Periodic Table & Periodicity

The correct order of atomic radii in group 13 elements is, $\mathrm{Tl}>\mathrm{In}>\mathrm{Al}>\mathrm{Ga}$.

Usually radii increases down the group, due to addition of shell, but Ga has smaller radii than Al due to poor shielding effect of $d$-electrons.

Amphoteric oxide:

$\mathrm{SnO}, \mathrm{PbO}, \mathrm{SnO}_2, \mathrm{PbO}_2, \mathrm{Ga}_2 \mathrm{O}_3$

Acidic oxide : $\mathrm{CO}_2, \mathrm{~B}_2 \mathrm{O}_3$

Neutral : CO

1. Determine the number of electrons for each ion:

   • $\mathrm{N}^{3-}$: Neutral N has 7 electrons. Gaining 3 electrons makes it 7 + 3 = 10 electrons.

   • $\mathrm{O}^{2-}$: Neutral O has 8 electrons. Gaining 2 electrons makes it 8 + 2 = 10 electrons.

   • $\mathrm{F}^{-}$: Neutral F has 9 electrons. Gaining 1 electron makes it 9 + 1 = 10 electrons.

   • $\mathrm{Na}^{+}$: Neutral Na has 11 electrons. Losing 1 electron makes it 11 - 1 = 10 electrons.

   • $\mathrm{Mg}^{2+}$: Neutral Mg has 12 electrons. Losing 2 electrons makes it 12 - 2 = 10 electrons.

   • $\mathrm{Al}^{3+}$: Neutral Al has 13 electrons. Losing 3 electrons makes it 13 - 3 = 10 electrons.

   All these ions are isoelectronic, meaning they all have the same electron configuration (like Neon, [Ne]).

2. Determine the nuclear charge (number of protons) for each ion:

   • $\mathrm{N}^{3-}$: Z = 7

   • $\mathrm{O}^{2-}$: Z = 8

   • $\mathrm{F}^{-}$: Z = 9

   • $\mathrm{Na}^{+}$: Z = 11

   • $\mathrm{Mg}^{2+}$: Z = 12

   • $\mathrm{Al}^{3+}$: Z = 13

3. Apply the rule for isoelectronic species: For isoelectronic ions, the ionic radius decreases as the nuclear charge (number of protons) increases. A higher nuclear charge means the electrons are pulled more strongly towards the nucleus, resulting in a smaller size.

   Ordering the ions by increasing nuclear charge (and thus decreasing size):

   • $\mathrm{N}^{3-}$ (Z=7) - Largest size

   • $\mathrm{O}^{2-}$ (Z=8)

   • $\mathrm{F}^{-}$ (Z=9)

   • $\mathrm{Na}^{+}$ (Z=11)

   • $\mathrm{Mg}^{2+}$ (Z=12)

   • $\mathrm{Al}^{3+}$ (Z=13) - Smallest size

Therefore, the ion with the largest size is $\mathrm{N}^{3-}$ and the ion with the smallest size is $\mathrm{Al}^{3+}$.

The final answer is $\boxed{\text{D}}$.

The cation with a greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus.

Thus,

$ \begin{aligned} & Q^{a+}=\mathrm{Al}^{3+}, X^{b+}=\mathrm{Mg}^{2+}, Y^{c+}=\mathrm{Si}^{4+} \\ & Z^{d+}=\mathrm{Na}^{+} \end{aligned} $

The order of first ionisation enthalpy is incorrect in option (c). The correct order is $\mathrm{C}<\mathrm{O}<\mathrm{N}$

The order of negative electrons gain enthalpy is

$ \mathrm{Cl}>\mathrm{S}>\mathrm{Li}>\mathrm{Na} $

Electron gain enthalpy becomes more negative with increase in the atomic number across a period. While E.G.E. becomes less negative as we go down the group because the size of the atom increases and the added $e^{-}$are farther away from the nucleus.

