General Characteristics Questions in English

(B) The standard reduction potential $(E^o)$ for the reaction $Ce^{+4} + e^- \rightarrow Ce^{+3}$ is given as $1.74 \ V$.
Since the value of $E^o$ is highly positive,it indicates that $Ce^{+4}$ has a strong tendency to gain an electron and get reduced to $Ce^{+3}$.
Therefore,$Ce^{+4}$ acts as a strong oxidizing agent.
Even though $Ce^{+4}$ has a noble gas configuration $([Xe])$,the high positive reduction potential indicates that it is unstable in the $+4$ oxidation state and readily converts to the more stable $+3$ oxidation state.

(D) The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For a magnetic moment of $5.92 \ BM$,we have $\sqrt{n(n+2)} = 5.92$,which implies $n = 5$.
$Fe^{3+}$ $([Ar] 3d^5)$ has $5$ unpaired electrons.
$Mn^{2+}$ $([Ar] 3d^5)$ has $5$ unpaired electrons.
$Cr^{+}$ $([Ar] 3d^5)$ has $5$ unpaired electrons.
Since all the given species have $5$ unpaired electrons,they all exhibit a magnetic moment of $5.92 \ BM$.

(C) The electronic configuration of the given elements is as follows:
$Ce$ $(Z=58)$: $[Xe] 4f^1 5d^1 6s^2$. Here,the $(n-2)f$ subshell is $4f$,which contains $1$ electron.
$U$ $(Z=92)$: $[Rn] 5f^3 6d^1 7s^2$. Here,the $(n-2)f$ subshell is $5f$,which contains $3$ electrons.
$Th$ $(Z=90)$: $[Rn] 5f^0 6d^2 7s^2$. Here,the $(n-2)f$ subshell is $5f$,which contains $0$ electrons.
Therefore,$Th$ is the element where the number of electrons in the $(n-2)f$ subshell is zero.

(D) The second ionization energy $(IE_2)$ refers to the energy required to remove an electron from a unipositive ion $(M^{+})$. The stability of the resulting $M^{2+}$ ion and the electronic configuration of the $M^{+}$ ion play a crucial role.
The electronic configurations of the $M^{+}$ ions are:
$Cr^{+} (3d^5)$ - This is a stable half-filled configuration.
$Mn^{+} (3d^5 4s^1)$
$Fe^{+} (3d^6 4s^1)$
$Co^{+} (3d^7 4s^1)$
Due to the extra stability of the $3d^5$ configuration in $Cr^{+}$,removing the second electron from $Cr^{+}$ requires significantly higher energy compared to the others. The experimental values for $IE_2$ are approximately:
$Mn^{+} \approx 1509 \, kJ/mol$
$Fe^{+} \approx 1563 \, kJ/mol$
$Co^{+} \approx 1644 \, kJ/mol$
$Cr^{+} \approx 1591 \, kJ/mol$
Comparing these values,the order is $Mn^{+} < Fe^{+} < Cr^{+} < Co^{+}$.

(A) Most compounds of transition metals are coloured due to $d-d$ transitions.
The aqueous colours of the given ions are:
$Ni^{+2} = \text{Green}$
$V^{+4} = \text{Blue}$
$Ti^{+3} = \text{Purple}$
$Co^{+2} = \text{Pink}$
$Fe^{+3} = \text{Yellow}$
$Cu^{+2} = \text{Blue}$
Therefore,the pair of ions that exhibit a blue colour in an aqueous state is $V^{+4}$ and $Cu^{+2}$.

(C) The atomic number of Chromium $(Cr)$ is $24$.
Its ground state electronic configuration is $[Ar] 3d^5 4s^1$.
To form the chromic ion $(Cr^{3+})$,three electrons are removed: one from the $4s$ orbital and two from the $3d$ orbital.
The electronic configuration of $Cr^{3+}$ becomes $[Ar] 3d^3$.
In the $3d^3$ configuration,the three electrons occupy three separate $d$-orbitals singly according to Hund's rule.
Therefore,the number of unpaired electrons in $Cr^{3+}$ is $3$.

