General Characteristics Questions in English

(A) German silver is an alloy of copper $(Cu)$,zinc $(Zn)$,and nickel $(Ni)$.
The typical composition is $60\% \ Cu$,$20\% \ Ni$,and $20\% \ Zn$.
It is named for its silver-white appearance,although it does not contain any silver $(Ag)$.

(D) In the $3d$ transition series,as we move from left to right,the effective nuclear charge $(Z_{eff})$ increases,which causes the atomic size to decrease.
Although $Zn$ shows a slight increase in size due to electron-electron repulsions,the element with the largest atomic radius in the $3d$ series is $Sc$ $(Z=21)$,as it has the lowest effective nuclear charge among the given elements.

(D) $1$. $Sc$ $(Z=21)$ has the configuration $[Ar] 3d^1 4s^2$. By losing three electrons,it attains the stable noble gas configuration $[Ar]$,making $+3$ the most stable oxidation state.
$2$. $Zn$ $(Z=30)$ has a completely filled $3d^{10} 4s^2$ configuration. Removing an electron from this stable configuration requires a very high amount of energy,hence it has a very high first ionization energy compared to other $3d$ elements.
$3$. In the $3d$ series,the metallic radius generally decreases from $Sc$ to $Cr$. However,$Mn$ has a slightly larger radius than expected due to half-filled $d^5$ configuration leading to weak inter-atomic metallic bonding. Thus,the metallic radius of $Mn$ is indeed greater than that of $Fe$.
$4$. Since all statements are correct,the correct option is $D$.

(A) Transition elements form complex compounds because they have small size,high nuclear charge,and the presence of vacant $d$-orbitals of appropriate energy to accept lone pairs of electrons from ligands.

(A) The transition elements exhibit variable oxidation states. Among the given options,$Os$ (Osmium) belongs to group $8$ and can exhibit a maximum oxidation state of $+8$ (e.g.,in $OsO_4$).
$Mn$ (Manganese) shows a maximum oxidation state of $+7$ (e.g.,in $KMnO_4$).
$Cr$ (Chromium) shows a maximum oxidation state of $+6$ (e.g.,in $K_2Cr_2O_7$).
$Co$ (Cobalt) typically shows $+2$ and $+3$ oxidation states.
Therefore,$Os$ shows the maximum oxidation state.

(D) The ionic size of trivalent ions generally increases down a group due to the addition of new shells.
$Ti^{3+}$ belongs to the $3d$ series,$Zr^{3+}$ to the $4d$ series,and $Hf^{3+}$ to the $5d$ series.
$La^{3+}$ is a lanthanide ion with a larger atomic radius compared to the transition metal ions listed.
Among the given options,$La^{3+}$ has the largest ionic radius due to its position in the $6th$ period and the lanthanide contraction effect being less pronounced for $La^{3+}$ compared to the subsequent $Hf^{3+}$ ion.

(D) The atomic number of Platinum $(Pt)$ is $78$.
The electronic configuration of $Pt$ is $[Xe] 4f^{14} 5d^9 6s^1$.
To find the total number of $d$ electrons,we look at the inner shells as well:
$1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^6 4f^{14} 5d^9 6s^1$.
Total $d$ electrons = $3d^{10} + 4d^{10} + 5d^9 = 10 + 10 + 9 = 29$.

(B) transition element is defined as an element which has incompletely filled $d$ orbitals in its ground state or in any one of its oxidation states.
$Zn$,$Cd$,and $Hg$ have completely filled $d^{10}$ configuration in their ground state as well as in their common oxidation states.
Therefore,$Zn$ is not considered a transition element,although it belongs to the $d$-block.

(A) The general electronic configuration of $d-$block elements is $(n-1)d^{1-10} ns^{0-2}$.
Here,the penultimate shell is the $(n-1)$ shell.
The electrons present in the $(n-1)$ shell are $(n-1)s^2$,$(n-1)p^6$,and $(n-1)d^{1-10}$.
Summing these,we get $2 + 6 + (1 \text{ to } 10) = 9$ to $18$ electrons.

