General Characteristics Questions in English

(B) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 1.73 \ BM$,we have $1.73 = \sqrt{n(n+2)}$.
Squaring both sides,$3 = n(n+2)$,which gives $n^2 + 2n - 3 = 0$.
Solving for $n$,we get $(n+3)(n-1) = 0$,so $n = 1$ (since $n$ cannot be negative).
Vanadium $(V)$ has the atomic number $23$ and electronic configuration $[Ar] 3d^3 4s^2$.
For $n=1$,the vanadium ion must have one unpaired electron.
In $VCl_4$,vanadium is in the $+4$ oxidation state $(V^{4+})$.
The configuration of $V^{4+}$ is $[Ar] 3d^1$,which contains $1$ unpaired electron.
Thus,the formula is $VCl_4$.

(D) The $E^{\circ}$ value for $Ce^{4+} / Ce^{3+}$ is $+1.74 \, V$. This high positive value indicates that $Ce^{4+}$ has a strong tendency to gain an electron to form $Ce^{3+}$.
The most stable oxidation state of lanthanide series elements is $+3$. Therefore,$Ce^{3+}$ is more stable than $Ce^{4+}$.
Thus,Statement $I$ is correct and Statement $II$ is incorrect.

(A) The element with atomic number $24$ is Chromium $(Cr)$.
The electronic configuration of $Cr$ is $[Ar] \, 4s^{1} \, 3d^{5}$.
Chromium exhibits a range of common oxidation states from $+2$ to $+6$ in its various compounds.

(B) $Mn_{3}O_{4}$ is a mixed oxide of $MnO$ and $Mn_{2}O_{3}$.
It exhibits ferrimagnetism due to the presence of unpaired electrons in the $d$-orbitals of manganese ions,which makes it magnetic.

(D) The electrode potential $E^{\Theta}$ for the reduction of $M^{2+}$ to $M$ in the $3d$-series is generally negative,except for copper.
Copper $(Cu)$ has a positive standard electrode potential $(E^{\Theta} = +0.34 \ V)$ because the high energy required to transform $Cu(s)$ to $Cu^{2+}(aq)$ (sum of enthalpy of atomization and ionization enthalpy) is not compensated by its hydration enthalpy.
Therefore,the correct option is $D$.

(A) The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$. Therefore,$Cr^{+}$ is $[Ar] 3d^5$.
The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$. Therefore,$Mn^{2+}$ is $[Ar] 3d^5$.
Both ions have the same outermost electronic configuration of $3d^5$.

(D) $T \ell I_{3}$ is isomorphous to $CsI_{3}$,which means it contains the triiodide ion $I_{3}^{\ominus}$.
Thus,$T \ell I_{3}$ is formulated as $T \ell^{\oplus} I_{3}^{\ominus}$,where $T \ell$ is in the $+1$ oxidation state. So,Assertion $A$ is correct.
The electronic configuration of $T \ell$ $(Z=81)$ is $[Xe] 4f^{14} 5d^{10} 6s^{2} 6p^{1}$.
It contains fourteen $4f$ electrons. So,Reason $R$ is correct.
However,the presence of $14f$ electrons is a general feature of the $T \ell$ atom and is not the direct reason why $T \ell I_{3}$ exists as $T \ell^{\oplus} I_{3}^{\ominus}$ (which is due to the inert pair effect and the stability of the $T \ell^{\oplus}$ ion). Therefore,$R$ is not the correct explanation of $A$.

(A) The electronic configuration of neutral Gadolinium $(Gd)$ with atomic number $64$ is $[Xe] 4f^7 5d^1 6s^2$.
When $Gd$ forms a $Gd^{2+}$ ion,it loses two electrons from the outermost $6s$ orbital.
Therefore,the electronic configuration of $Gd^{2+}$ becomes $[Xe] 4f^7 5d^1 6s^0$.
Thus,the number of $4f$ electrons in the ground state electronic configuration of $Gd^{2+}$ is $7$.

