Mix Examples - Classification of Elements and Periodicity in Properties Questions in English

(B) In the periodic table,as we move from left to right across a period,the effective nuclear charge increases and the atomic size decreases.
$1$. Ionization Potential $(IP)$ generally increases from $B$ to $O$.
$2$. Electronegativity $(EN)$ increases from $B$ to $O$ $(B=2.0, C=2.5, N=3.0, O=3.5)$.
$3$. Atomic size decreases from $B$ to $O$,so the order $B < C < N < O$ is incorrect for size.
$4$. Electron Affinity $(EA)$ does not follow a strictly increasing trend from $B$ to $O$ due to stable configurations (e.g.,$N$ has a half-filled $p$-orbital).
Both $IP$ and $EN$ follow the increasing trend $B < C < N < O$.

(C) The correct order of bond energy for halogens is $Cl_2 > Br_2 > F_2 > I_2$. The bond dissociation energy of $F_2$ is lower than $Cl_2$ and $Br_2$ due to high inter-electronic repulsion between the lone pairs of the small $F$ atoms. Therefore,the statement $F_2 > Cl_2 > Br_2 > I_2$ is incorrect.

$(B)$ $Be$ and $Mg$ belong to Group $2$, which are alkaline earth metals. This statement is correct.
$(b)$ $K^{+}$ ($18$ electrons) and $Ca^{2+}$ ($18$ electrons) are isoelectronic. For isoelectronic species, the ionic radius decreases as the nuclear charge $(Z)$ increases. Since $Z$ of $Ca$ $(20)$ is greater than $K$ $(19)$, $Ca^{2+}$ has a smaller radius than $K^{+}$. Thus, $K^{+}$ has a larger radius than $Ca^{2+}$. This statement is correct.
$(c)$ Transition elements are defined as elements having partially filled $d$-orbitals in their ground state or any common oxidation state. Elements like $Zn$, $Cd$, and $Hg$ have fully filled $d^{10}$ configurations and are not considered transition elements, even though they are $d$-block elements. This statement is incorrect.
$(d)$ The ionic radius of $H^{-}$ is approximately $208 \ pm$, while the ionic radius of $F^{-}$ is approximately $133 \ pm$. Thus, $H^{-}$ is larger than $F^{-}$. This statement is correct.
Therefore, only statement $(c)$ is incorrect.

(B) For element $W$,the jump in $IE$ occurs between $IE_3$ and $IE_4$ ($24.9 \ eV$ to $79.8 \ eV$),indicating that $W$ has $3$ valence electrons. Thus,$W$ exhibits a valency of $+3$.
For element $X$,the jump in $IE$ occurs between $IE_2$ and $IE_3$ ($14.8 \ eV$ to $78.9 \ eV$),indicating that $X$ has $2$ valence electrons. Thus,$X$ exhibits a valency of $+2$.
Element $Y$ has configuration $ns^2 np^4$,requiring $2$ electrons to complete its octet,so its valency is $-2$.
Element $Z$ has configuration $ns^2 np^5$,requiring $1$ electron to complete its octet,so its valency is $-1$.
Possible compounds:
$W$ $(+3)$ and $Y$ $(-2)$ form $W_2Y_3$.
$X$ $(+2)$ and $Y$ $(-2)$ form $XY$.
$W$ $(+3)$ and $Z$ $(-1)$ form $WZ_3$.
$X$ $(+2)$ and $Z$ $(-1)$ form $XZ_2$.
Comparing with the given options: $(a) W_2Y_3$ is possible,$(b) X_2Y_3$ is not possible,$(c) WZ_2$ is not possible,$(d) XZ_2$ is possible.
Therefore,the compounds that are not possible are $(b)$ and $(c)$.

(C) $I.$ Correct. An anion is formed by the gain of electrons,which increases electron-electron repulsion and decreases the effective nuclear charge per electron,leading to a larger ionic radius compared to the parent atom.
$II.$ Correct. Across a period,the atomic size decreases due to an increase in effective nuclear charge,making it harder to remove an electron,thus increasing the ionization energy.
$III.$ Incorrect. Electronegativity is defined as the tendency of an atom in a chemical compound to attract a shared pair of electrons,not an isolated atom.