(C) To find the number of unpaired electrons,we write the electronic configuration for each ion:
$1$. $Fe^{2+}$ $(Z=26)$: $[Ar] 3d^6$. In $3d^6$,there are $4$ unpaired electrons.
$2$. $Co^{2+}$ $(Z=27)$: $[Ar] 3d^7$. In $3d^7$,there are $3$ unpaired electrons.
$3$. $Mn^{2+}$ $(Z=25)$: $[Ar] 3d^5$. In $3d^5$,there are $5$ unpaired electrons.
$4$. $Cr^{2+}$ $(Z=24)$: $[Ar] 3d^4$. In $3d^4$,there are $4$ unpaired electrons.
Comparing the number of unpaired electrons: $Mn^{2+}$ has the maximum number of unpaired electrons,which is $5$.

(B) Paramagnetic substances are those that contain one or more unpaired electrons.
$1$. $Zn^{2+}$ $(Z=30)$: Electronic configuration is $[Ar] 3d^{10}$. It has $0$ unpaired electrons (diamagnetic).
$2$. $Ni^{2+}$ $(Z=28)$: Electronic configuration is $[Ar] 3d^8$. It has $2$ unpaired electrons (paramagnetic).
$3$. $Cu^{+}$ $(Z=29)$: Electronic configuration is $[Ar] 3d^{10}$. It has $0$ unpaired electrons (diamagnetic).
Therefore,$Ni^{2+}$ is paramagnetic.

(C) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
For $V^{3+}$ $([Ar] 3d^2)$: $n = 2$,$\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \text{ BM}$.
For $Mn^{3+}$ $([Ar] 3d^4)$: $n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \text{ BM}$.
For $Fe^{3+}$ $([Ar] 3d^5)$: $n = 5$,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92 \text{ BM}$.
For $Cu^{2+}$ $([Ar] 3d^9)$: $n = 1$,$\mu = \sqrt{1(3)} = \sqrt{3} \approx 1.73 \text{ BM}$.
Thus,$Fe^{3+}$ has the maximum number of unpaired electrons $(n=5)$,resulting in the maximum magnetic moment.

(C) Ions are diamagnetic if all the electrons are paired in their electronic configuration.
Electronic configuration of $Zn^{2+}$ is $[Ar] 3d^{10}$.
Since all $10$ electrons in the $3d$ subshell are paired,$Zn^{2+}$ is diamagnetic.
In contrast,$Cr^{3+} ([Ar] 3d^3)$,$Fe^{3+} ([Ar] 3d^5)$,and $Mn^{2+} ([Ar] 3d^5)$ all contain unpaired electrons and are therefore paramagnetic.

(A, D) An ion is coloured if it has unpaired electrons in its $d$-orbitals,allowing for $d-d$ transitions.
$1$. $Fe^{2+}$ $([Ar] 3d^6)$: Has $4$ unpaired electrons. It is coloured.
$2$. $Cu^{+}$ $([Ar] 3d^{10})$: Has no unpaired electrons. It is colourless.
$3$. $Sc^{3+}$ $([Ar] 3d^0)$: Has no unpaired electrons. It is colourless.
$4$. $Mn^{2+}$ $([Ar] 3d^5)$: Has $5$ unpaired electrons. It is coloured.
Note: Both $Fe^{2+}$ and $Mn^{2+}$ are coloured. However,in many standard contexts,$Mn^{2+}$ is specifically noted for its distinct pale pink colour,and $Fe^{2+}$ is green. Given the options,both $A$ and $D$ are technically correct. If a single choice is required,$Mn^{2+}$ is often the standard answer for $d^5$ systems.

(B) The central metal ions are $Mn^{2+}$,$Fe^{2+}$,$Ni^{2+}$,and $Cu^{2+}$.
Electronic configurations are $3d^5$,$3d^6$,$3d^8$,and $3d^9$ respectively.
The number of unpaired electrons $(n)$ for each is: $n=5$ for $Mn^{2+}$,$n=4$ for $Fe^{2+}$,$n=2$ for $Ni^{2+}$,and $n=1$ for $Cu^{2+}$.
Magnetic moment $\mu$ is directly proportional to the number of unpaired electrons $(n)$.
Therefore,the decreasing order of magnetism is $I > II > III > IV$.
The increasing order of magnetism is $IV < III < II < I$.

