MHT CET 2024 Chemistry Question Paper with Answer and Solution

(C) To find the oxidation state of chlorine $(Cl)$ in the given compounds,we use the rule that the sum of oxidation states of all atoms in a neutral molecule is $0$. Let the oxidation state of $Cl$ be $x$. The oxidation state of hydrogen $(H)$ is $+1$ and oxygen $(O)$ is $-2$.
$1.$ For $HClO_4$: $1 + x + 4(-2) = 0 \implies x - 7 = 0 \implies x = +7$.
$2.$ For $HClO_2$: $1 + x + 2(-2) = 0 \implies x - 3 = 0 \implies x = +3$.
$3.$ For $HClO_3$: $1 + x + 3(-2) = 0 \implies x - 5 = 0 \implies x = +5$.
$4.$ For $HCl$: $1 + x = 0 \implies x = -1$.
Thus,$HClO_3$ contains chlorine in the $+5$ oxidation state.

(A) The chemical formula of mustard gas is $(ClCH_2CH_2)_2S$.
Each molecule of mustard gas contains $1$ sulfur atom.
Therefore,$n$ moles of mustard gas molecules contain $n \times 1 = n$ moles of sulfur atoms.

(A) The reactivity of halogens towards alkanes is determined by the bond dissociation energy and the electronegativity of the halogen atoms.
Fluorine is the most reactive due to its high electronegativity and the low bond dissociation energy of the $F-F$ bond.
As we move down the group,the reactivity decreases.
Therefore,the correct order of reactivity is $F_2 > Cl_2 > Br_2 > I_2$.

(C) The elements of group $16$ (such as $O, S$,etc.) have a valence shell configuration of $ns^2 np^4$.
They require $2$ more electrons to complete their octet $(ns^2 np^6)$,which corresponds to the stable electronic configuration of the nearest noble gas.
Therefore,group $16$ elements achieve noble gas configuration by gaining $2$ electrons.

(A) To determine the number of electrons transferred,we calculate the change in the oxidation state of the central atom:
$1$. In $MnO_4^{-}$,the oxidation state of $Mn$ is $+7$. In $Mn^{2+}$,it is $+2$. The change is $7 - 2 = 5$ electrons.
$2$. In $CrO_4^{2-}$,the oxidation state of $Cr$ is $+6$. In $Cr^{3+}$,it is $+3$. The change is $6 - 3 = 3$ electrons.
$3$. In $NO_3^{-}$,the oxidation state of $N$ is $+5$. In $NH_4^{+}$,it is $-3$. The change is $5 - (-3) = 8$ electrons.
$4$. In $Cr_2O_7^{2-}$,the oxidation state of $Cr$ is $+6$. In $2Cr^{3+}$,the total charge change for two $Cr$ atoms is $2 \times (6 - 3) = 6$ electrons.
Thus,the change involving the transfer of $5$ electrons is $MnO_4^{-} \longrightarrow Mn^{2+}$.

(A) In the given redox reaction,the oxidation state of $Cr$ changes from $+6$ in $Cr_2O_7^{2-}$ to $+3$ in $Cr^{3+}$.
Since the oxidation state of $Cr$ decreases,$Cr$ is reduced.
The oxidation state of $I$ changes from $-1$ in $I^{-}$ to $0$ in $I_2$.
Since the oxidation state of $I$ increases,$I$ is oxidized.
Therefore,the element reduced is $Cr$.

(D) To find the oxidation state of chlorine $(Cl)$ in each compound,we use the rules for oxidation numbers:
$1$. In $KCl$,the oxidation state of $K$ is $+1$. Let the oxidation state of $Cl$ be $x$. So,$1 + x = 0$,which gives $x = -1$.
$2$. In $HClO$,the oxidation state of $H$ is $+1$ and $O$ is $-2$. Let the oxidation state of $Cl$ be $x$. So,$1 + x - 2 = 0$,which gives $x = +1$.
$3$. In $HClO_2$,the oxidation state of $H$ is $+1$ and $O$ is $-2$. Let the oxidation state of $Cl$ be $x$. So,$1 + x + 2(-2) = 0$,which gives $1 + x - 4 = 0$,so $x = +3$.
$4$. In $HClO_4$,the oxidation state of $H$ is $+1$ and $O$ is $-2$. Let the oxidation state of $Cl$ be $x$. So,$1 + x + 4(-2) = 0$,which gives $1 + x - 8 = 0$,so $x = +7$.
Comparing the values: $-1, +1, +3, +7$. The highest oxidation state is $+7$ in $HClO_4$.

(D) To find the oxidation number of iodine $(I)$ in each compound:
$1$. In $KIO_3$: $1 + x + 3(-2) = 0 \implies x = +5$
$2$. In $KI$: $1 + x = 0 \implies x = -1$
$3$. In $IF_5$: $x + 5(-1) = 0 \implies x = +5$
$4$. In $KIO_4$: $1 + x + 4(-2) = 0 \implies x = +7$
Comparing the values: $+5, -1, +5, +7$. The highest oxidation number is $+7$ in $KIO_4$.

(A) Let the oxidation state of $S$ in $SO_4^{2-}$ be $x$.
The sum of oxidation states of all atoms in an ion is equal to the charge on the ion.
$x + 4(-2) = -2$
$x - 8 = -2$
$x = -2 + 8$
$x = +6$
Therefore,the oxidation state of $S$ in $SO_4^{2-}$ is $+6$.

(C) Let $x$ be the oxidation state of $I$ in $I_2Cl_6$.
Since the oxidation state of chlorine $(Cl)$ in this compound is $-1$,we can write the equation:
$2x + 6 \times (-1) = 0$
$2x - 6 = 0$
$2x = 6$
$x = +3$
Therefore,the oxidation state of iodine in $I_2Cl_6$ is $+3$.

(C) Let the oxidation state of $C$ be $x$ in $K_2C_2O_4$.
The sum of oxidation states of all atoms in a neutral molecule is $0$.
$2(+1) + 2(x) + 4(-2) = 0$
$2 + 2x - 8 = 0$
$2x - 6 = 0$
$2x = 6$
$x = +3$
Therefore,the oxidation number of carbon is $+3$.