(A) The diagonal relationship between $Be$ and $Al$ arises due to their similar ionic potential (charge/size ratio) and similar electronegativity values.
The ionization energy of $Be$ $(9.32 \ eV)$ is very close to that of $Al$ ($5.98 \ eV$ for first,but the overall properties are governed by charge density).
Actually,the similarity in their charge-to-size ratio (ionic potential) is the primary reason for their diagonal relationship,which is directly related to their ionization energies and electronegativities.
Therefore,both the assertion and the reason are correct,and the reason explains the assertion.

(C) Electron gain enthalpy is the measure of the tendency of an isolated gaseous atom to accept an electron to form an anion,and it is an energy value measured in $kJ \ mol^{-1}$.
Electronegativity is the measure of the tendency of an atom in a chemical compound to attract a shared pair of electrons towards itself,and it is a relative,dimensionless value.

(D) The electronic configuration of an element having $5$ electrons in its outermost $p$-subshell is $ns^{2} np^{5}$. This corresponds to the halogen group $(Group \ 17)$. Examples include $F, Cl, Br, I$.
$(b)$ An element with $2$ valence electrons $(ns^{2})$ tends to lose two electrons to attain a stable noble gas configuration. This is characteristic of $Group \ 2$ (alkaline earth metals) such as $Mg$ or $Ca$.
$(c)$ An element with $6$ valence electrons $(ns^{2} np^{4})$ needs $2$ more electrons to complete its octet. This is characteristic of the oxygen family $(Group \ 16)$ such as $O$ or $S$.
$(d)$ $Group \ 17$ (halogens) contains elements in different states at room temperature: $F$ and $Cl$ are gases,$Br$ is a liquid,and $I$ and $At$ are solids (non-metals/metalloids).

(C) The chemistry of an element is primarily determined by its valence electrons.
$(a)$ The principal quantum number $(n)$ determines the energy and size of the valence shell.
$(b)$ The nuclear charge $(Z)$ determines the effective nuclear charge experienced by the valence electrons.
$(d)$ The number of core electrons determines the shielding effect,which influences the effective nuclear charge.
$(c)$ Nuclear mass depends on the number of protons and neutrons in the nucleus. While it affects the physical properties like density or isotopic behavior,it does not significantly influence the electronic configuration or the chemical behavior of the valence shell.

(N/A) Generally,ionization enthalpy refers to the first ionization enthalpy.
The first ionization enthalpies of elements with atomic numbers up to $60$ are plotted in the graph.
The periodicity of the graph is quite striking.
Maxima are observed at the noble gases,which have closed electron shells and very stable electron configurations.
Conversely,minima occur at the alkali metals,and their low ionization enthalpies can be correlated with their high reactivity.
Within a period,as we move from left to right,the ionization enthalpy generally increases.
Within a group,as we move from top to bottom,the ionization enthalpy generally decreases.

(N/A)

Electron gain enthalpy	Electronegativity
$1$. It can be measured.	$1$. It cannot be measured.
$2$. The change in enthalpy when $1$ electron is added to a gaseous atom is known as electron gain enthalpy.	$2$. The capacity to attract a shared pair of electrons in a covalent bond is known as electronegativity.
$3$. It represents the enthalpy change of the reaction $X_{(g)} + e^{-} \rightarrow X^{-}_{(g)}$.	$3$. Its value is relative and denoted by scales like Pauling's scale (electronegativity of fluorine is $4.0$).
$4$. It indicates the energy released or absorbed when an electron is added.	$4$. It indicates the relative tendency of an atom to attract shared electrons.
$5$. Its value is a property of an isolated gaseous atom.	$5$. Electronegativity is not constant; it depends on the chemical environment and the atom it is bonded to.

(N/A) The trends in periodic properties are as follows:
$1$. Across a period (left to right):
- Atomic radius decreases.
- Ionization enthalpy,electron gain enthalpy,and electronegativity generally increase.
- Non-metallic character increases,while metallic character decreases.
$2$. Down a group (top to bottom):
- Atomic radius increases.
- Ionization enthalpy,electron gain enthalpy,and electronegativity generally decrease.
- Metallic character increases,while non-metallic character decreases.