(B) compound is coloured if it contains transition metal ions with partially filled $d$-orbitals (i.e.,unpaired electrons) due to $d-d$ transitions.
$Ag_2SO_4$ contains $Ag^+$ $(4d^{10})$,which has no unpaired electrons,making it colourless.
$CuF_2$ contains $Cu^{2+}$ $(3d^9)$,which has $1$ unpaired electron,making it coloured.
$MgF_2$ contains $Mg^{2+}$ $(2p^6)$,which has no unpaired electrons,making it colourless.
$CuCl$ contains $Cu^+$ $(3d^{10})$,which has no unpaired electrons,making it colourless.
Therefore,the correct option is $B$.

(D) Interstitial compounds are formed when small atoms like $H$,$C$,$N$,or $B$ are trapped inside the crystal lattice of transition metals.
These compounds have the following properties:
$1$. They are extremely hard.
$2$. They have high melting points,higher than those of pure metals.
$3$. They retain metallic conductivity.
$4$. They become chemically inert.
$5$. Due to the presence of interstitial atoms,the mobility of metal layers is restricted,which makes the metal less ductile and less malleable.
Therefore,all the given options are correct.

(D) $1$. The stability of the $+3$ oxidation state generally decreases across the $3d$ series as the nuclear charge increases,making it harder to remove the third electron.
$2$. In the $3d$ series,from $Cr$ $(Z=24)$ to $Cu$ $(Z=29)$,the atomic radii remain almost constant because the increase in nuclear charge is balanced by the shielding effect of $d$-electrons.
$3$. The stability of the $+2$ oxidation state increases across the period because the ionization energy required to remove two electrons increases,but the hydration energy or lattice energy compensates for this,making the $+2$ state more common for later elements like $Ni^{2+}$ and $Cu^{2+}$.

(D) The ionisation energy of transition elements is higher than that of $s-$ block elements due to the smaller atomic size and higher effective nuclear charge.
However,it is lower than that of $p-$ block elements because $p-$ block elements have higher effective nuclear charge and smaller atomic radii compared to transition elements.
Therefore,the ionisation energies of transition elements are intermediate between $s-$ block and $p-$ block elements.
Thus,the correct statement is that they are more than $s-$ block elements and less than $p-$ block elements.

(A) Transition elements exhibit strong metallic bonding due to the presence of unpaired electrons in their $(n-1)d$ orbitals.
These unpaired electrons participate in covalent bonding in addition to the metallic bonding,which results in high enthalpies of atomization and makes them more metallic and harder compared to representative elements.

(D) An amalgam is an alloy of mercury $(Hg)$ with another metal.
Most metals form amalgams with mercury,but some metals like iron $(Fe)$,platinum $(Pt)$,tungsten $(W)$,and tantalum $(Ta)$ do not form amalgams.
Therefore,both iron $(Fe)$ and platinum $(Pt)$ do not form amalgams with mercury.
Thus,the correct option is $(d)$.

(D) Transition metals exhibit several characteristic properties:
$1$. Due to $d-d$ transitions,compounds containing ions of transition elements are usually coloured.
$2$. Transition metals exhibit variable oxidation states,and $+2$ is often a very common state,though $+3$ is also common. However,the statement 'All of the above' is generally accepted in the context of these properties being characteristic of transition metals.
$3$. They have a high charge-to-size ratio and vacant $d$-orbitals,which allow them to easily form complexes.

(D) The highest oxidation state shown by any transition metal is $+8$,which is exhibited by $Ru$ and $Os$ in their respective compounds like $RuO_4$ and $OsO_4$.
Therefore,both options $(b)$ and $(c)$ are correct,making option $(d)$ the most appropriate choice.

(B) The electronic configuration of $Zn$ $(Z=30)$ is $[Ar] 3d^{10} 4s^2$.
In transition elements,variable oxidation states arise due to the participation of both $(n-1)d$ and $ns$ electrons in bonding.
Since $Zn$ has a completely filled $3d$ subshell $(3d^{10})$,the $d$-electrons are stable and do not participate in chemical bonding.
Therefore,$Zn$ only exhibits a $+2$ oxidation state by losing its two $4s$ electrons,failing to show variable oxidation states.

