General Characteristics Questions in English

(A) Substances like iron $(Fe)$,cobalt $(Co)$,nickel $(Ni)$,gadolinium $(Gd)$,and $CrO_2$ are attracted very strongly by a magnetic field. Such substances are called ferromagnetic substances. Besides strong attraction,these substances can be permanently magnetized.

(B) $Gadolinium$ is ferromagnetic in nature.
$Oxygen$ is paramagnetic.
$Water$ and $Benzene$ are diamagnetic.

(B) The electronic configuration of $Cr$ is $[Ar] \ 3d^5 \ 4s^1$.
For $Cr^{2+}$,two electrons are removed,resulting in the configuration $[Ar] \ 3d^4$.
The number of unpaired electrons $(n)$ is $4$.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n=4$: $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.

(D) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
$1.$ $Sc$ $(Z=21)$: $[Ar] 3d^1 4s^2$. In the $+3$ state,$Sc^{3+}$ is $[Ar] 3d^0$. Here,$n=0$,so $\mu = 0 \text{ BM}$.
$2.$ $Ti$ $(Z=22)$: $[Ar] 3d^2 4s^2$. In the $+3$ state,$Ti^{3+}$ is $[Ar] 3d^1$. Here,$n=1$,so $\mu = \sqrt{3} \text{ BM}$.
$3.$ $V$ $(Z=23)$: $[Ar] 3d^3 4s^2$. In the $+3$ state,$V^{3+}$ is $[Ar] 3d^2$. Here,$n=2$,so $\mu = \sqrt{8} \text{ BM}$.
$4.$ $Cr$ $(Z=24)$: $[Ar] 3d^5 4s^1$. In the $+3$ state,$Cr^{3+}$ is $[Ar] 3d^3$. Here,$n=3$,so $\mu = \sqrt{15} \text{ BM}$.
Therefore,$Sc$ does not exhibit a spin-only magnetic moment in the $+3$ state because it has no unpaired electrons.

(C) The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$.
For $Ni^{2+}$,the configuration is $[Ar] 3d^8$.
In the $3d$ subshell,there are $8$ electrons,which results in $2$ unpaired electrons $(n=2)$.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n=2$: $\mu = \sqrt{2(2+2)} = \sqrt{2 \times 4} = \sqrt{8} \approx 2.83 \ BM$.
Thus,the value is approximately $2.8 \ BM$.

(B) The spin only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
For a species to have no spin only magnetic moment,it must have $n = 0$ (i.e.,all electrons are paired).
Let us analyze the electronic configurations:
$1. Cu^{2+} (Z=29): [Ar] 3d^9$. It has $1$ unpaired electron.
$2. Zn^{2+} (Z=30): [Ar] 3d^{10}$. It has $0$ unpaired electrons.
$3. Ti^{3+} (Z=22): [Ar] 3d^1$. It has $1$ unpaired electron.
$4. V^{3+} (Z=23): [Ar] 3d^2$. It has $2$ unpaired electrons.
Since $Zn^{2+}$ has $0$ unpaired electrons,its spin only magnetic moment is $0$.

(B) The atomic number $27$ corresponds to Cobalt $(Co)$.
The electronic configuration of neutral $Co$ $(Z=27)$ is $[Ar] \ 3d^7 \ 4s^2$.
When $Co$ is in the $+2$ oxidation state,it loses two electrons from the $4s$ orbital.
The resulting electronic configuration is $[Ar] \ 3d^7$.
Therefore,the number of electrons in the $d$-subshell is $7$.

(B) The enthalpy of atomization of transition elements depends on the strength of metallic bonding,which is directly related to the number of unpaired electrons in the $d$-orbital.
More unpaired electrons lead to stronger metallic bonding and higher enthalpy of atomization.
Let us analyze the number of unpaired electrons in the given elements:
$Sc$ $(Z=21)$: $[Ar] 3d^1 4s^2$ ($1$ unpaired electron)
$Fe$ $(Z=26)$: $[Ar] 3d^6 4s^2$ ($4$ unpaired electrons)
$Cu$ $(Z=29)$: $[Ar] 3d^{10} 4s^1$ ($1$ unpaired electron)
$Zn$ $(Z=30)$: $[Ar] 3d^{10} 4s^2$ ($0$ unpaired electrons)
Among the given options,$Fe$ has the maximum number of unpaired electrons $(4)$,resulting in the strongest metallic bonding and the highest enthalpy of atomization.

