Definition of Isoelectronic Species

Isoelectronic species are defined as atoms, molecules, or ions that possess the same total number of electrons. Despite having different nuclear charges (different atomic numbers), these species exhibit identical electronic configurations in terms of the total count of their extranuclear electrons [Per the Principles of Atomic Structure].

Step 1: Analyzing the Target Ion $Mg^{2+}$

To identify an isoelectronic species for $Mg^{2+}$, we must first determine the total number of electrons in the target ion.

• Atomic number of Magnesium ($Z$): $12$
• Neutral Magnesium ($Mg$) electron count: $12$ electrons
• Ionization state: The $2+$ charge indicates the loss of $2$ electrons from the neutral atom.
• Calculation: $12 - 2 = 10 \text{ electrons}$.

Step 2: Identifying Potential Isoelectronic Candidates

We require a species (atom or ion) that also contains exactly $10$ electrons. We examine elements with atomic numbers near $10$.

Candidate: Neon ($Ne$)

• Atomic number of Neon ($Z$): $10$
• Electron count: $10$ electrons.
• Status: $Ne$ is isoelectronic with $Mg^{2+}$.

Candidate: Fluoride ion ($F^-$)

• Atomic number of Fluorine ($Z$): $9$
• Ionization state: The $1-$ charge indicates the gain of $1$ electron.
• Calculation: $9 + 1 = 10 \text{ electrons}$.
• Status: $F^-$ is isoelectronic with $Mg^{2+}$.

Candidate: Sodium ion ($Na^+$)

• Atomic number of Sodium ($Z$): $11$
• Ionization state: The $1+$ charge indicates the loss of $1$ electron.
• Calculation: $11 - 1 = 10 \text{ electrons}$.
• Status: $Na^+$ is isoelectronic with $Mg^{2+}$.

Visual Representation of Isoelectronic Configuration

Comparative Data Table

Final Solution: A species isoelectronic with $Mg^{2+}$ is the Sodium ion ($Na^+$), the Fluoride ion ($F^-$), or the Neon atom ($Ne$), as all possess a total of $10$ electrons.

Step 1: Theoretical Basis of Atomic Radii

The radius of an atom is defined as the distance from the center of the nucleus to the outermost shell containing electrons. When an atom loses an electron to become a positively charged ion (cation), the structural configuration of the atom undergoes a significant transition governed by the principles of electrostatic attraction and electron shielding [Per the Bohr-Sommerfeld model and principles of quantum mechanics].

Step 2: Analysis of Force Distribution

When an electron is removed, the remaining electrons experience a higher effective nuclear charge ($Z_{eff}$). This occurs because the total number of protons ($Z$) remains constant while the number of electrons decreases. Consequently, the remaining electrons are pulled more strongly toward the nucleus by the electrostatic force of attraction, defined by Coulomb’s Law:

$F = k \cdot \frac{q_1 \cdot q_2}{r^2}$

Where $q_1$ is the nuclear charge and $q_2$ is the charge of the electron shell. As the ratio of protons to electrons increases, $Z_{eff}$ increases, resulting in a stronger contraction of the electron cloud.

Step 3: Reduction in Electron-Electron Repulsion

The removal of an electron reduces the overall magnitude of electron-electron repulsion within the valence shell and between the valence shell and inner shells. With decreased inter-electronic repulsion, the orbital contraction is further facilitated, drawing the electron density closer to the nucleus [Based on Slater’s Rules for shielding constants].

Step 4: Geometric Visualization of Cationic Shrinkage

Step 5: Comparative Summary

Final Solution: The radius of an atom decreases when it loses an electron because the reduction in the number of electrons leads to a higher effective nuclear charge ($Z_{eff}$) acting on the remaining electrons, combined with decreased inter-electronic repulsion, which results in the contraction of the electron cloud toward the nucleus.

Step 1: Theoretical Foundation of Ionization Enthalpy

The ionization enthalpy ($IE$) is defined as the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. Mathematically, for a hydrogen atom, this corresponds to the energy required to transition an electron from the ground state ($n=1$) to an infinite distance from the nucleus ($n=\infty$), where the potential energy of the electron is defined as zero.

The energy change ($\Delta E$) is expressed as:

$\Delta E = E_{\infty} - E_{ground}$

[Per the Bohr model of the hydrogen atom, where $E_{\infty} = 0$ J]

Step 2: Calculation of Energy for One Mole of Atoms

Given the energy of an electron in the ground state of a single hydrogen atom ($E_1$):

$E_1 = -2.18 \times 10^{-18} \, J$

The energy required to ionize one hydrogen atom is:

$\Delta E = 0 - (-2.18 \times 10^{-18} \, J) = +2.18 \times 10^{-18} \, J$

[Justification: By the Principle of Conservation of Energy, the energy absorbed must equal the difference between the final and initial states.]

