Periodic Table & Periodicity

Metallic character increases down a group and decreases from left to right across a period (NCERT trend).

Given elements:

• Period 2: $N,\ O,\ F$

• Period 3: $P,\ S,\ Cl$

1) Most metallic element

In period 3, metallic character is higher than in period 2, and within period 3 it decreases left to right:

$P > S > Cl$

So, $P$ is the most metallic among the given elements.

Valence electrons in $P$ (Group 15) $= 5$.

2) Least metallic element

Least metallic means most non-metallic, which is highest towards the top-right of the periodic table. Among the given, $F$ is the most non-metallic.

Valence electrons in $F$ (Group 17) $= 7$.

Final Answer

Number of valence electrons (most metallic, least metallic) $= (5,\ 7)$.

Option C: 5 and 7

In period 4 (from $\mathrm{K}$ to $\mathrm{Kr}$), the atomic radius decreases as we move from left to right because:

• Nuclear charge increases ($Z$ increases),

• Electrons are added in the same principal shell ($n = 4$),

• So the effective nuclear charge on valence electrons increases and pulls them closer.

Hence:

• The largest atomic radius is for the first element of the period: $\boxed{\mathrm{K}}$.

• The smallest atomic radius (as per NCERT trend, minimum at halogen) is for the halogen in that period: $\boxed{\mathrm{Br}}$.

So, the elements with highest and lowest atomic radii respectively are:

$\boxed{\mathrm{K} \ \& \ \mathrm{Br}}$

Correct option: C

To compare second ionization potential configuration of mono-cation is observed

$ \begin{array}{|l|l|l|l|} \hline \mathrm{C}^{+} & \mathrm{N}^{+} & \mathrm{O}^{+} & \mathrm{F}^{+} \\ \hline[\mathrm{He}] & {[\mathrm{He}]} & {[\mathrm{He}] 2 \mathrm{~s}^2 2 \mathrm{p}^3} & {[\mathrm{He}] 2 \mathrm{~s}^2 2 \mathrm{p}^4} \\ 2 \mathrm{~s}^2 \mathrm{sp}^1 & 2 \mathrm{~s}^2 2 \mathrm{p}^2 & \begin{array}{l} \text { Half-filled } \\ \text { stable. } \end{array} & \\ \hline \end{array} $

$ \begin{aligned} &2^{\text {nd }} \text { IE order }\\ &\mathrm{O}>\mathrm{F}>\mathrm{N}>\mathrm{C} \end{aligned} $

Metallic character means the tendency of an element to lose electrons and form a positive ion. In general (as per NCERT trends), metallic character is more in s-block elements than in p-block elements.

So, elements like $\mathrm{K}$ and $\mathrm{Mg}$ (s-block) show more metallic character than $\mathrm{Al}$ and $\mathrm{B}$ (p-block).

Now, compare atomic radius and ionic radius: if an atom forms a cation, its ionic radius becomes smaller than the atomic radius. If an atom forms an anion, its ionic radius becomes larger than the atomic radius.

Therefore, ionic radius is not always smaller than atomic radius. Specifically, anionic radius is greater than atomic radius, but cationic radius is always less than atomic radius for any element.

Statement I:

Na: $1s^2\,2s^2\,2p^6\,3s^1$

After first ionisation, $\mathrm{Na}^+$ becomes $1s^2\,2s^2\,2p^6$ (noble gas configuration).

So, the second ionisation enthalpy of Na means removing an electron from a stable noble gas core, which needs very high energy.

Mg: $1s^2\,2s^2\,2p^6\,3s^2$

After first ionisation, $\mathrm{Mg}^+$ becomes $1s^2\,2s^2\,2p^6\,3s^1$.

So, the second ionisation enthalpy of Mg removes the remaining valence $3s$ electron, which needs less energy than removing from a noble gas core.

Hence, $IE_2(\mathrm{Na}) > IE_2(\mathrm{Mg})$.

So, Statement I is true.

Statement II:

$\mathrm{O}^{2-}$ has $8 + 2 = 10$ electrons, and $\mathrm{F}^-$ has $9 + 1 = 10$ electrons.

So, $\mathrm{O}^{2-}$ and $\mathrm{F}^-$ are isoelectronic (same number of electrons).

In an isoelectronic series, ionic radius decreases as nuclear charge ($Z$) increases, because higher $Z$ pulls electrons more strongly.

Here, $Z(\mathrm{O}) = 8$ and $Z(\mathrm{F}) = 9$.

So, $\mathrm{F}^-$ (higher $Z$) will be smaller than $\mathrm{O}^{2-}$.

Therefore, $r(\mathrm{O}^{2-}) > r(\mathrm{F}^-)$.

So, Statement II is true.

Correct option: Option C

Both Statement I and Statement II are true.

As per the NCERT textbook, electronegativity is the ability of an atom to attract a shared pair of electrons. For Group 15, the electronegativity decreases as we go down the group.

Let's look at their approximate electronegativity values (on the Pauling scale):

• $EN(N) = 3.0$

• $EN(P) = 2.1$

• $EN(As) = 2.0$

• $EN(Sb) = 1.9$

• $EN(Bi) = 1.9$

Notice a very important point: The drop in electronegativity from Nitrogen (N) to Phosphorus (P) is very large ($3.0 - 2.1 = 0.9$). However, for the elements below Phosphorus, the values decrease very slowly.

The question states: "The difference between the electronegativity values of 'X' and phosphorus is higher than that of the difference between phosphorus and 'Y'."

Let's write this mathematically. Let $EN$ stand for electronegativity. The absolute difference is what matters.

$ |EN(X) - EN(P)| > |EN(P) - EN(Y)| $

This means we are looking for an element 'X' whose electronegativity is very different from Phosphorus, and an element 'Y' whose electronegativity is very close to Phosphorus.

Now, let's check each option using the electronegativity values.

• Option A: X = As & Y = Bi

  • Difference 1: $|EN(X) - EN(P)| = |EN(As) - EN(P)| = |2.0 - 2.1| = 0.1$

  • Difference 2: $|EN(P) - EN(Y)| = |EN(P) - EN(Bi)| = |2.1 - 1.9| = 0.2$

  • Is $0.1 > 0.2$? No. So, this option is incorrect.

• Option B: X = Bi & Y = N

  • Difference 1: $|EN(X) - EN(P)| = |EN(Bi) - EN(P)| = |1.9 - 2.1| = 0.2$

  • Difference 2: $|EN(P) - EN(Y)| = |EN(P) - EN(N)| = |2.1 - 3.0| = 0.9$

  • Is $0.2 > 0.9$? No. So, this option is incorrect.