(D) The electronic configuration of the element in the $+3$ oxidation state is $[Ar] 3d^5$.
To find the neutral atom,we add $3$ electrons back to the configuration: $[Ar] 3d^5 + 3e^- = [Ar] 3d^6 4s^2$.
This corresponds to the element Iron $(Fe)$,which has an atomic number of $26$.

(C) The atomic and ionic radii of $Zr$ $(Z=40)$ and $Hf$ $(Z=72)$ are very similar.
This phenomenon is known as the lanthanoid contraction.
As we move from $4d$ series to $5d$ series elements in the same group,the filling of $4f$ orbitals occurs.
The poor shielding effect of $4f$ electrons leads to a decrease in atomic size,which almost perfectly compensates for the expected increase in size due to the addition of a new shell.

(B) For an ion to be coloured and paramagnetic,it must have unpaired electrons in its $d$-orbitals.
$Cu^{2+}: [Ar] 3d^{9}$ (one unpaired $e^{-}$,coloured and paramagnetic).
$Cr^{3+}: [Ar] 3d^{3}$ (three unpaired $e^{-}$,coloured and paramagnetic).
$Sc^{+}: [Ar] 3d^{1} 4s^{1}$ (two unpaired $e^{-}$,coloured and paramagnetic).
Therefore,the set containing $Cu^{2+}, Cr^{3+}, Sc^{+}$ consists of ions that are both coloured and paramagnetic.

(A) Calculate the oxidation state of the metal in each oxide:
$(a)$ In $CrO_3$: $x + 3(-2) = 0 \implies x = +6$
$(b)$ In $Fe_2O_3$: $2x + 3(-2) = 0 \implies 2x = +6 \implies x = +3$
$(c)$ In $MnO_2$: $x + 2(-2) = 0 \implies x = +4$
$(d)$ In $V_2O_5$: $2x + 5(-2) = 0 \implies 2x = +10 \implies x = +5$
$(e)$ In $Cu_2O$: $2x + 1(-2) = 0 \implies 2x = +2 \implies x = +1$
Comparing the oxidation states: $(a) (+6) > (d) (+5) > (c) (+4) > (b) (+3) > (e) (+1)$.
Thus,the correct order is $(a) > (d) > (c) > (b) > (e)$.

(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.
For $Ti^{3+}$ $(Z=22)$: Electronic configuration is $[Ar] 3d^{1}$. Here,$n = 1$. Thus,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
For $V^{2+}$ $(Z=23)$: Electronic configuration is $[Ar] 3d^{3}$. Here,$n = 3$. Thus,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
For $Sc^{3+}$ $(Z=21)$: Electronic configuration is $[Ar] 3d^{0}$. Here,$n = 0$. Thus,$\mu = 0 \ BM$.
Therefore,the magnetic moments are $1.73, 3.87, 0$ respectively.

(A) The atomic number of Holmium $(Ho)$ is $67$.
The electronic configuration of neutral $Ho$ is $[Xe] 4f^{11} 6s^{2}$.
When $Ho$ forms a $Ho^{3+}$ ion,it loses three electrons (two from the $6s$ orbital and one from the $4f$ orbital).
Therefore,the electronic configuration of $Ho^{3+}$ is $[Xe] 4f^{10}$.
Thus,the number of electrons present in the $4f$ orbital is $10$.

(N/A) This is due to the lanthanoid contraction.
Due to the lanthanoid contraction, the atomic radii of $Zr$ $(160 \text{ pm})$ and $Hf$ $(159 \text{ pm})$ are almost identical.
Because of these similar atomic radii, they exhibit similar chemical properties, making them very difficult to separate.

(N/A) In the beginning of the actinoid series,when $5f$-orbitals begin to be occupied,they penetrate less into the inner core of electrons compared to $4f$-orbitals in lanthanoids.
Consequently,$5f$-electrons are more effectively shielded from the nuclear charge than $4f$-electrons.
Due to this,the outer electrons in actinoids are less tightly held by the nucleus and are more easily available for chemical bonding.
Therefore,$Th$,$Pa$ and $U$ have lower ionization enthalpies compared to their lanthanoid counterparts $Ce$,$Pr$ and $Nd$.

