General Characteristics Questions in English

(A) The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$. This configuration is half-filled,which provides extra stability.
The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$.
Oxidation to the $+3$ state involves the removal of an electron from the $d$-subshell. For $Mn^{2+}$,removing an electron disrupts the stable half-filled $d^5$ configuration,requiring a very high third ionization energy.
For $Fe^{2+}$,removing an electron from the $3d^6$ configuration results in a stable $3d^5$ configuration,which requires less energy.
Therefore,$Mn^{2+}$ is more resistant to oxidation to the $+3$ state compared to $Fe^{2+}$.

(D) The atomic number of $Gd$ is $64$.
The electronic configuration of neutral $Gd$ is $[Xe] 4f^7 5d^1 6s^2$.
When forming the $Gd^{2+}$ ion,electrons are removed from the outermost shell first,which is the $6s$ orbital.
Removing two electrons from the $6s$ orbital results in the configuration $[Xe] 4f^7 5d^1$.

(C) The paramagnetism of transition elements in a given series depends on the number of unpaired electrons in the $d$-orbitals.
As we move from left to right across a transition series,the number of unpaired electrons increases until the $d^5$ configuration is reached.
Consequently,the paramagnetism increases to a maximum value at the $d^5$ configuration.
Beyond $d^5$,the electrons begin to pair up in the $d$-orbitals ($d^6$ to $d^{10}$),which leads to a decrease in the number of unpaired electrons and thus a decrease in paramagnetism.
Therefore,the paramagnetism first increases to a maximum and then decreases.

(C) The electronic configuration of mercury ($Hg$,$Z=80$) is $[Xe] 4f^{14} 5d^{10} 6s^{2}$.
Since the $5d$ subshell is completely filled,the electrons are held tightly by the nucleus and do not participate in metallic bonding through $d-d$ overlapping.
This weak metallic bonding results in a low melting point,making mercury a liquid at room temperature.

(B) In the periodic table,transition metals are commonly used as catalysts because they provide a large surface area and have variable oxidation states.
These metals belong to the $d$-block.
Examples include $Ni$,$Pt$,$V_2O_5$,and $Fe$.

(B) The atomic number of Chromium $(Cr)$ is $24$.
Its ground state electronic configuration is $[Ar] 3d^{5} 4s^{1}$.
To form the $Cr^{3+}$ ion,we remove $3$ electrons: one from the $4s$ orbital and two from the $3d$ orbital.
Therefore,the electronic configuration of $Cr^{3+}$ is $[Ar] 3d^{3} 4s^{0}$.

(B) The spin-only magnetic moment $(\mu_{spin})$ is calculated using the formula $\mu_{spin} = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
| Electronic Configuration | No. of Unpaired Electrons $(n)$ | Magnetic Moment $(\mu_{spin})$ |
| $3d^7$ | $3$ | $\sqrt{3(3+2)} = \sqrt{15} \text{ BM}$ |
| $3d^5$ | $5$ | $\sqrt{5(5+2)} = \sqrt{35} \text{ BM}$ |
| $3d^8$ | $2$ | $\sqrt{2(2+2)} = \sqrt{8} = 2\sqrt{2} \text{ BM}$ |
| $3d^2$ | $2$ | $\sqrt{2(2+2)} = \sqrt{8} = 2\sqrt{2} \text{ BM}$ |
Comparing the values,the $3d^5$ configuration has the highest number of unpaired electrons $(n=5)$,resulting in the highest magnetic moment of $\sqrt{35} \text{ BM}$.

(C) The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] \ 3d^{5} \ 4s^{2}$.
For $Mn^{4+}$ ion,we remove four electrons,resulting in the configuration $[Ar] \ 3d^{3}$.
This ion contains $n=3$ unpaired electrons.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n=3$,we get $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
Rounding to the nearest integer,the value is $4 \ BM$.

(C) The correct matches are as follows:
$A$. Titanium $(Ti)$ is a $d$-block element,which is a transition metal $(II)$.
$B$. Fluorine $(F)$ is a halogen,which is a non-metal $(I)$.
$C$. Tellurium $(Te)$ is a metalloid $(IV)$.
$D$. Dysprosium $(Dy)$ is a $4f$-block element,which is a lanthanoid $(III)$.
Therefore,the correct matching is $A-II, B-I, C-IV, D-III$.