• Option C: X = N & Y = As

  • Difference 1: $|EN(X) - EN(P)| = |EN(N) - EN(P)| = |3.0 - 2.1| = 0.9$

  • Difference 2: $|EN(P) - EN(Y)| = |EN(P) - EN(As)| = |2.1 - 2.0| = 0.1$

  • Is $0.9 > 0.1$? Yes. This matches the condition given in the question.

• Option D: X = As & Y = Sb

  • Difference 1: $|EN(X) - EN(P)| = |EN(As) - EN(P)| = |2.0 - 2.1| = 0.1$

  • Difference 2: $|EN(P) - EN(Y)| = |EN(P) - EN(Sb)| = |2.1 - 1.9| = 0.2$

  • Is $0.1 > 0.2$? No. So, this option is incorrect.

Therefore, X is Nitrogen (N) and Y is Arsenic (As).

In general on moving from left to right in a period ionization energy increases as $\mathbf{Z}_{\text {eff }}$ increases.

Statement I (First ionization enthalpy)

Across a period, first ionization enthalpy generally increases due to increasing nuclear charge and decreasing atomic size.

But there is an important NCERT exception:

• Nitrogen ($\mathrm{N}$) has a half-filled configuration $2p^3$, which is extra stable.

• Oxygen ($\mathrm{O}$) has configuration $2p^4$, where one $p$-orbital contains a pair of electrons, causing extra electron–electron repulsion, so it is easier to remove one electron.

So,

$\mathrm{IE}_1(\mathrm{O}) < \mathrm{IE}_1(\mathrm{N})$

Also, in general:

$\mathrm{C} < \mathrm{O} < \mathrm{N} < \mathrm{F}$

✅ Statement I is true.

Statement II (Magnitude of electron gain enthalpy)

Electron gain enthalpy usually becomes less negative down a group due to increasing size (incoming electron feels less attraction).

But for group 16, NCERT notes:

• Oxygen is very small, so the incoming electron experiences strong repulsion in the compact $2p$ subshell.

• Hence oxygen has less negative electron gain enthalpy than sulphur.

• Sulphur has the most negative (highest magnitude) electron gain enthalpy in this group.

Therefore, the magnitude order is:

$\mathrm{S} > \mathrm{Se} > \mathrm{Te} > \mathrm{Po} > \mathrm{O}$

✅ Statement II is true.

Correct option: D

✅ Both Statement I and Statement II are true.

Statement I:

The order given is:

Al > Mg > Mg²⁺ > Al³⁺

We are comparing atomic and ionic radii.

• Mg (Z = 12) → configuration: [Ne]3s²

• Al (Z = 13) → configuration: [Ne]3s²3p¹

Across a period (from Mg to Al), the effective nuclear charge increases, so atomic radius decreases.

Therefore, Mg > Al (not Al > Mg).

For the ions:

• Mg²⁺ and Al³⁺ are isoelectronic (both have 10 electrons, same as Ne).

  Among isoelectronic species, the greater the nuclear charge, the smaller the radius.

  Hence, Al³⁺ < Mg²⁺.

So the correct order should be:

Mg > Al > Mg²⁺ > Al³⁺

Therefore, Statement I is false.

Statement II:

Electron gain enthalpy (EGE) refers to the energy change when an atom gains an electron.

The more negative the EGE, the greater the electron affinity.

The general trend for electron affinity (magnitude):

Cl > Br > S > O

Reasoning:

• Across a period and down a group, EGE usually becomes less negative (magnitude decreases down group 17).

• Halogens have high (negative) EGEs.

• Among main-group nonmetals, oxygen and sulfur have less negative EGE due to smaller size and electron–electron repulsion in the compact 2p orbital of oxygen.

Thus, Cl > Br > S > O is the correct order in magnitude.

So Statement II is true.

✅ Final Answer:

Option C: Statement I is false but Statement II is true.

Ionisation Energy (Option A)

• Across a period (left ➜ right) atoms get smaller and the nucleus pulls more strongly, so the first ionisation energy (IE) goes up.
• Down a group (top ➜ bottom) atoms get bigger, so the IE goes down.

$ \mathrm{F}>\mathrm{P}>\mathrm{S}>\mathrm{B} \text { (IE order) } $

Metallic Character (Option C)

• Metallic character means how easily an atom loses electrons.
• Across a period it falls because atoms hold their electrons more tightly.
• Down a group it rises because outer electrons are farther from the nucleus.

$ \mathrm{K}>\mathrm{Mg}>\mathrm{Al}>\mathrm{B} \text { (Metallic character order) } $

Basic Nature of Oxides (Option D)

• Metal oxides are basic; non-metal oxides are acidic.
• As metallic character increases down a group, basic strength of the oxides also increases.

$ \mathrm{K}_2 \mathrm{O}>\mathrm{Na}_2 \mathrm{O}>\mathrm{MgO}>\mathrm{Al}_2 \mathrm{O}_3 $

Electron Affinity (Option B)

• Electron affinity (EA) usually becomes more negative from left to right because atoms want to gain electrons.
• Among the main-group elements, EA follows Group 17 > Group 16 > Group 15.

$ \mathrm{Cl}>\mathrm{F}>\mathrm{S}>\mathrm{P} $

To determine which element has the lowest first ionization enthalpy among the given options, we look at the atomic numbers:

Atomic number 32 corresponds to Germanium (Ge).

Atomic number 35 corresponds to Bromine (Br).

Atomic number 87 corresponds to Francium (Fr).

Atomic number 19 corresponds to Potassium (K).

The element with the lowest first ionization energy is Francium (Fr) with atomic number 87. This can be represented by its electron configuration: $\text{Fr} - [\text{Rn}] 7s^1$. Francium, being a part of the alkali metal group, has the most loosely bound outer electron compared to the other elements listed, resulting in a lower first ionization enthalpy.

To determine the relative acidity and bond enthalpy of H₂Se and H₂Te, let's examine both:

Acidic Strength:

The acidity of hydrides increases as we move down the group in the periodic table. This is because the bond strength between hydrogen and the central atom decreases, making it easier for the hydrogen ions to dissociate. Therefore, H₂Te, being further down the group than H₂Se, is more acidic:

$ \mathrm{H}_2\mathrm{Se} < \mathrm{H}_2\mathrm{Te} $

Bond Enthalpy:

Bond enthalpy refers to the energy required to dissociate a bond. The bond enthalpy decreases down the group due to weaker bonds forming as the atomic size increases. Hence, H₂Se has a higher bond enthalpy compared to H₂Te:

$ \Delta_{\text{dis}} \mathrm{H}: \mathrm{H}_2\mathrm{Se} > \mathrm{H}_2\mathrm{Te} $

$ \text{H}_2\mathrm{Se} \, [276 \, \text{KJ/mol}] > \text{H}_2\mathrm{Te} \, [238 \, \text{KJ/mol}] $

Thus, H₂Se is indeed less acidic than H₂Te, but it has a higher bond enthalpy for dissociation.