(N/A) The elements of the second period ($Z=3$ to $10$) and their corresponding first ionisation enthalpies $(IE_1)$ in $kcal \, mol^{-1}$ are as follows:
$1$. $Li$ $(Z=3)$: $520$
$2$. $Be$ $(Z=4)$: $899$
$3$. $B$ $(Z=5)$: $801$
$4$. $C$ $(Z=6)$: $1086$
$5$. $N$ $(Z=7)$: $1402$
$6$. $O$ $(Z=8)$: $1314$
$7$. $F$ $(Z=9)$: $1681$
$8$. $Ne$ $(Z=10)$: $2080$
General Trend: As we move from left to right across a period,the atomic radius decreases and the effective nuclear charge increases,leading to a general increase in ionisation enthalpy.
Exceptions:
- $Be$ $(2s^2)$ has a higher $IE_1$ than $B$ $(2s^2 2p^1)$ because the $2s$ orbital is fully filled and more stable.
- $N$ $(2s^2 2p^3)$ has a higher $IE_1$ than $O$ $(2s^2 2p^4)$ because the $2p$ subshell is half-filled,providing extra stability.

(A) The elements are placed in the periodic table as follows:

$2^{nd}$ Period	$B$ (Group-$13$),$C$ (Group-$14$)
$3^{rd}$ Period	$Al$ (Group-$13$),$Si$ (Group-$14$)

$(a)$ Ionisation enthalpy increases from left to right across a period due to a decrease in atomic size and increases from bottom to top within a group. Among the given elements,$C$ is the furthest to the right and in the $2^{nd}$ period,making it have the highest first ionisation enthalpy.
$(b)$ Metallic character decreases from left to right across a period and increases from top to bottom within a group. $Al$ is in the $3^{rd}$ period and Group-$13$,which makes it the most metallic element among the four.

(N/A) The electronic configuration of ${}_{7}N$ is $1s^{2}, 2s^{2}, 2p^{3}$. Nitrogen has a stable half-filled $p$-orbital configuration,which makes the addition of an extra electron unfavorable,resulting in a positive electron gain enthalpy.
The electronic configuration of ${}_{8}O$ is $1s^{2}, 2s^{2}, 2p^{4}$. Adding an electron to oxygen leads to a more stable half-filled $p$-orbital $(2p^{3})$,making the process exothermic (negative electron gain enthalpy).
Regarding ionisation enthalpy,oxygen $(1s^{2}, 2s^{2}, 2p^{4})$ has a lower value than nitrogen $(1s^{2}, 2s^{2}, 2p^{3})$ because removing an electron from oxygen results in a stable half-filled $2p^{3}$ configuration. Conversely,nitrogen has a stable half-filled configuration,making the removal of an electron more difficult.

(N/A) The positions of the elements in the periodic table are as follows:

Period	Group $15$ and $16$
$2$nd period	$N$ (Group $15$),$O$ (Group $16$)
$3$rd period	$P$ (Group $15$),$S$ (Group $16$)

$(i)$ Ionisation enthalpy of nitrogen $(N: 1s^2, 2s^2, 2p^3)$ is greater than oxygen $(O: 1s^2, 2s^2, 2p^4)$ due to the extra stability of the exactly half-filled $2p$-orbitals. Similarly,the ionisation enthalpy of phosphorus $(P: 1s^2, 2s^2, 2p^6, 3s^2, 3p^3)$ is greater than sulphur $(S: 1s^2, 2s^2, 2p^6, 3s^2, 3p^4)$.
On moving down a group,ionisation enthalpy decreases due to an increase in atomic size. Therefore,the order of increasing first ionisation enthalpy is: $S < P < O < N$.
$(ii)$ Non-metallic character increases across a period (left to right) and decreases down a group. Therefore,the order of increasing non-metallic character is: $P < S < N < O$.

(N/A) As we move from left to right across a period,the atomic number increases,which means the nuclear charge increases while the number of shells remains the same. This leads to a stronger attraction between the nucleus and the valence electrons,causing the atomic radius to decrease and the electronegativity to increase.
$(b)$ As we move down a group,the number of shells increases,which increases the atomic radius. The valence electrons are placed at a greater distance from the nucleus,and the shielding effect of inner electrons increases. Consequently,the effective nuclear charge experienced by the valence electrons decreases,making it easier to remove an electron,which results in a decrease in ionisation enthalpy.