(C) The atomic number of palladium $(Pd)$ is $46$.
Following the Aufbau principle and considering the stability of the filled $d$-subshell,the electrons fill the $4d$ orbital completely.
The electronic configuration of $Pd$ is $[Kr] 4 d^{10} 5 s^0$.
Therefore,the outer electronic configuration is $4 d^{10} 5 s^0$.

(B) The atomic number of $Kr$ (Krypton) is $36$. The total number of electrons in the given configuration $[Kr] \ 4d^{10} \ 5s^0$ is $36 + 10 + 0 = 46$.
The element with atomic number $46$ is Palladium $(Pd)$.
According to the provided data:
- $Ag$ $(Z=47)$: $[Kr] \ 4d^{10} \ 5s^1$
- $Pd$ $(Z=46)$: $[Kr] \ 4d^{10} \ 5s^0$
- $Rh$ $(Z=45)$: $[Kr] \ 4d^8 \ 5s^1$
- $Tc$ $(Z=43)$: $[Kr] \ 4d^5 \ 5s^2$
Thus,the correct element is $Pd$.

(D) The electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$ and that of $Cu$ is $[Ar] 3d^{10} 4s^1$.
Their second ionisation enthalpies are higher than expected because after the removal of one electron,$Cr^+$ attains a stable half-filled $3d^5$ configuration,and $Cu^+$ attains a stable fully-filled $3d^{10}$ configuration.
These configurations are significantly more stable than the configurations of their neighbors.
Therefore,it requires much more energy to remove the second electron from these ions.
Thus,option $(D)$ is correct.

$(A)$ The atomic radius of $Zn$ is $137 \ pm$.
The atomic radius of $Fe$ is $126 \ pm$.
The ionic radius of $Fe^{2+}$ is $77 \ pm$.
The ionic radius of $Fe^{3+}$ is $63 \ pm$.
Comparing these values: $137 \ pm > 126 \ pm > 77 \ pm > 63 \ pm$.
Therefore, the correct order of size is $Zn > Fe > Fe^{2+} > Fe^{3+}$.
Hence, the correct option is $A$.

(A) The standard electrode potentials $(E_{M^{2+}/ M}^{\ominus})$ for the given transition metals are as follows:
$Ni^{2+}/Ni = -0.25 \ V$
$Mn^{2+}/Mn = -1.18 \ V$
$Fe^{2+}/Fe = -0.44 \ V$
$Cr^{2+}/Cr = -0.91 \ V$
Matching these values:
$A(Ni) - III (-0.25)$
$B(Mn) - I (-1.18)$
$C(Fe) - IV (-0.44)$
$D(Cr) - II (-0.91)$
Therefore,the correct sequence is $A-III, B-I, C-IV, D-II$.

(D) Let us analyze each statement:
$1$. $CrO$ is a basic oxide,while $Cr_2O_3$ is amphoteric. This statement is correct.
$2$. In an acidic medium,$KMnO_4$ oxidizes nitrite $(NO_2^-)$ to nitrate $(NO_3^-)$. This statement is correct.
$3$. The Wacker process involves the oxidation of ethylene to acetaldehyde using $PdCl_2$ as a catalyst and $CuCl_2$ as a co-catalyst. This statement is correct.
$4$. The reactivity of the earlier members of the lanthanide series is similar to that of calcium $(Ca)$,not aluminium $(Al)$. Lanthanides are highly electropositive metals,and their chemical behavior is comparable to alkaline earth metals like calcium. Therefore,this statement is incorrect.

(B) Interstitial compounds are formed when small atoms like $H, C, N$ or $B$ are trapped inside the crystal lattice of transition metals.
These compounds exhibit the following properties:
$1$. They have high melting points,higher than those of pure metals.
$2$. They are very hard.
$3$. They retain metallic conductivity.
$4$. They are chemically inert.
Therefore,the statement that they lose electrical conductivity is incorrect,as they retain their metallic conductivity.

