General Characteristics Questions in English

(A) The electronic configurations and the number of unpaired electrons are as follows:
$Sc (Z=21): [Ar] 4s^2 3d^1 \rightarrow 1 \text{ unpaired electron}$
$Ti (Z=22): [Ar] 4s^2 3d^2 \rightarrow 2 \text{ unpaired electrons}$
$V (Z=23): [Ar] 4s^2 3d^3 \rightarrow 3 \text{ unpaired electrons}$
$Mn (Z=25): [Ar] 4s^2 3d^5 \rightarrow 5 \text{ unpaired electrons}$
$Cr (Z=24): [Ar] 4s^1 3d^5 \rightarrow 6 \text{ unpaired electrons}$
Increasing order of unpaired electrons: $Sc(1) < Ti(2) < V(3) < Mn(5) < Cr(6)$,which corresponds to $A < D < C < E < B$.

(C) Statement $I$ is true: For transition elements,the stability of higher oxidation states increases down the group (e.g.,$W(VI)$ is more stable than $Cr(VI)$),whereas in $p-$block elements,the stability of higher oxidation states decreases down the group due to the inert pair effect.
Statement $II$ is true: Copper has a positive standard reduction potential $(E^{\circ}_{Cu^{2+}/Cu} = +0.34 \ V)$,which is higher than that of hydrogen $(E^{\circ}_{H^{+}/H_2} = 0.00 \ V)$. Therefore,copper cannot displace hydrogen from acids.

(B) The second ionization enthalpy $(IE_2)$ involves removing an electron from the $M^{+}$ ion. For $Cr$ $(Z=24)$,the electronic configuration is $[Ar] 3d^5 4s^1$. The $Cr^{+}$ ion has the configuration $[Ar] 3d^5$. Removing the second electron from this stable half-filled $d^5$ subshell requires a very high amount of energy.
Thus,$M = Cr$.
The $Cr^{+}$ ion has the configuration $[Ar] 3d^5$,which means it has $n = 5$ unpaired electrons.
The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \ BM$.
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
Rounding to the nearest integer,we get $6 \ BM$.

(B) The standard electrode potential $E^{\circ}$ for the $Mn^{3+} / Mn^{2+}$ couple is significantly higher because the reduction of $Mn^{3+}$ to $Mn^{2+}$ involves a transition from a $3d^4$ configuration to a more stable $3d^5$ configuration.
$Mn^{3+} ([Ar] 3d^4) + e^- \rightarrow Mn^{2+} ([Ar] 3d^5)$
The $3d^5$ configuration is particularly stable due to the half-filled $d$-orbital subshell,which provides extra stability through exchange energy.
In contrast,the reduction of $Cr^{3+} (3d^3)$ to $Cr^{2+} (3d^4)$ or $Fe^{3+} (3d^5)$ to $Fe^{2+} (3d^6)$ does not result in such a significant gain in stability.

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

Ion and Configuration	No. of unpaired electrons $(n)$
$Ti^{3+}: 3d^{1}$	$1$
$Cr^{2+}: 3d^{4}$	$4$
$Mn^{2+}: 3d^{5}$	$5$
$Fe^{2+}: 3d^{6}$	$4$
$Sc^{3+}: 3d^{0}$	$0$

Since $Cr^{2+}$ and $Fe^{2+}$ both have $n = 4$ unpaired electrons,they have the same 'spin only' magnetic moment.

(A, B, C, D) $(1)$ $Cr^{2+}$ $(d^4)$ is a reducing agent because it gets oxidized to $Cr^{3+}$ $(d^3)$,which has a stable half-filled $t_{2g}$ subshell.
$(2)$ $Mn^{3+}$ $(d^4)$ is an oxidizing agent because it gets reduced to $Mn^{2+}$ $(d^5)$,which has a stable half-filled $d$-orbital configuration.
$(3)$ Both $Cr^{2+}$ and $Mn^{3+}$ have $d^4$ electronic configuration.
$(4)$ Since all statements $(A), (B), (C),$ and $(D)$ are correct,the correct option is $(A, B, C, D)$ (Note: Given the provided options,all statements are factually correct).

(B) The electronic configuration of Niobium $(Nb, Z=41)$ is $[Kr] \ 4d^4 \ 5s^1$. Thus,the number of electrons in the $4d$ orbital is $x = 4$.
The electronic configuration of Ruthenium $(Ru, Z=44)$ is $[Kr] \ 4d^7 \ 5s^1$. Thus,the number of electrons in the $4d$ orbital is $y = 7$.
Therefore,the value of $x+y = 4+7 = 11$.