The element with the highest electronegativity in the periodic table is fluorine (F). It is located in period 2 and group 17.

The elements of given electronic configuration are $(A) \mathrm{Na},(B) \mathrm{Al}(C) \mathrm{Mg}$ and $(D)$ Si. Thus, the correct order of first ionisation enthalpy of these elements is $D>C>B>A$

Ionisation enthalpy generally increases across a period due to increase nuclear charge and decreasing atomic radius.

The correct order of non-metallic character is

$ \mathrm{F}>\mathrm{N}>\mathrm{C}>\mathrm{B}>\mathrm{Si} $

Non-metallic character increases across the period and decreases down the group.

Among the given orders, order of ionisation enthalpy is incorrect. The correct order is

$ \mathrm{Ge}>\mathrm{Pb}>\mathrm{Sn} $

The correct order of atomic radii is,

$ \mathrm{C}<\mathrm{S}<\mathrm{Al} $

As we go left to right in period, radii decreases. As we go down in group radii increases.

The order of negative electron gain enthalpy is incorrect the correct order is,

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

$ \begin{aligned} &\text { The match is A-III, B-II, C-I, D-IV }\\ &\begin{aligned} & 112-[\mathrm{Rn}] 5 f^{14} 6 d^{10} 7 s^2 \\ & 116-[\mathrm{Rn}] 5 f^{14} 6 d^{10} 7 s^2 7 p^4 \\ & 100-[\mathrm{Rn}] 5 f^{12} 7 s^2 \\ & 88-[\mathrm{Rn}] 7 s^2 \end{aligned} \end{aligned} $

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.

The third ionization enthalpy refers to the energy required to remove the third electron from a di-positive ion ($M^{2+}$) to form a tri-positive ion ($M^{3+}$). In the case of transition metals, the energies involved in removing the third electron are typically higher than for the first and second electrons due to an increasing effective nuclear charge and the electrons being removed from closer energy levels to the nucleus or from a half-filled or full-filled d-subshell which is relatively more stable.

Let us look at the electronic configurations of the ions of each metal listed to determine which would require the highest third ionisation enthalpy. Since we are talking about $3^{\text{rd}}$ ionization, we would be considering the removal of three electrons, ending up with a $+3$ oxidation state for each metal.

• Manganese ($\mathrm{Mn}$): [Ar] $3d^5 4s^2$


• Iron ($\mathrm{Fe}$): [Ar] $3d^6 4s^2$


• Chromium ($\mathrm{Cr}$): [Ar] $3d^5 4s^1$


• Vanadium ($V$): [Ar] $3d^3 4s^2$

After removing two electrons (assuming they come from the 4s orbital and the next available d orbital), the configurations would be:

• $\mathrm{Mn^{2+}}$: [Ar] $3d^5$


• $\mathrm{Fe^{2+}}$: [Ar] $3d^6$


• $\mathrm{Cr^{2+}}$: [Ar] $3d^4$


• $V^{2+}$: [Ar] $3d^3$

Now we look at the energy required to remove the third electron from the $M^{2+}$ ions:

• For $\mathrm{Mn^{2+}}$, removing the third electron would break the half-filled $3d^5$ configuration and require significant energy.
• For $\mathrm{Fe^{2+}}$, removing the third electron would form half-filled $3d^5$ configuration, which forms stable configuration.
• For $\mathrm{Cr^{2+}}$, the third electron removal will disrupt the $3d^4$ configuration, which is not particularly stable.
• For $V^{2+}$, the third electron removal will disrupt the $3d^3$ configuration.

In terms of stability, the half-filled $d^5$ subshell in manganese ($\mathrm{Mn^{2+}}$) provide extra stability. So, the $3^{\text{rd}}$ ionisation enthalpy of $\mathrm{Mn^{2+}}$ is the highest because it involves breaking a half-filled configuration which is more stable.