(D) $1. Fe$ electronic configuration: $[Ar] 3d^6 4s^2$. The $(n-1)$ shell is $3d$,which has $6$ electrons.
$2. Mn$ electronic configuration: $[Ar] 3d^5 4s^2$. The $(n-1)$ shell is $3d$,which has $5$ electrons.
$3. Zn$ electronic configuration: $[Ar] 3d^{10} 4s^2$. The $(n-1)$ shell is $3d$,which has $10$ electrons.
$4. Sc$ electronic configuration: $[Ar] 3d^1 4s^2$. The $(n-1)$ shell is $3d$,which has $1$ electron.
$5. K$ electronic configuration: $[Ar] 4s^1$. The $(n-1)$ shell is $3s^2 3p^6$,which has $8$ electrons.
$6. Cr$ electronic configuration: $[Ar] 3d^5 4s^1$. The $(n-1)$ shell is $3d$,which has $5$ electrons.
Comparing the options:
$A. Fe (6), Mn (5)$
$B. Zn (10), Fe (6)$
$C. K (8), Sc (1)$
$D. Mn (5), Cr (5)$
Thus,$Mn$ and $Cr$ have the same number of electrons in the $(n-1)$ shell.

(D) The transition metal with the highest melting point is $W$ (Tungsten).
This is because $W$ has a large number of unpaired electrons in its $d$-orbitals, which leads to strong interatomic metallic bonding.
Stronger metallic bonding results in a higher melting point.

(C) The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$.
To form $Cr^{3+}$,we remove three electrons (one from $4s$ and two from $3d$).
Thus,the configuration of $Cr^{3+}$ is $[Ar] 3d^3$.
Therefore,the number of $d$-electrons is $3$.

(A) The transition metals and their compounds are known for their catalytic activity.
This activity is ascribed to their ability to adopt multiple oxidation states and to form complexes.
Catalysts at a solid surface involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst (first row transition metals utilise $3d$ and $4s$ electrons for bonding).
This has the effect of increasing the concentration of the reactants at the catalyst surface and also weakening of the bonds in the reacting molecules,which results in the lowering of the activation energy.
Thus,the reason correctly explains why transition metals show catalytic activity.

(A) Transition elements are defined as elements that have partially filled $(n-1)d$ orbitals in their ground state or in any of their common oxidation states.
$Zn$,$Cd$,and $Hg$ have a $d^{10}$ configuration in their ground state and also in their common $+2$ oxidation state.
Since they do not have partially filled $d$-orbitals,they are not considered transition elements.
$Cu$ and $Ag$ are considered transition elements because they exhibit a $d^9$ configuration in their $+2$ oxidation state.
Electronic configurations:
$Zn: [Ar] 3d^{10} 4s^2$
$Cd: [Kr] 4d^{10} 5s^2$
$Hg: [Xe] 4f^{14} 5d^{10} 6s^2$

(A) Transition elements exhibit high enthalpies of atomization due to the presence of a large number of unpaired electrons in their $(n-1)d$ orbitals.
These unpaired electrons facilitate strong interatomic interactions,leading to strong metallic bonding between the atoms.
Consequently,more energy is required to break these bonds,resulting in higher enthalpies of atomization,as well as high melting and boiling points.
Therefore,both $(A)$ and $(R)$ are correct,and $(R)$ is the correct explanation of $(A)$.

(C) $CuCl$ contains $Cu^+$ ion with $3d^{10}$ configuration. All electrons are paired,so it is colorless.
$ScCl_3$ contains $Sc^{3+}$ ion with $3d^0$ configuration. It has no unpaired electrons,so it is colorless.
$CuCl_2$ contains $Cu^{2+}$ ion with $3d^9$ configuration. It has one unpaired electron,which allows for $d-d$ transitions,making it colored.
$TiCl_4$ contains $Ti^{4+}$ ion with $3d^0$ configuration. It has no unpaired electrons,so it is colorless.
Therefore,$CuCl_2$ is the colored compound.