In a period (row on the periodic table), the atomic size generally decreases as you move from left to right. This is because additional electrons are being added to the same outer shell while the number of protons in the nucleus increases, pulling the electron cloud closer to the nucleus and reducing the atomic radii.

To determine the number of valence electrons in the metal with the lowest enthalpy of atomization among Chromium (Cr), Cobalt (Co), Iron (Fe), and Nickel (Ni), we need to examine their respective electron configurations.

Chromium has the electron configuration:

$ \text{Cr} = [\text{Ar}] 3d^5 4s^1 $

Based on this configuration, Chromium has a total of six valence electrons (1 from the 4s orbital and 5 from the 3d orbital).

Chromium is noted for having the lowest enthalpy of atomization among the given metals. Thus, the number of valence electrons in Chromium, the metal with the lowest enthalpy of atomization, is 6.

Given the ionization energy values for elements $A$ and $B$ in group 14, where $A$ has an ionization energy of 708 kJ/mol and $B$ has 715 kJ/mol, these values are the lowest among their group. This helps identify the elements as tin (Sn) and lead (Pb), respectively. In this context:

The $\mathrm{A}^{2+}$ ion corresponds to $\mathrm{Sn}^{2+}$, which acts as a reducing agent.

The $\mathrm{B}^{4+}$ ion corresponds to $\mathrm{Pb}^{4+}$, which acts as an oxidizing agent.

Therefore, Sn in the +2 oxidation state tends to donate electrons (reduce), while Pb in the +4 oxidation state tends to accept electrons (oxidize), underscoring their respective roles as reducing and oxidizing agents.

$\mathrm{Mn}^{2+}:[\mathrm{Ar}] 3 \mathrm{~d}^5$

Half filled stability

More 1E than $\mathrm{Fe}^{2+}$

$\mathrm{Fe}^{2+}:[\mathrm{Ar}] 3 \mathrm{~d}^6$

$\begin{aligned} &\text { IE order }\\ &\mathrm{B}>\mathrm{T} \ell>\mathrm{Ga}>\mathrm{Al}>\mathrm{In} \end{aligned}$

$\begin{array}{lc} \text { Radius Order } & \mathrm{T} \ell>\mathrm{In}>\mathrm{A} \ell>\mathrm{Ga}>\mathrm{B} \\ \text { EN Order } & \mathrm{B}>\mathrm{T} \ell>\mathrm{In}>\mathrm{Ga}>\mathrm{Al} \\ \text { Density Order } & \mathrm{T} \ell>\mathrm{In}>\mathrm{Ga}>\mathrm{A} \ell>\mathrm{B} \\ \mathrm{IE}_1 \text { Order } & \mathrm{B}>\mathrm{T} \ell>\mathrm{Ga}>\mathrm{A} \ell>\mathrm{In} \end{array}$

To correctly match the elements with their respective families and symbols, consider the following associations:

Pnictogen is matched with Moscovium (Mc), which has the atomic number 115.

Chalcogen corresponds to Livermorium (Lv), with the atomic number 116.

Halogen is associated with Tennessine (Ts), with atomic number 117.

Noble gas correlates with Oganesson (Og), which has an atomic number of 118.

This alignment clarifies the classification of elements based on their family groups and chemical symbols.

(A) The process of adding an $\mathrm{e}^{-}$to a neutral gaseous atom is not always exothermic it may be exothermic or endothermic.

$\begin{array}{cc}\text { (C) } \mathrm{Be} & \mathrm{B} \\ 1 \mathrm{~s}^2 2 s^2 & 1 s^2 2 s^2 2 p^1\end{array}$

In Be 2s subshell in fully filled

So, need high energy to remove $\mathrm{e}^{-}$as compared to B .

due to partially positive charge $z_{\text {eff }} \uparrow, \mathrm{EN} \uparrow$

So, EN of $\mathrm{C} \Rightarrow \mathrm{CCl}_4>\mathrm{CH}_4$

(E) Cs is most electropositive.

The electronic configurations of four elements, A, B, C, and D, are presented as follows:

(A) $1s^2 2s^2 2p^3$

(B) $1s^2 2s^2 2p^4$

(C) $1s^2 2s^2 2p^5$

(D) $1s^2 2s^2 2p^2$

To determine the correct order of increasing electronegativity according to Pauling's scale, analyze the elements as follows:

Nitrogen ($1s^2 2s^2 2p^3$): Electronegativity = 3.0

Oxygen ($1s^2 2s^2 2p^4$): Electronegativity = 3.5

Fluorine ($1s^2 2s^2 2p^5$): Electronegativity = 4.0

Carbon ($1s^2 2s^2 2p^2$): Electronegativity = 2.55

The correct order of increasing electronegativity is:

Carbon (D) < Nitrogen (A) < Oxygen (B) < Fluorine (C)

$\Rightarrow$ The metallic radius order of $\mathrm{Al} \& \mathrm{Ga}$ is

(due to poor shielding of d-subshell electrons)

$\Rightarrow$ The ionic radius order of $\mathrm{Al}^{+3} \& \mathrm{Ga}^{+3}$ is

$\mathrm{B}^{+3}<\mathrm{Al}^{+3}<\mathrm{Ga}^{+3}<\mathrm{In}^{+3}<\mathrm{T} \ell^{+3}$

The order of first four elements of group-17 are as follows.

$\mathrm{F}<\mathrm{Cl}<\mathrm{Br}<\mathrm{I} \text { (Covalent radius) }$

$\mathrm{Cl}>\mathrm{F}>\mathrm{Br}>\mathrm{I}$ (Electron affinity)

$\mathrm{F}<\mathrm{Cl}^{-}<\mathrm{Br}^{-}<\mathrm{I}$ (Ionic radius)

$\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$ ( $\mathrm{I}^{\mathrm{st}}$ ionization energy)

Electron affinity order is irregular.

The elements given are

Li, Na, Be, Mg, B and Al

2$^{nd}$ period elements $\to$ Li, Be, B

Order of atomic radii $\to$ Li > Be > B

3$^{rd}$ period elements $\to$ Na, Mg, Al

Order of atomic radii $\to$ Na > Mg > Al

In a period, atomic radii decreases on moving from left to right.

So, the element with least atomic radii in 2$_{nd}$ period (Li, Be, B) is B. The element with least atomic radii 3$_{rd}$ period (Na, Mg, Al) is Al.

In B and Al $\Rightarrow$ B and Al are same group elements. Down the group, radii increases. So, B has least radii.

B is the element with least atomic radius to form the oxide $A_2O_3$.

Oxidation states of the A in oxides :

${A_0} \to + 2\,\,{A_{{0_2}}} \to + 2\,\,{A_{{2^0}}} \to + 1\,\,{A_{2_3^0}} \to + 3$

B outer configuration is s$^2$P$^1$. 3 valence electrons and form A$_2$O$_3$ type oxide.