(A) The electronic configuration of Gadolinium ($Gd$,$Z=64$) is $[Xe] 4f^7 5d^1 6s^2$.
In this configuration,one electron enters the $5d$ orbital due to the stability provided by the half-filled $4f^7$ subshell.
For other lanthanoids like $Eu$ $(Z=63)$,$Tb$ $(Z=65)$,and $Dy$ $(Z=66)$,the $5d$ orbital remains empty in their ground state configurations as electrons fill the $4f$ subshell.

(A) Transition elements are defined as elements that have incompletely filled $d$-orbitals in their ground state or in any of their oxidation states.
$Zinc$ ($Zn$,atomic number $30$) has the electronic configuration $[Ar] 3d^{10} 4s^2$.
In its common oxidation state of $+2$,it forms $Zn^{2+}$ with the configuration $[Ar] 3d^{10}$.
Since the $d$-orbital is completely filled in both the ground state and the $+2$ oxidation state,$Zinc$ is not considered a transition element and does not exhibit variable oxidation states.

(D) The correct answer is $D$.
Due to the lanthanoid contraction,the atomic radii of elements in the $5d$ series are nearly the same as those of the corresponding elements in the $4d$ series.
Therefore,the statement that the atomic sizes of the $5d$ series are greater than the $4d$ series is incorrect.

(A) The similarity in the properties of $Zr$ and $Hf$ is primarily due to the lanthanoid contraction. Due to the filling of $4f$ orbitals in the elements between $La$ and $Hf$, the atomic radius of $Hf$ $(159 \text{ pm})$ is almost identical to that of $Zr$ $(160 \text{ pm})$. This phenomenon is known as the lanthanoid contraction, which results in similar atomic and ionic radii for elements of the same group in the $4d$ and $5d$ series. Therefore, the correct answer is $A$.

(D) The magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 5.92 \ BM$,we have $\sqrt{n(n+2)} = 5.92$,which implies $n \approx 5$.
For a divalent ion $(M^{2+})$ to have $5$ unpaired electrons,the electronic configuration must be $3d^5$.
Among the given elements,$Mn$ $(Z=25)$ has the ground state configuration $[Ar] 3d^5 4s^2$.
Upon forming the $Mn^{2+}$ ion,it loses two $4s$ electrons,resulting in the configuration $[Ar] 3d^5$.
This configuration contains $5$ unpaired electrons,consistent with the magnetic moment of $5.92 \ BM$.

(B) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $Fe^{3+}$ $(Z=26)$: The electronic configuration is $[Ar] 3d^5$. The number of unpaired electrons $(n)$ = $5$.
For $Mn^{2+}$ $(Z=25)$: The electronic configuration is $[Ar] 3d^5$. The number of unpaired electrons $(n)$ = $5$.
Since both $Fe^{3+}$ and $Mn^{2+}$ have $5$ unpaired electrons,they have the same magnetic moment of $\sqrt{5(5+2)} = \sqrt{35} \ BM$.

(B) The formation of an alloy requires the atomic radii of the two elements to differ by no more than $15\%$. In this case, the difference in radii is $(187 - 115) = 72 \ pm$. The percentage difference is $\frac{72}{115} \times 100 \approx 62.6\%$. Since this difference is significantly greater than $15\%$, the elements cannot form an alloy. Therefore, option $B$ is the correct reason.

(C) An alloy is a homogeneous mixture of two or more metals or a metal and a non-metal.
Transition metals,due to their similar atomic sizes,readily form alloys with each other.
$Fe$,$Ni$,and $Cr$ are all transition metals with comparable atomic radii,allowing them to form a stable solid solution known as stainless steel.
Therefore,the correct mixture is $Fe, Ni, Cr$.