(B) $Mn$ $(Z=25)$ has the electronic configuration $[Ar] 3d^5 4s^2$.
Due to the presence of $5$ unpaired electrons in the $3d$ orbital and $2$ electrons in the $4s$ orbital,it can exhibit a wide range of oxidation states from $+2$ to $+7$ (e.g.,in $MnO$,$Mn_2O_3$,$MnO_2$,$MnO_4^{2-}$,and $MnO_4^-$).
$Fe$ $(Z=26)$ typically shows $+2$ and $+3$.
$Cr$ $(Z=24)$ shows oxidation states from $+1$ to $+6$.
$V$ $(Z=23)$ shows oxidation states from $+2$ to $+5$.
Therefore,$Mn$ exhibits the greatest number of oxidation states among the given options.

(B) The electronic configurations of the given elements in their ground state are:
${}_{25}Mn = [Ar] 3d^5 4s^2$ (Number of unpaired electrons = $5$)
${}_{24}Cr = [Ar] 3d^5 4s^1$ (Number of unpaired electrons = $6$)
${}_{28}Ni = [Ar] 3d^8 4s^2$ (Number of unpaired electrons = $2$)
${}_{26}Fe = [Ar] 3d^6 4s^2$ (Number of unpaired electrons = $4$)
Comparing these,${}_{24}Cr$ has the highest number of unpaired electrons,which is $6$.

(C) The colour of transition metal compounds is generally due to the presence of unpaired electrons in the $d$-orbitals,which allow for $d-d$ transitions.
In $CuCl_2$,copper exists as $Cu^{2+}$ ions with a $3d^9$ electronic configuration.
Due to the presence of $1$ unpaired electron in the $3d$-orbital,$CuCl_2$ exhibits colour.
In contrast,$Zn^{2+}$ $(3d^{10})$,$Mg^{2+}$ $(2p^6)$,and $Ag^+$ $(4d^{10})$ have fully filled orbitals and no unpaired electrons,making their compounds colourless.
Therefore,$CuCl_2$ is the expected coloured compound.

(B) In both chromate $(CrO_4^{2-})$ and dichromate $(Cr_2O_7^{2-})$ ions,$Cr$ is present in $+6$ oxidation state,which corresponds to a $d^0$ electronic configuration.
Since there are no electrons in the $d$-orbitals,$d-d$ transitions are impossible.
The intense colours (yellow for $CrO_4^{2-}$ and orange for $Cr_2O_7^{2-}$) arise due to ligand-to-metal charge transfer $(LMCT)$,where electrons from oxygen are transferred to the vacant $d$-orbitals of $Cr(VI)$.

(C) Transition metals commonly exhibit a $+2$ oxidation state.
This is because the two electrons in the outermost $s$-orbital are relatively easy to remove.
Removing additional electrons from the inner $d$-subshell requires significantly higher energy,which limits the most common stable oxidation state to $+2$ for many transition elements.
Hence,the correct option is $(C)$.

(B) The electronic configuration of $Zn$ is $[Ar] 3d^{10} 4s^2$.
After the removal of two electrons,the configuration becomes $[Ar] 3d^{10}$.
Since $3d^{10}$ is a completely filled stable configuration,removing the third electron requires a very high amount of energy.
Therefore,$Zn$ has the highest third ionisation enthalpy among the $3d$-series elements.
Thus,the correct option is $B$.

(A) Gallium $(Ga)$ has a very low melting point of approximately $29.76 \ ^\circ C$ $(302.91 \ K)$.
Since the summer temperature in many regions often exceeds this value,$Ga$ exists in a liquid state during the summer season.

(C) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$1$. $Mn^{2+}$ $(3d^5)$: $n = 5$,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92 \ BM$.
$2$. $Fe^{2+}$ $(3d^6)$: $n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \ BM$.
$3$. $Ti^{2+}$ $(3d^2)$: $n = 2$,$\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \ BM$.
$4$. $V^{2+}$ $(3d^3)$: $n = 3$,$\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87 \ BM$.
$5$. $Cr^{2+}$ $(3d^4)$: $n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \ BM$.
$6$. $Co^{2+}$ $(3d^7)$: $n = 3$,$\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87 \ BM$.
Comparing the values,$V^{2+}$ and $Co^{2+}$ both have $n = 3$ and thus the same magnetic moment.