(A) The electronic configuration is $A = [Rn]\,6d^2\,7s^2$.
Since the last electron enters the $d-$ orbital,it belongs to the $d-$ block.
The principal quantum number of the outermost shell is $n = 7$,so the period is $7$.
The group number for $d-$ block elements is calculated as $(n-1)d + ns$ electrons.
Here,$2 + 2 = 4$ electrons,which corresponds to group $4$ or $IVB$ in the old $IUPAC$ notation.
Therefore,the element is in the $d-$ block,$IVB$ group,and $7^{th}$ period.

(B) species is diamagnetic if it has no unpaired electrons.
$1$. $Fe^{+2}$ $([Ar] 3d^6)$ has $4$ unpaired electrons; $Fe^{+3}$ $([Ar] 3d^5)$ has $5$ unpaired electrons.
$2$. $Hg_2^{+2}$ contains two $Hg^{+}$ ions $([Xe] 4f^{14} 5d^{10} 6s^1)$ bonded together,resulting in a paired electron configuration $([Xe] 4f^{14} 5d^{10} 6s^0)$,making it diamagnetic. $Hg^{+2}$ $([Xe] 4f^{14} 5d^{10})$ has a completely filled $d$-orbital,making it diamagnetic.
$3$. $Cu^{+}$ $([Ar] 3d^{10})$ is diamagnetic,but $Cu^{+2}$ $([Ar] 3d^9)$ has $1$ unpaired electron.
Therefore,the pair $Hg_2^{+2}, Hg^{+2}$ is diamagnetic.

(C) Across the $3d$ transition series from $Sc$ $(Z=21)$ to $Zn$ $(Z=30)$,the atomic radius generally decreases due to the increase in effective nuclear charge,which pulls the electrons closer to the nucleus.
However,the decrease is not uniform. From $Sc$ to $Cr$,the size decreases.
From $Mn$ to $Ni$,the atomic radii remain almost constant because the increase in nuclear charge is balanced by the shielding effect of $d$-electrons.
Towards the end of the series,specifically at $Zn$ $(3d^{10} 4s^2)$,the atomic radius increases due to the strong inter-electronic repulsions between the $d$-electrons,which outweighs the effect of the increased nuclear charge.
Therefore,$Zn$ has the largest atomic radius in the $3d$ series.

(B) $KMnO_4$ is a purple-coloured salt due to charge transfer.
$BaSO_4$ is a white (colourless) salt as $Ba^{2+}$ and $SO_4^{2-}$ ions have a $d^0$ configuration and no $d-d$ transitions occur.
$Na_2CrO_4$ is yellow due to charge transfer in the chromate ion.
$CoCl_2$ is blue (in anhydrous form) or pink (in hydrated form) due to $d-d$ transitions.

(D) The elements of group $11$ are Copper $(Cu)$,Silver $(Ag)$,and Gold $(Au)$.
These elements belong to the $d$-block of the periodic table.
The general electronic configuration for group $11$ elements is $(n-1)d^{10} ns^1$,where $n$ is the outermost shell.
For example,Copper $(Cu)$ has the configuration $[Ar] 3d^{10} 4s^1$.

(A) $Pd$ (Palladium) belongs to Group $10$ and Period $5$ of the periodic table.
To find the elements in the same group,we use the magic numbers for the $d$-block elements.
The element above $Pd$ is $Ni$ $(Z = 28)$ in Period $4$.
The element below $Pd$ is $Pt$ $(Z = 78)$ in Period $6$.
Thus,the atomic numbers are $28$ and $78$.

(B) In the $d$-block elements,specifically the first transition series ($3d$ series),the atomic radii of elements like $Fe$ $(26)$,$Co$ $(27)$,and $Ni$ $(28)$ are very similar.
This is because the increase in nuclear charge is almost perfectly balanced by the screening effect of the $d$-electrons,leading to a very small change in atomic size across these elements.

(D) pseudo inert gas configuration is defined as having $18$ electrons in the outermost shell,represented by the electronic configuration $ns^2 np^6 nd^{10}$.
$Zn^{2+}$ $(Z=30)$: The electronic configuration of $Zn$ is $[Ar] 3d^{10} 4s^2$. Removing $2$ electrons gives $Zn^{2+} = [Ar] 3d^{10}$,which has $18$ electrons in the outermost shell.
$Cu^{+}$ $(Z=29)$: The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$. Removing $1$ electron gives $Cu^{+} = [Ar] 3d^{10}$,which has $18$ electrons in the outermost shell.
$Pd$ $(Z=46)$: The electronic configuration of $Pd$ is $[Kr] 4d^{10} 5s^0$,which also exhibits a pseudo inert gas configuration.
Therefore,all the given species possess a pseudo inert gas configuration.