(B) $CuCl_{2}$ is more stable in aqueous solution.
This is because the high negative hydration enthalpy of $Cu^{2+}$ ion compensates for the energy required to remove the second electron from $Cu^{+}$ to $Cu^{2+}$,making $Cu^{2+}$ more stable than $Cu^{+}$ in aqueous medium.

(N/A) In transition elements,the $(n-1)d$-electrons are involved in interatomic metallic bonding in addition to $ns$ electrons. This increased number of electrons participating in metallic bonding leads to stronger metallic bonds,which results in high melting points.

(A) The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$ and $Zn$ $(Z=30)$ is $[Ar] 3d^{10} 4s^2$.
In $Cr$,the electron is removed from the $4s$ orbital,which is relatively easier due to the stability gained by the resulting $3d^5$ half-filled configuration.
In $Zn$,the electron is removed from a completely filled $4s^2$ orbital,and the $3d^{10}$ configuration is already very stable,requiring more energy to remove an electron.
Therefore,the first ionisation enthalpy of $Cr$ is lower than that of $Zn$.

(B) The $E^{\circ}$ value for $Mn^{2+}$ is more negative because of the extra stability of the half-filled $d^5$ configuration.
For $Zn^{2+}$,the $E^{\circ}$ value is more negative due to the stability of the fully-filled $d^{10}$ configuration.
For $Ni^{2+}$,the large negative $E^{\circ}$ value is attributed to its very high negative hydration enthalpy.

(A) The electronic configuration of $Gd$ $(Z=64)$ is $[Xe] 4f^7 5d^1 6s^2$.
Upon formation of $Gd^{2+}$,the configuration becomes $[Xe] 4f^7 5d^1$.
The third ionisation enthalpy involves the removal of the $5d^1$ electron.
After the removal of this electron,the remaining $4f^7$ configuration is half-filled,which is highly stable due to high exchange energy.
Therefore,the energy required for the third ionisation is relatively low.

(B) The spin-only magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$, where $n$ is the number of unpaired electrons.
To find the lowest magnetic moment, we need to identify the ion with the minimum number of unpaired electrons $(n)$.
$1$. $V^{2+}$ $(Z=23)$: Electronic configuration is $[Ar] 3d^3$. Number of unpaired electrons $(n)$ = $3$.
$2$. $Ni^{2+}$ $(Z=28)$: Electronic configuration is $[Ar] 3d^8$. Number of unpaired electrons $(n)$ = $2$.
$3$. $Cr^{2+}$ $(Z=24)$: Electronic configuration is $[Ar] 3d^4$. Number of unpaired electrons $(n)$ = $4$.
$4$. $Fe^{2+}$ $(Z=26)$: Electronic configuration is $[Ar] 3d^6$. Number of unpaired electrons $(n)$ = $4$.
Since $Ni^{2+}$ has the lowest number of unpaired electrons $(n=2)$, it has the lowest spin-only magnetic moment value.

(B) Cerium $(Ce)$ has an atomic number of $58$. Its electronic configuration is $[Xe] 4f^1 5d^1 6s^2$.
In the $+4$ oxidation state,$Ce^{4+}$ has the configuration $[Xe]$,which is a stable noble gas configuration.
However,$Ce^{4+}$ has a strong tendency to gain an electron to return to the more stable $+3$ oxidation state $(Ce^{3+})$.
The reduction reaction is: $Ce^{4+} + e^- \rightarrow Ce^{3+} \quad E^0 = +1.61 \ V$.
Since it readily accepts an electron,it acts as a strong oxidizing agent.

(C) The standard electrode potential $(E^{\circ})$ for $M^{2+}/M$ in the $3d$ series depends on the sum of enthalpy of sublimation,ionization enthalpy,and hydration enthalpy.
$Cr^{2+}/Cr \rightarrow -0.90 \, V$
$Fe^{2+}/Fe \rightarrow -0.44 \, V$
$Cu^{2+}/Cu \rightarrow +0.34 \, V$
$Zn^{2+}/Zn \rightarrow -0.76 \, V$
Among the given options,$Cu$ has the highest positive value $(+0.34 \, V)$,which is due to its high enthalpy of atomization and low hydration enthalpy,making it the only metal in the $3d$ series with a positive $E^{\circ}$ value for the $M^{2+}/M$ couple.