(A) $I$. $V^{2+}$ is violet,$Cr^{3+}$ is violet,and $Mn^{3+}$ is violet. All three ions exhibit a violet colour in aqueous solution.
$II$. $Zn^{2+}$ is colourless,$V^{3+}$ is green,and $Fe^{3+}$ is yellow.
$III$. $Ti^{4+}$ is colourless,$V^{4+}$ is blue,and $Mn^{2+}$ is pink.
$IV$. $Sc^{3+}$ is colourless,$Ti^{3+}$ is purple,and $Cr^{2+}$ is blue.
Therefore,the set with the same colour is $V^{2+}, Cr^{3+}, Mn^{3+}$.

(B) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$, where $n$ is the number of unpaired electrons.
$A$. $Ti^{3+}$ $(3d^1)$: $n = 1$, $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ B.M.$ $(III)$
$B$. $V^{2+}$ $(3d^3)$: $n = 3$, $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ B.M.$ $(I)$
$C$. $Ni^{2+}$ $(3d^8)$: $n = 2$, $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84 \ B.M.$ $(IV)$
$D$. $Sc^{3+}$ $(3d^0)$: $n = 0$, $\mu = \sqrt{0(0+2)} = 0.00 \ B.M.$ $(II)$
Therefore, the correct matching is $A-III, B-I, C-IV, D-II$.

(B) The structure of $ClF_3$ is $T$-shaped with two lone pairs occupying the equatorial positions to minimize repulsion.
Thus,$n = 2$.
We need to identify ions with $2$ unpaired electrons:
$(A) V^{3+}: [Ar] 3d^2$ (Unpaired electrons = $2$)
$(B) Ti^{3+}: [Ar] 3d^1$ (Unpaired electrons = $1$)
$(C) Cu^{2+}: [Ar] 3d^9$ (Unpaired electrons = $1$)
$(D) Ni^{2+}: [Ar] 3d^8$ (Unpaired electrons = $2$)
$(E) Ti^{2+}: [Ar] 3d^2$ (Unpaired electrons = $2$)
Therefore,ions with $n = 2$ unpaired electrons are $A, D,$ and $E$.

(C) The melting points of transition metals depend on the number of unpaired electrons and the strength of metallic bonding.
For the $3d$ series, $Mn$ has a lower melting point than $Fe$ because $Mn$ has a stable $d^5$ configuration, which results in weaker metallic bonding compared to $Fe$.
For the $4d$ and $5d$ series, the trend is $Tc < Ru$ and $Os < Re$ because the number of unpaired electrons available for metallic bonding increases as we move from $Tc$ to $Ru$ and decreases from $Re$ to $Os$.
Thus, the correct order is $Mn < Fe$, $Tc < Ru$, and $Os < Re$.

(B) Statement-$I$ is true because $CrO_3$ is a stronger oxidizing agent than $MoO_3$.
Statement-$II$ is false because $Cr(VI)$ is less stable than $Mo(VI)$.
As we move down the group in the $d$-block,the stability of the highest oxidation state increases. Therefore,$Mo(VI)$ is more stable than $Cr(VI)$.
Due to the lower stability of $Cr(VI)$,$CrO_3$ acts as a stronger oxidizing agent.

(D)

Element	Enthalpy of Atomisation $(\text{kJ/mol})$
$Sc$	$326$
$Mn$	$281$
$Co$	$425$
$Cu$	$339$

The element with the highest enthalpy of atomisation is $Co$ $(425 \text{ kJ/mol})$.
The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.
In its $+2$ oxidation state,$Co^{2+}$ has the configuration $[Ar] 3d^7$.
The number of unpaired electrons $(n)$ in $3d^7$ is $3$.
The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \text{ BM}$.
$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM}$.
The nearest integer value is $4$.

(A) To determine the number of unpaired electrons,we write the electronic configuration of each ion:

$Ion$	$Electronic \ Configuration$	$Unpaired \ e^-$
$V^{2+}$	$[Ar] 3d^3$	$3$
$Co^{2+}$	$[Ar] 3d^7$	$3$
$Ti^{2+}$	$[Ar] 3d^2$	$2$
$Fe^{3+}$	$[Ar] 3d^5$	$5$
$Cr^{2+}$	$[Ar] 3d^4$	$4$
$Ti^{3+}$	$[Ar] 3d^1$	$1$
$Mn^{2+}$	$[Ar] 3d^5$	$5$

Comparing the pairs:
$A. V^{2+} (3)$ and $Co^{2+} (3)$ have the same number of unpaired electrons.
Thus,the correct option is $A$.