(A) In the reactant $H_2S$,the oxidation number of $S$ is calculated as follows: $2(+1) + x = 0$,which gives $x = -2$.
In the product $S$ (elemental state),the oxidation number of $S$ is $0$.
Therefore,the change in the oxidation number of $S$ is from $-2$ to $0$.

(A) Let the oxidation state of $S$ be $x$ in $SO_3^{2-}$.
The sum of oxidation states of all atoms in an ion is equal to the charge on the ion.
$x + 3(-2) = -2$
$x - 6 = -2$
$x = -2 + 6$
$x = +4$
Therefore,the oxidation number of $S$ in $SO_3^{2-}$ is $+4$.

(A) The ionic radius of alkali metals increases down the group as the number of shells increases.
$Li^+$ has $2$ electrons in $1$ shell.
$Na^+$ has $10$ electrons in $2$ shells.
$K^+$ has $18$ electrons in $3$ shells.
$Rb^+$ has $36$ electrons in $4$ shells.
Since $Rb^+$ has the highest number of shells,it has the largest ionic radius.
Therefore,the correct option is $A$.

(A) In $NO_3^{-}$,let the oxidation number of $N$ be $x$. Since the oxidation number of $O$ is $-2$,we have: $x + 3(-2) = -1 \implies x - 6 = -1 \implies x = +5$.
In $NH_4^{+}$,let the oxidation number of $N$ be $y$. Since the oxidation number of $H$ is $+1$,we have: $y + 4(+1) = +1 \implies y + 4 = +1 \implies y = -3$.
Therefore,the oxidation number of nitrogen changes from $+5$ to $-3$.

(C) Let the oxidation state of phosphorus be $x$.
In the phosphate ion $(PO_4^{3-})$,the sum of oxidation states of all atoms equals the charge on the ion.
$x + 4 \times (-2) = -3$
$x - 8 = -3$
$x = -3 + 8$
$x = +5$
Therefore,the oxidation state of phosphorus is $+5$.

(C) The given reaction is: $x CuO + y NH_3 \rightarrow x Cu + N_2 + x H_2 O$.
Step $1$: Balance the atoms other than $O$ and $H$. Since $x$ is the coefficient for $CuO$ and $Cu$,the $Cu$ atoms are balanced.
Step $2$: Balance $N$ atoms. There are $2$ nitrogen atoms in $N_2$,so we need $2$ $NH_3$ molecules to provide $2$ $N$ atoms. Thus,$y = 2$.
Step $3$: Balance $O$ atoms. There are $x$ oxygen atoms in $x CuO$,so we need $x$ $H_2 O$ molecules. This matches the equation.
Step $4$: Balance $H$ atoms. There are $3y$ hydrogen atoms in $y NH_3$ and $2x$ hydrogen atoms in $x H_2 O$. Therefore,$3y = 2x$.
Step $5$: Substitute $y = 2$ into the equation $3y = 2x$: $3(2) = 2x$ $\Rightarrow 6 = 2x$ $\Rightarrow x = 3$.
Thus,$x = 3$ and $y = 2$. The balanced equation is $3 CuO + 2 NH_3 \rightarrow 3 Cu + N_2 + 3 H_2 O$.

(C) Alkali metals dissolve in liquid ammonia to form a deep blue coloured solution.
This colour is due to the presence of ammoniated electrons,which absorb energy in the visible region of the electromagnetic spectrum.

(C) In the Solvay process,the recovery of ammonia is achieved by treating the ammonium chloride $(NH_4Cl)$ solution with slaked lime $(Ca(OH)_2)$.
The chemical reaction is as follows:
$2NH_4Cl + Ca(OH)_2 \rightarrow CaCl_2 + 2H_2O + 2NH_3$
As shown in the reaction,calcium chloride $(CaCl_2)$ is obtained as a by-product.

(B) Barium tests are used to examine the digestive tract using a white powder called barium sulphate $(BaSO_4)$.
This powder is insoluble in water and is not absorbed by the body,making it safe for internal use.
For a barium meal,the barium sulphate powder is mixed with water and swallowed.
Since barium is a heavy element,it absorbs $X$-rays effectively,allowing the digestive tract to be clearly visible on $X$-ray images.

(D) $Be$ (Beryllium) does not react with water even at high temperatures because of the formation of a stable,protective oxide layer on its surface.

(C) Beryllium $(Be)$ is the only alkaline earth metal that does not react with water or steam,even at red heat.
This is due to its exceptionally small atomic size and high ionization energy compared to other elements in the group $(Group \ 2)$.

(B) In the reaction $3 \ Mg + N_2 \longrightarrow Mg_3N_2$,the oxidation state of $Mg$ changes from $0$ (in elemental form) to $+2$ (in $Mg_3N_2$).
Since there is an increase in the oxidation state of $Mg$,it undergoes oxidation.
Similarly,the oxidation state of $N$ changes from $0$ (in $N_2$) to $-3$ (in $Mg_3N_2$),which means $N_2$ undergoes reduction.
Therefore,$Mg$ is oxidised.

(B) The $second$ group consists of alkaline earth metals,which are located in Group $2$ of the periodic table.
The $fifth$ period corresponds to the row where the element is located.
Strontium $(Sr)$ has an atomic number of $38$ and its electronic configuration is $[Kr] 5s^2$.
Therefore,it belongs to Group $2$ and Period $5$ of the periodic table.

(D) Beryllium oxide $(BeO)$ is amphoteric in nature,meaning it reacts with both acids and bases.
$1$. Reaction with aqueous $HCl$:
$BeO + 2HCl_{(aq)} \longrightarrow BeCl_2 + H_2O$
The products are beryllium chloride $(BeCl_2)$ and water.
$2$. Reaction with aqueous $NaOH$:
$BeO + 2NaOH_{(aq)} \longrightarrow Na_2BeO_2 + H_2O$
The products are sodium beryllate $(Na_2BeO_2)$ and water.
Therefore,the products obtained are $BeCl_2$ and $Na_2BeO_2$.

(C) In the Solvay process,$NH_4Cl$ is treated with slaked lime $(Ca(OH)_2)$ to recover ammonia $(NH_3)$.
The chemical reaction is as follows:
$2NH_4Cl + Ca(OH)_2 \rightarrow CaCl_2 + 2H_2O + 2NH_3$
Thus,$NH_3$ is recovered.