Step 3: Conversion to Molar Enthalpy

To calculate the ionization enthalpy in terms of $J \, mol^{-1}$, we must multiply the energy per atom by the Avogadro constant ($N_A$), which represents the number of atoms in one mole ($6.022 \times 10^{23} \, mol^{-1}$).

$IE = \Delta E \times N_A$

$IE = (2.18 \times 10^{-18} \, J \cdot atom^{-1}) \times (6.022 \times 10^{23} \, atom \cdot mol^{-1})$

Step 4: Algebraic Evaluation

Executing the multiplication of the coefficients and the exponents:

• Coefficient product: $2.18 \times 6.022 \approx 13.12796$
• Exponent sum: $10^{-18} \times 10^{23} = 10^{5}$

$IE = 13.12796 \times 10^5 \, J \, mol^{-1}$

Normalizing to scientific notation:

$IE = 1.312796 \times 10^6 \, J \, mol^{-1}$

Final Solution: The ionization enthalpy of atomic hydrogen is $1.31 \times 10^6 \, J \, mol^{-1}$.

Step 1: Identification of Element 71

The element with atomic number $Z = 71$ is Lutetium ($Lu$). [Per the Periodic Table of Elements, Lutetium is the final element in the Lanthanide series].

Step 2: Electronic Configuration Analysis

The ground state electronic configuration of Lutetium ($Lu$) is $[Xe] 4f^{14} 5d^1 6s^2$. [By the Aufbau Principle, the filling of the $4f$ subshell is complete at Ytterbium ($Z=70$), and the third electron enters the $5d$ orbital for Lutetium].

Given the configuration, Lutetium predominantly exhibits a stable oxidation state of $+3$ ($Lu^{3+}$) by losing the three electrons in the $5d^1 6s^2$ orbitals to achieve a noble gas configuration consistent with the $[Xe]$ core.

Step 3: Identification of Fluorine and its Valency

Fluorine ($F$) is a Group 17 (Halogen) element with an atomic number of $9$. Its electronic configuration is $1s^2 2s^2 2p^5$.

Fluorine requires one electron to achieve the stable octet configuration of Neon ($1s^2 2s^2 2p^6$). Consequently, Fluorine typically forms the fluoride ion, $F^-$, exhibiting a valency of $1$.

Step 4: Determination of the Chemical Formula (Criss-Cross Method)

To form a stable binary compound, the total positive charge must neutralize the total negative charge to maintain electrical neutrality [Per the Law of Constant Proportions].

Using the criss-cross method: The magnitude of the valency of $Lu$ ($3$) becomes the subscript for $F$, and the magnitude of the valency of $F$ ($1$) becomes the subscript for $Lu$.

Formula = $Lu_1 F_3$

Visual Representation of the Bonding Mechanism

Final Solution: The formula of the stable binary compound formed by Lutetium and Fluorine is $LuF_3$ (Lutetium(III) fluoride).

Theoretical Foundation: Atomic and Ionic Radius Dynamics

The radius of an atom when it loses an electron undergoes a process known as the formation of a cation. The change in the atomic radius is governed by the interplay between nuclear charge ($Z$) and the total number of electrons ($e^-$), which dictates the effective nuclear charge experienced by the valence shell.

Step 1: Analysis of the Initial State

In a neutral atom, the nuclear charge $Z$ (the number of protons) is exactly balanced by the number of electrons. The atomic radius is defined as the distance from the center of the nucleus to the outermost shell containing electrons [Per the definition of covalent or metallic radius].

Step 2: The Mechanism of Electron Loss

When an atom loses one or more electrons to form a positive ion ($M \rightarrow M^+ + e^-$), two primary factors facilitate a reduction in size:

• Increased Effective Nuclear Charge ($Z_{eff}$): Because the number of protons remains constant while the number of electrons decreases, the remaining electrons experience a stronger electrostatic attraction toward the nucleus [Per Coulomb's Law, $F = k\frac{q_1q_2}{r^2}$].
• Reduction in Electron-Electron Repulsion: With fewer electrons in the valence shell, the inter-electronic repulsion is significantly diminished, allowing the remaining electron cloud to be pulled closer to the nucleus.

Step 3: Structural Visualization

Step 4: Comparative Summary

Furthermore, in many cases, the loss of electrons may result in the complete removal of the outermost shell (principal quantum number $n$ decreases), leading to a drastic contraction in the ionic radius compared to the atomic radius [Per the shell model of the atom].