(C) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $Cr^{3+}$ $(Z=24)$: Electronic configuration is $[Ar] 3d^3$. Number of unpaired electrons $(n)$ = $3$. Magnetic moment $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
For $Mn^{2+}$ $(Z=25)$: Electronic configuration is $[Ar] 3d^5$. Number of unpaired electrons $(n)$ = $5$. Magnetic moment $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
For $Fe^{3+}$ $(Z=26)$: Electronic configuration is $[Ar] 3d^5$. Number of unpaired electrons $(n)$ = $5$. Magnetic moment $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
Thus,the order is $Cr^{3+} < Mn^{2+} = Fe^{3+}$.

(C) The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
$1$. For $V^{3+}$ $([Ar] 3d^2)$: $n = 2$,$\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \text{ BM}$.
$2$. For $Mn^{3+}$ $([Ar] 3d^4)$: $n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \text{ BM}$.
$3$. For $Fe^{3+}$ $([Ar] 3d^5)$: $n = 5$,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92 \text{ BM}$.
$4$. For $Cu^{3+}$ $([Ar] 3d^8)$: $n = 2$,$\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \text{ BM}$.
Thus,$Fe^{3+}$ has the maximum number of unpaired electrons $(n=5)$,resulting in the maximum magnetic moment.

(A) The basicity of lanthanoid oxides $(Ln_2O_3)$ decreases as the ionic radius of the $Ln^{3+}$ ion decreases across the lanthanoid series due to lanthanoid contraction.
As we move from $Pr$ $(Z=59)$ to $Lu$ $(Z=71)$,the ionic radius decreases.
Therefore,the basic character follows the order: $Pr_2O_3 > Sm_2O_3 > Gd_2O_3 > Lu_2O_3$.
Thus,$Pr_2O_3$ has the maximum basicity.

(B) 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 $Cr^{3+}$ $([Ar] 3d^3)$: $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
$2$. For $Fe^{3+}$ $([Ar] 3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
$3$. For $Ti^{3+}$ $([Ar] 3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
$4$. For $Co^{3+}$ $([Ar] 3d^6)$: In high spin,$n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
Comparing these values,$Fe^{3+}$ has the maximum number of unpaired electrons $(n=5)$,hence it has the maximum theoretical magnetic moment.

(A) transition element is defined as an element which has an incompletely filled $d$-orbital in its ground state or in any of its oxidation states.
$Au$ (Gold) has the ground state electronic configuration $[Xe] 4f^{14} 5d^{10} 6s^1$. However,in its $+3$ oxidation state,it has the configuration $[Xe] 4f^{14} 5d^8$,which has an incompletely filled $d$-orbital.
$Zn$,$Cd$,and $Hg$ belong to group $12$ and have the general electronic configuration $(n-1)d^{10} ns^2$. They do not have incompletely filled $d$-orbitals in their ground state or common oxidation states $(+2)$,and thus are not considered transition elements.
Therefore,$Au$ is the correct answer.

(A) The basic strength of metallic hydroxides depends on the ionic character of the $M-OH$ bond.
As the size of the metal cation increases,the ionic character of the $M-OH$ bond increases,leading to higher basicity.
$Al(OH)_3$ is amphoteric and is the least basic among the given compounds.
Among the lanthanoids,the basic strength decreases as the ionic radius decreases due to lanthanoid contraction $(Ce^{3+} > Lu^{3+})$.
$Ca(OH)_2$ is a strong base compared to the lanthanoid hydroxides.
Therefore,the correct order of increasing basic strength is $Al(OH)_3 < Lu(OH)_3 < Ce(OH)_3 < Ca(OH)_2$.

(B) The stability of $Cu^{2+}(aq)$ ions is primarily due to their very high hydration enthalpy,which more than compensates for the energy required to remove the second electron from the $Cu^+(g)$ ion. Therefore,$Cu^{2+}$ is more stable than $Cu^+$ in aqueous solutions. The correct option is $B$.

(D) German silver is an alloy consisting of $Copper$ $(Cu)$,$Zinc$ $(Zn)$,and $Nickel$ $(Ni)$.
Despite its name,it does not contain any $Silver$ $(Ag)$.
Therefore,the correct composition is $Zinc$,$Nickel$,and $Copper$.