(A) The volatile metal group refers to Group $12$ of the periodic table,which includes $Zn$,$Cd$,$Hg$,and $Cn$ (Copernicium).
The last element of this group is $Cn$ (atomic number $Z = 112$).
The electronic configuration of $Cn$ is $[Rn] \ 5f^{14} \ 6d^{10} \ 7s^2$.

(A) The melting point of transition elements depends on the strength of metallic bonding,which is determined by the number of unpaired electrons available for delocalization.
In the $3d$ series,$Zn$ has the electronic configuration $[Ar] 3d^{10} 4s^2$.
Since all $d$-orbitals are completely filled,there are no unpaired electrons available for metallic bonding.
Consequently,the interatomic metallic bonding in $Zn$ is relatively weak,leading to a lower melting point compared to other $3d$ transition elements.

(D) Transition metals exhibit catalytic activity primarily because they can adopt multiple oxidation states and form complexes.
They form unstable intermediate compounds with reactants,which provide an alternative reaction path with lower activation energy.
This reduction in activation energy increases the reaction rate.
Finally,these intermediates decompose to yield the product,and the catalyst is regenerated.

(B) The electronic configuration of $La$ $(Z = 57)$ is $[Xe] \, 5d^1 \, 6s^2$.
Following this,the $4f$ orbitals are filled.
For $Eu$ $(Z = 63)$,the configuration is $[Xe] \, 4f^7 \, 6s^2$,which represents a stable half-filled $f$-subshell.
For the next element,$Gd$ $(Z = 64)$,the additional electron enters the $5d$ orbital instead of the $4f$ orbital to maintain the stability of the half-filled $4f^7$ configuration.
Thus,the electronic configuration of $Gd$ $(Z = 64)$ is $[Xe] \, 4f^7 \, 5d^1 \, 6s^2$.

(D) $Zn$ has a $3d^{10} 4s^2$ electronic configuration.
Because the $3d$ orbitals are completely filled,the $d-$electrons do not participate in metallic bonding.
This results in weaker interatomic forces compared to other transition metals,leading to a lower melting point.

(B) The electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$.
In an aqueous medium,the $Cr^{3+}$ ion is the most stable oxidation state.
This is because the $Cr^{3+}$ ion has a $t_{2g}^3$ configuration,which corresponds to a half-filled $t_{2g}$ subshell,providing extra stability due to crystal field stabilization energy in octahedral complexes.

(A) The magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $A$,$\mu = 2 \sqrt{2} = \sqrt{8}$,so $n = 2$,corresponding to $Mn^{5+}$.
For $B$,$\mu = \sqrt{15}$,so $n = 3$,corresponding to $Mn^{4+}$.
For $C$,$\mu = \sqrt{35}$,so $n = 5$,corresponding to $Mn^{2+}$.
For $D$,$\mu = \sqrt{24}$,so $n = 4$,corresponding to $Mn^{3+}$.
Ionization energy $(IE)$ is directly proportional to the positive charge on the ion. Therefore,$Mn^{5+}$ has the highest ionization energy.

(A) The third ionisation enthalpy $(IE_3)$ corresponds to the energy required to remove an electron from the $M^{2+}$ ion.
The electronic configurations of $M^{2+}$ ions are:
$Ti^{2+} ([Ar] 3d^2)$
$V^{2+} ([Ar] 3d^3)$
$Cr^{2+} ([Ar] 3d^4)$
$Mn^{2+} ([Ar] 3d^5)$
Removing an electron from $Mn^{2+}$ $(3d^5)$ is difficult due to the stable half-filled $d$-orbital configuration.
Removing an electron from $Cr^{2+}$ $(3d^4)$ is relatively easier than $Mn^{2+}$,but $Cr^{2+}$ has a higher $IE_3$ than $V^{2+}$ and $Ti^{2+}$ due to the exchange energy and stability factors.
The correct order of decreasing $IE_3$ is $Mn > Cr > V > Ti$.

(D) Transition elements are defined by the presence of partially filled $d$-orbitals in their ground state or any of their oxidation states.
They are known for forming a large number of complexes due to their small size,high nuclear charge,and availability of vacant $d$-orbitals.
They exhibit variable oxidation states because the energy difference between $(n-1)d$ and $ns$ orbitals is small.
They also form organometallic compounds.
However,they do not have a fixed group oxidation state of $(+3)$; their oxidation states vary significantly across the series.