(C) $1$. Metals that have low ionization enthalpies are used in photoelectric cells.
$2$. The most popular metal used in photoelectric cells is Caesium $(Cs)$,which has the atomic number $55$.
$3$. Caesium belongs to group $1$ (alkali metals) and period $6$.
$4$. The electronic configuration of Caesium is $[Xe] 6s^1$.
$5$. Caesium is used widely in photocells as it can easily convert solar energy into electrical energy.
$6$. When Caesium is exposed to sunlight,it starts emitting electrons,which induces electric flow and generates electricity.
$7$. Thus,Caesium is the most preferred metal used in photoelectric cells.

(C) $Be$ is the first element of group $2$ which shows a diagonal relationship with $Al$,the second element of group $13$.

(A) Dipole-induced dipole interaction occurs between polar and non-polar molecules.
When a polar molecule approaches a non-polar molecule,it distorts the electron cloud of the non-polar molecule,inducing a temporary dipole in it.
This results in an attractive force known as dipole-induced dipole interaction.

(A) dipole-induced dipole interaction occurs between a polar molecule (having a permanent dipole) and a non-polar molecule (which becomes polarized due to the presence of the polar molecule).
$NH_3$ is a polar molecule with a permanent dipole moment.
$C_6H_6$ (benzene) is a non-polar molecule.
Therefore,the interaction between $NH_3$ and $C_6H_6$ is a dipole-induced dipole interaction.
In contrast,$NaCl + H_2O$ involves ion-dipole interactions,$CH_4 + C_2H_6$ involves London dispersion forces,and $HF + H_2O$ involves hydrogen bonding.

(D) The dissociation of $CaCO_3$ is given by: $CaCO_{3(s)} \rightleftharpoons Ca^{2+}_{(aq)} + CO^{2-}_{3(aq)}$.
Let the solubility be $S = 7 \times 10^{-5} \ mol \ dm^{-3}$.
The solubility product expression is $K_{sp} = [Ca^{2+}][CO^{2-}_{3}] = S \times S = S^2$.
Substituting the value of $S$:
$K_{sp} = (7 \times 10^{-5})^2 = 49 \times 10^{-10} = 4.9 \times 10^{-9}$.

(B) The degree of dissociation $(\alpha)$ is given as $4.2 \times 10^{-2}$.
To find the percent dissociation,multiply the degree of dissociation by $100$.
$\text{Percent dissociation} = \alpha \times 100$
$\text{Percent dissociation} = 4.2 \times 10^{-2} \times 100 = 4.2 \%$

(B) The law of multiple proportions states that when two elements combine to form more than one compound,the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.
$CuO$ and $Cu_2O$ consist of $Cu$ and $O$.
$CO$ and $CO_2$ consist of $C$ and $O$.
$N_2O_4$ and $N_2O_5$ consist of $N$ and $O$.
$NaNO_3$ and $CaCO_3$ are different compounds containing different sets of elements ($Na, N, O$ vs $Ca, C, O$).
Therefore,they do not satisfy the criteria for the law of multiple proportions.

(D) The statement "$A$ given compound always contains the same proportion of elements by mass" refers to the $Law \ of \ definite \ proportions$.
This law states that a chemical compound always consists of the same elements combined together in the same fixed proportion by mass,regardless of its source or method of preparation.
For example,water $(H_2O)$ always contains hydrogen and oxygen in a mass ratio of $1:8$.

(C) The correct answer is $C$ (Law of Multiple Proportions).
Explanation:
The Law of Multiple Proportions states that when two elements combine to form more than one compound,the different masses of one element that combine with a fixed mass of the other element are in simple whole number ratios.
- In $H_2O$,$2 \ g$ of hydrogen combines with $16 \ g$ of oxygen.
- In $H_2O_2$,$2 \ g$ of hydrogen combines with $32 \ g$ of oxygen.
- The masses of oxygen ($16 \ g$ and $32 \ g$) that combine with a fixed mass of hydrogen $(2 \ g)$ are in the ratio $16:32$,which simplifies to $1:2$,a simple whole number ratio.
- Therefore,these compounds illustrate the Law of Multiple Proportions.

(D) The law of multiple proportions states that when two elements combine to form two or more compounds,the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.
This law is applicable only when the same two elements form different compounds.
In the pair $Na_2S$ and $NaF$,the elements involved are different ($S$ and $F$),so they cannot demonstrate the law of multiple proportions.

(A) $\text{Number of moles} = 5 = \frac{\text{Given mass (g)}}{\text{Molar mass}}$
$\text{For acetic acid, } M = 60 \ g \ mol^{-1}$
$\text{Given mass (g)} = 5 \ mol \times 60 \ g \ mol^{-1} = 300 \ g$
$\text{Mass in } kg = \frac{300 \ g}{1000} = 0.3 \ kg$

(A) Given mass of carbon $= 3.6 \ kg = 3600 \ g$.
Molar mass of carbon $= 12 \ g/mol$.
The number of moles is calculated as:
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{3600 \ g}{12 \ g/mol} = 300 \ mol$.
This can be expressed as $3.0 \times 10^2 \ mol$.

(C) The chemical formula of cytosine is $C_4H_5N_3O$.
From the chemical formula,it is evident that one molecule of cytosine contains $3$ nitrogen atoms.
Therefore,one mole of cytosine contains $3$ moles of nitrogen atoms.

(D) The molar mass of water $(H_2O)$ is $18 \times 10^{-3} \ kg \ mol^{-1}$.
Number of moles $(n) = \frac{\text{Given mass}}{\text{Molar mass}}$.
$n = \frac{9.10 \times 10^{-2} \ kg}{18 \times 10^{-3} \ kg \ mol^{-1}}$.
$n = \frac{91 \times 10^{-3}}{18 \times 10^{-3}} = 5.055 \approx 5.0 \ mol$.