(C) The elements belonging to the $s$-block and $p$-block are collectively known as representative elements or main group elements.

(N/A) $(i)$ $6s^2 4f^3$: The last electron enters the $4f$ orbital,so it belongs to the $f$-block.
$(ii)$ $3s^2 3p^4$: The last electron enters the $3p$ orbital,so it belongs to the $p$-block.
$(iii)$ $3s^1$: The last electron enters the $3s$ orbital,so it belongs to the $s$-block.
$(iv)$ $3d^2 4s^2$: The last electron enters the $3d$ orbital,so it belongs to the $d$-block.

(N/A) The elements of the second period,such as $Li$,$Be$,and $B$,show similarities in properties with the elements of the third period,such as $Mg$,$Al$,and $Si$,respectively. This similarity in properties between elements placed diagonally in the periodic table is known as the $diagonal \ relationship$.

(A) The first element of each group differs in many respects from the other members of its group.
For example,the first element of group-$1$ is Lithium $(Li)$.
The first element of group-$2$ is Beryllium $(Be)$.
Similarly,the first elements of groups-$13$ to $17$ (from Boron to Fluorine) show anomalous properties compared to the rest of their group members.

(N/A) The first element of a group differs from the subsequent elements due to its small size,high electronegativity,and the absence of $d$-orbitals in its valence shell. The differences are summarized below:

First element of the group	Other elements of the group
Forms covalent compounds.	Forms ionic compounds.
Small atomic/ionic size.	Larger atomic/ionic size.
High charge/radius ratio.	Low charge/radius ratio.
High electronegativity.	Low electronegativity.
Only $4$ orbitals available in the valence shell ($s$ and $p$).	$9$ orbitals available in the valence shell ($s, p,$ and $d$).

(N/A) The first elements of a group (e.g.,$Li, Be, B, C, N, O, F$) exhibit anomalous properties compared to the rest of the elements in their respective groups due to their small size,high electronegativity,and absence of $d$-orbitals.

$1$. First element of the group	$2$. Subsequent elements of the group
They can form a maximum of $4$ bonds (e.g.,$[BF_4]^-$).	They can form more than $4$ covalent bonds (e.g.,$[Al(H_2O)_6]^{3+}$).
They form multiple bonds with themselves and other elements (e.g.,$C=C, N=N, C\equiv C, N\equiv N, C=O, C\equiv N, N=O$).	They do not typically form multiple bonds.

(N/A) $(i)$ Isoelectronic species
$(ii)$ Ionization enthalpy
$(iii)$ Shielding effect
$(iv)$ Electron gain enthalpy

(N/A) $(i)$ Pauling scale
$(ii)$ Diagonal relationship
$(iii)$ Amphoteric or neutral

(A) $(i)$ True. Higher ionization enthalpy implies stronger nuclear attraction,which is often associated with a lower screening effect.
$(ii)$ True. $Be$ $(1s^2 2s^2)$ has a fully filled $2s$ orbital,which is more stable than the $2p^1$ configuration of $B$.
$(iii)$ False. The shielding effect remains approximately constant as we move across a period because electrons are added to the same shell.
$(iv)$ True. The order $B < Be < O < N$ is correct due to the stability of fully filled and half-filled orbitals.

(A) $(i)$ True: Across a period,the effective nuclear charge increases,which pulls the electrons closer,making it harder to remove an electron,thus increasing ionization enthalpy.
$(ii)$ True: Metallic character is associated with the ability to lose electrons,which correlates with larger atomic radii as one moves down a group.
$(iii)$ True: $Mn$ has the electronic configuration $[Ar] 3d^5 4s^2$. The principal quantum number $n = 4$ indicates it belongs to the $4^{th}$ period.