(B) The only $d-$block element that exists as a liquid at room temperature is Mercury $(Hg)$.
Mercury exhibits a high specific heat capacity compared to other metals.
It is less reactive than hydrogen,meaning it does not displace hydrogen from dilute acids.
Mercury$(II)$ chloride $(HgCl_2)$ is a covalent compound that sublimes (is volatile) upon heating.

(D) $Cu$,$Ag$,and $Au$ are group $11$ elements belonging to the $d$-block.
They possess partially filled $d$-orbitals in their common oxidation states,which is the characteristic feature of transition elements.
Group $11$ metals are also known as coinage metals due to their historical usage in minting coins.

(D) Interstitial compounds are those in which small atoms like $H, C, N$ or $B$ are trapped inside the crystal lattice of transition metals.
Transition metals like $Fe, Co, Ni$ and others have large atomic sizes and empty interstitial sites in their crystal structures,which allow them to form these compounds.
Since $Fe, Co$,and $Ni$ are all transition metals capable of forming such structures,the correct answer is all of the above.

(A) $Molybdenum$ $(Mo)$ is a transition metal with a high melting point and good thermal conductivity.
It is commonly used as the target material in $X$-ray tubes because it can withstand the intense heat generated by the electron bombardment required to produce $X$-rays.

(D) Transition elements exhibit catalytic activity due to several factors:
$1$. They show $variable \ oxidation \ states$ and can form complexes,which allow them to provide alternative pathways with lower activation energy.
$2$. They provide a large $surface \ area$ for the adsorption of reactants,which increases the concentration of reactants on the surface and facilitates the reaction.
$3$. Therefore,all the given factors contribute to their catalytic properties.

(A) The higher oxidation states of transition elements are found in combination with $F$ and $O$ because $F$ and $O$ are the most electronegative elements.
For example,$Os$ shows the highest oxidation state of $+8$ in $OsO_4$.
Similarly,$Mn$ shows the highest oxidation state of $+7$ in $KMnO_4$.

(D) According to Fajan's rule,the covalent character of a compound increases with an increase in the oxidation state (charge) of the metal cation.
Given that the oxidation states $x$ and $y$ differ by two units,let us assume $x < y$.
Since $y > x$,the compound with oxidation state $y$ will have more covalent character,and the compound with oxidation state $x$ will have more ionic character.
Therefore,if $x < y$,compounds in oxidation state $x$ are more ionic,and compounds in oxidation state $y$ are more covalent.
Thus,both statements $(b)$ and $(c)$ are correct.

(D) The magnetic moment (spin only) is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$Sc^{3+}$ $([Ar] 3d^0)$: $n = 0$,so $\mu = 0 \ BM$.
$Ti^{3+}$ $([Ar] 3d^1)$: $n = 1$,so $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
$Cu^{2+}$ $([Ar] 3d^9)$: $n = 1$,so $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
Since both $Ti^{3+}$ and $Cu^{2+}$ have $n=1$,they possess a definite non-zero magnetic moment.

(D) An ion is diamagnetic if all its electrons are paired in its electronic configuration.
$Ce^{+4}$ has the electronic configuration $[Xe] 4f^0 5d^0 6s^0$. Since all orbitals are empty,there are no unpaired electrons,making it diamagnetic.
$Pa^{+4}$ has the configuration $[Rn] 5f^1$,which has one unpaired electron.
$Yb^{+3}$ has the configuration $[Xe] 4f^{13}$,which has one unpaired electron.
$U^{+3}$ has the configuration $[Rn] 5f^3$,which has three unpaired electrons.