(A) The maximum oxidation state of a transition element is determined by the total number of electrons in the $(n-1)d$ and $ns$ orbitals.
For option $A$: $(n-1)d^5 ns^2$,total electrons = $5 + 2 = 7$.
For option $B$: $(n-1)d^8 ns^2$,total electrons = $8 + 2 = 10$ (but the maximum stable oxidation state is usually limited by the number of unpaired electrons and stability of the configuration).
For option $C$: $(n-1)d^5 ns^1$,total electrons = $5 + 1 = 6$.
For option $D$: $(n-1)d^3 ns^2$,total electrons = $3 + 2 = 5$.
Among the given configurations,the configuration $(n-1)d^5 ns^2$ (e.g.,$Mn$ in $3d^5 4s^2$) allows for a maximum oxidation state of $+7$. While $d^8 s^2$ has more electrons,it does not exhibit an oxidation state of $+10$; transition metals typically show a maximum oxidation state corresponding to the sum of $ns$ and $(n-1)d$ electrons up to $Mn$ $(+7)$. Thus,$(n-1)d^5 ns^2$ is the correct choice for the highest common oxidation state.

(A) The $d-d$ transition is possible only in ions that have partially filled $d$-orbitals ($d^1$ to $d^9$ configuration).
$1$. $Ti^{4+}$: The electronic configuration is $[Ar] 3d^0$. Since there are no electrons in the $d$-orbital,$d-d$ transition is not possible.
$2$. $Cr^{3+}$: The electronic configuration is $[Ar] 3d^3$. It has partially filled $d$-orbitals,so $d-d$ transition is possible.
$3$. $Mn^{2+}$: The electronic configuration is $[Ar] 3d^5$. It has partially filled $d$-orbitals,so $d-d$ transition is possible.
$4$. $Cu^{2+}$: The electronic configuration is $[Ar] 3d^9$. It has partially filled $d$-orbitals,so $d-d$ transition is possible.
Therefore,the correct option is $A$.

(A) The magnetic moment (spin-only) is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$1$. For $Fe^{3+}$ $(Z=26)$: Electronic configuration is $[Ar] 3d^5$. Number of unpaired electrons $(n)$ = $5$.
$2$. For $Cr^{3+}$ $(Z=24)$: Electronic configuration is $[Ar] 3d^3$. Number of unpaired electrons $(n)$ = $3$.
$3$. For $Co^{3+}$ $(Z=27)$: Electronic configuration is $[Ar] 3d^6$. Number of unpaired electrons $(n)$ = $4$.
$4$. For $Ti^{3+}$ $(Z=22)$: Electronic configuration is $[Ar] 3d^1$. Number of unpaired electrons $(n)$ = $1$.
Since $Fe^{3+}$ has the highest number of unpaired electrons $(n=5)$,it has the highest magnetic moment.

(D) Transition elements are characterized by the presence of partially filled $d$-orbitals,which allow them to exhibit variable oxidation states.
However,$Scandium$ ($Sc$,atomic number $21$) has an electronic configuration of $[Ar] 3d^1 4s^2$.
In its compounds,it loses all three valence electrons to form the $Sc^{3+}$ ion,which has a stable noble gas configuration $([Ar])$.
Because it does not have other stable oxidation states,$Scandium$ is the only transition element among the choices that does not exhibit variable oxidation states.
Therefore,the correct option is $D$.

(D) transition element is defined as an element which has an incompletely filled $d$-orbital in its ground state or in any of its oxidation states.
$Zn$ $([Ar] 3d^{10} 4s^2)$,$Cd$ $([Kr] 4d^{10} 5s^2)$,and $Hg$ $([Xe] 4f^{14} 5d^{10} 6s^2)$ have completely filled $d$-orbitals in their ground state as well as in their common oxidation states $(+2)$.
$Cu$ $([Ar] 3d^{10} 4s^1)$ has a completely filled $d$-orbital in its ground state,but in its $+2$ oxidation state $(Cu^{2+})$,it has a $3d^9$ configuration,which is incompletely filled.
Therefore,$Cu$ is considered a transition element.