Final Solution: The radius of an atom decreases upon the loss of an electron because the increase in effective nuclear charge pulls the remaining electrons closer to the nucleus, and the reduction in inter-electronic repulsion allows for a more compact electron cloud. Consequently, the ionic radius of a cation is always smaller than the atomic radius of its parent atom.

1. Theoretical Foundation of Non-Metallic Character

Non-metallic character is defined as the tendency of an element to gain electrons to form anions or to share electrons to form covalent bonds. This property is directly proportional to electronegativity and electron gain enthalpy, and inversely proportional to atomic radius [Per Periodic Trends of Elements].

In the periodic table:

• Across a Period (Left to Right): Non-metallic character increases as effective nuclear charge ($Z_{eff}$) increases and atomic size decreases.
• Down a Group (Top to Bottom): Non-metallic character decreases as atomic size increases and valence electrons are further from the nucleus, reducing electronegativity.

2. Analysis of Given Elements

We are analyzing Boron ($B$), Carbon ($C$), Nitrogen ($N$), Fluorine ($F$), and Silicon ($Si$).

3. Sequential Comparison

Step A: Compare elements in Period 2 ($B, C, N, F$)
Following the periodic trend across a period, electronegativity increases from left to right. Therefore, the order of non-metallic character for Period 2 elements is $B < C < N < F$.

Step B: Compare Carbon ($C$) and Silicon ($Si$)
Both $C$ and $Si$ belong to Group 14. Silicon is below Carbon. Per the group trend, non-metallic character decreases down a group due to the increase in atomic size and shielding effect. Thus, $Si < C$.

Step C: Integration
Since $Si < C$ and $B < C < N < F$, we integrate these observations. We know that Silicon is a metalloid and less non-metallic than Carbon. Boron is also a metalloid. Comparing $Si$ and $B$, Silicon is more metallic than Boron [due to larger size]. Combining these: $Si < B < C < N < F$.

4. Visual Representation of Trends

5. Conclusion

By evaluating the electronegativity values ($F: 4.0, N: 3.0, C: 2.5, B: 2.0, Si: 1.8$), the quantitative trend confirms the qualitative periodic arrangement [Pauling Scale reference].

Final Solution: The correct order of non-metallic character is $Si < B < C < N < F$.

Definition of Isoelectronic Species

Isoelectronic species are defined as atoms, molecules, or ions that possess the same total number of electrons. Despite having identical electron counts, these species exhibit different chemical properties due to variations in nuclear charge ($Z$), which affects the attraction exerted by the nucleus on the electron cloud [Per Atomic Theory and periodic trends].

Step 1: Analyzing the Target Species ($Ar$)

The target species is the noble gas Argon ($Ar$).
Atomic number of $Ar$ ($Z$) = $18$.
Since $Ar$ is a neutral atom, the number of electrons is equal to the atomic number: $18$ electrons.

Step 2: Identifying Isoelectronic Counterparts

To find species isoelectronic with $Ar$, we must identify ions that result in a total of $18$ electrons. We use the formula for ion electron count: $e^- = Z - \text{charge}$.

• For a Cation ($X^{n+}$): The species must have $18 + n$ protons.
  • Consider Potassium ($K$, $Z=19$): $K^+ \rightarrow 19 - 1 = 18$ electrons.
  • Consider Calcium ($Ca$, $Z=20$): $Ca^{2+} \rightarrow 20 - 2 = 18$ electrons.
• For an Anion ($Y^{n-}$): The species must have $18 - n$ protons.
  • Consider Chlorine ($Cl$, $Z=17$): $Cl^- \rightarrow 17 + 1 = 18$ electrons.
  • Consider Sulfur ($S$, $Z=16$): $S^{2-} \rightarrow 16 + 2 = 18$ electrons.

Step 3: Comparative Analysis of Electron Configuration

All identified species share the same electronic configuration: $1s^2 2s^2 2p^6 3s^2 3p^6$.

Visual Representation of Isoelectronic Orbitals

Final Solution: Examples of species isoelectronic with $Ar$ ($18$ electrons) include $Cl^-$, $K^+$, $S^{2-}$, and $Ca^{2+}$.

Theoretical Foundation: The Quantum Mechanical Basis of Periodicity

The modern periodic table is constructed on the principle of the Aufbau process, which governs the filling of atomic orbitals based on increasing energy levels. In this structural arrangement, the horizontal rows are defined as periods, while the vertical columns are defined as groups. [Per the Bohr-Bury scheme and the Pauli Exclusion Principle].

Step 1: Identifying the Quantum Number Association

The position of an element in the periodic table is determined by its electronic configuration. Specifically, the period number corresponds to the principal quantum number ($n$) of the valence shell (outermost shell) being filled for that specific element.