(A) The lanthanide elements that exhibit the $+4$ oxidation state are $Ce$ $(Z=58)$,$Pr$ $(Z=59)$,$Nd$ $(Z=60)$,$Tb$ $(Z=65)$,and $Dy$ $(Z=66)$.
Among the given elements:
$1$. $Ce$ $(4f^1 5d^1 6s^2)$ forms $Ce^{4+}$ $(4f^0)$ which is stable due to noble gas configuration.
$2$. $Pr$ $(4f^3 6s^2)$ can exhibit $+4$ state.
$3$. $Nd$ $(4f^4 6s^2)$ can exhibit $+4$ state.
$4$. $Tb$ $(4f^9 6s^2)$ forms $Tb^{4+}$ $(4f^7)$ which is stable due to half-filled $f$-orbital.
$5$. $Dy$ $(4f^{10} 6s^2)$ can exhibit $+4$ state.
Thus,there are $5$ elements in the given list that exhibit the $+4$ oxidation state.

(B) The ionic radii of lanthanoids decrease as the atomic number increases due to lanthanide contraction.
As we move from $Pr$ $(Z=59)$ to $Yb$ $(Z=70)$,the effective nuclear charge increases,causing the ionic radii to decrease.
The correct order of ionic radii is: $Yb^{3+} < Dy^{3+} < Sm^{3+} < Pr^{3+}$.

(D) The atomic number of Gadolinium $(Gd)$ is $64$.
Its electronic configuration follows the Aufbau principle with an exception due to the stability of the half-filled $4f$ subshell.
The ground state electronic configuration is $[Xe] 4f^7 5d^1 6s^2$.
In the $+3$ oxidation state,it loses three electrons (two from $6s$ and one from $5d$),resulting in $[Xe] 4f^7$.

(C) The electronic configurations are as follows:
$(I)$ Lanthanum ($La$,$Z=57$): $[Xe]_{54} \cdot 5d^1 \cdot 6s^2$ (No $f$-electrons)
$(II)$ Uranium ($U$,$Z=92$): $[Rn]_{86} \cdot 5f^3 \cdot 6d^1 \cdot 7s^2$ (Has $f$-electrons)
$(III)$ Lawrencium ($Lr$,$Z=103$): $[Rn]_{86} \cdot 5f^{14} \cdot 6d^1 \cdot 7s^2$ (Has $f$-electrons)
$(IV)$ Thorium ($Th$,$Z=90$): $[Rn]_{86} \cdot 6d^2 \cdot 7s^2$ (No $f$-electrons)
$(V)$ Actinium ($Ac$,$Z=89$): $[Rn]_{86} \cdot 6d^1 \cdot 7s^2$ (No $f$-electrons)
$(VI)$ Cerium ($Ce$,$Z=58$): $[Xe]_{54} \cdot 4f^1 \cdot 5d^1 \cdot 6s^2$ (Has $f$-electrons)
Based on the electronic configurations,the elements $La$,$Ac$,and $Th$ do not possess $f$-electrons.
Therefore,option $(c)$ is the correct answer.

(B) Option $B$ is incorrect because $MnO_{4}^{2-}$ (manganate ion) is stable in alkaline medium and is not a strong oxidizing agent compared to $MnO_{4}^{-}$ (permanganate ion). $CrO_{4}^{2-}$ (chromate ion) is also not a strong oxidizing agent. Both are oxyanions in the $+6$ oxidation state.
$A$ is correct: $E^{0}$ for $Fe^{3+}/Fe^{2+}$ is $+0.77 \ V$,while for $Mn^{3+}/Mn^{2+}$ it is $+1.51 \ V$.
$C$ is correct: Due to lanthanoid contraction,the atomic radii of the second and third transition series are very similar.
$D$ is correct: $Mn^{3+}$ is a strong oxidizing agent because $Mn^{2+}$ has a stable $d^{5}$ configuration,making the $E^{0}$ value for $Mn^{3+}/Mn^{2+}$ $(+1.51 \ V)$ much more positive than for $Cr^{3+}/Cr^{2+}$ $(-0.41 \ V)$.