(A) The atomic number of $Pt$ is $78$.
Following the Aufbau principle,the expected configuration is $[Xe]\, 4f^{14}\, 5d^{8}\, 6s^{2}$.
However,$Pt$ exhibits an exceptional electronic configuration due to the stability of the $d^{9}s^{1}$ arrangement.
Therefore,the correct electronic configuration is $[Xe]\, 4f^{14}\, 5d^{9}\, 6s^{1}$.

(A) The standard electrode potential $E^{0}_{M^{3+}/M^{2+}}$ is negative only for the $Cr^{3+}/Cr^{2+}$ pair $(E^{0} = -0.41 \, V)$.
The electronic configuration of $Cr^{3+}$ is $[Ar] 3d^{3}$.
The number of unpaired electrons $(n)$ in $Cr^{3+}$ is $3$.
The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \, B.M.$
$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, B.M.$
The nearest integer value is $4$.

(B) Gallium $(Ga)$ is a metal with an exceptionally low melting point $(29.76^\circ C)$.
In the periodic table,it is located in Group $13$ and is adjacent to the metalloid Germanium $(Ge)$.

(D) The oxidation states of $Mn$ in $MnF_{4}$,$MnF_{3}$,and $MnF_{2}$ are $+4$,$+3$,and $+2$ respectively.
The electronic configurations are:
$MnF_{4} (Mn^{4+}) = [Ar] 3d^{3}$
$MnF_{3} (Mn^{3+}) = [Ar] 3d^{4}$
$MnF_{2} (Mn^{2+}) = [Ar] 3d^{5}$
Among these,$Mn^{3+}$ is the strongest oxidizing agent because it can easily gain an electron to form the stable $d^{5}$ configuration $(Mn^{2+})$.
For $Mn^{3+}$ $(3d^{4})$,the number of unpaired electrons $(n)$ is $4$.
The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ B.M.$
$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.89 \ B.M.$
The nearest integer value is $5$.

(D) Isoelectronic species are those that have the same number of electrons.
$1. Sm^{2+}: 62 - 2 = 60 \text{ electrons}; Er^{3+}: 68 - 3 = 65 \text{ electrons}$. Since $60 \neq 65$,they are not isoelectronic.
$2. Yb^{2+}: 70 - 2 = 68 \text{ electrons}; Lu^{3+}: 71 - 3 = 68 \text{ electrons}$. Since $68 = 68$,they are isoelectronic.
$3. Tb^{2+}: 65 - 2 = 63 \text{ electrons}; Tm^{4+}: 69 - 4 = 65 \text{ electrons}$. Since $63 \neq 65$,they are not isoelectronic.
Therefore,both pairs $(A)$ and $(C)$ are not isoelectronic.

(B) The enthalpy of hydration $(\Delta_{hyd}H)$ is directly proportional to the charge density of the ion.
Charge density is defined as $\frac{\text{charge}}{\text{size}}$.
For the given $3d$-metal ions $(Cr^{2+}, Mn^{2+}, Fe^{2+}, Co^{2+})$,the charge is the same $(+2)$.
Therefore,the ion with the largest ionic radius will have the lowest charge density and consequently the lowest enthalpy of hydration.
According to periodic trends,the ionic radius decreases across the $3d$ series due to increasing effective nuclear charge.
Among the given ions,$Mn^{2+}$ has the largest ionic radius.
Thus,$Mn^{2+}$ has the lowest enthalpy of hydration.

(B) The most abundant transition metal in the human body is iron $(Fe)$.
Iron is a crucial component of haemoglobin.
It is the iron-containing oxygen transport metalloprotein found in red blood cells.
Haemoglobin in the blood carries oxygen from the lungs to the rest of the body.