Step 2: Analysis of the Relationship

• For any element, the principal quantum number ($n$) represents the main energy level.
• The periodic table is arranged such that all elements within a single horizontal row share the same highest principal quantum number ($n_{max}$) for their valence electrons.
• Example: For an element like Sodium ($Z=11$), the configuration is $1s^2 2s^2 2p^6 3s^1$. The highest value of $n$ is $3$, which places Sodium in Period 3.

Step 3: Visual Representation of Quantum Shells

Step 4: Comparative Summary

The period number in the modern periodic table indicates the value of the principal quantum number ($n$) for the valence shell of the elements in that period.

Final Solution: The period indicates the value of the principal quantum number ($n$) of the outermost or valence shell of the atom.

Step 1: Substitution of Principal Quantum Number

Given the general outer electronic configuration for the element as $(n-1)d^{2}ns^{2}$ and the specified value of $n=4$, we perform the substitution:

• For $n = 4$:
• $(n-1)d = (4-1)d = 3d$
• $ns = 4s$

Step 2: Determination of Atomic Number and Identification

To identify the element, we determine the full electronic configuration by filling the preceding inner orbitals ($1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}$): $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{2}$ Summing the electrons: $2 + 2 + 6 + 2 + 6 + 2 + 2 = 22$. The element with atomic number $Z = 22$ is Titanium (Ti).

Step 3: Determination of Group and Period

We classify the element based on its orbital occupancy:

• Period: The highest principal quantum number ($n$) in the configuration is 4. Therefore, the element belongs to Period 4. [Rule: The principal quantum number of the valence shell corresponds to the period number].
• Group: For $d$-block elements, the group number is calculated as the sum of electrons in $(n-1)d$ and $ns$ orbitals. Group Number = (Number of $d$-electrons) + (Number of $s$-electrons) Group Number = $2 + 2 = 4$. Therefore, the element belongs to Group 4. [Rule: For transition elements, Group = $(n-1)d + ns$ electrons].

Step 4: Verification via Visualization

Summary of Classification

Final Solution: The element is Titanium (Ti), situated in Period 4, Group 4 of the periodic table.

Step 1: Identifying the Foundational Property

Dmitri Mendeleev formulated his periodic classification based on the atomic mass of the elements. His central thesis, known as the Mendeleev’s Periodic Law, states: "The properties of the elements are a periodic function of their atomic masses."

Mendeleev arranged elements in increasing order of their atomic masses, observing that elements with similar physical and chemical properties recur at regular intervals [Per Mendeleev's Periodic Law].

Step 2: Analysis of Adherence to the Principle

Mendeleev did not strictly adhere to the increasing order of atomic masses in all instances. He demonstrated significant scientific flexibility to ensure that elements with similar properties were grouped in the same family (Group). When a strict adherence to the atomic mass sequence conflicted with the chemical properties of an element, he prioritized the chemical properties.

Step 3: Evidence of Deviations

The following instances exemplify his departure from the strict atomic mass sequence:

• Inversion of Masses: In several cases, an element with a higher atomic mass was placed before an element with a lower atomic mass to maintain horizontal periodicity and vertical group similarity. For example, Tellurium (atomic mass $127.6\text{ u}$) was placed before Iodine (atomic mass $126.9\text{ u}$) because the properties of Tellurium resembled those of Group 16 elements (Oxygen, Sulfur), and Iodine resembled Group 17 elements (Fluorine, Chlorine).
• Predictive Gaps: He left gaps for elements that were undiscovered at the time (e.g., Eka-aluminium, later discovered as Gallium), predicting their properties based on the trends of the surrounding elements rather than forcing an existing, incompatible element into those spots [By Periodic Law Application].

Step 4: Comparative Summary of Classification Logic

Final Solution: Mendeleev used atomic mass as the primary property for classification; however, he did not stick to it strictly. Whenever the atomic mass order conflicted with the requirement that elements with similar properties be placed in the same group, he prioritized the chemical and physical properties (valency and reactivity) over the strict numerical sequence of atomic mass.

Step 1: Theoretical Framework for Metallic Reactivity

The chemical reactivity of a metal is directly proportional to its ability to lose valence electrons, a process quantified by the first ionization enthalpy ($\Delta_{i}H_{1}$). A lower $\Delta_{i}H_{1}$ indicates that the valence electron is less tightly bound to the nucleus and can be removed with less energy input [Per Periodic Trends in Metallic Character].