(A) $(1-D), (2-A), (3-B), (4-C)$
$1$. $He$ (Helium) is a noble gas with a stable electronic configuration,resulting in a very high ionization enthalpy $(D)$.
$2$. $Cl$ (Chlorine) is a halogen with a high tendency to gain an electron,resulting in a high electron gain enthalpy $(A)$.
$3$. $Ca$ (Calcium) is an alkaline earth metal,which is electropositive in nature $(B)$.
$4$. $Li$ (Lithium) is an alkali metal and acts as a strong reducing agent $(C)$.

(A) $(1)$ Chlorine $(Cl)$ is a highly reactive non-metal with a very high negative electron gain enthalpy.
$(2)$ Sodium $(Na)$ is a soft alkali metal with a low first ionization enthalpy.
$(3)$ Antimony $(Sb)$ is a metalloid that forms oxides of the type ${M_2}{O_3}$.
$(4)$ Helium $(He)$ is a noble gas and is chemically inert.
Therefore,the correct matching is $(1-D, 2-A, 3-B, 4-C)$.

(A) $(i)$ The most reactive non-metal has a high negative electron gain enthalpy. Element $B$ has $\Delta_{eg}H = -328 \ kJ/mol$,which is the most negative.
$(ii)$ The most reactive metal has the lowest first ionization enthalpy $(\Delta_{i}H_1)$. Element $A$ has $\Delta_{i}H_1 = 419 \ kJ/mol$,which is the lowest.
$(iii)$ The least reactive element is a noble gas,which has very high ionization enthalpies and a positive electron gain enthalpy. Element $D$ has $\Delta_{i}H_1 = 2372 \ kJ/mol$ and $\Delta_{eg}H = +48 \ kJ/mol$.
$(iv)$ Metals forming dihalides are alkaline earth metals. Element $C$ has $\Delta_{i}H_1 = 738 \ kJ/mol$ and $\Delta_{i}H_2 = 1451 \ kJ/mol$,which is characteristic of group $2$ elements.

(N/A) The main properties associated with individual particles of matter are:
$(i)$ Atomic size
$(ii)$ Ionization enthalpy
$(iii)$ Electron charge density
$(iv)$ Molecular shape
$(v)$ Molecular polarity,etc.

(D) Across a period,the effective nuclear charge $(Z_{eff})$ increases,which leads to a decrease in the atomic radius.
Conversely,properties like ionization enthalpy,electron gain enthalpy,and electronegativity increase across a period due to the stronger attraction between the nucleus and the valence electrons.
Therefore,atomic radius shows an opposite trend compared to the other three properties.

(D) The pairs $Li-Mg$,$B-Si$,and $Be-Al$ exhibit a diagonal relationship in the periodic table.
$Li$ and $Na$ belong to the same group $(Group \ 1)$ and do not show a diagonal relationship,as they are placed vertically relative to each other.

(B) The element $E$ with configuration $4s^2 4p^1$ is Gallium $(Ga)$,which belongs to period $4$ and group $13$.
Diagonal relationship exists between elements of period $2$ and $3$,and period $3$ and $4$. However,the question asks for an element in period $5$ placed diagonally to $E$ $(Ga)$.
In the periodic table,the element placed diagonally to $Ga$ $(Z=31)$ in the next period (period $5$) is Tin ($Sn$,$Z=50$).
$Sn$ is a group $14$ element in period $5$.
The electronic configuration of $Sn$ $(Z=50)$ is $[Kr] 4d^{10} 5s^2 5p^2$.

(D) To determine the block of an element from its atomic number,we look at the last orbital filled:
$1$. Atomic number $37$ is Rubidium $(Rb)$,which belongs to the $s$-block (Group $1$). Thus,$A-IV$.
$2$. Atomic number $78$ is Platinum $(Pt)$,which is a transition metal belonging to the $d$-block. Thus,$B-II$.
$3$. Atomic number $52$ is Tellurium $(Te)$,which belongs to the $p$-block (Group $16$). Thus,$C-I$.
$4$. Atomic number $65$ is Terbium $(Tb)$,which is a lanthanoid belonging to the $f$-block. Thus,$D-III$.
Therefore,the correct matching is $A-IV, B-II, C-I, D-III$.