Thus, out of the given options, manganese ($\mathrm{Mn}$) would have the highest third ionization enthalpy. Therefore, the correct answer is:

Option A : $\mathrm{Mn}$

To answer this question, let's investigate both statements in detail.

Statement (I): Both metals and non-metals exist in p and d-block elements.

The periodic table is divided into blocks based on the electron configuration of the elements. These blocks are labeled s, p, d, and f. The p-block contains a mix of metals, non-metals, and metalloids. For example, the p-block contains nonmetals such as oxygen (O) and nitrogen (N), as well as metals like aluminum (Al) and lead (Pb).

The d-block, also known as the transition metals block, primarily consists of metals. However, the d-block does not typically contain elements that are traditionally classified as non-metals. The elements in the d-block have a wide range of properties but are generally characterized by their metallic traits such as conductivity and malleability.

Therefore, Statement I is not entirely true because while both metals and nonmetals exist in the p-block of the periodic table, the d-block is principally composed of metals.

Statement (II): Non-metals have higher ionisation enthalpy and higher electronegativity than the metals.

Ionization enthalpy (or ionization energy) is the energy required to remove an electron from a gaseous atom or ion. Non-metals generally have higher ionization energies compared to metals because non-metals have more tightly bound electrons to their nucleus. This is partly due to the fact that non-metals tend to have higher electronegativities and smaller atomic radii, meaning that their outer electrons are closer to the nucleus and more strongly attracted to it.

Electronegativity is a measure of an atom's ability to attract and bond with electrons. Non-metals have higher electronegativities typically because they are more eager to gain electrons to achieve a full valence shell, reflecting their position on the right side of the periodic table. Metals, on the other hand, are more inclined to lose electrons and form positive ions, indicating their lower electronegativities.

Hence, Statement II is true as non-metals indeed have higher ionisation enthalpy and higher electronegativity than metals.

Given the analysis:

• Statement I is not true.
• Statement II is true.

Thus, the most appropriate answer is:

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

In general along the period from left to right, size decreases and metallic character decrease.

In general down the group, size increases and metallic character increases.

$\begin{aligned} & \mathrm{B}^{\prime}<\mathrm{A}^{\prime}(\text { size }) \quad \mathrm{C}^{\prime}>\mathrm{A}^{\prime}(\text { size }) \\ & \mathrm{D}^{\prime}<\mathrm{C}^{\prime}(\text { size }) \quad \mathrm{D}^{\prime}>\mathrm{B}^{\prime}(\text { size }) \\ & \mathrm{B}^{\prime}<\mathrm{A}^{\prime}(\text { metallic character }) \\ & \mathrm{D}^{\prime}<\mathrm{C}^{\prime}(\text { metallic character }) \\ & \mathrm{B}^{\prime+}<\mathrm{A}^{\prime+}(\text { size }) \\ & \mathrm{D}^{\prime+}<\mathrm{C}^{\prime+}(\text { size }) \\ & \therefore \text { C statement is incorrect. } \end{aligned}$

$\begin{array}{ll} \text { Element } & \Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{kJ} / \mathrm{mol}) \\ \mathrm{F} & -333 \\ \mathrm{~S} & -200 \\ \mathrm{Br} & -325 \\ \mathrm{Ar} & +96 \end{array}$

Due to lower Bond dissociation enthalpy of $\mathrm{H}_2 \mathrm{Te}$ it ionizes to give $\mathrm{H}^{+}$ more easily than $\mathrm{H}_2 \mathrm{S}$.

Chemical reactivity of elements decreases along the period therefore statement - I is false.

Group - 1 elements from basic nature oxides while group - 17 elements form acidic oxides therefore statement - II is true.