(C) Enthalpy of atomisation is the energy required to break all the bonds in one mole of a substance to obtain atoms in the gaseous phase.
Transition elements exhibit high enthalpies of atomisation due to strong metallic bonding resulting from the presence of unpaired electrons in their $(n-1)d$ orbitals.
As we move across a $3d$ series,the number of unpaired electrons increases up to $d^5$ configuration,leading to stronger metallic bonding and higher enthalpy of atomisation.
Among the given options,$V$ $(3d^3 4s^2)$ has more unpaired electrons than $Sc$ $(3d^1 4s^2)$ and $Mn$ ($3d^5 4s^2$ is an exception due to stable half-filled configuration,but $V$ generally shows higher values in specific contexts or comparisons),while $Zn$ $(3d^{10} 4s^2)$ has no unpaired electrons,resulting in the lowest enthalpy of atomisation.
Thus,$V$ is expected to have a high enthalpy of atomisation.

(C) The given electronic configuration is $(n - 1)s^2 (n - 1)p^6 (n - 1)d^x ns^2$.
Given $n = 4$ and $x = 5$.
Substituting these values,we get the configuration: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2$.
The total number of electrons is $2 + 2 + 6 + 2 + 6 + 5 + 2 = 25$.
Since the atom is neutral,the number of protons equals the number of electrons,which is $25$.

(C) The electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^3 4s^2$,which can be rearranged as $[Ar] 3d^3 4s^2$.
For $d$-block elements,the group number is determined by the sum of $(n-1)d$ electrons and $ns$ electrons.
Here,the number of electrons in the $(n-1)d$ subshell is $3$ and in the $ns$ subshell is $2$.
Total electrons = $3 + 2 = 5$.
Therefore,the element belongs to the $5^{th}$ group.

(C) Ionisation energy depends on the electronic configuration and the stability of the resulting ion.
For $Mn^{2+}$,$Ni^{2+}$,and $Ca^{2+}$:
$Mn^{2+} = [Ar] 3d^5$ (Half-filled stable configuration,high $IE$).
$Ni^{2+} = [Ar] 3d^8$.
$Ca^{2+} = [Ar]$ (Noble gas configuration,but removing an electron from a stable shell requires high energy,however,in the context of transition metals,$Mn^{2+}$ has a stable $d^5$ configuration,making its $IE$ higher than $Ni^{2+}$ and $Ca^{2+}$).
Comparing the options:
Option $C$ is correct because $Mn^{2+}$ has a stable $d^5$ configuration,leading to a higher ionisation energy compared to $Ni^{2+}$ and $Ca^{2+}$.

(C) The atomic number of $Uub$ (Copernicium,$Cn$) is $112$.
Its electronic configuration is $[Rn] \ 5f^{14} \ 6d^{10} \ 7s^2$.
Since the valence electrons enter the $d$-orbital,it belongs to the $d$-block.
The group number for $d$-block elements is calculated as $(n-1)d + ns$ electrons.
Group number $= 10 + 2 = 12$.
The highest principal quantum number $(n)$ is $7$,so the period is $7$.
Therefore,the correct set is $12, 7$.

(D) Option $A$ is incorrect because the conversion of chromate to dichromate occurs in acidic medium $(pH < 7)$,not basic. The correct reaction is $2CrO_4^{2-} + 2H^+ \to Cr_2O_7^{2-} + H_2O$.
Option $B$ is incorrect because $I^-$ is oxidized to $I_2$ by $KMnO_4$ in acidic medium,not $IO_3^-$.
Option $C$ is incorrect because Misch metal is an alloy consisting of a lanthanoid metal (typically $Ce$ $\approx 50\%$,$La$ $\approx 25\%$,$Nd$ $\approx 18\%$,$Pr$ $\approx 5\%$) and iron ($Fe$ $\approx 5\%$).
Option $D$ is correct because $Ce$ and $Tb$ are known to exhibit the $+2$ oxidation state in addition to their common $+3$ state.

(C) The electronic configuration of $Cr^{3+}$ is $[Ar] 3d^3$.
In an octahedral field,this corresponds to a $t_{2g}^3 e_g^0$ configuration.
This configuration represents a half-filled $t_{2g}$ subshell,which provides extra stability due to exchange energy.
Therefore,$Cr^{3+}$ is the most stable cation among the given options in an aqueous medium.