(C) The electronic configurations of the given ions are:
$Ti^{+} (Z=22): [Ar] 3d^2 4s^1$
$V^{+} (Z=23): [Ar] 3d^3 4s^1$
$Cr^{+} (Z=24): [Ar] 3d^5 4s^0$
$Mn^{+} (Z=25): [Ar] 3d^5 4s^1$
Ionisation enthalpy is the energy required to remove an electron from the outermost shell.
$Cr^{+}$ has a stable half-filled $3d^5$ configuration with no electrons in the $4s$ orbital.
Removing an electron from a stable,half-filled $d$-subshell requires significantly more energy compared to removing an electron from the $4s$ orbital of the other ions.
Therefore,$Cr^{+}$ has the highest ionisation enthalpy.

(D) The colour of aqueous solutions of transition metal ions depends on the presence of unpaired $d$-electrons,which allow for $d-d$ transitions.
$1$. $Cu^{2+}$ has $3d^9$ configuration ($1$ unpaired electron),so it is blue.
$2$. $Mn^{2+}$ has $3d^5$ configuration ($5$ unpaired electrons),so it is pink.
$3$. $Ni^{2+}$ has $3d^8$ configuration ($2$ unpaired electrons),so it is green.
$4$. $Zn^{2+}$ has $3d^{10}$ configuration ($0$ unpaired electrons). Since there are no unpaired electrons,$d-d$ transitions are not possible,making the solution colourless.
Therefore,the correct option is $D$.

(C) The colour of transition metal ions is due to $d-d$ transitions,which require unpaired electrons in the $d$-orbitals.
$Zn^{2+}$ has the electronic configuration $[Ar] 3d^{10}$,meaning all its $d$-orbitals are completely filled.
Since there are no unpaired electrons,no $d-d$ transitions occur,making its aqueous solution colourless.

(C) For a $d-d$ transition to occur,the metal ion must have a partially filled $d$-orbital (i.e.,$d^1$ to $d^9$ configuration).
$1$. $Cu^{+}$: Electronic configuration is $[Ar] 3d^{10}$. It has a fully filled $d$-orbital,so no $d-d$ transition is possible.
$2$. $Zn^{2+}$: Electronic configuration is $[Ar] 3d^{10}$. It has a fully filled $d$-orbital,so no $d-d$ transition is possible.
$3$. $Mn^{3+}$: Electronic configuration is $[Ar] 3d^4$. It has a partially filled $d$-orbital,allowing for $d-d$ transition.
$4$. $Sc^{3+}$: Electronic configuration is $[Ar] 3d^0$. It has an empty $d$-orbital,so no $d-d$ transition is possible.
Therefore,the correct option is $C$.

(C) $Sc^{3+}$ has completely empty $d$-orbitals ($d^0$ configuration); hence,it forms a colourless solution in aqueous medium.
$Sc \ (21): [Ar] \ 3d^1 \ 4s^2$
$Sc^{3+}: [Ar] \ 3d^0 \ 4s^0$
Since there are no unpaired electrons to undergo $d-d$ transitions,the solution is colourless.

(C) The electronic configuration of the element with atomic number $21$ $(Sc)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^1$.
Since this element contains a partially filled $d$-orbital in its ground state,it is classified as a $d$-block element.
$d$-block elements are also known as transition elements.

(B) The electronic configurations and the number of unpaired electrons are as follows:
$Zn (Z=30): [Ar] \ 3d^{10} 4s^{2} \implies Zn^{2+}: [Ar] \ 3d^{10}$ (Number of unpaired electrons = $0$)
$Fe (Z=26): [Ar] \ 3d^{6} 4s^{2} \implies Fe^{2+}: [Ar] \ 3d^{6}$ (Number of unpaired electrons = $4$)
$Ni (Z=28): [Ar] \ 3d^{8} 4s^{2} \implies Ni^{3+}: [Ar] \ 3d^{7}$ (Number of unpaired electrons = $3$)
$Cu (Z=29): [Ar] \ 3d^{10} 4s^{1} \implies Cu^{+}: [Ar] \ 3d^{10}$ (Number of unpaired electrons = $0$)
Thus,$Fe^{2+}$ has the maximum number of unpaired '$d$' electrons.