(D) The correct option is $D$.
We calculate the oxidation state $(x)$ for the metal in each compound:
$(i)$ In $K_2CrO_4$: $2(+1) + x + 4(-2) = 0 \implies 2 + x - 8 = 0 \implies x = +6$.
The oxidation state of $Cr$ is $+6$.
$(ii)$ In $NbCl_5$: $x + 5(-1) = 0 \implies x - 5 = 0 \implies x = +5$.
The oxidation state of $Nb$ is $+5$.
$(iii)$ In $MnO_2$: $x + 2(-2) = 0 \implies x - 4 = 0 \implies x = +4$.
The oxidation state of $Mn$ is $+4$.
Comparing the values $(+6, +5, +4)$,$Cr$ has the highest oxidation state.

(B)
Isoelectronic species are those that have the same number of electrons.
$(a)$ $Sc^{2+}$ ($Z=21$,$21-2=19$ electrons) and $V^{3+}$ ($Z=23$,$23-3=20$ electrons).
$(b)$ $Mn^{2+}$ ($Z=25$,$25-2=23$ electrons) and $Fe^{3+}$ ($Z=26$,$26-3=23$ electrons).
$(c)$ $Mn^{3+}$ ($Z=25$,$25-3=22$ electrons) and $Fe^{2+}$ ($Z=26$,$26-2=24$ electrons).
$(d)$ $Ni^{3+}$ ($Z=28$,$28-3=25$ electrons) and $Fe^{2+}$ ($Z=26$,$26-2=24$ electrons).
Thus,$Mn^{2+}$ and $Fe^{3+}$ have the same number of electrons and form an isoelectronic pair. The correct option is $(b)$.

(C) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
For $\mu = 3.87 \ B.M.$,we have $\sqrt{n(n+2)} = 3.87$,which implies $n(n+2) \approx 15$,so $n = 3$.
$Cr^{2+}: [Ar] 3d^4, n=4, \mu = \sqrt{4(6)} = 4.89 \ B.M.$
$Mn^{2+}: [Ar] 3d^5, n=5, \mu = \sqrt{5(7)} = 5.91 \ B.M.$
$V^{2+}: [Ar] 3d^3, n=3, \mu = \sqrt{3(5)} = 3.87 \ B.M.$
$Ti^{2+}: [Ar] 3d^2, n=2, \mu = \sqrt{2(4)} = 2.82 \ B.M.$
Thus,the metal ion is $V^{2+}$.

(B) The spin-only magnetic moment is given by the formula $\mu_{s} = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.

Ion Configuration	$n$
$V^{3+}: [Ar] 3d^{2}$	$2$
$Cr^{3+}: [Ar] 3d^{3}$	$3$
$Fe^{2+}: [Ar] 3d^{6}$	$4$
$Ni^{3+}: [Ar] 3d^{7}$	$3$

Comparing the values of $n$,we see that $Cr^{3+}$ and $Ni^{3+}$ both have $n = 3$,meaning they have the same spin-only magnetic moment. Thus,there are $2$ such ions.

(C) The first ionization enthalpy of $3d$ series elements is generally comparable to or lower than that of group $2$ metals (alkaline earth metals) for the early members of the series. For example,the first ionization energy of $Mg$ is $737 \ kJ/mol$,while for $Sc$ it is $631 \ kJ/mol$ and for $Ti$ it is $656 \ kJ/mol$. Thus,Assertion $(A)$ is false.
Reason $(R)$ states that in the $3d$ series of elements,successive filling of $d$-orbitals takes place. This is a correct statement as the electronic configuration of $3d$ series elements involves the filling of the $3d$ subshell.
Therefore,$(A)$ is false but $(R)$ is true.

(B) The stability of $Cu^{2+}(aq)$ in aqueous solution is higher than that of $Cu^{+}(aq)$.
This is because the enthalpy of hydration $(\Delta_{hyd}H)$ for $Cu^{2+}$ is much more negative (higher magnitude) than that of $Cu^{+}$.
This large negative hydration energy of $Cu^{2+}$ more than compensates for the energy required for the second ionization enthalpy of $Cu$,making $Cu^{2+}$ more stable in water.
Therefore,Assertion $(A)$ is correct,but Reason $(R)$ is incorrect because the enthalpy of hydration for $Cu^{2+}$ is much more negative (greater in magnitude) than that of $Cu^{+}$,not less.