(D) Ionisation enthalpy $(IE)$ is the energy required to remove an electron from a gaseous atom or ion.
$Mn^{2+}$ has a stable $d^5$ configuration $([Ar] 3d^5)$,which makes it very difficult to remove another electron,resulting in a very high $IE$.
$Fe^{2+}$ has a $d^6$ configuration $([Ar] 3d^6)$.
Since $Mn^{2+}$ is more stable than $Fe^{2+}$,the energy required to remove an electron from $Mn^{2+}$ is higher than that for $Fe^{2+}$.
Therefore,the relationship $Mn^{2+} < Fe^{2+}$ is incorrect; the correct relationship is $Mn^{2+} > Fe^{2+}$.

(C) The enthalpy of atomisation depends on the number of unpaired electrons available for metallic bonding. Among the given transition metals ($Cr$,$Co$,$Fe$,$Ni$),$Cr$ has the lowest enthalpy of atomisation because it has a stable $d^5$ configuration,which limits the number of electrons participating in metallic bonding compared to others.
For $Cr$ $(Z=24)$,the electronic configuration is $[Ar] 3d^5 4s^1$.
The total number of valence electrons is the sum of electrons in the outermost $s$ and $d$ orbitals.
Valence electrons = $5 (d) + 1 (s) = 6$.

(C) The spin only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$Cu^{+} : [Ar] 3d^{10}$,$n = 0$,$\mu = 0 \ BM$.
$Cu^{2+} : [Ar] 3d^{9}$,$n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
$Cr^{2+} : [Ar] 3d^{4}$,$n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
$Cr^{3+} : [Ar] 3d^{3}$,$n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
Comparing the values: $4.90 > 3.87 > 1.73 > 0$.
Therefore,the decreasing order is $Cr^{2+} > Cr^{3+} > Cu^{2+} > Cu^{+}$.

(C) Statement $I$: Ferromagnetism is indeed considered an extreme form of paramagnetism because,like paramagnetism,it arises from the presence of unpaired electrons,but the interaction between them is much stronger,leading to permanent magnetic moments even in the absence of an external magnetic field. This statement is true.
Statement $II$: For $Cr^{2+}$ $(Z=24)$,the electronic configuration is $[Ar] 3d^4$. It has $4$ unpaired electrons.
For $Nd^{3+}$ $(Z=60)$,the electronic configuration is $[Xe] 4f^3$. It has $3$ unpaired electrons.
Since $4 \neq 3$,the number of unpaired electrons is not the same. This statement is false.
Therefore,Statement $I$ is true but Statement $II$ is false.

(B) $p-$block elements form acidic,basic,and amphoteric oxides. Therefore,the statement that they form only acidic oxides is incorrect.
Additionally,for $p-$block elements,the oxidation states differ by $2$ units due to the inert pair effect (e.g.,$Pb^{+2}$ and $Pb^{+4}$).

(B) . Misch metal is a well-known alloy which consists of a lanthanoid metal $(Ln)$ $(\approx 95\%)$,iron $(Fe)$ $(\approx 5\%)$,and traces of sulfur $(S)$,carbon $(C)$,silicon $(Si)$,aluminum $(Al)$,and calcium $(Ca)$.
Therefore,the most appropriate option representing the composition is $Ln + Fe$.

(A) The electronic configuration of mercury $(Hg)$ is $[Xe] 4f^{14} 5d^{10} 6s^2$.
Statement-$I$ is correct because mercury is the only metal that exists in a liquid state at room temperature due to its unique electronic configuration.
Statement-$II$ is also correct. Because the $5d$ subshell is completely filled $(5d^{10})$,there are no unpaired electrons available for metallic bonding. The $6s$ electrons are also tightly held by the nucleus due to relativistic effects,resulting in very weak metallic bonding,which explains its low melting point and liquid state.