(B) The molar mass of water $(H_2O)$ is $18 \,g/mol$.
$1 \,mol$ of water contains $6.022 \times 10^{23}$ molecules.
Mass of $6.022 \times 10^{23}$ molecules of water $= 18 \,g$.
Therefore, the mass of $1$ molecule of water $= \frac{18 \,g}{6.022 \times 10^{23}} = 2.988 \times 10^{-23} \,g$.
Given that $\text{density} = \frac{\text{mass}}{\text{volume}}$, the volume occupied by $1$ molecule is $\text{Volume} = \frac{\text{mass}}{\text{density}}$.
$\text{Volume} = \frac{2.988 \times 10^{-23} \,g}{1 \,g \,cm^{-3}} = 2.988 \times 10^{-23} \,cm^3 \approx 2.98 \times 10^{-23} \,cm^3$.

(D) Let the relative abundance of $^{35}Cl$ be $x\%$ and $^{37}Cl$ be $(100-x)\%$.
Average atomic mass = $\frac{(35 \times x) + (37 \times (100-x))}{100} = 35.5$
$35x + 3700 - 37x = 3550$
$-2x = -150$
$x = 75$
Abundance of $^{35}Cl = 75\%$ and $^{37}Cl = 25\%$.
Ratio = $75:25 = 3:1$.

(B) Let the percentage abundance of $^{35}Cl$ be $x$ and that of $^{37}Cl$ be $(100 - x)$.
Average atomic mass is given by the formula:
$\text{Average atomic mass} = \frac{(\text{mass of } ^{35}Cl \times x) + (\text{mass of } ^{37}Cl \times (100 - x))}{100}$
Substituting the values:
$35.5 = \frac{35x + 37(100 - x)}{100}$
$3550 = 35x + 3700 - 37x$
$3550 = 3700 - 2x$
$2x = 3700 - 3550$
$2x = 150$
$x = 75$
Therefore,the abundance of $^{35}Cl$ is $75 \%$ and the abundance of $^{37}Cl$ is $100 - 75 = 25 \%$.

(A) The chemical equation for the combustion of carbon is: $C + O_2 \rightarrow CO_2$
From the stoichiometry of the reaction,$12 \ g$ of $C$ produces $44 \ g$ of $CO_2$.
Therefore,$0.6 \ g$ of $C$ will produce: $\frac{44 \times 0.6}{12} = 2.2 \ g$ of $CO_2$.
The molar mass of $CO_2$ is $44 \ g/mol$.
Number of moles of $CO_2 = \frac{2.2 \ g}{44 \ g/mol} = 0.05 \ mol$.
Number of molecules of $CO_2 = \text{moles} \times N_A = 0.05 \times 6.022 \times 10^{23} = 3.011 \times 10^{22}$ molecules.

(A) The chemical formula of carnallite is $KCl \cdot MgCl_2 \cdot 6 H_2O$.
From the formula,it is evident that $1$ mole of carnallite contains $6$ moles of water molecules.

(C) The number of moles is calculated using the formula: $\text{Moles} = \frac{\text{Mass in grams}}{\text{Molar mass}}$
Given mass $= 6.9 \times 10^{-2} \ kg = 6.9 \times 10^{-2} \times 10^3 \ g = 69 \ g$
Molar mass of sodium $= 23 \ g \ mol^{-1}$
Number of moles $= \frac{69 \ g}{23 \ g \ mol^{-1}} = 3 \ mol$

(B) The chemical formula of ammonia is $NH_3$.
The molar mass of $NH_3 = 14 + (3 \times 1) = 17 \ g \ mol^{-1}$.
Mass in grams $= \text{moles} \times \text{molar mass} = 2.5 \ mol \times 17 \ g \ mol^{-1} = 42.5 \ g$.
To convert the mass into $kg$,divide by $1000$:
Mass in $kg = \frac{42.5 \ g}{1000} = 0.0425 \ kg = 4.25 \times 10^{-2} \ kg$.

(D) At $STP$,$1 \ mol$ of any gas occupies $22.4 \ dm^3$ volume.
The molar mass of $CO_2$ is $12 + (2 \times 16) = 44 \ g/mol$.
Thus,$22.4 \ dm^3$ of $CO_2$ at $STP$ has a mass of $44 \ g$.
Therefore,the mass of $4.48 \ dm^3$ of $CO_2$ at $STP$ is calculated as:
$\text{Mass} = \frac{44 \ g}{22.4 \ dm^3} \times 4.48 \ dm^3 = 8.8 \ g$.
To convert the mass into $kg$,we divide by $1000$:
$8.8 \ g = 8.8 \times 10^{-3} \ kg$.

(C) The balanced chemical equation for the combustion of methane is:
$CH_4(g) + 2O_2(g) \longrightarrow CO_2(g) + 2H_2O(l)$
According to the stoichiometry of the reaction,$1 \ mole$ of $CH_4$ requires $2 \ moles$ of $O_2$.
At $\text{S.T.P.}$,$1 \ mole$ of any gas occupies $22.4 \ dm^3$.
Therefore,$2 \ moles$ of $O_2$ occupy $2 \times 22.4 \ dm^3 = 44.8 \ dm^3$.
For $0.25 \ mole$ of $CH_4$,the volume of $O_2$ required is:
$V = 0.25 \times 44.8 \ dm^3 = 11.2 \ dm^3$.

(D) At $\text{STP}$, $1 \ mol$ of any ideal gas occupies a volume of $22.4 \ dm^3$.
Therefore, the volume occupied by $2.5 \ mol$ of ammonia gas is calculated as:
$V = n \times 22.4 \ dm^3 \ mol^{-1}$
$V = 2.5 \ mol \times 22.4 \ dm^3 \ mol^{-1} = 56.0 \ dm^3$.

(D) The number of oxidation states exhibited by transition elements depends on the number of unpaired electrons in their $d$-orbitals.
$Sc$ $([Ar] 3d^1 4s^2)$ shows $+2, +3$ oxidation states.
$Ti$ $([Ar] 3d^2 4s^2)$ shows $+2, +3, +4$ oxidation states.
$Cu$ $([Ar] 3d^{10} 4s^1)$ shows $+1, +2$ oxidation states.
$Zn$ $([Ar] 3d^{10} 4s^2)$ has a completely filled $d$-subshell and shows only $+2$ oxidation state.
Therefore,$Zn$ exhibits the lowest number of different oxidation states.