Step 2: Analysis of Given Data

We evaluate the provided elements based on their $\Delta_{i}H_{1}$ values (in $\text{kJ mol}^{-1}$):

Step 3: Comparative Evaluation

• Element I: $\Delta_{i}H_{1} = 520 \text{ kJ mol}^{-1}$
• Element II: $\Delta_{i}H_{1} = 419 \text{ kJ mol}^{-1}$
• Element III: $\Delta_{i}H_{1} = 1681 \text{ kJ mol}^{-1}$
• Element IV: $\Delta_{i}H_{1} = 1008 \text{ kJ mol}^{-1}$
• Element V: $\Delta_{i}H_{1} = 2372 \text{ kJ mol}^{-1}$
• Element VI: $\Delta_{i}H_{1} = 738 \text{ kJ mol}^{-1}$

By comparing the values, Element II possesses the lowest first ionization enthalpy ($\Delta_{i}H_{1} = 419 \text{ kJ mol}^{-1}$). A lower value signifies the highest ease of ionization, which is the defining characteristic of a highly reactive metal [By Definition of Chemical Reactivity in Group 1 Metals].

Step 4: Visualizing Ionization Energy Trends

Final Solution: Element II is the most reactive metal because it requires the least amount of energy to lose its valence electron, as evidenced by its lowest first ionization enthalpy of $419 \text{ kJ mol}^{-1}$.

1. Conceptual Foundation

The first ionization enthalpy ($\Delta_iH$) is defined as the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. The process is represented by the equation:

$X(g) \rightarrow X^+(g) + e^-$

2. Defining Isotopes

Isotopes are atoms of the same element that possess an identical number of protons ($Z$) and electrons ($e^-$), but a differing number of neutrons ($N$). Consequently, they possess identical electronic configurations, which is the primary determinant of chemical behavior and energy levels [Per the Pauli Exclusion Principle and Aufbau Principle].

3. Analysis of Variables Affecting Ionization Enthalpy

The energy required to remove an electron depends on the effective nuclear charge ($Z_{eff}$) and the distance of the electron from the nucleus ($r$):

• Electronic Configuration: Since isotopes have the same atomic number ($Z$), they have the same number of electrons arranged in identical shells and subshells.
• Nuclear Attraction: The electrostatic force of attraction between the nucleus and the valence electron is governed by Coulomb's Law: $F = k \frac{q_1 q_2}{r^2}$. Because $q_1$ (number of protons) and $q_2$ (electron charge) are identical for isotopes, and the orbital radii ($r$) are defined by the quantum principal number $n$, the force remains effectively identical.

4. Comparative Evaluation

5. Justification

While isotopes differ in mass due to the number of neutrons, neutrons do not participate in electronic interactions or contribute to the electrostatic screening of the valence electrons. Although there exists a negligible "nuclear mass effect" (related to the slight difference in nuclear motion), it is orders of magnitude smaller than the electronic transition energies involved in ionization. Therefore, in the context of standard chemical periodicity, the first ionization enthalpies are considered identical [Per the Born-Oppenheimer approximation, which separates electronic and nuclear motion].

Final Solution: The first ionization enthalpies for two isotopes of the same element are expected to be the same because they possess identical electronic configurations and effective nuclear charges.

Step 1: Theoretical Analysis of Binary Halide Formation

A stable binary halide of the formula $MX_2$ implies that the metal $M$ typically exhibits an oxidation state of $+2$. For an element to form a stable $MX_2$ compound, the energy required to remove two electrons (the sum of the first and second ionization enthalpies, $\Delta_i H_1 + \Delta_i H_2$) must be low enough to be compensated by the lattice energy of the resulting ionic crystal or the bond dissociation energy in a covalent network [Per Born-Haber Cycle principles].

Step 2: Evaluating the Energetics of Potential Candidates

We analyze the sum of the first two ionization enthalpies ($\Sigma\Delta_i H = \Delta_i H_1 + \Delta_i H_2$) for the provided elements to identify which are most likely to form divalent cations ($M^{2+}$):

Step 3: Justification for Element VI

• Elements I and V: The jump between $\Delta_i H_1$ and $\Delta_i H_2$ is massive (I: $7300 - 520 = 6780$; V: $5251 - 2372 = 2879$), indicating that these are likely Alkali Metals (Group 1), which form $M^+$ ions, not $M^{2+}$.
• Element VI: The $\Sigma\Delta_i H$ value is the lowest in the table ($2189$ kJ $mol^{-1}$). A relatively low $\Delta_i H_2$ suggests that the second electron is removed from the same shell or subshell without a massive energy penalty, which is characteristic of Alkaline Earth Metals (Group 2, e.g., Magnesium). Group 2 elements characteristically form stable $MX_2$ halides (e.g., $MgCl_2$).
• Comparison: While Element IV also has a relatively low $\Sigma\Delta_i H$, Element VI fits the periodic trend of an Alkaline Earth Metal more precisely, where the configuration allows for the stable loss of two valence $s$-electrons.