(D) The electronic configuration of $Zn$ is $[Ar] 3d^{10} 4s^2$.
Because the $3d$ subshell is completely filled,the $d$-electrons do not participate in the formation of metallic bonds.
Metallic bonding in transition elements depends on the number of unpaired electrons in the $d$-orbitals.
Since $Zn$ has zero unpaired electrons $(U.P.E.)$,the metallic bonding is weak,resulting in a lower melting point compared to other $3d$ series elements.

(A) $i$. $Os$ $(\text{Osmium})$ is a transition metal that exhibits a maximum oxidation state of $+8$ in $OsO_4$.
$ii$. $Mn$ $(\text{Manganese})$ is a $3d$ series element that shows oxidation states ranging from $+2$ to $+7$ (e.g.,in $KMnO_4$).
$iii$. $Cr$ $(\text{Chromium})$ has a relatively high melting point among $3d$ series elements due to the involvement of more electrons in metallic bonding.
Therefore,the correct matching is $i-c, ii-a, iii-b$.

(C) Transition metals exhibit variable valency because the energy difference between the $ns$ and $(n-1)d$ orbitals is very small.
This allows electrons from both the $ns$ and $(n-1)d$ subshells to participate in bond formation.
Therefore,the Assertion is True,but the Reason is False because the energy difference is small,not large.

(D) In the first transition series ($3d$ series),the element with the maximum melting point is $Cr$ due to the presence of a large number of unpaired electrons in the $d$-orbitals,which leads to strong metallic bonding.
In the second transition series ($4d$ series),the element with the minimum melting point is $Cd$ because it has a completely filled $d^{10}$ configuration,resulting in weak metallic bonding.

(A) In $d$-block elements,the stability of the highest oxidation state increases down the group. $Cr(VI)$ is a strong oxidising agent because it is unstable and tends to get reduced to $Cr(III)$.
Conversely,$Mo(VI)$ and $W(VI)$ are much more stable than $Cr(VI)$,which makes $MoO_3$ and $WO_3$ poor oxidising agents.
Therefore,both the Assertion and the Reason are true,and the Reason correctly explains why $Cr(VI)$ acts as a strong oxidising agent while $Mo(VI)$ and $W(VI)$ do not.

(D) The correct answer is $(d)$.
Transition elements (also known as transition metals) are elements that have partially filled $d$ orbitals.
$IUPAC$ defines transition elements as an element having a $d$ subshell that is partially filled with electrons,or an element that has the ability to form stable cations with an incompletely filled $d$ orbital.
In general,any element which corresponds to the $d$-block of the modern periodic table (which consists of groups $3-12$) is considered to be a transition element.

(B) Each series of transition elements ($d$-block elements) corresponds to the filling of the $(n-1)d$ orbitals.
Since a $d$-subshell can accommodate a maximum of $10$ electrons,each transition series contains $10$ elements.

(A) 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.
However,there are irregularities due to stable electronic configurations (like half-filled or fully-filled $d$-orbitals).
The electronic configurations are: $Sc ([Ar] 3d^1 4s^2)$,$Cr ([Ar] 3d^5 4s^1)$,$Fe ([Ar] 3d^6 4s^2)$,and $Zn ([Ar] 3d^{10} 4s^2)$.
Comparing their $(IE_1)$ values,the order is $Zn > Fe > Cr > Sc$.

(C) The electronic configuration of $Zn$ is $[Ar] 3d^{10} 4s^2$.
Due to the completely filled $3d$ and $4s$ orbitals,$Zn$ exhibits high stability.
Consequently,it requires the highest amount of energy to remove an electron compared to the other elements listed ($Ti$,$Sc$,$Ni$) in the $3d$ series.

(B) The elements $Pr$ $(Z=59)$,$Gd$ $(Z=64)$,$Er$ $(Z=68)$,and $Yb$ $(Z=70)$ belong to the lanthanoid series.
In the lanthanoid series,as the atomic number $(Z)$ increases,the atomic size decreases due to lanthanoid contraction.
Therefore,the correct decreasing order of atomic size is $Pr > Gd > Er > Yb$.

(B) The electronic configuration of $Zn$ $(Z=30)$ is $[Ar] 3d^{10} 4s^2$.
In the $+2$ oxidation state,$Zn^{2+}$ has the configuration $[Ar] 3d^{10}$.
Since the $3d$ subshell is completely filled,there are no unpaired electrons.
Therefore,$Zn^{2+}$ compounds are colourless due to the absence of $d-d$ transitions.

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

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

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

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

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

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

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

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

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

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

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

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

(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.