Apalite family - A group of phosphate minerals that includes fluorapatite, chlorapatite, and hychoxyapatite. Chemical formula $C{a_{10}}{(P{O_4})_6}{(OH,F,Cl)_2}$

$\left. \matrix{ Fluorapatite \to C{a_{10}}{(P{O_4})_6}{F_2} \hfill \cr Chlorapatite \to C{a_{10}}{(P{O_4})_6}C{l_2} \hfill \cr Hydroxyapatite \to C{a_{10}}{(P{O_4})_6}{(OH)_2} \hfill \cr} \right\}\matrix{ {Main\,components:} \cr {Calcium\,and\,phosphate} \cr }$

So, the element considered is phosphorus.

The first four group 15 elements are Nitrogen (N), Phosphorus (P), Arsenic (As) and Antimony (Sb). The outer configuration of phosphorus is half filled and hence stable $({s^2}{p^3})$.

For a stable configuration, ionization enthalpy is high. Ionization enthalpy - Energy required to remove an electron from the gaseous atom.

Nitrogen has the highest first ionization enthalpy in group 15.

Down the group, ionization enthalpy decreases.

Down the group, atomic size increases, valence electrons are less attracted to the nucleus. And here, electron removal will become easier. So, ionization enthalpy is also decreases.

The highest value of ionization enthalpy 1402 KJ mol$^{-1}$ corresponds to nitrogen element.

So, the first ionization enthalpy of the element that is a main component of apatite family is option D) 1012 kJ mol$^{-1}$.

Element $E \rightarrow$ Ionisation enthalpy value $\rightarrow 374 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Element $A \rightarrow$ Electron gain enthalpy value $\rightarrow-328 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Element $C \rightarrow$ Electron gain enthalpy value $\rightarrow-325 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}$

Element $D \rightarrow$ Electron gain enthalpy value $\rightarrow-295 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Ionization enthalpy is the energy required to remove an election from an atom or ion in the gas phase. An element with lower ionization enthalpy indicate easier cation formation and this lead to higher ionic character. Electron gain enthalpy is the energy released when an election is added to an atom or ion in the gas phase. A highly negative election gain enthalpy Indicate that the atom readily gains electrons, favouring the formation of anions. This leads to stronger ionic character in compounds. High ionic character generally occurs when the difference in ionisation enthalpy and electron gain enthalpy between the two atoms is large.

For $E A$,

Difference between ionisation enthalpy and electron gain enthalpy

$\begin{aligned} & =374 \mathrm{kJmol}^{-1}-\left(-328 \mathrm{~kJ} \mathrm{~mol}^{-1}\right) \\ & =702 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$

For $E B$,

$\begin{aligned} &\text {Difference between ionisation enthalpy and elation gain enthalpy } \\ & =374 \mathrm{~kJ} \mathrm{~mol}^{-1}-\left(-349 \mathrm{~kJ~mol}^1\right) \\ & =723 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$

For $E C$,

$\begin{aligned} & \text { Difference between ionisation enthalpy and election gain enthalpy } \\ & =374 \mathrm{~kJ} \mathrm{~mol}^{-1}-\left(-325 \mathrm{~kJ} \mathrm{~mol}^{-1}\right) \\ & =699 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$

For $ED$,

$\begin{aligned} & =374 \mathrm{~kJ} \mathrm{~mol}^{-1}-\left(-295 \mathrm{~kJ} \mathrm{~mol}^{-1}\right) \\ & =669 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$

Higher difference between ionisation enthalpy and election gain enthalpy is for compound EB. So, EB has higher ionic character.

The Lower value of difference between ionisation enthalpy and election gain enthalpy is for compound ED. So, ED has lower ionic character. The increasing order of compounds for the difference between ionisation enthalpy and electron gain enthalpy $\to$

$E B< E C < E A< E B$

So, increasing. order of compounds for ionic character:

$E D< E C < E A< E B$

In the decreasing order,

$E B>E A>E C>E D$

Correct option (1)

Statement I : Correct

The radii of isoelectronic species increases in the order $\mathrm{Mg^{2+} < Na^+ < F^- < O^{2-}}$ (number of electrons = 10)

In general, radius of cation is smaller than anion $\mathrm{A^+ < A^-}$ cation with positive charge have higher effective nuclear charge.

Reason : In cations, number of electrons decreases, but number of protons remains the same. This causes a stronger pull on the remaining electrons by the nucleus, resulting in a higher effective nuclear charge.

With increase in effective nuclear charge, radii decreases so, increase in positive charge indicate more number of electron removal and hence increase in effective nuclear charge. As a result, radii decreases. So, Mg$^{2+}$ has lower radii than Na$^+$.

For anion, number of electrons are increased. The electron-electron repulsion weakens the attraction between the nucleus and the electrons, resulting in a decrease in effective nuclear charge. So, effective nuclear charge of O^{2-}$ is lower than F^-$.

So, O$^{2-}$ with lower effective nuclear charge causes longer radii of it.

So, ${F^ - } < {O^{2 - }}$

So, $M{g^{2 + }} < N{a^ + } < {F^ - } < {O^{2 - }}$

Ionic radii of isoelectronic species decreases as nuclear charge increases. The cation with greater +ve charge will have a smaller radius and the anion with the greater $-$ve charge will have a bigger radius for isoelectronic species.

Statement II : Correct

The magnitude of electron gain enthalpy of halogen decreases in the order Cl > F > Br > I.

Electron gain enthalpy is the amount of energy released when an atom accepts the electron from any neutral isolated gaseous atom to form an anion.

Generally, electron gain enthalpy decreases down the group. The exception is for F and cl. Electron gain enthalpy of F is less than Cl. It is due to the comparatively small size of F atom.

For Cl, Br and I, the electron gain enthalpy decreases as go from Cl to I.

So, both the statements are correct.

Law of octaves was arranged in the increasing order of their atomic weight.

Lothar Meyer plotted the physical properties such as atomic volume, melting point and boiling point against atomic weight.

Let's analyze the options based on the periodic trends in atomic radii. Recall two key trends:

Atomic radii decrease as you move from left to right in a period due to increasing effective nuclear charge.

Atomic radii increase as you move down a group because additional electron shells are added.

We'll review each option:

Option A: $\mathrm{Si} > \mathrm{P} > \mathrm{Cl} > \mathrm{F}$

Silicon (Si), phosphorus (P), and chlorine (Cl) are in the same period, with a gradual decrease in atomic radius from left to right.

Fluorine (F) is in an earlier period (it has only 2 electron shells) and is much smaller.

Hence, the order follows the expected trend.