(B) Statement $A$ is incorrect because the electron gain enthalpy of $Cl$ is more negative than that of $F$ due to the small size and interelectronic repulsion in $F$.
Statement $B$ is correct because ionization enthalpy generally decreases down a group due to an increase in atomic size and shielding effect.
Statement $C$ is incorrect because electronegativity is a property of an atom in a molecule and it depends on the hybridization and the nature of the atoms bonded to it.
Statement $D$ is incorrect because $Al_2O_3$ is amphoteric,but $NO$ is a neutral oxide.
Therefore,statements $A, C,$ and $D$ are incorrect.

(B) Based on the provided image,the elements are arranged as follows:
Row $1$: $A^{\prime}, B^{\prime}$ (left to right)
Row $2$: $C^{\prime}, D^{\prime}$ (left to right)
$1$. Atomic Radii: In a period,size decreases from left to right. In a group,size increases from top to bottom.
Thus,$B^{\prime} < A^{\prime}$ and $D^{\prime} < C^{\prime}$. Also,elements in the second row are larger than those in the first row ($A^{\prime} < C^{\prime}$ and $B^{\prime} < D^{\prime}$).
The correct order is $B^{\prime} < A^{\prime} < D^{\prime} < C^{\prime}$. Statement $A$ is true.
$2$. Metallic Character: Metallic character decreases from left to right in a period and increases from top to bottom in a group.
Thus,$B^{\prime} < A^{\prime}$ and $D^{\prime} < C^{\prime}$. Also,$A^{\prime} < C^{\prime}$ and $B^{\prime} < D^{\prime}$.
The correct order is $B^{\prime} < A^{\prime} < D^{\prime} < C^{\prime}$. Statement $B$ is true.
$3$. Size of the element: As established,$B^{\prime} < A^{\prime} < D^{\prime} < C^{\prime}$. Statement $C$ is false.
$4$. Ionic Radii: For cations,the trend follows the atomic radii trend. Thus,$B^{\prime+} < A^{\prime+} < D^{\prime+} < C^{\prime+}$. Statement $D$ is true.
Therefore,statements $A, B,$ and $D$ are true.

(A) Statement $I$: The first ionization enthalpy $(IE_1)$ increases across a period and decreases down a group. The values are: $Na$ $(496 \ kJ/mol)$,$Li$ $(520 \ kJ/mol)$,$Cl$ $(1256 \ kJ/mol)$,$F$ $(1681 \ kJ/mol)$. Thus,the order $Na < Li < Cl < F$ is correct.
Statement $II$: The negative electron gain enthalpy $(-\Delta_{eg}H)$ generally increases across a period and decreases down a group. However,due to the small size of $F$,the electron-electron repulsion is high,making its electron gain enthalpy less negative than that of $Cl$. The values are: $Na$ $(-53 \ kJ/mol)$,$Li$ $(-60 \ kJ/mol)$,$F$ $(-328 \ kJ/mol)$,$Cl$ $(-349 \ kJ/mol)$. Thus,the order $Na < Li < F < Cl$ is correct.
Both statements are true.

(C) $A. Cl, S$ are elements with the highest negative electron gain enthalpy $(IV)$.
$B. Ge, As$ are metalloids,showing properties of both metals and non-metals $(III)$.
$C. Fr, Ra$ are elements with the largest atomic size in their respective groups $(II)$.
$D. F, O$ are elements with the highest electronegativity $(I)$.
Therefore,the correct matching is $A-IV, B-III, C-II, D-I$.

(C) . $Al^{3+} < Mg^{2+} < Na^{+} < F^{-}$ represents the order of ionic radii for isoelectronic species,where size increases as nuclear charge decreases. Thus,$A-IV$.
$B$. $B < C < O < N$ represents the order of first ionisation enthalpy,where $N$ has a higher value than $O$ due to stable half-filled $p$-orbitals. Thus,$B-I$.
$C$. $B < Al < Mg < K$ represents the order of metallic character,which increases down a group and decreases across a period. Thus,$C-II$.
$D$. $Si < P < S < Cl$ represents the order of electronegativity,which increases across a period. Thus,$D-III$.
Therefore,the correct match is $A-IV, B-I, C-II, D-III$.