(B) The electronic configuration of Scandium $(Sc)$ is $[Ar] 3d^1 4s^2$. It typically exhibits a $+3$ oxidation state.
Zinc $(Zn)$ has the electronic configuration $[Ar] 3d^{10} 4s^2$. It can only lose two electrons from the $4s$ orbital to show a $+2$ oxidation state.
Both $Sc$ and $Zn$ do not show variable valency because they do not have partially filled $d$ orbitals that allow for multiple oxidation states in the same way as other transition metals.
Therefore,$Sc$ and $Zn$ are the elements in the $3d$ series that do not exhibit variable valency.

(B) Let us analyze the given electronic configurations:
$(i) \; [Xe] \; 6s^1$: This is Cesium $(Cs)$,an alkali metal. It has a low ionization enthalpy and is a strong reducing agent. This statement is true.
$(ii) \; [Xe] \; 4f^{14}5d^16s^2$: This is Lutetium $(Lu)$. Although it has a $5d^1$ electron,it is the last element of the lanthanoid series ($4f$ block). It is considered an $f-$block element,not a $d-$block element. This statement is false.
$(iii) \; [Ne] \; 3s^23p^5$: This is Chlorine $(Cl)$,a halogen. It has a high negative electron gain enthalpy. This statement is true.
$(iv) \; [Ar] \; 3d^74s^2$: This is Cobalt $(Co)$,a transition metal. Transition metals typically show variable oxidation states. This statement is true.
Therefore,the incorrect statement is $(ii)$.

(C) In the $3d$ transition series,the atomic radii generally decrease from $Sc$ to $Mn$ due to an increase in effective nuclear charge.
However,from $Fe$ to $Cu$,the atomic radii remain almost constant because the increase in nuclear charge is balanced by the shielding effect of $d$-electrons.
For $Zn$,the atomic radius increases due to inter-electronic repulsions among $d$-electrons,which causes the electron cloud to expand.
Comparing $Cr$ $(3d^5 4s^1)$ and $Mn$ $(3d^5 4s^2)$,the atomic radius of $Cr$ is larger than $Mn$ $(Cr > Mn)$.
Comparing $Cu$ $(3d^{10} 4s^1)$ and $Zn$ $(3d^{10} 4s^2)$,the atomic radius of $Zn$ is larger than $Cu$ $(Cu < Zn)$.
Therefore,both statements $(A)$ and $(B)$ are correct.

(C) Transition metals are characterized by the following properties:
$I$: They exhibit variable oxidation states due to the participation of $(n-1)d$ and $ns$ electrons in bonding.
$II$: They form coloured ions in solution due to $d-d$ transitions.
$IV$: They act as effective catalysts due to their ability to adopt multiple oxidation states and provide a large surface area for adsorption.
However,statement $III$ is incorrect because most transition metals do not burn vigorously in oxygen; they typically form stable oxide layers on their surfaces (passivation) or react slowly under standard conditions.
Therefore,statements $I, II,$ and $IV$ are correct.

(B) For an ion to be paramagnetic and coloured,it must have unpaired electrons in its $4f$ orbital.
$1$. $Ce^{+4}$ $([Xe] 4f^0)$: Diamagnetic and colourless.
$2$. $Yb^{+2}$ $([Xe] 4f^{14})$: Diamagnetic and colourless.
$3$. $Gd^{+3}$ $([Xe] 4f^7)$: Paramagnetic and coloured.
$4$. $Nd^{+3}$ $([Xe] 4f^3)$: Paramagnetic and coloured.
$5$. $La^{+3}$ $([Xe] 4f^0)$: Diamagnetic and colourless.
$6$. $Lu^{+3}$ $([Xe] 4f^{14})$: Diamagnetic and colourless.
$7$. $Lr^{+3}$ $([Rn] 5f^{14})$: Diamagnetic and colourless.
Therefore,$Gd^{+3}$ and $Nd^{+3}$ are both paramagnetic and coloured.

(A) The lanthanoid contraction is primarily due to the poor shielding effect of $4f$ electrons.
Because the $4f$ orbitals have a diffuse shape,they are ineffective at shielding the outer electrons from the increasing nuclear charge as we move across the lanthanoid series.
Consequently,the effective nuclear charge increases,causing the atomic and ionic radii to decrease steadily.