(B) The colour of a transition metal ion depends on the presence of unpaired electrons in its $d$-orbitals,which allows for $d-d$ transitions.
| Metal Ion | Electronic Configuration | Unpaired Electrons |
| :--- | :--- | :--- |
| $Zn^{2+}$ | $4s^0 3d^{10}$ | $0$ |
| $Cu^{+}$ | $4s^0 3d^{10}$ | $0$ |
| $Fe^{2+}$ | $4s^0 3d^6$ | $4$ |
| $Sc^{3+}$ | $4s^0 3d^0$ | $0$ |
Since $Fe^{2+}$ has $4$ unpaired electrons,it can undergo $d-d$ transitions and thus forms coloured compounds.

(B) The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$,which contains a half-filled $d$-subshell $(d^5)$.
The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$,which also contains a half-filled $d$-subshell $(d^5)$.
Therefore,the pair of elements with half-filled $d$-orbitals is $Mn$ and $Cr$.

(C) The magnetic moment $\mu$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$1$. For $Fe^{2+}$ $(3d^6)$: $n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.89 \ BM$.
$2$. For $Cr^{2+}$ $(3d^4)$: $n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.89 \ BM$.
$3$. For $Mn^{2+}$ $(3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.91 \ BM$.
$4$. For $Ni^{2+}$ $(3d^8)$: $n = 2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.
Comparing the values,$Mn^{2+}$ has the highest number of unpaired electrons $(n=5)$,and therefore exhibits the highest magnetic moment.

(C) The $5d$ series of transition elements includes elements from $Z=72$ $(Hf)$ to $Z=80$ $(Hg)$.
Lanthanum ($La$,$Z=57$) is a $d$-block element but is often considered the start of the lanthanide series.
The statement in option $C$ is false because the $5d$ series does not include all elements from $La$ to $Hg$ as a continuous transition series,and the range provided is inaccurate regarding the definition of the $5d$ transition series.

(D) The magnetic moment of an ion is given by the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
For an ion to exhibit a magnetic moment,it must have at least one unpaired electron $(n > 0)$.
Let us analyze the electronic configuration of the given elements in their $+2$ oxidation state:
$1$. $Mn^{2+}$: $[Ar]3d^5$ $(n = 5)$
$2$. $Co^{2+}$: $[Ar]3d^7$ $(n = 3)$
$3$. $Fe^{2+}$: $[Ar]3d^6$ $(n = 4)$
$4$. $Zn^{2+}$: $[Ar]3d^{10}$ $(n = 0)$
Since $Zn^{2+}$ has a completely filled $3d$ subshell,it has $0$ unpaired electrons. Therefore,it does not exhibit a magnetic moment.

(C) The oxidation states of the given transition elements are as follows:
$Fe$ $(3d^6 4s^2)$: $+2, +3, +4, +5, +6$
$Cr$ $(3d^5 4s^1)$: $+2, +3, +4, +5, +6$
$Mn$ $(3d^5 4s^2)$: $+2, +3, +4, +5, +6, +7$
$Ni$ $(3d^8 4s^2)$: $+2, +3, +4$
Comparing these,$Mn$ exhibits the highest number of oxidation states (from $+2$ to $+7$).

(B) The first ionization enthalpy $(IE_1)$ of the $3d$ transition series elements generally increases from left to right across the period due to the increase in effective nuclear charge.
Among the given elements ($Sc$,$Mn$,$Cu$,$Zn$),$Sc$ $(Z=21)$ is the first element of the $3d$ series.
Since $IE_1$ increases as we move from left to right,the element at the beginning of the series will have the lowest ionization enthalpy.
Therefore,$Sc$ has the lowest $IE_1$ value.

(C) Transition elements are defined as elements that have partially filled $d-$orbitals in their ground state or in any of their common oxidation states.
$Hg$ has an electronic configuration of $[Xe] 4f^{14} 5d^{10} 6s^2$.
Since $Hg$ has a completely filled $d-$orbital $(d^{10})$ in its ground state and does not form ions with partially filled $d-$orbitals,it is not considered a transition element.
Similarly,$Zn$ and $Cd$ are also not considered transition elements for the same reason.

(A) The $5d$ transition series consists of elements from Lanthanum ($La$,atomic number $57$) to Mercury ($Hg$,atomic number $80$).
This series comprises a portion of the $d$-block elements,specifically those where the $5d$ electron shell is being filled progressively.
$Hg$ (Mercury) is the last element of this series,with an atomic number of $80$.
Its electron configuration is $[Xe] 4f^{14} 5d^{10} 6s^2$.
Mercury is well-known for being the only metal that is liquid at room temperature.

(A) Nichrome,an alloy of nickel and chromium in the ratio $80:20$,is specifically used for the construction of gas turbine engines due to its high heat resistance and oxidation resistance.
Titanium alloys are also used in aerospace applications for their ability to withstand stress at high temperatures.

(C) Transition elements are metals and possess metallic properties.
They are generally hard,malleable,and ductile.
They can form alloys with other metals.
Transition elements are good conductors of both heat and electricity.
Therefore,the statement that they do not conduct heat is incorrect.

(C) Stainless steels are used in the construction of the outer fuselage of ultra-high-speed aircraft due to their high strength and resistance to corrosion at elevated temperatures.

(C) To determine the number of unpaired electrons,we look at the electronic configuration of each ion:
$Fe^{3+} = [Ar] 3d^5$ ($5$ unpaired electrons)
$Mn^{2+} = [Ar] 3d^5$ ($5$ unpaired electrons)
$Co^{2+} = [Ar] 3d^7$ ($3$ unpaired electrons)
$Sc^{3+} = [Ar] 3d^0$ ($0$ unpaired electrons)
$Zn^{2+} = [Ar] 3d^{10}$ ($0$ unpaired electrons)
$Mn^{3+} = [Ar] 3d^4$ ($4$ unpaired electrons)
Comparing the pairs:
Option $A$: $Co^{2+}$ $(3)$ and $Mn^{3+}$ $(4)$ - Different
Option $B$: $Mn^{2+}$ $(5)$ and $Zn^{2+}$ $(0)$ - Different
Option $C$: $Fe^{3+}$ $(5)$ and $Mn^{2+}$ $(5)$ - Same
Option $D$: $Sc^{3+}$ $(0)$ and $Co^{2+}$ $(3)$ - Different
Therefore,the pair with the same number of unpaired electrons is $Fe^{3+}$ and $Mn^{2+}$.

