge of the $Ln^{3+}$ ion.
Thus,the general formula for lanthanoid hydroxide is $Ln(OH)_3$.

(B) When lanthanoids $(Ln)$ are heated with carbon at elevated temperatures,they form carbides of the general formula $LnC_2$.
These carbides are typically ionic in nature and are known as acetylides,as they react with water to produce acetylene gas $(C_2H_2)$.

(D) The first ionisation enthalpy $(IE_1)$ of lanthanoids generally increases with an increase in atomic number due to the lanthanoid contraction,which leads to a decrease in atomic size and stronger attraction of the nucleus for the valence electrons.
Among the given elements,$Ce$ $(Z=58)$,$La$ $(Z=57)$,$Gd$ $(Z=64)$,and $Yb$ $(Z=70)$,the element $Yb$ has the highest atomic number.
$Yb$ has a stable electronic configuration of $[Xe] 4f^{14} 6s^2$. Due to the completely filled $4f$ subshell and the highest effective nuclear charge among the options,it requires the most energy to remove the first electron.
Therefore,$Yb$ has the highest $IE_1$.

(C) In the lanthanoid series,the ionic radii of $Ln^{3+}$ ions decrease regularly with an increase in atomic number due to lanthanoid contraction.
Since $La$ $(Z=57)$ has the lowest atomic number among the given elements $(La, Pr, Sm, Yb)$,its $La^{3+}$ ion will have the largest ionic size.
The order of ionic size is: $La^{3+} > Pr^{3+} > Sm^{3+} > Yb^{3+}$.

(B) The effective magnetic moment $(\mu_{eff})$ is calculated using the formula $\mu_{eff} = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For a lanthanoid to exhibit zero effective magnetic moment in the $+3$ state,it must have zero unpaired electrons $(n=0)$.
$Lu$ (Lutetium) has the atomic number $71$ and its electronic configuration is $[Xe] \ 4f^{14} \ 5d^1 \ 6s^2$.
In the $+3$ oxidation state,$Lu^{3+}$ loses three electrons to form $[Xe] \ 4f^{14}$.
Since the $4f$ subshell is completely filled,there are no unpaired electrons $(n=0)$,resulting in a magnetic moment of $0 \ BM$.

(D) The first ionisation enthalpy $(IE_1)$ of lanthanoids generally increases across the series due to the lanthanoid contraction,which leads to a decrease in atomic size and an increase in effective nuclear charge.
Among the given options,$Yb$ (Ytterbium,$Z=70$) has the highest atomic number.
$Yb$ has a stable electronic configuration of $[Xe] 4f^{14} 6s^2$.
Due to the completely filled $4f$ subshell and the high effective nuclear charge,$Yb$ exhibits a very high first ionisation enthalpy compared to the other listed lanthanoids ($Nd$,$Dy$,$Tm$).

(A) In the lanthanoid series,as we move from $La$ $(Z = 57)$ to $Lu$ $(Z = 71)$,the ionic radius of the $M^{3+}$ ions decreases regularly. This phenomenon is known as lanthanoid contraction.
Lanthanoid contraction occurs due to the poor shielding effect of $4f$ electrons,which causes the effective nuclear charge to increase,pulling the electrons closer to the nucleus.
Therefore,the ionic size decreases as the atomic number increases.
Among the given elements,$Lu$ $(Z = 71)$ has the highest atomic number,so $Lu^{3+}$ has the smallest ionic size.

(C) The electronic configuration of Cerium ($Ce$,atomic number $58$) is $[Xe] 4f^1 5d^1 6s^2$.
When $Ce$ loses four electrons to form $Ce^{4+}$,it loses the two $6s$,one $5d$,and one $4f$ electron.
The resulting configuration of $Ce^{4+}$ is $[Xe] 4f^0$,which is a stable noble gas configuration.
Therefore,$Ce$ is the lanthanoid that exhibits a $+4$ oxidation state with an $f^0$ configuration.