(D) Paramagnetic substances are those that contain one or more unpaired electrons.
$Cu^{+2}$ has electronic configuration $[Ar] 3d^9$,which contains $1$ unpaired electron.
$Fe^{+3}$ has electronic configuration $[Ar] 3d^5$,which contains $5$ unpaired electrons.
$Cr^{+3}$ has electronic configuration $[Ar] 3d^3$,which contains $3$ unpaired electrons.
$MnO$ contains $Mn^{+2}$ ion,which has electronic configuration $[Ar] 3d^5$. However,in the solid state,$MnO$ is an antiferromagnetic substance because the magnetic moments of the $Mn^{+2}$ ions are aligned in a way that they cancel each other out.

(B) To determine if an ion is paramagnetic,we check for the presence of unpaired electrons in its electronic configuration.
$1$. $O_2^{2-}$ (Peroxide ion): Total electrons = $8 + 8 + 2 = 18$. Configuration: $(\sigma 1s)^2, (\sigma^* 1s)^2, (\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2 = (\pi 2p_y)^2, (\pi^* 2p_x)^2 = (\pi^* 2p_y)^2$. All electrons are paired,so it is diamagnetic.
$2$. $Cr^{3+}$: Atomic number of $Cr$ is $24$. Configuration of $Cr$ is $[Ar] 3d^5 4s^1$. $Cr^{3+}$ loses $3$ electrons: $[Ar] 3d^3$. It has $3$ unpaired electrons,so it is paramagnetic.
$3$. $Na^+$: Atomic number $11$. Configuration: $1s^2 2s^2 2p^6$. All electrons are paired,so it is diamagnetic.
$4$. $Cu^+$: Atomic number $29$. Configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$. $Cu^+$ loses $1$ electron: $[Ar] 3d^{10}$. All electrons are paired,so it is diamagnetic.
Therefore,$Cr^{3+}$ is the paramagnetic ion.

(D) The element $Lr$ (Lawrencium) is a synthetic (man-made) element,which is a radioactive transuranic element with atomic number $103$.
$Ra$ (Radium),$Fr$ (Francium),and $Rn$ (Radon) are naturally occurring radioactive elements.

(D) The electronic configuration of $Cu$ $(Z=29)$ is $[Ar] 3d^{10} 4s^1$.
For $Cu^+$ $(Cu^+)$,the configuration is $[Ar] 3d^{10}$,which has no unpaired electrons.
For $Cu^{2+}$ $(Cu^{2+})$,the configuration is $[Ar] 3d^9$,which has one unpaired electron.
Therefore,the Assertion is incorrect because $Cu^+$ has no unpaired electrons,while $Cu^{2+}$ has one.
Regarding the Reason,$Cu^+$ is indeed colourless due to the absence of unpaired electrons ($d^{10}$ configuration),and $Cu^{2+}$ is blue due to $d-d$ transitions facilitated by the presence of an unpaired electron.
Since the Assertion is incorrect but the Reason is correct,the correct option is $D$.

(B) The composition of Portland cement typically includes the following major constituents:
$1$. Tricalcium silicate $(Ca_3SiO_5)$: $50-70\%$
$2$. Dicalcium silicate $(Ca_2SiO_4)$: $20-30\%$
$3$. Tricalcium aluminate $(Ca_3Al_2O_6)$: $5-10\%$
$4$. Tetracalcium aluminoferrite $(Ca_4Al_2Fe_2O_{10})$: $5-15\%$
Comparing the given options,$Ca_3SiO_5$ (tricalcium silicate) is present in the highest amount.

(A) Gallium $(Ga)$ is a soft,silvery metal.
Its melting point is $30\,^oC$.
This metal expands by $3.1\%$ when it solidifies and hence,it should not be stored in glass or metal containers.