Step 4: Verification of Periodic Properties

The ionization energy profile of Element VI ($738$ kJ $mol^{-1}$ for $\Delta_i H_1$ and $1451$ kJ $mol^{-1}$ for $\Delta_i H_2$) matches the expected trend for a metal located in Group 2 of the Periodic Table. By the Law of Conservation of Energy and Lattice Energy stability, the formation of $M^{2+}$ is exothermic or energetically favorable for such metals when forming halides [Per Born-Haber Cycle].

Final Solution: Element VI is the metal which can form a stable binary halide of the formula $MX_2$.

Analysis of Statements Regarding the Modern Periodic Table

To determine the incorrect statement, we must evaluate the fundamental principles governing the Modern Periodic Law and the electronic configuration of elements.

The Modern Periodic Law states: "The physical and chemical properties of the elements are periodic functions of their atomic numbers." [Per Moseley's Law].

Step 1: Evaluation of Core Theoretical Principles

Step 2: Technical Justification of the Incorrect Statement

The period number corresponds to the principal quantum number ($n$) of the valence shell. For the first period:

• The electron configuration involves the $1s$ orbital.
• Maximum capacity $N = 2n^2$ is not applicable here as a simple sum; rather, it is limited by the $1s$ subshell capacity.
• $1s$ subshell holds a maximum of $2$ electrons.
• Therefore, the first period contains exactly $2$ elements: $H$ ($1s^1$) and $He$ ($1s^2$).
• The assertion that it contains $8$ elements is mathematically and physically inconsistent with the Aufbau Principle.

Visual Representation of Period 1 Configuration

Final Solution: The incorrect statement is "The first period contains 8 elements," as the first period consists of only 2 elements ($H$ and $He$) corresponding to the filling of the $1s$ orbital.

Step 1: Application of the Aufbau Principle

To locate the element with atomic number $Z = 114$ in the periodic table, we must determine its electronic configuration by filling orbitals in increasing order of energy, as dictated by the Aufbau Principle [Per the $(n+l)$ rule].

The sequence of orbital filling is: $1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p$.

Distributing $114$ electrons:

$1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 6s^2, 4f^{14}, 5d^{10}, 6p^6, 7s^2, 5f^{14}, 6d^{10}, 7p^2$

Summing the electrons: $2+2+6+2+6+2+10+6+2+10+6+2+14+10+6+2+14+10+2 = 114$.

Step 2: Identification of Period and Group

• Period Identification: The principal quantum number $n$ represents the period. The highest value of $n$ in the valence shell configuration is $7$. Therefore, the element belongs to Period 7.
• Group Identification: For a $p$-block element, the group number is calculated as $10 + (\text{number of electrons in } ns + \text{number of electrons in } np)$ [Per IUPAC Periodic Table Classification rules].
• Calculation: $10 + (2 + 2) = 14$. Therefore, the element belongs to Group 14.

Visual Representation of Valence Shell Position

Analytical Summary

Final Solution: The element with atomic number $Z = 114$ is located in Period 7 and Group 14 of the periodic table (also known as Flerovium, Fl).

Step 1: Historical Context of Synthetic Elements

The discovery and naming of transuranium elements (elements with atomic numbers $Z > 92$) were primarily conducted by research groups at the Lawrence Berkeley National Laboratory (LBNL). The group led by Glenn T. Seaborg was instrumental in the synthesis of several transuranic elements, specifically those in the actinide series and the initial transactinide series.

Step 2: Identifying the Elements Synthesized by Seaborg's Group

Seaborg's group at the University of California, Berkeley, was responsible for the synthesis of the following elements:

• Americium ($Z = 95$)
• Curium ($Z = 96$)
• Berkelium ($Z = 97$)
• Californium ($Z = 98$)
• Einsteinium ($Z = 99$)
• Fermium ($Z = 100$)
• Mendelevium ($Z = 101$)
• Nobelium ($Z = 102$)
• Seaborgium ($Z = 106$) [Named in honor of Glenn T. Seaborg]

Step 3: Answering the Specific Identification

In the context of the IUPAC nomenclature and historical periodic table questions, Seaborgium (element 106) is the element explicitly named after Glenn T. Seaborg. However, if the question refers to the element synthesized by his group that historically completed the actinide series, it is Lawrencium ($Z = 103$), or the element Seaborgium ($Z = 106$), which carries the namesake of the group leader.

