Periodic Table & Periodicity

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.