(D) Statement $I$ is incorrect because $Cr^{2+}$ $(3d^4)$ is a reducing agent as it changes to $Cr^{3+}$ ($3d^3$,stable $t_{2g}^3$ configuration),and $Mn^{3+}$ $(3d^4)$ is an oxidising agent as it changes to $Mn^{2+}$ ($3d^5$,stable half-filled $d^5$ configuration).
Statement $II$ is correct because $Sc^{3+}$ has a $3d^0$ configuration,meaning it has no unpaired electrons. Therefore,it is diamagnetic and is repelled by an applied magnetic field.

(D) The formula for chromyl chloride is $CrO_2Cl_2$.
In $CrO_2Cl_2$,the oxidation state of $Cr$ is calculated as: $x + 2(-2) + 2(-1) = 0$,which gives $x = +6$.
The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$. Thus,$Cr(VI)$ is $[Ar] 3d^0$.
Now,let us check the $d$-electron count for the options:
$Ti(III)$ $(Z=22)$: $[Ar] 3d^1$
$Fe(III)$ $(Z=26)$: $[Ar] 3d^5$
$V(IV)$ $(Z=23)$: $[Ar] 3d^1$
$Mn(VII)$ $(Z=25)$: $[Ar] 3d^0$
Since $Cr(VI)$ and $Mn(VII)$ both have $0$ $d$-electrons,the correct option is $D$.

(D) The stability of $Cu^{2+}_{(aq)}$ is more than $Cu^{+}_{(aq)}$ is due to the much more negative $\Delta_{hyd}H^{\circ}$ of $Cu^{2+}_{(aq)}$ than $Cu^{+}_{(aq)}$,which more than compensates for the second ionisation enthalpy of $Cu$.
$\Delta_{hyd}H^{\circ}$ of $Cu^{2+}_{(aq)} = -2121 \ kJ \ mol^{-1}$
$\Delta_{i}H_{1}^{\circ}$ of $Cu = +745 \ kJ \ mol^{-1}$
$\Delta_{i}H_{2}^{\circ}$ of $Cu = +1960 \ kJ \ mol^{-1}$

(D) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.

$3d$ configuration	Unpaired electrons $(n)$
$3d^7$	$3$
$3d^8$	$2$
$3d^3$	$3$
$3d^6$	$4$

Comparing the values:
For $3d^7$: $\mu = \sqrt{3(3+2)} = \sqrt{15} \ BM \approx 3.87 \ BM$
For $3d^8$: $\mu = \sqrt{2(2+2)} = \sqrt{8} \ BM \approx 2.83 \ BM$
For $3d^3$: $\mu = \sqrt{3(3+2)} = \sqrt{15} \ BM \approx 3.87 \ BM$
For $3d^6$: $\mu = \sqrt{4(4+2)} = \sqrt{24} \ BM \approx 4.90 \ BM$
Thus,the configuration $3d^6$ has the highest number of unpaired electrons $(n=4)$ and consequently the highest magnetic moment.

(B) The electronic configurations of the given elements are:
$Cu (Z=29): [Ar] 3d^{10} 4s^1$
$Zn (Z=30): [Ar] 3d^{10} 4s^2$
$Cd (Z=48): [Kr] 4d^{10} 5s^2$
$Ag (Z=47): [Kr] 4d^{10} 5s^1$
All these elements have a completely filled $d$-orbital ($d^{10}$ configuration) in their ground state.
Therefore,option $B$ is correct.

(A) $Zn, Cd$ and $Hg$ have fully filled $d$-orbitals,which results in weak metallic bonding and thus they exhibit the lowest enthalpy of atomization in their respective transition series.
$(B)$ $Zn$ and $Cd$ show only $+II$ oxidation state,whereas $Hg$ shows $+I$ and $+II$ oxidation states.
$(C)$ Since they have a fully filled $d^{10}$ configuration,their compounds are diamagnetic in nature.
$(D)$ Due to weak metallic bonding,$Zn, Cd$ and $Hg$ are considered soft metals.
Therefore,statements $B$ and $D$ are correct.