$\begin{gathered} { }_{24} \mathrm{Cr} \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^1 ; \mathrm{Cr}^{2+} \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^4 \\ { }_{25} \mathrm{Mn} \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{s}^2 ; \mathrm{Mn}^{+} \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{s}^1 \\ { }_{28} \mathrm{Ni} \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{s}^2 ; \mathrm{Ni}^{2+} \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^8 \\ { }_{23} \mathrm{V} \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^3 4 \mathrm{s}^2 ; \mathrm{V}^{+} \rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^3 4 \mathrm{s}^1 \end{gathered}$

Statement- 1 is false because chlorine has most negative electron gain enthalpy in its group.

$\mathrm{Al}<\mathrm{Si}<\mathrm{C}<\mathrm{N} ; \mathrm{IE}_1 \text { order. }$

Fact Based.

First ionisation energy increases along the period. Along the period $\mathrm{Z}$ increases which outweighs the shielding effect.

$\begin{array}{ll} \text { Gr-14 } & \text { EN } \\ \mathrm{C} & 2.5 \\ \mathrm{Si} & 1.8 \\ \mathrm{Ge} & 1.8 \\ \mathrm{Sn} & 1.8 \\ \mathrm{~Pb} & 1.9 \end{array}$

The electronegativity values for elements from $\mathrm{Si}$ to $\mathrm{Pb}$ are almost same. So Statement $\mathrm{I}$ is false.

Statement I is correct because the 4f and 5f series elements, commonly referred to as the lanthanides and actinides, respectively, have unique properties and configurations that distinguish them from other elements. Placing them separately in the Periodic table helps to maintain the orderly classification based on atomic numbers and avoids disrupting the pattern established by the periodic law.

Statement II is incorrect because s-block elements, which include Group 1 (alkali metals) and Group 2 (alkaline earth metals), are indeed highly reactive. They typically do not exist in their pure form in nature due to their reactivity. Instead, they are mostly found in compound forms, combined with other elements. For instance, sodium (Na) and potassium (K) from Group 1 react with water and are usually found as salts, not in their metallic, pure state.

The electronic configurations of the given elements are as follows:

Na (11): [Ne] 3s¹

Mg (12): [Ne] 3s²

Al (13): [Ne] 3s² 3p¹

Si (14): [Ne] 3s² 3p²

The ionization energy for Mg is greater than that for Na due to its fully filled 3s orbitals, which offer greater stability. In contrast, Al has a lower ionization energy than Mg because it has one electron more than a stable configuration, making it easier to remove. However, Al's ionization energy is still less than that of Si because Si has a higher effective nuclear charge, which increases its ionization enthalpy.

Thus, the correct order of their first ionization enthalpies is:

$ \text{Na} < \text{Mg} > \text{Al} < \text{Si} $

Among the options provided, only option (A) $ 575 \, \text{kJ/mol} $ conforms to this order.

$ \mathrm{Kr}, \mathrm{Xe}, \mathrm{Rn} $ has positive electron gain enthalpies. All of above gases are noble gases. All the occupied subshell of noble gases are completely filled. Therefore, the additional electron has to be placed in an orbital of next higher shell. We can also say that energy has to be supplied to add an additional electron and hence, electron gain enthalpy of all noble gases are positive.

A is not correct but R is correct.

The correct form of A is $ \mathrm{Na}^{+} $has smaller size than $ \mathrm{F}^{-} $because after loosing an electron Na will has less electron cloud and more effective nuclear charge whereas F after gaining an electron will have less effective nuclear charge than the parent atom.

Hence, $ \mathrm{F}^{-} $size increases.

$ \mathrm{Na}^{+} $has smaller size than $ \mathrm{F}^{-} $. It is because after losing an electron Na will have more effective nuclear charge whereas $ F $ after gaining an electron will have less effective nuclear charge than the parent atom so the size increases.

In the given species the ratio of $s$-electrons and $p$-electrons are same in $\mathrm{K}^{+}, \mathrm{Cr}^{3+}$.