Visual Representation of Actinide Synthesis

Analytical Summary

The naming of elements following the transuranium synthesis often reflects the geographical location (e.g., Berkelium, Californium) or the scientists involved. Per the IUPAC recommendations regarding the naming of elements:

Final Solution: The element specifically named for the leader of the group is Seaborgium ($Z = 106$), though the group was responsible for the synthesis of elements 95 through 102, and 106.

Step 1: Electronic Configuration Analysis

To determine the relative ionization enthalpies ($\Delta_{i}H$) of Nitrogen ($N$) and Oxygen ($O$), we must first analyze their ground-state electronic configurations in the second period of the periodic table.

• Nitrogen ($N$, $Z=7$): $1s^{2} 2s^{2} 2p^{3}$
• Oxygen ($O$, $Z=8$): $1s^{2} 2s^{2} 2p^{4}$

[Per the Aufbau Principle and Hund’s Rule of Maximum Multiplicity]

Step 2: Analysis of Subshell Stability

The stability of an atom is significantly influenced by the occupancy of its valence subshells. Specifically, half-filled and fully-filled orbitals possess extra stability due to symmetry and exchange energy.

The $2p$ subshell consists of three degenerate orbitals ($p_{x}, p_{y}, p_{z}$).

• In Nitrogen, the $2p^{3}$ configuration implies one electron in each of the three $p$-orbitals. This is a symmetrical, half-filled configuration, which provides additional stability. [Per Hund's Rule]
• In Oxygen, the $2p^{4}$ configuration implies one orbital is doubly occupied ($p_{x}^{2} p_{y}^{1} p_{z}^{1}$). This introduces inter-electronic repulsion between the two electrons sharing the same $p$-orbital.

Step 3: Energetic Considerations for Ionization

Ionization enthalpy is the energy required to remove the most loosely bound electron from an isolated gaseous atom. The process can be represented as:

$N(g) \rightarrow N^{+}(g) + e^{-}$; $\Delta_{i}H = 1402 \text{ kJ/mol}$

$O(g) \rightarrow O^{+}(g) + e^{-}$; $\Delta_{i}H = 1314 \text{ kJ/mol}$

Although Oxygen has a higher effective nuclear charge ($Z_{eff}$) than Nitrogen—which would normally pull electrons closer and increase $\Delta_{i}H$—the stability factor dominates here.

• Removing an electron from Nitrogen requires breaking the stable, half-filled $2p$ configuration.
• Removing an electron from Oxygen actually relieves the inter-electronic repulsion present in the doubly occupied $2p$ orbital, making the removal energetically more favorable than expected.

Step 4: Comparative Orbital Visualization

Step 5: Conclusion on Periodic Trends

Regarding Fluorine ($F$), it possesses a $2p^{5}$ configuration. Despite the inter-electronic repulsions, the significantly higher $Z_{eff}$ of Fluorine compared to Nitrogen creates a stronger electrostatic attraction between the nucleus and the valence electrons, which overrides the stability factor of Nitrogen. Therefore, the $\Delta_{i}H$ of $N$ is lower than $F$ due to the atomic size and $Z_{eff}$ increase, but Oxygen is lower than Nitrogen due to the destabilization caused by electron-electron repulsion in the $2p^{4}$ orbital.

Final Solution: Oxygen exhibits a lower ionization enthalpy than Nitrogen because Nitrogen's half-filled $2p^{3}$ subshell configuration provides extra electronic stability, whereas the removal of an electron from Oxygen's $2p^{4}$ configuration reduces the inter-electronic repulsions existing in the doubly occupied $p$-orbital.

Step 1: Theoretical Basis of Non-Metallic Character

Non-metallic character is defined by the tendency of an element to accept electrons to form anions. This ability is directly proportional to the element's electronegativity and inversely proportional to its atomic radius. Across a period (from left to right), the effective nuclear charge ($Z_{eff}$) increases, and atomic size decreases, leading to an increase in non-metallic character. Down a group, atomic size increases due to the addition of new shells, which decreases the effective nuclear charge felt by valence electrons, thus decreasing non-metallic character [Per Periodic Trends in Electronegativity].

Step 2: Position in the Periodic Table

We map the given elements within the periodic framework:

Step 3: Comparative Analysis

1. Intra-period Comparison (Period 2): $B, C, N,$ and $F$ all belong to Period 2. Following the periodic law, non-metallic character increases as we move from left to right (Group 13 to 17). Thus, the order for these elements is $B < C < N < F$.

2. Inter-group/Period Comparison ($Si$ vs $C$): $Si$ is located in the same group as $C$ (Group 14) but resides in Period 3. Because non-metallic character decreases down a group due to increased shielding and atomic radius, $Si$ is significantly less non-metallic than $C$.