(A) The standard reduction potential $E^o$ for $Mn^{3+}/Mn^{2+}$ $(+1.51 \ V)$ is much higher than that for $Cr^{3+}/Cr^{2+}$ $(-0.41 \ V)$.
This is because $Mn^{2+}$ has a stable $d^5$ configuration,making the reduction of $Mn^{3+}$ to $Mn^{2+}$ highly favorable.
Conversely,the third ionization energy of $Mn$ is very high because it involves removing an electron from the stable $d^5$ configuration of $Mn^{2+}$.
Since the third ionization energy is a measure of the energy required to remove an electron from the $M^{2+}$ ion,the high value for $Mn$ explains why $Mn^{3+}$ is a strong oxidizing agent.
Therefore,both Assertion and Reason are correct,and the Reason is the correct explanation for the Assertion.

(D) The colour in transition metal compounds is due to the presence of unpaired $d-$electrons,which allow for $d-d$ transitions.
$Ti^{3+} = [Ar] \, 3d^1$ (one unpaired electron).
$Fe^{3+} = [Ar] \, 3d^5$ (five unpaired electrons).
$Co^{2+} = [Ar] \, 3d^7$ (three unpaired electrons).
Since all these ions contain unpaired $d-$electrons,all the given compounds are coloured.

(B) The stability of oxidation states in $d$-block elements is often determined by the electronic configuration.
$Mn^{2+}$ has a $3d^5$ configuration,which is a half-filled stable configuration.
$Mn^{3+}$ has a $3d^4$ configuration,which is less stable than the half-filled $3d^5$ state.
Therefore,$Mn^{2+} > Mn^{3+}$ is the correct representation of stability.

(D) The electronic configuration of $Cu^{+}$ is $[Ar] 3d^{10}$. Since all $d-$orbitals are completely filled,there are no unpaired electrons,making it colourless.
The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$. It contains one unpaired electron in the $d-$orbital,which allows for $d-d$ transitions,making it coloured.

(A) The first ionization energy $(IE_1)$ generally increases with an increase in atomic number across a period due to the increase in effective nuclear charge.
However,the trend is irregular among $d-$block elements due to shielding effects and stable electronic configurations.
Comparing the given elements:
$Zn (3d^{10} 4s^2)$ has a completely filled $d-$orbital and $s-$orbital,making it very stable.
$Fe (3d^6 4s^2)$ has a relatively high $IE_1$.
$Cu (3d^{10} 4s^1)$ has a filled $d-$subshell but a single $s-$electron.
$Cr (3d^5 4s^1)$ has a half-filled $d-$subshell.
Based on experimental data for the first ionization potential,the order is $Zn > Fe > Cu > Cr$.

(D) The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.
For $Cu^{+}$ (cuprous ion),the configuration is $[Ar] 3d^{10}$,which has no unpaired electrons.
For $Cu^{2+}$ (cupric ion),the configuration is $[Ar] 3d^9$,which has one unpaired electron.
Therefore,the Assertion is incorrect because it states the opposite.
$Cu^{+}$ is colourless due to the absence of unpaired electrons,while $Cu^{2+}$ is blue in aqueous solution due to $d-d$ transitions in the presence of water ligands. Thus,the Reason is correct.
The correct option is $D$.

(C) The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] \, 3d^5 \, 4s^1$.
In this configuration,there are $5$ unpaired electrons in the $3d$ subshell and $1$ unpaired electron in the $4s$ subshell,totaling $6$ unpaired electrons.
The stability of the $Cr$ atom is due to the half-filled $d$ subshell $(3d^5)$,not the $s$ orbital.
Therefore,the Assertion is correct,but the Reason is incorrect.

(C) The assertion is correct because transition metals have partially filled $d$-orbitals,allowing them to lose electrons from both the $ns$ and $(n-1)d$ subshells.
The reason is incorrect because the energy difference between the $ns$ and $(n-1)d$ electrons is actually very small,which is precisely why both sets of electrons can participate in bond formation,leading to variable valency.

(B) The borax bead test is used to identify transition metal ions by the formation of characteristic coloured metaborates. $Al^{3+}$ ions do not form coloured metaborates,which is why the test is not suitable for $Al(III)$.
$Al_2O_3$ is indeed insoluble in water,but this property is unrelated to the failure of the borax bead test for aluminium. Therefore,both statements are correct,but the reason does not explain the assertion.