(B) The correct match is $I \to (ii), II \to (iii), III \to (i)$.
- $Zn^{2+} (Z=30)$: The electronic configuration is $[Ar] 3d^{10}$. Since all $d$-orbitals are fully filled,there are no unpaired electrons,making it colourless.
- $Cu^{2+} (Z=29)$: The electronic configuration is $[Ar] 3d^9$. It has one unpaired electron. The magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ \text{BM}$.
- $Ni^{2+} (Z=28)$: The electronic configuration is $[Ar] 3d^8$. Thus,it has a $d^8$ configuration.

(D) The magnetic moment $(\mu)$ is calculated using the formula: $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $Ti^{3+}$ $(3d^1)$,$n = 1$. Thus,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
For $V^{3+}$ $(3d^2)$,$n = 2$. Thus,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \ BM$.
For $Cr^{3+}$ $(3d^3)$,$n = 3$. Thus,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
For $Fe^{3+}$ $(3d^5)$,$n = 5$. Thus,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.91 \ BM$.
Therefore,$Ti^{3+}$ has a magnetic moment of approximately $1.73-1.75 \ BM$.

(C) The formation of a coloured solution is possible when the metal ion in the compound contains unpaired electrons,which allow for $d-d$ transitions.
For example:
$Cu^{+}: 3d^{10} 4s^{0}$ (no unpaired electrons,colourless)
$Cu^{2+}: 3d^{9} 4s^{0}$ (one unpaired electron,blue)

(A) Transition metal ions are colorless if they do not have unpaired electrons in their $d$-orbitals,meaning they have $d^{0}$ or $d^{10}$ configurations,as they cannot undergo $d-d$ transitions.
$Sc^{3+}$: Atomic number $21$,configuration $[Ar] 3d^{0} 4s^{0}$ ($d^{0}$ configuration,colorless).
$Zn^{2+}$: Atomic number $30$,configuration $[Ar] 3d^{10} 4s^{0}$ ($d^{10}$ configuration,colorless).
Therefore,both $Sc^{3+}$ and $Zn^{2+}$ are colorless.

(D) The general electronic configuration for the elements in the given groups is $(n-1)d^{10} ns^2$.
$Zn$ $([Ar] 3d^{10} 4s^2)$,$Cd$ $([Kr] 4d^{10} 5s^2)$,and $Hg$ $([Xe] 4f^{14} 5d^{10} 6s^2)$ follow this configuration.
$Cu$ has the configuration $[Ar] 3d^{10} 4s^1$.
$Ag$ has the configuration $[Kr] 4d^{10} 5s^1$.
In the pair $Cu, Zn$,$Cu$ does not have the $(n-1)d^{10} ns^2$ configuration.
In the pair $Ag, Cu$,neither element has the $(n-1)d^{10} ns^2$ configuration.
However,the question asks for the pair where both elements do not have this configuration. Since $Ag$ and $Cu$ both have $ns^1$ instead of $ns^2$,this is the correct pair.

(D) The general electronic configuration of group $12$ elements $(Zn, Cd, Hg, Cn)$ is $(n-1)d^{10}ns^2$.
Copper $(Cu)$ belongs to group $11$ and has an electronic configuration of $3d^{10}4s^1$.
Zinc $(Zn)$ belongs to group $12$ and has an electronic configuration of $3d^{10}4s^2$.
In option $D$,$Cu$ does not have the $(n-1)d^{10}ns^2$ configuration,while $Zn$ does.
Therefore,the pair where both elements do not have this configuration is $Cu, Zn$.

(C) The property stated against option $(C)$ is incorrect.
The electronic configurations of the given ions are:
$Cr^{2+} = [Ar] 3d^4$ ($4$ unpaired electrons)
$Mn^{2+} = [Ar] 3d^5$ ($5$ unpaired electrons)
$Fe^{2+} = [Ar] 3d^6$ ($4$ unpaired electrons)
Paramagnetic behaviour is directly proportional to the number of unpaired electrons.
Since $Mn^{2+}$ has $5$ unpaired electrons,it shows the maximum paramagnetic behaviour among the three.
Therefore,the correct order of paramagnetic behaviour is $Cr^{2+} = Fe^{2+} < Mn^{2+}$.