Option B: $\mathrm{Mg} > \mathrm{Al} > \mathrm{C} > \mathrm{O}$

Magnesium (Mg) and aluminum (Al) are period 3 elements, whereas carbon (C) and oxygen (O) belong to period 2.

Period 3 elements generally have larger radii than period 2 counterparts.

The order is consistent: Mg and Al are larger than C and O, and within each period, radii decrease from left to right.

Option C: $\mathrm{Al} > \mathrm{B} > \mathrm{N} > \mathrm{F}$

Aluminum (Al) is in period 3 and is expected to be larger than boron (B), nitrogen (N), and fluorine (F) from period 2.

Within period 2 (B, N, F), the atomic radii decrease from left to right.

Thus, this order is correct as well.

Option D: $\mathrm{Be} > \mathrm{Mg} > \mathrm{Al} > \mathrm{Si}$

Beryllium (Be) is in period 2, while magnesium (Mg), aluminum (Al), and silicon (Si) are in period 3.

Since atomic radii increase with the number of electron shells, Be (with only 2 shells) should be smaller than Mg.

This means that stating $\mathrm{Be} > \mathrm{Mg}$ is incorrect.

Thus, the incorrect decreasing order of atomic radii is given in Option D.

Among $\mathrm{Al}, \mathrm{In}, \mathrm{Tl}, \mathrm{Ge}, \mathrm{Sn}, \mathrm{Pb}$, the metal having highest $\mathrm{IE}_1$ is Ge and lowest $\mathrm{IE}_1$ is $\mathrm{In}$.

Most stable oxidation state of Ge is +4 and In is +3 .

We can determine the number of valence electrons by looking for a large jump in successive ionization energies. Here's how we can analyze the data:

The ionization energies given are:

$IE_1 = 800 \, \text{kJ/mol}$

$IE_2 = 2427 \, \text{kJ/mol}$

$IE_3 = 3658 \, \text{kJ/mol}$

$IE_4 = 25024 \, \text{kJ/mol}$

$IE_5 = 32824 \, \text{kJ/mol}$

Notice that:

The first three ionization energies are relatively low.

There is a significant jump between the third and fourth ionization energy.

What does the gap indicate?

The large jump from $IE_3$ to $IE_4$ means that after removing the first three electrons (which are the valence electrons), the next electron to be removed (a core electron) is much more strongly bound.

This implies that the element has 3 valence electrons.

Which group does an element with 3 valence electrons belong to?

Elements with 3 valence electrons are found in Group 13 of the periodic table.

Thus, the element most likely belongs to Group 13.

Correct Answer: Option C (Group 13).

Order of I.E. (in KJ/mol) :

$\matrix{ C & > & {Si} & > & {Ge} & > & {Sn} & > & {Pb} \cr {1086} & {} & {786} & {} & {761} & {} & {708} & {} & {715} \cr }$

Let's analyze each statement one by one:

$\textbf{A. The properties of elements are function of atomic weights.}$

Originally, elements were arranged by atomic weights, but it was later understood that their properties depend on the atomic number (the number of protons) rather than the atomic weight.

Thus, this statement is NOT true.

$\textbf{B. The properties of elements are function of atomic numbers.}$

The periodic table is arranged by increasing atomic number, and an element’s properties directly relate to its number of protons and the corresponding electron configuration.

This statement is true.

$\textbf{C. Elements having similar outer electronic configurations are arranged in same period.}$

Elements that have similar outer electronic configurations are actually grouped in the same column (group) rather than the same row (period).

Therefore, this statement is NOT true.

$\textbf{D. An element's location reflects the quantum numbers of the last filled orbital.}$

The arrangement of elements corresponds to the electron configurations determined by quantum numbers, especially for the last electron added.

This statement is true.

$\textbf{E. The number of elements in a period is same as the number of atomic orbitals available in energy level that is being filled.}$

In an energy level, the number of atomic orbitals does not directly equal the number of elements in a period. For example, period 2 has 1s filled, and then the 2s and 2p orbitals are filled (a total of 4 orbitals), yet there are 8 elements because each orbital can accommodate 2 electrons.

Hence, this statement is NOT true.

So, the statements that are NOT true about the periodic table are A, C, and E.

Thus, the correct answer is:

Option A: A, C and E Only.

Order of M.P. of group 14: $\mathrm{C}>\mathrm{Si}>\mathrm{Ge}>\mathrm{Pb}>\mathrm{Sn}$

Let's analyze the periods for each element:

Platinum (Pt) has an atomic number of 78, which places it in period 6.

Osmium (Os) has an atomic number of 76, also in period 6.

Iridium (Ir) has an atomic number of 77, which is in period 6 as well.

Palladium (Pd) has an atomic number of 46, placing it in period 5.

Since Palladium is in period 5 while the other elements are in period 6, the element that does not belong to the same period is:

$\textbf{Option D: Palladium.}$

First, let us re-state the two statements clearly:

Statement (I): An element in the extreme left of the periodic table forms acidic oxides.

Statement (II): Acid is formed during the reaction between water and oxide of a reactive element present in the extreme right of the periodic table.

Analyzing Statement (I)

The extreme left of the periodic table corresponds to the alkali metals (Group 1) and alkaline earth metals (Group 2).

These metals typically form basic (or occasionally amphoteric, in the case of some Group 2 elements) oxides, not acidic oxides.

For example, Na₂O, K₂O, MgO, CaO, etc., all form basic solutions (e.g., Na₂O + H₂O → 2 NaOH).

Hence, Statement (I)—that an element in the extreme left forms acidic oxides—is false.

Analyzing Statement (II)

The extreme right of the periodic table corresponds to the nonmetals in Groups 15, 16, 17 (and noble gases in Group 18).

Nonmetal oxides (such as those of sulfur, phosphorus, chlorine) are generally acidic.

When these oxides dissolve in water, they typically form acids.

Example: SO₃ + H₂O → H₂SO₄ (sulfuric acid)

Example: P₂O₅ + 3 H₂O → 2 H₃PO₄ (phosphoric acid)

Example: Cl₂O₇ + H₂O → 2 HClO₄ (perchloric acid)

Thus, Statement (II)—that acid is formed when water reacts with an oxide of a reactive element in the extreme right—is true.

Conclusion

Statement (I) is false.

Statement (II) is true.

Therefore, the correct choice is:

(D) Statement (I) is false but Statement (II) is true.

(A) $\mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{F}^{-}$

All four ions ($\mathrm{Al}^{3+},\mathrm{Mg}^{2+},\mathrm{Na}^{+},\mathrm{F}^{-}$) are isoelectronic (each has 10 electrons). For isoelectronic species, ionic radii decrease as the positive nuclear charge increases, which is why the smallest ion here is $\mathrm{Al}^{3+}$ (Z=13) and the largest is $\mathrm{F}^{-}$ (Z=9). Thus, this ordering is one of increasing ionic radius.