(B) Let us analyze each option:
$1$. $Mn < Cr > Zn \, (IP)$: The first ionization potential $(IP)$ of $Cr$ $(3d^5 4s^1)$ is lower than $Mn$ $(3d^5 4s^2)$ due to the stability of the half-filled $d$-subshell in $Mn$. Thus,the order is incorrect.
$2$. $La > Sc < Zn \, (size)$: Atomic size increases down a group. $Sc$ $(3d)$ is smaller than $La$ $(5d)$. $Zn$ is smaller than $Sc$ due to the increase in effective nuclear charge across the period. The order $La > Sc < Zn$ is correct.
$3$. $F > P < N \, (IP)$: $N$ $(2p^3)$ has a higher $IP$ than $F$ $(2p^5)$ due to the stable half-filled configuration. The order is incorrect.
$4$. $Cl < F > Br \, (EA)$: The electron affinity $(EA)$ of $Cl$ is higher than $F$ due to inter-electronic repulsion in the small $2p$ orbital of $F$. Thus,$Cl > F$. The order is incorrect.

(D) Transition elements are characterized by the presence of partially filled $d$-orbitals.
Due to their small size,high nuclear charge,and availability of $d$-orbitals,they exhibit variable oxidation states and have a high tendency to form complex compounds.
They also form a variety of organometallic compounds.
However,transition elements generally show a significant degree of covalent character in their compounds due to high charge density and polarizing power,rather than having a maximum tendency to form purely ionic compounds.
Therefore,the statement that they have a maximum tendency to form ionic compounds is not a characteristic feature.

(C) The atomic number of Gadolinium $(Gd)$ is $64$.
Following the Aufbau principle and the stability of half-filled orbitals,the $4f$ subshell is half-filled at Europium ($Eu$,$Z = 63$) with the configuration $[Xe] 4f^7 6s^2$.
For the next element,Gadolinium $(Z = 64)$,the $4f$ subshell is already stable (half-filled),so the next electron enters the $5d$ orbital instead of the $4f$ orbital.
Therefore,the electronic configuration of $Gd$ is $[Xe] 4f^7 5d^1 6s^2$.

(C) For transition elements,the group number is determined by the sum of $(n-1)d$ and $ns$ electrons.
Given configuration: $4d^3$ $5s^2$.
Number of electrons in $(n-1)d$ orbital = $3$.
Number of electrons in $ns$ orbital = $2$.
Total electrons = $3 + 2 = 5$.
Therefore,the element belongs to group $V$ $B$.

(C) Zirconium $(Zr)$ belongs to the $4d$ transition series and Hafnium $(Hf)$ belongs to the $5d$ transition series.
They exhibit similar physical and chemical properties primarily because they possess similar atomic radii.
This similarity in atomic radii is a direct consequence of the lanthanoid contraction,which occurs due to the poor shielding effect of $4f$ electrons in the $5d$ series elements.

(D) $4f$ and $5f$ orbitals are not equally shielded.
$5f$ orbitals provide poorer shielding than $4f$ orbitals because $5f$ orbitals are more diffused and extend further from the nucleus.
Thus,the statement that $4f$ and $5f$ orbitals are equally shielded is incorrect.

(B) The standard electrode potential ($E^o$ $M^{2+}/M$) values for the first transition series generally become less negative as we move from left to right,with some exceptions.
$Mn^{2+}$ has a stable $d^5$ configuration,which makes its reduction potential more negative than expected.
The electronic configurations of the $M^{2+}$ ions are:
$Cr^{2+}$: $[Ar] 3d^4$
$Mn^{2+}$: $[Ar] 3d^5$ (half-filled,extra stable)
$Fe^{2+}$: $[Ar] 3d^6$
$Co^{2+}$: $[Ar] 3d^7$
Due to the extra stability of the $Mn^{2+}$ ion,the $E^o$ value for $Mn^{2+}/Mn$ is more negative than that of $Cr^{2+}/Cr$. The overall order of the magnitude of negative values is $Mn > Cr > Fe > Co$.