3. Synthesis: Since $Si$ is less non-metallic than $C$, it must also be less non-metallic than $B$, as $B$ is positioned to the left of $C$ in Period 2 but is a metalloid, whereas $Si$ is a metalloid with even lower electronegativity relative to the small, compact atoms of Period 2. The combined relative non-metallic order is $Si < B < C < N < F$.

Visual Representation of Periodic Trend

Step 4: Final Conclusion

By assessing the electronegativity values ($F \approx 4.0, N \approx 3.04, C \approx 2.55, B \approx 2.04, Si \approx 1.90$ on the Pauling scale), we confirm the logical progression [Per Pauling Scale Data].

Final Solution: The correct order of non-metallic character is $Si < B < C < N < F$.

Step 1: Mendeleev’s Classification Criterion

Dmitri Mendeleev formulated his periodic table based on the atomic mass (or atomic weight) of the elements as the fundamental property. He proposed the Periodic Law, which states that the physical and chemical properties of the elements are a periodic function of their atomic masses [Based on the historical Periodic Law of 1869].

Step 2: Analysis of Adherence to the Principle

Mendeleev did not strictly adhere to the rule of increasing atomic mass. He demonstrated significant scientific intuition by prioritizing the chemical properties (valency, oxide/hydride formation) over the strict sequential order of atomic masses in several instances. Notable deviations include:

• Inversion of pairs: In certain positions, he placed an element with a slightly higher atomic mass before an element with a lower atomic mass to maintain the integrity of groups based on similar chemical properties.
• Example (Tellurium and Iodine): Tellurium ($Te$, atomic mass $127.6$) was placed before Iodine ($I$, atomic mass $126.9$) because Tellurium resembles the elements in Group VI, while Iodine resembles elements in Group VII [Per Mendeleev’s principle of chemical periodicity].
• Gaps for undiscovered elements: Mendeleev left intentional gaps in his table (e.g., Eka-boron, Eka-aluminum, and Eka-silicon), predicting that their properties would be consistent with their group members despite their missing atomic masses.

Step 3: Visualizing the Logic of Periodic Trends

Step 4: Comparative Summary Table

Final Solution: Mendeleev used atomic mass as the primary criterion for classifying elements. He did not strictly adhere to this principle, as he purposefully inverted the order of certain elements—such as Tellurium ($127.6$) and Iodine ($126.9$)—whenever the chemical properties necessitated a deviation from the sequential arrangement based on atomic mass.

Step 1: Theoretical Basis of Oxidizing Property

The oxidizing property of an element is defined by its ability to gain electrons, which is synonymous with its electron affinity and electronegativity. In the periodic table, oxidizing power generally increases across a period (from left to right) due to increasing effective nuclear charge ($Z_{eff}$) and decreases down a group due to increased shielding and atomic radius [Per Periodic Trends in Chemistry].

Step 2: Analysis of Given Elements

• Fluorine (F): Atomic number $Z=9$. Located in Period 2, Group 17. It possesses the highest electronegativity ($4.0$ on the Pauling scale) and a high electron gain enthalpy, making it the strongest oxidizing agent.
• Oxygen (O): Atomic number $Z=8$. Located in Period 2, Group 16. It is more electronegative than Nitrogen and Chlorine, but less so than Fluorine.
• Nitrogen (N): Atomic number $Z=7$. Located in Period 2, Group 15. Due to its half-filled $p$-orbital ($2p^3$), it has a lower tendency to gain electrons compared to O and F.
• Chlorine (Cl): Atomic number $Z=17$. Located in Period 3, Group 17. Although it is a halogen, its oxidizing power is lower than Fluorine due to its larger atomic size and smaller $Z_{eff}$ relative to F and O.

Step 3: Comparative Electronegativity & Oxidizing Power

The oxidizing power follows the order of electronegativity values: $F > O > Cl > N$.

Note: While Cl and N have similar electronegativities, Cl acts as a better oxidizing agent in aqueous solutions due to its ability to attain a stable octet more effectively under standard conditions compared to the chemically inert characteristics often observed with $N_2$.

Step 4: Visualization of Periodic Trend

Step 5: Final Logical Deduction

Based on the periodic trends [Per Periodic Law], the oxidizing power relies on the ease of gaining electrons. Fluorine, being the most electronegative element, leads the sequence. Oxygen follows, and while Chlorine is highly reactive, its position beneath Fluorine in the halogen group designates it as less powerful than F and O. Nitrogen, having the lowest electronegativity among these, concludes the sequence.

Final Solution: The correct order of oxidizing property is $F > O > Cl > N$.