$ \boxed{ (A) \;\longrightarrow\; \text{(IV) Ionic radii} } $

(B) $\mathrm{B}<\mathrm{C}<\mathrm{O}<\mathrm{N}$

Check first ionization enthalpies ($\mathrm{IE}_1$):

B: $\approx 801$\,kJ/mol

C: $\approx 1086$\,kJ/mol

O: $\approx 1314$\,kJ/mol

N: $\approx 1402$\,kJ/mol

Hence, the order of increasing $\mathrm{IE}_1$ is

$ B < C < O < N. $

This matches the given sequence exactly.

$ \boxed{ (B) \;\longrightarrow\; \text{(I) Ionisation enthalpy} } $

(C) $\mathrm{B}<\mathrm{Al}<\mathrm{Mg}<\mathrm{K}$

Consider metallic character (the tendency to lose electrons easily, show metallic properties). Across a period (left to right), metallic character decreases; down a group, it increases.

B (metalloid) has the least metallic character here.

Al (group 13 metal) is more metallic than B.

Mg (group 2 metal) is typically more metallic than Al.

K (group 1 metal) is the most metallic among these.

Thus, $\mathrm{B} < \mathrm{Al} < \mathrm{Mg} < \mathrm{K}$ is an order of increasing metallic character.

$ \boxed{ (C) \;\longrightarrow\; \text{(II) Metallic character} } $

(D) $\mathrm{Si}<\mathrm{P}<\mathrm{S}<\mathrm{Cl}$

Check electronegativities:

Si: $\approx 1.90$

P: $\approx 2.19$

S: $\approx 2.58$

Cl: $\approx 3.16$

They increase in the order

$ \mathrm{Si}<\mathrm{P}<\mathrm{S}<\mathrm{Cl}, $

which matches the given sequence for increasing electronegativity.

$ \boxed{ (D) \;\longrightarrow\; \text{(III) Electronegativity} } $

Final Matching

$ (A) \to (IV),\quad (B) \to (I),\quad (C) \to (II),\quad (D) \to (III). $

Looking at the choices given:

Option A: $(A)-(IV), (B)-(I), (C)-(II), (D)-(III)$

This is exactly what we found.

Answer: Option A

Let’s compare the known electronegativities (Pauling scale) of the elements in each option:

F ~ 3.98

O ~ 3.44

N ~ 3.04

Cl ~ 3.16

C ~ 2.55

B ~ 2.04

S ~ 2.58 (sometimes listed as 2.5–2.58)

Si ~ 1.90

Al ~ 1.61

Be ~ 1.57

Mg ~ 1.31

Now, check each statement:

Option A: $\mathrm{S}<\mathrm{Cl}<\mathrm{O}<\mathrm{F}$

S (2.58) < Cl (3.16) < O (3.44) < F (3.98)

This order is correct.

Option B: $\mathrm{Al}<\mathrm{Si}<\mathrm{C}<\mathrm{N}$

Al (1.61) < Si (1.90) < C (2.55) < N (3.04)

This order is correct.

Option C: $\mathrm{Al}<\mathrm{Mg}<\mathrm{B}<\mathrm{N}$

Actual electronegativities are Al (1.61) and Mg (1.31).

The statement says “Al < Mg,” which would mean 1.61 < 1.31, which is wrong.

Therefore, Option C is incorrect.

Option D: $\mathrm{Mg}<\mathrm{Be}<\mathrm{B}<\mathrm{N}$

Mg (1.31) < Be (1.57) < B (2.04) < N (3.04)

This order is correct.

Answer: Option C is the incorrect order.

To match each element with its correct electronic configuration, let's consider the electronic configurations of each element based on their positions in the periodic table:

• Nitrogen (N): It is in the 2nd period and belongs to the p-block, specifically group 15. The electronic configuration of nitrogen is $1\text{s}^2 2\text{s}^2 2\text{p}^3$. Looking at the options in List II, this matches with the electronic configuration labeled 'III' ($[\mathrm{He}] 2 \mathrm{s}^2 2 \mathrm{p}^3$), which reflects the electronic configuration of nitrogen starting after the helium core.


• Sulfur (S): Sulfur is in the 3rd period, also in the p-block, specifically group 16. Its electronic configuration is $1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^4$. In the given list, this corresponds to 'II' ($[\mathrm{Ne}] 3 \mathrm{s}^2 3 \mathrm{p}^4$), which captures the configuration of sulfur starting after the neon core.


• Bromine (Br): Bromine is located in the 4th period and in the p-block, belonging to group 17. Its electronic configuration is $1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^{10} 4\text{s}^2 4\text{p}^5$, which is correctly represented by 'I' ($[\mathrm{Ar}] 3 \mathrm{d}^{10} 4 \mathrm{s}^2 4 \mathrm{p}^5$), showing the configuration after the argon core.


• Krypton (Kr): Krypton falls in the 4th period and is a noble gas situated in the p-block at group 18. Its electronic configuration is $1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^{10} 4\text{s}^2 4\text{p}^6$, which aligns with 'IV' ($[\mathrm{Ar}] 3 \mathrm{d}^{10} 4 \mathrm{s}^2 4 \mathrm{p}^6$), clearly depicting krypton's electronic configuration post-argon.

Therefore, the correct match is:

1. A-III (Nitrogen matches with $[\mathrm{He}] 2 \mathrm{s}^2 2 \mathrm{p}^3$)
2. B-II (Sulfur matches with $[\mathrm{Ne}] 3 \mathrm{s}^2 3 \mathrm{p}^4$)
3. C-I (Bromine matches with $[\mathrm{Ar}] 3 \mathrm{d}^{10} 4 \mathrm{s}^2 4 \mathrm{p}^5$)
4. D-IV (Krypton matches with $[\mathrm{Ar}] 3 \mathrm{d}^{10} 4 \mathrm{s}^2 4 \mathrm{p}^6$)

Thus, the correct answer is Option C: A-III, B-II, C-I, D-IV.

Melting point : $\mathrm{B}>\mathrm{A} \ell>\mathrm{T} \ell>\mathrm{In}>\mathrm{Ga}$

Ionic radius $(\mathrm{M}^{+3} / \mathrm{pm}): \mathrm{T} \ell>\mathrm{In}>\mathrm{Ga}>\mathrm{A} \ell>\mathrm{B}$

$\left(\Delta_{\mathrm{IE}} \mathrm{H}\right)_1\left[\frac{\mathrm{kJ}}{\mathrm{mol}}\right]: \mathrm{B}>\mathrm{T} \ell>\mathrm{A} \ell \approx \mathrm{Ga}>\mathrm{In}$

Atomic radius (in $\mathrm{pm}$) : $\mathrm{T} \ell>\mathrm{In}>\mathrm{A} \ell>\mathrm{Ga}>\mathrm{B}$

To evaluate the given statements, we need to consider the concepts of oxidation states and bonding tendencies across different periods in the periodic table.