(C)

Element	$SEP \text{ } (M^{3+} / M^{2+})$
$V$	$V^{3+} / V^{2+} = -0.26 \ V$
$Cr$	$Cr^{3+} / Cr^{2+} = -0.4 \ V$
$Mn$	$Mn^{3+} / Mn^{2+} = +1.5 \ V$
$Fe$	$Fe^{3+} / Fe^{2+} = +0.77 \ V$

The standard electrode potential $(SEP)$ for the $Mn^{3+} / Mn^{2+}$ couple is the most positive $(+1.5 \ V)$ among the given elements.
This is because $Mn^{2+}$ has a stable $d^5$ electronic configuration,making the reduction of $Mn^{3+}$ to $Mn^{2+}$ highly favorable.

(B) Statement $a$ is correct: As the atomic number increases in the lanthanide series,the ionic radius decreases due to lanthanide contraction. This increases the covalent character of the $M-OH$ bond,thereby decreasing the basic strength. Thus,the order $Ce(OH)_3 > Gd(OH)_3 > Lu(OH)_3$ is correct.
Statement $b$ is incorrect: $O^{2-}$,$N^{3-}$,$F^{-}$,and $Na^{+}$ all have $10$ electrons,but the original statement listed $O^{-}$ and $Mg^{+}$,which are not isoelectronic with the others.
Statement $c$ is correct: Due to lanthanide contraction,the atomic radii of $Zr$ $(160 \ pm)$ and $Hf$ $(159 \ pm)$ are approximately the same.
Therefore,statements $a$ and $c$ are correct.

(C) The atomic number of gold $(Au)$ is $79$.
Following the Aufbau principle and considering the stability of the fully filled $d$-subshell,the electronic configuration is written as $[Xe] 4f^{14} 5d^{10} 6s^1$.

(C) $Lanthanum$ $(La)$ has an atomic number of $57$. Its expected electronic configuration based on the $(n+l)$ rule would be $[Xe] 4f^1 6s^2$.
However,the actual observed electronic configuration is $[Xe] 5d^1 6s^2$.
This occurs because the energy levels of the $4f$ and $5d$ orbitals are very close in energy,and the $5d$ orbital is slightly lower in energy than the $4f$ orbital for $La$,leading to a violation of the $Aufbau$ principle.

(B) The electronic configuration of $Sc^{3+}$ is $[Ar] 3d^0$.
Since there are no unpaired electrons in the $d$-orbital,$d-d$ transitions are not possible,making the ion colourless.
In contrast,$Ti^{3+}$ $(3d^1)$,$V^{4+}$ $(3d^1)$,and $Cr^{4+}$ $(3d^2)$ contain unpaired electrons,allowing for $d-d$ transitions,which result in the exhibition of colour.

(C) The enthalpy of atomization depends on the strength of metallic bonding,which is determined by the number of unpaired electrons in the $d$-orbital.
$Zn$ has a $3d^{10} 4s^2$ electronic configuration.
Due to the absence of unpaired electrons in the $d$-subshell,the metallic bonding in $Zn$ is the weakest among the given transition metals.
Therefore,$Zn$ has the lowest enthalpy of atomization.

(D) The electronic configurations of the given elements are as follows:
$A$. $Cd$ $(Z=48)$: $[Kr] 4d^{10} 5s^2$. It belongs to the $d$-block $(III)$.
$B$. $Eu$ $(Z=63)$: $[Xe] 4f^7 6s^2$. It belongs to the $f$-block $(I)$.
$C$. $Se$ $(Z=34)$: $[Ar] 3d^{10} 4s^2 4p^4$. It belongs to the $p$-block $(IV)$.
$D$. $Ba$ $(Z=56)$: $[Xe] 6s^2$. It belongs to the $s$-block $(II)$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(D) $\rightarrow (II), (B)$ $\rightarrow (I), (C)$ $\rightarrow (IV), (D)$ $\rightarrow (III)$
The electronic configuration of elements is as follows:
$(A)$ $Ra$ $(Z=88)$: The last electron enters the $7s$ orbital. Thus,it is an $s$-block element.
$(B)$ $Uuq$ ($Z=114$,Flerovium): The last electron enters the $7p$ orbital. Thus,it is a $p$-block element.
$(C)$ $Ds$ ($Z=110$,Darmstadtium): The last electron enters the $6d$ orbital. Thus,it is a $d$-block element.
$(D)$ $Fm$ ($Z=100$,Fermium): The last electron enters the $5f$ orbital. Thus,it is an $f$-block element.