(C) The elements from $Sc$ $(Z=21)$ to $Cu$ $(Z=29)$ are transition elements belonging to the $3d$ series.
For all these elements,the $3d$ subshell is partially filled in their ground state electronic configuration.
For example:
$Sc: [Ar] 3d^1 4s^2$
$Ti: [Ar] 3d^2 4s^2$
...
$Cu: [Ar] 3d^{10} 4s^1$
Since the $3d$ orbital is not completely filled for all elements from $Sc$ to $Cu$ in their ground state (except $Cu$,but the question asks for a general property of the series elements),option $(c)$ is the most appropriate description of the $3d$ series characteristics.

(D) Ions are coloured in aqueous solution if they have unpaired electrons in their $d$-orbitals (i.e.,$d^1$ to $d^9$ configuration).
$1$. $Sc^{3+}$ $(3d^0)$: Colourless (no unpaired electrons).
$2$. $Ti^{3+}$ $(3d^1)$: Coloured (one unpaired electron).
$3$. $Mn^{2+}$ $(3d^5)$: Coloured (five unpaired electrons).
$4$. $Ni^{2+}$ $(3d^8)$: Coloured (two unpaired electrons).
$5$. $Cu^{+}$ $(3d^{10})$: Colourless (no unpaired electrons).
$6$. $Ti^{4+}$ $(3d^0)$: Colourless (no unpaired electrons).
Comparing the options:
- Option $(A)$: $Sc^{3+}$ is colourless.
- Option $(B)$: $Ti^{4+}$ is colourless.
- Option $(C)$: $Cu^{+}$ is colourless.
- Option $(D)$: Both $Mn^{2+}$ and $Ti^{3+}$ have unpaired electrons and are coloured.
Therefore,the correct pair is $(Mn^{2+}, Ti^{3+})$.

(C) substance is paramagnetic if it contains at least one unpaired electron. Let us analyze the electronic configurations of the ions in option $C$:
$Ti$ $(Z=22)$: $[Ar] 4s^{2} 3d^{2}$. $Ti^{2+}$: $[Ar] 3d^{2}$ (contains $2$ unpaired electrons).
$Cu$ $(Z=29)$: $[Ar] 4s^{1} 3d^{10}$. $Cu^{2+}$: $[Ar] 3d^{9}$ (contains $1$ unpaired electron).
$Mn$ $(Z=25)$: $[Ar] 4s^{2} 3d^{5}$. $Mn^{3+}$: $[Ar] 3d^{4}$ (contains $4$ unpaired electrons).
Since all these ions contain unpaired electrons,they are paramagnetic.

(B) The correct answer is $(B)$.
$Zn$ (Zinc) has a completely filled $d$-orbital $(3d^{10}4s^2)$ and does not exhibit variable oxidation states; it only shows a $+2$ oxidation state.

(C) The electronic configuration of the ion $X^{3+}$ is given as $[Ar] 3d^5$.
To find the atomic number of the neutral element $X$,we add the number of electrons lost during oxidation to the number of electrons in the ion.
Number of electrons in $X^{3+} = 18 (Ar) + 5 (3d) = 23 \ e^-$.
Since the oxidation state is $+3$,the neutral element $X$ has $23 + 3 = 26 \ e^-$.
Therefore,the atomic number of element $X$ is $26$,which corresponds to Iron $(Fe)$.

(A) The magnetic moment of the metal ions is determined by the number of unpaired electrons $(n)$ using the formula $\mu_{spin} = \sqrt{n(n+2)} \ BM$.
$1$. For $Mn^{2+}$ $(3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \ BM \approx 5.92 \ BM$.
$2$. For $Cu^{2+}$ $(3d^9)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \ BM \approx 1.73 \ BM$.
$3$. For $Sc^{2+}$ $(3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \ BM \approx 1.73 \ BM$.
$4$. For $Cu^{+}$ $(3d^{10})$: $n = 0$,$\mu = 0 \ BM$.
Since $Mn^{2+}$ has the highest number of unpaired electrons $(n=5)$,it exhibits the maximum paramagnetic behaviour.