$\mathrm{K}^{+}$(18 electrons) : $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 \quad \ldots \text { (i) }$

$ \begin{gathered} s \text {-electrons }=6 ; p \text {-electron }=12 \\\\ s: p=6: 12 \text { or } 1: 2 \quad \ldots \text { (ii) } \end{gathered} $

Similarly,

$\mathrm{Cr}^{3+}$ (21 electrons) :

$ \begin{gathered} 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^3 \quad \ldots \text { (iii) } \\\\ s \text {-electrons }=6 ; p \text {-electron }=12 \\\\ s: p=6: 12 \text { or } 1: 2 \quad \ldots \text { (iv) } \end{gathered} $

Option D is correct:

A - II: Ra (Radium) belongs to the s-block as it is an alkaline earth metal.

B - I: Uuq (Ununquadium) is now known as Fl (Flerovium) and belongs to the p-block, as it is a part of the post-transition metals.

C - IV: Ds (Darmstadtium) belongs to the d-block, as it is a transition metal.

D - III: Fm (Fermium) belongs to the f-block, as it is an actinide element.

These elements are classified based on their electron configurations, which determine their position in the periodic table's blocks.

Electronic configuration of $P$

$ \Rightarrow \quad \begin{aligned} & 1 s^2 2 s^2 3 p^6 3 s^2 3 p^3 \\\\ & s_{\text {electron }}=6, p_{\text {electron }}=9 \\\\ & s: p \Rightarrow 6: 9 \text { or } 2: 3 \quad \ldots \text { (i) } \end{aligned} $

Electronic configuration of Ca

$ \begin{aligned} & 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 \\\\ & \Rightarrow \quad s_{\text {electron }}=8, p_{\text {electron }}=12 \\\\ & \quad s: p \Rightarrow 8: 12 \text { or } 2: 3 \quad \ldots \text { (ii) } \end{aligned} $

Hence, P and Ca have same ratio of $S_{\text {electron }}$ and $p_{\text {electron }}$.

Electron gain enthalpy generally becomes more negative across a period as we move from left to right. Within a group, electron gain enthalpy becomes less negative down the group. However adding an electron to the $2 p$-orbital leads to greater repulsion than adding an electron to the larger $3 p$-orbital. Hence, most electronegative enthalpy is chlorine and least electron gain enthalpy is oxygen.

Here’s how the elements line up by group number:

Moscovium (Mc, Z = 115) is in group 15 ⇒ III

Livermorium (Lv, Z = 116) is in group 16 ⇒ I

Flerovium (Fl, Z = 114) is in group 14 ⇒ IV

Tennessine (Ts, Z = 117) is in group 17 ⇒ II

That corresponds to Option C.

Statement I: "Nitrogen has more ionisation enthalpy and electronegativity than beryllium."

Ionisation Enthalpy: Nitrogen, being a non-metal and located in group 15, has a higher ionisation energy than beryllium, a metal in group 2. For example, the first ionisation energy of nitrogen is around $1402 \, \text{kJ/mol}$, while for beryllium it is around $900 \, \text{kJ/mol}$.

Electronegativity: Nitrogen is significantly more electronegative than beryllium. This reflects in their chemical behavior, as non-metals tend to attract electrons more strongly than metals.

Conclusion for Statement I: This statement is correct.

Statement II: "$\mathrm{CrO}_3, \mathrm{~B}_2 \mathrm{O}_3$ are acidic oxide."

Chromium Trioxide ($\mathrm{CrO}_3$): This oxide is acidic. It reacts with water to form chromic acid:

$\mathrm{CrO}_3 + \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2\mathrm{CrO}_4.$

Diboron Trioxide ($\mathrm{B}_2 \mathrm{O}_3$): This oxide is also acidic. It reacts with water to yield boric acid:

$\mathrm{B}_2 \mathrm{O}_3 + 3 \, \mathrm{H}_2 \mathrm{O} \rightarrow 2 \, \mathrm{H}_3 \mathrm{BO}_3.$

Conclusion for Statement II: This statement is also correct.

Since both statements are correct, the answer is:

Option A

Both Statement I and Statement II are correct.