(A) The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
$1. \ Co^{3+} (3d^6)$: $n = 4$,so $\mu = \sqrt{4(4+2)} = \sqrt{24} \ B.M.$ (Matches $iv$)
$2. \ Cr^{3+} (3d^3)$: $n = 3$,so $\mu = \sqrt{3(3+2)} = \sqrt{15} \ B.M.$ (Matches $v$)
$3. \ Fe^{3+} (3d^5)$: $n = 5$,so $\mu = \sqrt{5(5+2)} = \sqrt{35} \ B.M.$ (Matches $ii$)
$4. \ Ni^{2+} (3d^8)$: $n = 2$,so $\mu = \sqrt{2(2+2)} = \sqrt{8} \ B.M.$ (Matches $i$)
Therefore,the correct sequence is $(a)-(iv), (b)-(v), (c)-(ii), (d)-(i)$.

(C) The atomic radii of $Ag$ $(144 \text{ pm})$ and $Au$ $(144 \text{ pm})$ are nearly identical.
This phenomenon occurs due to lanthanide contraction, where the poor shielding effect of $4f$ electrons leads to an increase in effective nuclear charge, which offsets the expected increase in atomic size as one moves down the group from the $4d$ to the $5d$ series.

(A) The electronic configurations of the given elements are:
$Mn (Z=25): [Ar] 3d^5 4s^2$
$Fe (Z=26): [Ar] 3d^6 4s^2$
$Co (Z=27): [Ar] 3d^7 4s^2$
$Ni (Z=28): [Ar] 3d^8 4s^2$
After the removal of two electrons ($M^{2+}$ ions):
$Mn^{2+}: [Ar] 3d^5$
$Fe^{2+}: [Ar] 3d^6$
$Co^{2+}: [Ar] 3d^7$
$Ni^{2+}: [Ar] 3d^8$
The third ionization enthalpy involves the removal of an electron from the $3d$ orbital. For $Fe^{2+}$,removing one electron results in the stable half-filled $3d^5$ configuration. Since the removal of an electron from $Fe^{2+}$ leads to a more stable state,the energy required (third ionization enthalpy) is the lowest among the given options.

(N/A) $(i)$ Lawrencium $(Lr)$ with $Z=103$ and Berkelium $(Bk)$ with $Z=97$ were named by the Lawrence Berkeley Laboratory.
$(ii)$ Seaborgium $(Sg)$ with $Z=106$ was named by Seaborg's group.

(N/A) transition element is defined as an element which has incompletely filled $d$ orbitals in its ground state or in any one of its oxidation states.
Scandium $(Z=21)$ has the electronic configuration $[Ar] 3d^1 4s^2$. Since it has an incompletely filled $3d$ orbital,it is considered a transition element.
Zinc $(Z=30)$ has the electronic configuration $[Ar] 3d^{10} 4s^2$. In its ground state and its common oxidation state $(Zn^{2+}: [Ar] 3d^{10})$,the $d$ orbitals are completely filled. Therefore,it is not considered a transition element.

(N/A) Transition elements have a large number of unpaired electrons in their $(n-1)d$ orbitals.
These unpaired electrons lead to stronger interatomic interactions and metallic bonding between the atoms.
Due to this strong metallic bonding,a large amount of energy is required to break these bonds,resulting in higher enthalpies of atomisation.

(A) Scandium ($Sc$,$Z=21$) has the electronic configuration $[Ar] 3d^1 4s^2$. It only exhibits a $+3$ oxidation state by losing its three valence electrons,and therefore does not show variable oxidation states.

(N/A) $Cr^{2+}$ is reducing because its configuration changes from $d^{4}$ to $d^{3}$,where the $d^{3}$ configuration has a stable half-filled $t_{2g}$ level.
On the other hand,the change from $Mn^{3+}$ to $Mn^{2+}$ results in the $d^{5}$ configuration,which is half-filled and possesses extra stability.

(A) The $E^{\Theta } (M^{2+}/M)$ values are not regular because they depend on the sum of the enthalpy of sublimation,the enthalpy of ionization,and the enthalpy of hydration.
$1$. The irregular variation in ionization enthalpies $(\Delta_{i} H_{1} + \Delta_{i} H_{2})$ plays a significant role in these values.
$2$. The enthalpy of sublimation is also a key factor,which is relatively much lower for elements like $Mn$ and $V$,contributing to their more negative $E^{\Theta }$ values.
$3$. The stability of the $d^5$ configuration in $Mn^{2+}$ and the high hydration energy of $Cu^{2+}$ (which makes its $E^{\Theta }$ positive) are also critical factors.