(D) The lanthanoids are the elements with atomic numbers $57$ to $71$.
$Pm$ (Promethium,$Z=61$),$Er$ (Erbium,$Z=68$),and $Yb$ (Ytterbium,$Z=70$) are lanthanoids.
$Hf$ (Hafnium,$Z=72$) is a $d$-block transition element belonging to the $6th$ period and $4th$ group.

(B) $La$ (Lanthanum) has an atomic number of $57$.
In the $ 3$ oxidation state,its electronic configuration is $[Xe] 4f^0$.
Since there are no unpaired electrons $(n = 0)$,the effective magnetic moment $(\mu_{eff} = \sqrt{n(n 2)} \ BM)$ is $0 \ BM$.

(D) The phenomenon of lanthanide contraction is observed in $4f$ elements due to the increase in effective nuclear charge as we move across the series.
As a result,the atomic and ionic radii decrease continuously from lanthanum $(Z=57)$ to lutetium $(Z=71)$.
Since lutetium $(Lu)$ is the last element in the lanthanide series,it has the highest effective nuclear charge and therefore the smallest ionic radius in the $+3$ state.

(A) The electronic configuration of Gadolinium ($Gd$,$Z=64$) is $[Xe] 4f^7 5d^1 6s^2$.
In this configuration,the $4f$ subshell contains $7$ electrons,which corresponds to a half-filled $4f$ orbital ($f^{14}$ is fully filled,so $f^7$ is half-filled).

(A) $1$. These are bad conductors of heat and electricity: This statement is incorrect. Lanthanoids are metallic in nature and are generally good conductors of heat and electricity,similar to other metals.
$2$. These are soft metals: This is true. Lanthanoids are relatively soft metals,and their hardness increases with increasing atomic number.
$3$. Coordination number is usually greater than six: This is true. Due to their large ionic radii,lanthanoids often exhibit high coordination numbers,typically ranging from $8$ to $12$ in their complexes.
$4$. These are strongly paramagnetic: This is true. Lanthanoids possess unpaired $f$-electrons,which result in paramagnetic behavior. Many lanthanoid ions exhibit strong paramagnetism.

(A) The actinoids are the elements from atomic number $89$ to $103$.
Among the given options,$Cm$ (Curium,atomic number $96$) belongs to the actinoid series.
$Pm$ (Promethium),$Tm$ (Thulium),and $Sm$ (Samarium) are lanthanoids.

(C) The atomic number of $Lu$ is $71$.
Its electronic configuration is $[Xe] 4f^{14} 5d^1 6s^2$.
In this configuration,the $4f$ subshell is completely filled with $14$ electrons,which matches both the expected and observed electronic configurations.

(A) $(1)$ $Sm$ stands for Samarium,which is a rare earth element. Rare earth elements are a group of $17$ elements that include the $15$ lanthanides plus scandium $(Sc)$ and yttrium $(Y)$. Samarium $(Sm)$ is a member of the lanthanide series and is considered a rare earth element.
$(2)$ $Hg$ stands for Mercury,which is a transition metal,not a rare earth element.
$(3)$ $Zn$ stands for Zinc,which is a transition metal,not a rare earth element.
$(4)$ $W$ stands for Tungsten,which is a transition metal,not a rare earth element.

(C) The most stable and common oxidation state of lanthanoids $(Ln)$ is $+3$.
Therefore,when $Ln^{3+}$ reacts with hydroxide ions $(OH^-)$,the resulting compound is $Ln(OH)_3$.

(B) The ground state electronic configurations are as follows:
$Yb (Z=70): [Xe] 4f^{14} 6s^2$
$Eu (Z=63): [Xe] 4f^7 6s^2$
$Cd (Z=48): [Kr] 4d^{10} 5s^2$
$W (Z=74): [Xe] 4f^{14} 5d^4 6s^2$
Among the given options,$Yb$ has a completely filled $4f$ orbital $(4f^{14})$.

(A) The actinoid series consists of elements with atomic numbers from $89$ to $103$.
- $No$ (Nobelium) has an atomic number of $102$,which places it within the actinoid series.
- $Mo$ (Molybdenum) is a $d$-block element with atomic number $42$.
- $Co$ (Cobalt) is a $d$-block element with atomic number $27$.
- $Ce$ (Cerium) is a lanthanoid element with atomic number $58$.
Therefore,$No$ is the correct actinoid element.

(C) The electrolysis of molten $NaCl$ is a non-spontaneous process because the standard cell potential $(E^\circ_{cell})$ is negative.
Therefore,the decomposition of $NaCl$ into $Na_{(s)}$ and $Cl_{2(g)}$ is not spontaneous.
Electrical energy is required to drive this non-spontaneous reaction.
At the anode,$Cl^-$ ions are oxidized to $Cl_2$ gas.
At the cathode,$Na^+$ ions are reduced to $Na$ metal.

(A) The cell reaction is: $Pb_{(s)} + 2Ag^{+}_{(10 \ M)} \rightarrow Pb^{2+}_{(1 \ M)} + 2Ag_{(s)}$
Using the Nernst equation at $25^{\circ} C$ $(298 \ K)$:
$E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{n} \log_{10} Q$
Here,$n = 2$ (number of electrons transferred).
$Q = \frac{[Pb^{2+}]}{[Ag^{+}]^{2}} = \frac{1}{(10)^{2}} = 10^{-2}$
Substituting the values:
$E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{2} \log_{10} (10^{-2})$
$E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{2} \times (-2)$
$E_{cell} = E^{\circ}_{cell} + 0.0592 \ V$

(B) During the electrolysis of molten $NaCl$,the oxidation reaction at the anode is:
$2Cl^{-} \longrightarrow Cl_2 + 2e^{-}$
From the stoichiometry of the reaction,$1 \ mole$ of $Cl_2$ gas is produced by the transfer of $2 \ moles$ of electrons.
Therefore,to liberate $0.1 \ mole$ of $Cl_2$,the required number of moles of electrons is $0.1 \times 2 = 0.2 \ mole$.
The quantity of electricity $Q$ is given by $Q = n \times F$,where $n$ is the number of moles of electrons and $F$ is Faraday's constant $(96500 \ C \ mol^{-1})$.
$Q = 0.2 \ mole \times 96500 \ C \ mol^{-1} = 19300 \ C$.

