Periodic Table & Periodicity

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 ionization energy of an atom or molecule describes the minimum amount of energy required to remove an electron from the atom or molecule in the gaseous state. It generally increases across a period (from left to right) on the periodic table because the number of protons is increasing, making the nucleus more positively charged and therefore more strongly attracting the negatively charged electrons.

However, there are exceptions to this trend when half-filled or fully-filled sub-shells are present, which are particularly stable configurations. The increased stability of these configurations makes it harder to remove an electron, increasing the ionization energy.

The order of first ionization enthalpy for these elements, considering the above factors, should be:

Li < B < Be < C < O < N < F

This is because:

• Lithium (Li) has the lowest nuclear charge in this group and thus the lowest ionization energy.


• Beryllium (Be) has a fully filled 2s subshell, which increases its ionization energy above that of boron (B).


• Carbon (C) has more protons than boron, increasing its ionization energy.


• Nitrogen (N) has a half-filled 2p subshell, which increases its ionization energy above that of oxygen (O).


• Fluorine (F) has the highest nuclear charge in this group and thus the highest ionization energy.

So the correct answer is: $\mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}$

The correct answer is Option C: Both A and R are correct and R is the correct explanation of A.

• Assertion A: The energy required to form Mg²⁺ from Mg is much higher than that required to produce Mg⁺.

This is true because it takes more energy to remove two electrons from an atom than it does to remove one electron. The second electron is more tightly bound to the nucleus than the first electron, so it requires more energy to remove it.

• Reason R: Mg²⁺ is a small ion and carry more charge than Mg⁺.

This is also true. Mg²⁺ has a smaller radius than Mg⁺ because it has lost two electrons. The smaller radius means that the protons in the nucleus are more tightly packed together, which makes the charge of the nucleus more effective at attracting the electrons. This makes it more difficult to remove an electron from Mg²⁺ than it is to remove an electron from Mg⁺.

The reason is the correct explanation of the assertion because it explains why it takes more energy to form Mg²⁺ than it does to form Mg⁺. The smaller radius of Mg2+ and the higher charge of the nucleus make it more difficult to remove electrons from Mg²⁺, which requires more energy.

Electronegativity is a measure of an atom's ability to attract shared electrons to itself. On the periodic table, electronegativity generally increases as we move from left to right across a period due to increase in effective nuclear charge.

Here are the Pauling electronegativities for the elements:

• Carbon (C): 2.55


• Phosphorus (P): 2.19


• Astatine (At): 2.2


• Bromine (Br): 2.96

Based on these values, the correct order of electronegativity from highest to lowest is Br > C > At > P.

Therefore, Option B is the correct answer:

Br > C > At > P

The melting point of gallium is very low (about 29.76°C or 85.57°F). Therefore, gallium can melt just from the heat of one's hand. The boiling point of gallium, on the other hand, is quite high (2400°C or 4352°F).

On the contrary, water boils at 100°C (212°F) under normal atmospheric pressure. Therefore, inside boiling water, gallium would melt into a liquid but would not boil or evaporate, unlike the other elements listed. Lithium (Li), Bromine (Br), and Cesium (Cs) have boiling points below 100°C, so they would turn into a gas in boiling water.

The correct order of bond strength is

$\mathrm{H_2O > H_2S > H_2Se > H_2Te}$

Statement I is correct as the decrease in first ionization enthalpy from B to Al is indeed much larger than from Al to Ga. This is due to the shielding effect of the electron in the outermost shell and the increased effective nuclear charge as one moves from B to Al and from Al to Ga.

Statement II is also correct, as the d orbitals in Ga are indeed completely filled. This results in the shielding effect being maximum for Ga and thus, the decrease in first ionization enthalpy from Al to Ga is smaller compared to that from B to Al.

The given statements pertain to the concepts of atomic and ionic radii in chemical bonding and periodic properties. Let's analyze each statement individually for clarity.

Statement I: In $\mathrm{Cl}_{2}$ molecule the covalent radius is double of the atomic radius of chlorine.

The atomic radius of an element like chlorine refers to the size of its atoms, typically measured when the element is in its gas phase and not bonded to anything else. The covalent radius, on the other hand, is a measure of the size of an atom that forms part of a single covalent bond - it's essentially half the distance between two atoms bonded together.

In the case of a $\mathrm{Cl}_{2}$ molecule, the covalent bond is formed between two chlorine atoms. Therefore, the distance from the nucleus of one chlorine atom to the nucleus of the other (the bond length) is essentially twice the covalent radius of chlorine. This makes the statement true only if correctly interpreted: the covalent bond length is double the covalent radius, but it tends to be conflated with the concept of the atomic radius. However, it is important to clarify that the atomic radius and covalent radius are different measures. In a diatomic molecule like $\mathrm{Cl}_2$, saying the covalent radius is "double of the atomic radius" is inaccurate. The precise statement should be that the bond length (twice the covalent radius) is not directly twice the atomic radius but rather the distance between the nuclei of the bonding atoms. Therefore, this statement as phrased is misleading or incorrect.

Statement II: Radius of anionic species is always greater than their parent atomic radius.

This statement is based on the idea that when an atom gains electrons and becomes an anion, the increased electron-electron repulsions in the electron cloud cause it to expand. This is generally true across the periodic table. For example, when chlorine gains an electron to become $\mathrm{Cl}^{-}$, its radius increases due to the addition of an extra electron which increases repulsion among electrons and expands the electron cloud. This principle holds for virtually all anions compared to their parent atoms, so Statement II is correct.

Given the analysis, the correct option is:

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

Element $E$ is in period 4 and group 16.

The element in period 4 and group 16 is Selenium $(Se)$.

The element just above Selenium in the same group is Sulphur $(S)$.

Sulphur belongs to period 3, and its valence shell electronic configuration is:

$3s^2\,3p^4$

So, the correct option is:

Option A: $3s^2,\ 3p^4$