(C) The general electronic configuration $(n-1)d^{1-10}ns^2$ represents the transition elements (d-block elements).
Let us analyze the configurations of the elements in option $C$:
${}_{108}Hs \Rightarrow [Rn] 5f^{14} 6d^6 7s^2$
${}_{80}Hg \Rightarrow [Xe] 4f^{14} 5d^{10} 6s^2$
${}_{74}W \Rightarrow [Xe] 4f^{14} 5d^4 6s^2$
All these elements belong to the d-block and follow the general configuration $(n-1)d^{1-10}ns^2$.

(A) In the periodic table,as we move from left to right,the atomic size decreases due to an increase in effective nuclear charge.
For the $4f$-series (lanthanoids),the decrease in atomic size is less pronounced due to the poor shielding effect of $4f$-electrons,which leads to lanthanoid contraction.
Conversely,the $3d$-series elements show a more rapid decrease in atomic size compared to the $4f$-series because $3d$-electrons provide better shielding than $4f$-electrons,but the overall effective nuclear charge increase is more dominant in the $3d$ transition series.
Therefore,both Assertion $(A)$ and Reason $(R)$ are correct,and $(R)$ is the correct explanation of $(A)$.

(B) The atomic/ionic size depends on the effective nuclear charge $(Z_{eff})$ and the number of shells.
For ions,as the positive charge increases,the $Z_{eff}$ increases,leading to a decrease in ionic size. Thus,$Zn^{2+}$ is smaller than $Cu^{+}$.
For neutral atoms,the size increases down a group. Since $Ag$ is in the $5^{th}$ period and $Cu$ is in the $4^{th}$ period,$Ag > Cu$.
Comparing the species: $Zn^{2+}$ ($Z=30$,$28$ electrons),$Cu^{+}$ ($Z=29$,$28$ electrons),$Cu$ ($Z=29$,$29$ electrons),and $Ag$ ($Z=47$,$47$ electrons).
The correct order of size is $Zn^{2+} < Cu^{+} < Cu < Ag$.

$ (A) $ The group assignments for the given elements are as follows:
$A$. $Mn, Tc, Re$ belong to Group $7$ (Transition metals).
$B$. $Zn, Cd, Hg$ belong to Group $12$ (Post-transition metals).
$C$. $Ti, Zr, Hf$ belong to Group $4$ (Transition metals).
$D$. $Ga, In, Tl$ belong to Group $13$ (Post-transition metals).
Therefore, the correct matching is $A-IV, B-I, C-II, D-V$.

(A) $Ti = [Ar] 3d^2 4s^2$
$Ti^{3+} = [Ar] 3d^1 4s^0$
$Ti^{4+} = [Ar] 3d^0 4s^0$
Since $Ti^{4+}$ has a vacant $d$-orbital and it acquires the nearest noble gas configuration,it is more stable compared to $Ti^{3+}$ and $Ti^{2+}$.
$Cu^{2+} / Cu = +0.34 \ V$ has an exceptionally positive reduction potential among $3d$ series elements.
$Sc$ shows $+3$ oxidation state and $Zn$ shows $+2$ oxidation state. They do not exhibit $+1$ oxidation state.
Therefore,only $(i)$ and $(ii)$ are correct.

(C) The melting points of transition metals in a group generally increase as we move down the group due to the increase in the number of unpaired electrons and the strength of metallic bonding.
For the group $6$ elements ($Cr$,$Mo$,$W$),the melting points follow the order $W > Mo > Cr$.