(B) The reduction half-reaction is: $MnO_4^{-} + 8H^{+} + 5e^{-} \longrightarrow Mn^{2+} + 4H_2O$.
From the stoichiometry,$1 \ mol$ of $MnO_4^{-}$ requires $5 \ mol$ of electrons $(5 \ F)$ for reduction to $Mn^{2+}$.
Therefore,for $0.08 \ mol$ of $MnO_4^{-}$,the electricity required is $0.08 \times 5 \ F = 0.4 \ F$.
Since $1 \ F = 96500 \ C$,the total charge in coulombs is $0.4 \times 96500 \ C = 38600 \ C$.

(D) The cell reaction is: $Cd_{(s)} + 2Ag^{+}_{(aq)} \longrightarrow Cd^{2+}_{(aq)} + 2Ag_{(s)}$
For this reaction,the number of electrons transferred is $n = 2$.
The Nernst equation at $298 \ K$ is given by: $E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0592}{n} \log_{10} \frac{[Cd^{2+}]}{[Ag^{+}]^2}$
Given $[Cd^{2+}] = 1 \ M$ and $[Ag^{+}] = 1 \ M$,the reaction quotient $Q = \frac{1}{(1)^2} = 1$.
Substituting the values: $E_{\text{cell}} = 1.2 - \frac{0.0592}{2} \log_{10}(1)$
Since $\log_{10}(1) = 0$,we get $E_{\text{cell}} = 1.2 - 0 = 1.2 \ V$.

(D) The reduction reaction for magnesium is: $Mg^{2+} + 2e^{-} \longrightarrow Mg$.
According to the stoichiometry,$2 \ mol$ of electrons $(2 \ F)$ are required to deposit $1 \ mol$ of $Mg$.
Molar mass of $Mg = 24 \ g \ mol^{-1}$.
Number of moles of $Mg = \frac{4.8 \ g}{24 \ g \ mol^{-1}} = 0.2 \ mol$.
Electricity required $= 0.2 \ mol \times 2 \ F \ mol^{-1} = 0.4 \ F$.

(A) The cell reaction is: $Al(s) + 3H^+(aq) \rightarrow Al^{3+}(aq) + \frac{3}{2}H_2(g)$.
For this cell,the anode is $Al$ and the cathode is $Pt, H_2$.
The standard electrode potential of the standard hydrogen electrode $(SHE)$ is $E^{\circ}_{H^+/H_2} = 0.00 \ V$.
The standard cell potential is given by the formula: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
Substituting the values: $E^{\circ}_{cell} = E^{\circ}_{H^+/H_2} - E^{\circ}_{Al^{3+}/Al} = 0.00 \ V - (-1.66 \ V) = 1.66 \ V$.

(D) During the electrolysis of aqueous $NaCl$,the ions present in the solution are $Na^+$,$Cl^-$,$H^+$,and $OH^-$.
At the cathode,the reduction potential of water is higher than that of $Na^+$ ions.
Therefore,water is reduced in preference to $Na^+$ ions.
The reaction at the cathode is: $2H_2O_{(l)} + 2e^- \rightarrow H_{2(g)} + 2OH^-_{(aq)}$.

(B) During the electrolysis of fused (molten) $NaCl$,the electrolyte dissociates into $Na^+$ and $Cl^-$ ions.
At the cathode,$Na^+$ ions undergo reduction: $Na^+ + e^- \rightarrow Na_{(s)}$.
At the anode,$Cl^-$ ions undergo oxidation: $2Cl^- \rightarrow Cl_{2(g)} + 2e^-$.
Therefore,the product formed at the cathode is metallic sodium,$Na_{(s)}$.

(C) The reduction reaction at the cathode is: $Ca^{2+} + 2e^{-} \longrightarrow Ca_{(s)}$
From the reaction,$2 \ mol$ of electrons are required to deposit $1 \ mol$ of $Ca$.
The total charge passed $(Q)$ is given by $Q = I \times t = 0.8 \ A \times (60 \times 60 \ s) = 2880 \ C$.
The number of moles of electrons passed is $n = \frac{Q}{F} = \frac{2880}{96500} \approx 0.0298 \ mol$.
Since $2 \ mol$ of $e^{-}$ deposit $1 \ mol$ of $Ca$,the moles of $Ca$ deposited is $\frac{0.0298}{2} = 0.0149 \ mol$.
Mass of $Ca = \text{moles} \times \text{molar mass} = 0.0149 \ mol \times 40 \ g \ mol^{-1} \approx 0.597 \ g \approx 0.6 \ g$.

(A) The given cell is composed of a copper electrode as the anode and a silver electrode as the cathode.
The standard cell potential is given by the formula:
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
Substituting the given values:
$E^{\circ}_{cell} = E^{\circ}_{Ag^{+}|Ag} - E^{\circ}_{Cu^{2+}|Cu}$
$0.647 \ V = E^{\circ}_{Ag^{+}|Ag} - 0.153 \ V$
$E^{\circ}_{Ag^{+}|Ag} = 0.647 \ V + 0.153 \ V = 0.8 \ V$

(A) The higher the standard reduction potential $(E^\circ)$ value,the greater the tendency of the species to accept electrons and undergo reduction.
Order of $E^\circ_{red}: Ag^{+} (0.80 \ V) > Cu^{2+} (0.337 \ V) > Sn^{2+} (-0.136 \ V) > Cd^{2+} (-0.403 \ V)$.
Therefore,the decreasing order of deposition of metal on the electrode is $Ag > Cu > Sn > Cd$.