(B) The electronic configurations of the given species are as follows:
$1$. $Cr$ $(Z=24)$: $[Ar] 3d^5 4s^1$. Therefore,$Cr^{2+}$ is $[Ar] 3d^4$ (Matches $III$).
$2$. $Mn$ $(Z=25)$: $[Ar] 3d^5 4s^2$. Therefore,$Mn^{+}$ is $[Ar] 3d^5 4s^1$ (Matches $IV$).
$3$. $Ni$ $(Z=28)$: $[Ar] 3d^8 4s^2$. Therefore,$Ni^{2+}$ is $[Ar] 3d^8$ (Matches $I$).
$4$. $V$ $(Z=23)$: $[Ar] 3d^3 4s^2$. Therefore,$V^{+}$ is $[Ar] 3d^3 4s^1$ (Matches $II$).
Thus,the correct matching is $A-III, B-IV, C-I, D-II$.

(B) The $3^{rd}$ ionisation enthalpy $(IE_3)$ corresponds to the removal of an electron from the $M^{2+}$ ion.
Electronic configurations of $M^{2+}$ ions:
$V^{2+}: [Ar] 3d^3$
$Cr^{2+}: [Ar] 3d^4$
$Mn^{2+}: [Ar] 3d^5$
$Fe^{2+}: [Ar] 3d^6$
$Mn^{2+}$ has a stable half-filled $d$-orbital configuration $(3d^5)$. Removing an electron from this stable configuration requires a significantly higher amount of energy compared to the others.
Experimental values $(kJ/mol)$:
$V: 2833$
$Cr: 2990$
$Mn: 3260$
$Fe: 2962$
Thus,$Mn$ has the highest $3^{rd}$ ionisation enthalpy among the given elements.

(A) $Cr^{2+}$ $(d^4)$ is a strong reducing agent because it changes to $Cr^{3+}$ $(d^3)$,which has a stable half-filled $t_{2g}$ configuration.
$Mn^{3+}$ $(d^4)$ is an oxidizing agent because it changes to $Mn^{2+}$ $(d^5)$,which has a stable half-filled $d$-orbital configuration.
Both Assertion and Reason are true,and the Reason correctly explains the stability associated with half-filled configurations that drives these redox processes.

(B) $Sc$ (Scandium) has an electronic configuration of $[Ar] 3d^1 4s^2$.
It loses three electrons to form $Sc^{3+}$ ion,which has a stable noble gas configuration.
Therefore,$Sc$ exhibits only a $+3$ oxidation state in its compounds.
In contrast,$Co$,$Ti$,and $Ni$ are transition elements that exhibit variable oxidation states.

(D) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 3.86 \ BM$,we have $\sqrt{n(n+2)} = 3.86$.
Squaring both sides,$n(n+2) \approx 15$,which gives $n = 3$.
For a metal ion in the $+2$ oxidation state to have $3$ unpaired electrons,the electronic configuration must be $d^3$.
Checking the options:
${}_{22} Ti^{+2} = [Ar] 3d^2$ $(n=2)$
${}_{23} V^{+2} = [Ar] 3d^3$ $(n=3)$
${}_{25} Mn^{+2} = [Ar] 3d^5$ $(n=5)$
${}_{26} Fe^{+2} = [Ar] 3d^6$ $(n=4)$
Thus,the metal is Vanadium $(V)$ with atomic number $23$.

(B) $Mn$ (Manganese) exhibits the highest number of oxidation states among the $3d$ transition series metals,ranging from $+2$ to $+7$.
Specifically,it shows the maximum oxidation state of $+7$ in compounds like $KMnO_4$.

(A) The ion that acts as a strong oxidizing agent in aqueous solution is $Co^{3+}$ because it has a high reduction potential $(Co^{3+} + e^{-} \rightarrow Co^{2+})$.
The electronic configuration of $Co^{3+}$ $(Z = 27)$ is $[Ar] 3d^6$.
The number of unpaired electrons $(n)$ in $3d^6$ is $4$.
The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \ BM$.
$\mu = \sqrt{4(4+2)} = \sqrt{24} \ BM \approx 4.9 \ BM$.
The nearest integer value is $5$.