(A) The modern periodic law states that the properties of elements are a periodic function of their atomic numbers,making statement $A$ incorrect.
Elements with similar outer electronic configurations are placed in the same group,not the same period,making statement $C$ incorrect.
The number of elements in a period is determined by the number of electrons that can be accommodated in the subshells being filled,which is twice the number of available orbitals,making statement $E$ incorrect.
Statements $B$ and $D$ are true.
Therefore,statements $A, C,$ and $E$ are not true.

(B) The first ionisation enthalpy $(IE_1)$ generally decreases down a group and increases across a period. Among the given elements ($Al$,$In$,$Tl$,$Ge$,$Sn$,$Pb$),$Ge$ (Germanium) has the highest $IE_1$ and $In$ (Indium) has the lowest $IE_1$.
For $Ge$ (Group $14$),the most stable oxidation state is $+4$.
For $In$ (Group $13$),the most stable oxidation state is $+3$.
Therefore,the most stable oxidation states for the elements with the highest and lowest $IE_1$ are $+4$ and $+3$,respectively.

(A) The ionic character of a bond is related to the electronegativity difference between the bonded atoms.
In the formation of an ionic bond between $E$ and a non-metal,the ionic character increases as the electron gain enthalpy of the non-metal becomes more negative (i.e.,the non-metal becomes more electronegative).
Given electron gain enthalpy values:
$B (-349 \ kJ \ mol^{-1}) < A (-328 \ kJ \ mol^{-1}) < C (-325 \ kJ \ mol^{-1}) < D (-295 \ kJ \ mol^{-1})$.
Since $B$ has the most negative electron gain enthalpy,it is the most electronegative,resulting in the highest ionic character for $EB$.
Thus,the order of ionic character is $EB > EA > EC > ED$.

(D) $(i)$ For isoelectronic species,as the negative charge increases,the ionic radius increases due to a decrease in effective nuclear charge. Thus,the order $Mg^{2+} < Na^{+} < F^{-} < O^{2-}$ is correct.
$(ii)$ The magnitude of electron gain enthalpy for halogens is $Cl > F > Br > I$. Due to the small size of $F$,there is high inter-electronic repulsion,making its electron gain enthalpy less negative than that of $Cl$. Thus,the order is correct.

(A) Incorrect: The process of adding an $e^-$ to a neutral gaseous atom is not always exothermic; it can be endothermic for elements like noble gases or nitrogen.
$(B)$ Correct: Removing an electron requires energy to overcome the electrostatic attraction between the nucleus and the electron,making it always endothermic.
$(C)$ Correct: $Be$ $(1s^2 2s^2)$ has a fully filled $2s$ subshell,which is more stable than $B$ $(1s^2 2s^2 2p^1)$. Thus,the $1^{st}$ ionization energy of $Be > B$.
$(D)$ Incorrect: Electronegativity is not a constant value; it depends on the oxidation state and hybridization. In $CCl_4$,the carbon atom is bonded to highly electronegative $Cl$ atoms,which increases its effective nuclear charge and thus its electronegativity compared to $C$ in $CH_4$.
$(E)$ Incorrect: Electropositive character increases down the group. $Cs$ is the most electropositive element in group $I$.
Therefore,only statements $B$ and $C$ are correct.

(B) $A.$ $Ga$ (Gallium) has a low melting point $(303 \ K)$,while $Cs$ (Cesium) also has a low melting point $(302 \ K)$. This statement is false.
$B.$ On the Pauling scale,both $N$ and $Cl$ have an electronegativity value of $3.0$. This statement is false.
$C.$ $Ar$ $(18e^-)$,$K^+$ $(19-1=18e^-)$,$Cl^-$ $(17+1=18e^-)$,$Ca^{2+}$ $(20-2=18e^-)$,and $S^{2-}$ $(16+2=18e^-)$ all have $18$ electrons. They are isoelectronic. This statement is true.
$D.$ The correct order of first ionization enthalpy is $Si > Mg > Al > Na$. $Mg$ has a higher value than $Al$ due to its stable $3s^2$ configuration. This statement is false.
$E.$ Atomic radius increases down the group. Thus,$Cs > Rb > Li$. This statement is true.
Therefore,only $C$ and $E$ are true.