(A) The standard cell potential is given by the formula:
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
Here,the calomel electrode acts as the cathode and the zinc electrode acts as the anode.
$E^{\circ}_{cell} = E^{\circ}_{calomel} - E^{\circ}_{Zn^{2+}/Zn}$
Substituting the given values:
$1.007 \ V = 0.242 \ V - E^{\circ}_{Zn^{2+}/Zn}$
$E^{\circ}_{Zn^{2+}/Zn} = 0.242 \ V - 1.007 \ V$
$E^{\circ}_{Zn^{2+}/Zn} = -0.765 \ V$

(C) At equilibrium,the cell potential $E_{\text{cell}} = 0$ and the reaction quotient $Q = K$.
The Nernst equation is given by: $E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{0.0592}{n} \log_{10} Q$.
Substituting the equilibrium conditions: $0 = E_{\text{cell}}^{\circ} - \frac{0.0592}{n} \log_{10} K$.
Rearranging this gives: $E_{\text{cell}}^{\circ} = \frac{0.0592}{n} \log_{10} K$.

(A) The reduction reaction at the cathode is: $Al^{3+} + 3e^{-} \longrightarrow Al_{(s)}$
According to the stoichiometry,$1 \ mol$ of $Al$ $(27 \ g)$ requires $3 \ F$ of electricity.
Therefore,the number of faradays required to produce $0.18 \ g$ of $Al$ is:
$F = \frac{3 \times 0.18}{27} = \frac{0.54}{27} = 0.02 \ F$.

(B) During the discharging of a lead accumulator,the following overall cell reaction occurs:
$Pb_{(s)} + PbO_{2_{(s)}} + 2 H_2 SO_{4_{(aq)}} \rightarrow 2 PbSO_{4_{(s)}} + 2 H_2 O_{(l)}$
In this process,$Pb$ is oxidized at the anode and $PbO_2$ is reduced at the cathode.
As the reaction proceeds,$H_2 SO_4$ is consumed,leading to a decrease in the concentration of the electrolyte.

(C) In an $H_2-O_2$ fuel cell,the electrodes are typically porous carbon electrodes impregnated with catalysts like platinum,palladium,or nickel to facilitate the reactions.
The use of platinum wires as electrodes is not standard practice because:
$1$. Platinum is expensive,and using it as solid wires would not be cost-effective.
$2$. Porous electrodes increase the surface area for the reaction and improve efficiency.
Conclusion: The statement that platinum wires are used as anode and cathode is $NOT$ correct.

(A) The balanced half-reaction for the reduction of dichromate in acidic medium is:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O$
From the stoichiometry,$1 \ mol$ of $Cr_2O_7^{2-}$ requires $6 \ mol$ of electrons for reduction.
Therefore,for $1.1 \ mol$ of $Cr_2O_7^{2-}$,the number of moles of electrons required is $1.1 \times 6 = 6.6 \ mol$.
The total charge $Q$ is given by $Q = n \times F$,where $n$ is the number of moles of electrons and $F$ is Faraday's constant $(96500 \ C/mol)$.
$Q = 6.6 \ mol \times 96500 \ C/mol = 636900 \ C = 6.369 \times 10^5 \ C$.

(C) During the discharging of a lead storage battery (lead accumulator),the cathode is made of lead dioxide $(PbO_2)$.
At the cathode,$PbO_2$ undergoes reduction by gaining electrons in the presence of sulfuric acid $(H_2SO_4)$.
The balanced chemical equation for the reduction reaction at the cathode is:
$PbO_{2(s)} + 4 H_{(aq)}^{+} + SO_{4(aq)}^{2-} + 2 e^{-} \longrightarrow PbSO_{4(s)} + 2 H_2 O_{(l)}$
Therefore,the correct option is $(C)$.

(C) In a dry cell,the cathode is an inert graphite rod placed in the center,surrounded by an electrolyte paste.
The electrolyte consists of a moist paste of $NH_4Cl$ and $ZnCl_2$.
Starch is added to this paste to increase its viscosity,which prevents the leakage of the electrolyte from the cell.

(C) The conductivity $(\kappa)$ is given by the formula: $\kappa = \frac{\text{cell constant}}{R}$.
Given: Cell constant = $0.33 \ cm^{-1}$,Resistance $(R)$ = $30 \ \Omega$.
$\kappa = \frac{0.33 \ cm^{-1}}{30 \ \Omega} = 0.011 \ \Omega^{-1} \ cm^{-1}$.

(C) The relationship between conductivity $(k)$,cell constant $(G^*)$,and resistance $(R)$ is given by:
$k = \frac{G^*}{R}$
Therefore,the cell constant $(G^*)$ is:
$G^* = k \times R$
Substituting the given values:
$G^* = (1.64 \times 10^{-4} \ S \ cm^{-1}) \times (120 \ \Omega)$
$G^* = 0.01968 \ \text{cm}^{-1} \approx 0.0196 \ \text{cm}^{-1}$

(A) The formula for molar conductivity is $\Lambda_{m} = \frac{1000 \kappa}{C}$.
Given: $\Lambda_{m} = 410 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $C = 0.02 \ M$.
Rearranging the formula to find conductivity $(\kappa)$: $\kappa = \frac{\Lambda_{m} \times C}{1000}$.
Substituting the values: $\kappa = \frac{410 \times 0.02}{1000} \ \Omega^{-1} \ cm^{-1}$.
$\kappa = \frac{8.2}{1000} \ \Omega^{-1} \ cm^{-1} = 8.2 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$.

(A) The cell constant $(G^*)$ is defined as the ratio of the distance between the electrodes $(l)$ to the area of cross-section $(A)$:
$G^* = \frac{l}{A}$
Given:
$l = 0.92 \ cm$
$A = 1.2 \ cm^2$
Calculation:
$G^* = \frac{0.92 \ cm}{1.2 \ cm^2} = 0.7666... \ cm^{-1} \approx 0.767 \ cm^{-1}$

(B) The relationship between conductivity $(k)$,resistance $(R)$,and cell constant $(G^*)$ is given by the formula:
$k = \frac{1}{R} \times G^*$
Rearranging the formula to solve for the cell constant:
$G^* = k \times R$
Given values:
$k = 0.0015 \ \Omega^{-1} \ cm^{-1}$
$R = 600 \ \Omega$
Substituting the values:
$G^* = 0.0015 \ \Omega^{-1} \ cm^{-1} \times 600 \ \Omega = 0.90 \ cm^{-1}$