Statement (I) describes the oxidation state of an element in a compound as the hypothetical charge an atom would have if all bonds to atoms of different elements were 100% ionic. This consideration does involve looking at how electrons are shared or transferred between atoms, but it's not solely based on electron gain enthalpy. Rather, the oxidation state is determined by a set of rules which include, but are not limited to, electronegativity differences and known oxidation states of elements. Therefore, the statement is somewhat misleading in its specificity about electron gain enthalpy being the sole consideration. The electron gain enthalpy is a measure of the energy change when an electron is added to a neutral atom in the gaseous state to form a negative ion, which indirectly influences the discussion of oxidation states but is not the direct basis for their determination.

Statement (II) points out that $\mathrm{p}\pi-\mathrm{p}\pi$ bond formation is more prevalent among second period elements compared to elements in other periods. This is correct and can be attributed to the smaller size of the second period elements (such as carbon, nitrogen, and oxygen), which allows their p orbitals to overlap more effectively for $\mathrm{p}\pi-\mathrm{p}\pi$ bonding. As we move down the periods, the atomic size increases, which leads to less effective overlapping of p orbitals for $\mathrm{p}\pi-\mathrm{p}\pi$ bonding due to increased distance between the valence electrons and nucleus. This makes $\mathrm{p}\pi-\mathrm{p}\pi$ bonds less common and less strong in elements of higher periods compared to those in the second period.

Therefore, the most accurate assessment of the statements given the explanations above is:

• Statement I is not entirely accurate because it overemphasizes electron gain enthalpy in determining oxidation states.
• Statement II is correct as it accurately describes the prevalence of $\mathrm{p}\pi-\mathrm{p}\pi$ bonding among second period elements.

Hence, the correct option is:

Option C: Statement I is incorrect but Statement II is correct.

To match the elements from List I with their corresponding properties in List II, let's analyze each pair:

A. $\mathrm{Cl,S}$

These elements are known for having high negative electron gain enthalpy (energy released when an electron is added to an atom). Chlorine ($\mathrm{Cl}$) and sulfur ($\mathrm{S}$) are both elements that release a significant amount of energy when they gain an electron.

Hence, they match with IV: Elements with highest negative electron gain enthalpy.

B. $\mathrm{Ge,As}$

Germanium ($\mathrm{Ge}$) and arsenic ($\mathrm{As}$) are metalloids, meaning they show properties of both metals and non-metals.

Hence, they match with III: Elements which show properties of both metals and non-metal.

C. $\mathrm{Fr,Ra}$

Francium ($\mathrm{Fr}$) and radium ($\mathrm{Ra}$) are elements that have very large atomic sizes, being at the bottom of their respective groups on the periodic table.

Hence, they match with II: Elements with largest atomic size.

D. $\mathrm{F,O}$

Fluorine ($\mathrm{F}$) and oxygen ($\mathrm{O}$) are elements with the highest electronegativity, meaning they strongly attract electrons toward themselves.

Hence, they match with I: Elements with highest electronegativity.

Summarizing, we have:

A-IV, B-III, C-II, D-I

The correct option is:

Option D

A-IV, B-III, C-II, D-I

To determine which elements have negative electron affinity values, we'll analyze each option individually.

Understanding Electron Affinity

Electron Affinity (EA): The energy change when an electron is added to a neutral atom in the gas phase to form a negative ion.

Convention:

If energy is released (exothermic process), the electron affinity is considered positive.

If energy is absorbed (endothermic process), the electron affinity is considered negative.

Option A: $\mathrm{Be} \rightarrow \mathrm{Be}^{-}$

Electronic Configuration of Be: $1s^2\,2s^2$

Analysis:

Adding an electron to beryllium means placing it into the higher energy $2p$ orbital.

Beryllium has a filled $2s$ subshell, so the added electron experiences higher energy and electron-electron repulsion.

Result: Energy is absorbed; the process is endothermic.

Conclusion: Electron affinity of Be is negative.

Option B: $\mathrm{N} \rightarrow \mathrm{N}^{-}$

Electronic Configuration of N: $1s^2\,2s^2\,2p^3$

Analysis:

Nitrogen has a half-filled $2p$ subshell.

Adding an electron introduces repulsion in the half-filled orbital.

Result: Energy is absorbed; the process is endothermic.

Conclusion: Electron affinity of N is negative.

Option C: $\mathrm{O} \rightarrow \mathrm{O}^{2-}$

Electronic Configuration of O: $1s^2\,2s^2\,2p^4$

Analysis:

First Electron Affinity (O to O⁻): Exothermic (energy released).

Second Electron Affinity (O⁻ to O²⁻): Endothermic because adding an electron to a negative ion requires energy due to electron-electron repulsion.

Overall Process (O to O²⁻): The second step dominates, making the overall process endothermic.

Conclusion: Electron affinity for forming $\mathrm{O}^{2-}$ is negative.

Option D: $\mathrm{Na} \rightarrow \mathrm{Na}^{-}$

Electronic Configuration of Na: $1s^2\,2s^2\,2p^6\,3s^1$

Analysis:

Adding an electron fills the $3s$ orbital.

Sodium tends to lose an electron to form $\mathrm{Na}^+$, not gain one.

However, adding an electron is still an exothermic process due to the low energy of the $3s$ orbital.

Result: Energy is released; the process is exothermic.

Conclusion: Electron affinity of Na is positive.

Option E: $\mathrm{Al} \rightarrow \mathrm{Al}^{-}$

Electronic Configuration of Al: $1s^2\,2s^2\,2p^6\,3s^2\,3p^1$

Analysis:

Adding an electron to the $3p$ orbital.

The process releases energy due to the addition of an electron to a partially filled orbital.

Result: Energy is released; the process is exothermic.

Conclusion: Electron affinity of Al is positive.

Final Conclusion

Negative Electron Affinity Values (Endothermic Processes):

Option A: Beryllium ($\mathrm{Be}$)

Option B: Nitrogen ($\mathrm{N}$)

Option C: Oxygen forming $\mathrm{O}^{2-}$

Therefore, the correct answer is:

Option D: A, B, and C only

Answer: Option D

To determine the correctness of the given statements, let's analyze each one individually based on atomic and ionic radii concepts.

Statement I:

The metallic radius of $\mathrm{Na}$ is $1.86\, \text{Å}$ and the ionic radius of $\mathrm{Na}^+$ is lesser than $1.86\, \text{Å}$.

Analysis:

Metallic Radius of Sodium ($\mathrm{Na}$):

Sodium is a metal, and its metallic radius is indeed approximately $1.86\, \text{Å}$.

The metallic radius refers to half the distance between the nuclei of two adjacent atoms in a metallic lattice.