(B) Let us identify the elements based on their electronic configurations $:-$
$(I) \ [Xe] 6s^1$ is Cesium $(Cs)$,an alkali metal. Alkali metals are strong reducing agents.
$(II) \ [Xe] 4f^{14} 5d^1 6s^2$ is Lutetium $(Lu)$. Although it has a $5d^1$ electron,it is classified as an $f-$block element (lanthanoid) because the $4f$ subshell is completely filled.
$(III) \ [Ar] 4s^2 4p^5$ is Bromine $(Br)$,a halogen. Halogens have high electron affinity.
$(IV) \ [Ar] 3d^7 4s^2$ is Cobalt $(Co)$,a transition metal. Transition metals exhibit variable oxidation states.
Therefore,the statement that $(II)$ is a $d-$block element is incorrect,as it is an $f-$block element.

(B) First,identify the elements based on their electronic configurations:
$A = 1s^2 \ 2s^2 \ 2p^6$ (Neon,$Ne$,noble gas)
$B = 1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^5$ (Chlorine,$Cl$,halogen)
$C = 1s^2 \ 2s^2 \ 2p^5$ (Fluorine,$F$,halogen)
$D = 1s^2 \ 2s^2 \ 2p^6 \ 3s^1$ (Sodium,$Na$,alkali metal)
$1$. $EA$ (Electron Affinity): Fluorine $(C)$ has a high $EA$,but Chlorine $(B)$ has the highest $EA$ among all elements due to smaller size and inter-electronic repulsion in $F$.
$2$. $EN$ (Electronegativity): Fluorine $(C)$ is the most electronegative element in the periodic table.
$3$. $IP$ (Ionization Potential): Neon $(A)$ is a noble gas with a stable octet configuration,hence it has the highest $IP$ among these.
Thus,the order is $B, C, A$.

(C) Let us analyze each set:
$(a) \ IE \ (Na < Li < B < Be)$: The ionization energy order is $Na < Li < B < Be$. This is correct because $Na$ is in period $3$,while $Li, B, Be$ are in period $2$. Within period $2$,$Be$ has a higher $IE$ than $B$ due to its stable fully-filled $2s^2$ configuration.
$(b) \ EA \ (N < P < O < S)$: The electron affinity order is $N < P < O < S$. This is correct. $N$ has a very low $EA$ due to its stable half-filled $2p^3$ configuration. Generally,$EA$ increases across a period and decreases down a group,but $S > O$ and $P > N$ due to inter-electronic repulsions in smaller atoms.
$(c) \ EN \ (S < O < Cl < F)$: The electronegativity order is $S (2.58) < C (2.55)$ is incorrect,but here $S (2.58) < Cl (3.16) < O (3.44) < F (3.98)$. The given order $S < O < Cl < F$ is incorrect because $Cl$ is more electronegative than $S$,but $O$ is more electronegative than $Cl$. The correct order is $S < Cl < O < F$.
$(d) \ \text{Atomic radius} \ (F < O < N < Na)$: The atomic radius order is $F < O < N < Na$. This is correct as $F, O, N$ are in the same period $(2)$ where radius decreases from left to right $(N > O > F)$,and $Na$ is in period $3$,making it the largest.

(C) $1$. $S \rightarrow S^{-}$: Addition of an electron to $S$ releases energy (exothermic,$Q$) and results in a stable configuration.
$2$. $N^{-} \rightarrow N$: Removal of an electron from $N^{-}$ $(2p^4)$ to form $N$ $(2p^3)$ is exothermic $(Q)$ and results in a half-filled stable configuration $(S)$.
$3$. $O^{-} \rightarrow O^{2-}$: Addition of an electron to $O^{-}$ is endothermic $(P)$ due to inter-electronic repulsion,and $O^{2-}$ achieves a stable inert gas configuration $(R)$.
$4$. $P \rightarrow P^{-}$: Addition of an electron to $P$ $(3p^3)$ is exothermic $(Q)$ as it moves towards a more stable state,but does not reach inert gas or half-filled configurations.
Thus,the correct match is $A$ $\rightarrow Q, B$ $\rightarrow Q, S, C$ $\rightarrow P, R, D$ $\rightarrow Q$.