Ionic Radius of Sodium Ion ($\mathrm{Na}^+$):

When sodium loses an electron to form $\mathrm{Na}^+$, it loses its outermost electron shell (the 3s orbital).

This results in a significant decrease in size due to:

Decrease in Electron-Electron Repulsion: Fewer electrons mean less repulsion among them.

Unchanged Nuclear Charge: The number of protons remains the same, so the effective nuclear charge per electron increases, pulling the remaining electrons closer to the nucleus.

The ionic radius of $\mathrm{Na}^+$ is approximately $0.95\, \text{Å}$, which is significantly smaller than $1.86\, \text{Å}$.

Conclusion:

Statement I is correct.

Statement II:

Ions are always smaller in size than the corresponding elements.

Analysis:

General Trends:

Cations ($\text{Positive Ions}$):

Formed by the loss of one or more electrons.

Result: Cations are smaller than their parent atoms due to loss of electron(s) and decreased electron-electron repulsion.

Example: $\mathrm{Na} \rightarrow \mathrm{Na}^+$ (size decreases).

Anions ($\text{Negative Ions}$):

Formed by the gain of one or more electrons.

Result: Anions are larger than their parent atoms due to added electron(s) and increased electron-electron repulsion.

Example: $\mathrm{Cl} \rightarrow \mathrm{Cl}^-$ (size increases).

Exceptions to Statement II:

Since anions are larger than their corresponding neutral atoms, the statement that "ions are always smaller in size than the corresponding elements" is incorrect.

Conclusion:

Statement II is false.

Final Answer:

Statement I is correct but Statement II is false.

Statement I:

"In group 13, the stability of +1 oxidation state increases down the group."

Analysis:

Group 13 Elements:

Boron (B)

Aluminum (Al)

Gallium (Ga)

Indium (In)

Thallium (Tl)

Common Oxidation States:

The typical oxidation state for group 13 elements is +3.

However, due to the inert pair effect, the +1 oxidation state becomes more stable as we move down the group.

Inert Pair Effect:

The reluctance of the s-electrons (ns²) in the valence shell to participate in bonding.

This effect becomes more significant in heavier elements due to poor shielding by inner d and f orbitals.

Stability Trend:

Boron (B): Exhibits only the +3 oxidation state.

Aluminum (Al): Predominantly +3 oxidation state.

Gallium (Ga): +3 is more stable, but +1 oxidation state starts appearing.

Indium (In): +1 oxidation state becomes significant.

Thallium (Tl): The +1 oxidation state is more stable than the +3 oxidation state.

Conclusion for Statement I:

Correct.

The stability of the +1 oxidation state increases as we move down group 13 due to the inert pair effect.

Statement II:

"The atomic size of gallium is greater than that of aluminium."

Analysis:

Expected Trend:

Generally, atomic size increases down a group because each successive element has an additional electron shell.

Actual Atomic Radii:

Aluminum (Al): Approximately 143 picometers (pm)

Gallium (Ga): Approximately 135 pm

Anomalous Behavior:

**Gallium's atomic radius is *slightly smaller* than that of aluminum.**

This anomaly is due to the presence of 10 d-electrons in gallium's electron configuration (Ga: [Ar] 3d¹⁰ 4s² 4p¹).

Poor Shielding Effect:

3d electrons do not shield the nuclear charge effectively.

As a result, the effective nuclear charge increases, pulling the outer electrons closer to the nucleus.

This causes gallium to have a smaller atomic radius compared to aluminum.

Conclusion for Statement II:

Incorrect.

The atomic size of gallium is less than that of aluminum due to poor shielding by d-electrons.

Final Answer:

Statement I is correct, as the stability of the +1 oxidation state increases down group 13.

Statement II is incorrect, because gallium has a smaller atomic radius than aluminum.

Answer: Option B

To answer this question, let's address each statement individually.

(A) All are isoelectronic

An isoelectronic species is a group of ions or atoms which have the same number of electrons. The electron configuration for each species is as follows:

• $\mathrm{O}^{2-}$: Oxygen has 8 protons and normally 8 electrons. Gaining 2 electrons gives it a total of 10 electrons.
• $\mathrm{F}^{-}$: Fluorine has 9 protons and normally 9 electrons. Gaining 1 electron gives it a total of 10 electrons.
• $\mathrm{Na}^{+}$: Sodium has 11 protons and normally 11 electrons. Losing 1 electron leaves it with 10 electrons.
• $\mathrm{Mg}^{2+}$: Magnesium has 12 protons and normally 12 electrons. Losing 2 electrons leaves it with 10 electrons.

All of these ions have the same number of electrons (10), making them isoelectronic. Therefore, statement (A) is correct.

(B) All have the same nuclear charge

The nuclear charge refers to the total charge within the nucleus, which is determined by the number of protons. The nuclear charges are:

• $\mathrm{O}^{2-}$: 8 protons
• $\mathrm{F}^{-}$: 9 protons
• $\mathrm{Na}^{+}$: 11 protons
• $\mathrm{Mg}^{2+}$: 12 protons

Since the number of protons varies among these species, they do not have the same nuclear charge. Therefore, statement (B) is incorrect.

(C) $\mathrm{O}^{2-}$ has the largest ionic radii

In a series of isoelectronic ions, the ionic radius decreases with increasing nuclear charge because the greater the nuclear charge, the more strongly the electrons are pulled towards the nucleus, reducing the size of the ion. As $\mathrm{O}^{2-}$ has the lowest nuclear charge among the given ions, it will have the largest ionic radius. Therefore, statement (C) is correct.

(D) $\mathrm{Mg}^{2+}$ has the smallest ionic radii

Following the same logic as above, $\mathrm{Mg}^{2+}$, having the highest nuclear charge among the given isoelectronic species, will have the smallest ionic radius because its electrons are held most tightly by the nucleus. Thus, statement (D) is correct.

Given the evaluations above, the most appropriate answer is:

Option B (A), (C) and (D) only.

I. $\mathrm{E}_1$ of $\mathrm{TI}=589 \mathrm{~kJ} / \mathrm{mol}$

I. $\mathrm{E}_1$ of $\mathrm{Ga}=579 \mathrm{~kJ} / \mathrm{mol}$

I. $\mathrm{E_1}$ of $\mathrm{Al}=577 \mathrm{~kJ} / \mathrm{mol}$

First ionization enthalpy of $\mathrm{F}>\mathrm{Cl}$ as first IE decreases down the group.

Similarly, first IE of $\mathrm{Li}>\mathrm{Na}$.

So, statement-I is correct.

Electron gain enthalpy of given elements are negative. Considering the magnitude the given order is correct.

Thus, statement-II is correct

Correct ionization enthalpy order :

$\quad$B < Be < O < N < F

or E < C < A < B < D

N, F and O will not satisfy the given condition.