Class 11 Chemistry (Chemistry Part-I) Chapter 3 Classification of Elements and Periodicity in Properties

Key Concept: Periodic Table Anomalies, Real-World Application

b) Copper has a filled d-subshell providing extra stability and higher ionization enthalpy.

[Solution Description] Copper (Cu) has an electronic configuration of [Ar] $3d^{10}$ $4s^1$, whereas potassium (K) has an electronic configuration of [Ar] $4s^1$. The fully filled $3d^{10}$ subshell in Cu provides additional stability due to exchange energies, resulting in higher ionization enthalpy. Potassium, with only one electron in the outermost 4s shell, lacks this stability, leading to easier removal of the valence electron.

Key Concept: Naming Controversies

c) Instability and Minute Quantities

[Solution Description] The controversies in naming new elements usually arise due to their instability and being produced in minute quantities. These challenges make it difficult to confirm discoveries and assign official names.

Key Concept: Metallic Character

a) Magnesium (Mg)

[Solution Description] Metallic character refers to how easily an element can lose electrons. The periodic trend shows that metallic character decreases across a period from left to right. Among Mg, Al, Si, and Cl, magnesium is positioned furthest to the left in the periodic table, hence it has the highest metallic character.

Key Concept: Anomalous Behavior, Real-World Application

d) Fluorine

[Solution Description] Anomalies in the second period arise primarily due to small size and high electronegativity. Fluorine, for instance, has a very high electronegativity causing it to participate vigorously in hydrogen bonding, profoundly affecting its interactions in biological and chemical systems. Its small atomic radius also leads to exceptionally high hydration enthalpies, influencing solubility patterns significantly compared to other halogens.

Key Concept: Electronic Configuration

d) Their valence electrons are in s orbitals.

[Solution Description] Elements in the s-block have their outermost electron(s) in an s orbital. The general electronic configuration for s-block elements is $ns^1$ or $ns^2$, where n is the principal quantum number of the valence shell.

Key Concept: Ionization Enthalpy Trend

c) Ionization enthalpy increases due to decreasing atomic size and increasing nuclear charge

[Solution Description]

As we move from left to right across a period in the periodic table, the atomic radius decreases and nuclear charge increases. This results in higher attraction between the nucleus and valence electrons, leading to an increase in ionization enthalpy.

Key Concept: Quantum Numbers

c) Period 4

[Solution Description] The principal quantum number $n = 3$ indicates that the element belongs to the third period of the periodic table. However, since this element also involves 3d orbitals, it actually aligns with the beginning of the fourth period, as d-block elements start from period 4.

Key Concept: Periodic Trends

d) It increases steadily

[Solution Description] As we move from left to right across the periodic table, the nuclear charge increases due to more protons being added to the nucleus. This increased positive charge attracts electrons more strongly, making it harder to remove them and thus increasing the ionization enthalpy.

Key Concept: Oxidation State Calculation

b) +6

[Solution Description] In $SO_3$, oxygen has an oxidation state of -2. Let the oxidation state of sulfur be x. The equation is:

$x + 3(-2) = 0$

Simplifying gives:

$x – 6 = 0

x = +6$

Therefore, the oxidation state of sulfur in $SO_3$ is +6.

Key Concept: Advanced Configuration, Series Complexity

d) Nobelium (Z = 102)

[Solution Description] To solve this, consider the complexity introduced by multiple oxidation states and incomplete f orbitals typical of actinoids. Among actinoids, elements with higher atomic numbers tend to have more complex configurations due to additional electrons filling 5f orbitals and their interactions with other electrons. Hence, an element like Nobelium (Z = 102), characterized by a nearly filled 5f subshell, is expected to have intricate electron-electron interactions affecting its configuration.

Key Concept: Analytical Thinking, Advanced Comparison

c) Both imply recurrence of properties but differ in basis.

[Solution Description] Newlands proposed that every eighth element shares similar properties, aligning with modern electron configurations where elements in the same group have similar outer shell electron configurations. However, while Newlands focused on atomic weights, modern chemistry emphasizes atomic numbers and electron arrangements.

Key Concept: Electronegativity Trends

d) Cs

[Solution Description] Electronegativity decreases down a group as the atomic number increases because the increased distance from the nucleus makes it harder for the nucleus to attract electrons. Of Li (Lithium), Na (Sodium), K (Potassium), and Cs (Cesium), Cesium is at the bottom of this group, so it has the lowest electronegativity.

Key Concept: Systematic Naming

a) Ununquadium

[Solution Description]

The atomic number 114 can be broken down into digits: 1, 1, and 4.

Using the provided roots:

– 1 corresponds to “un”

– 1 corresponds to “un”

– 4 corresponds to “quad”

Therefore, combining these gives: Ununquadium.

Finally, add “ium” at the end to get the full name: Ununquadium.

Key Concept: Electronegativity Trends

c) Rubidium

[Solution Description] Electronegativity is the tendency of an atom to attract electrons towards itself. It increases across a period and decreases down a group. Among the given options, rubidium is located further down a group compared to other elements listed, hence it has the lowest electronegativity.

Key Concept: Analytical Thinking

c) By reordering elements according to atomic number instead of atomic mass

[Solution Description] Mendeleev had arranged elements by increasing atomic mass, but there were several instances where elements with lower atomic masses appeared after those with higher masses due to their chemical properties (such as iodine and tellurium). Moseley’s use of atomic number resolved these inconsistencies because atomic numbers are unique for each element and reflect their true underlying structure, aligning with both physical and chemical properties.

Key Concept: Oxidation State Calculation, Periodic Trends and Reactivity

c) +5

[Solution Description]

To find the oxidation state of phosphorus in $P_4O_{10}$, let the oxidation state of phosphorus be x.

The compound $P_4O_{10}$ consists of 4 phosphorus atoms and 10 oxygen atoms.

The oxidation state for oxygen is typically -2.

The sum of oxidation states should equal zero for a neutral compound:

$4x + 10(-2) = 0$

Solving this equation gives:

$4x – 20 = 0$

$4x = 20$

$x = +5$

Therefore, the oxidation state of P in $P_4O_{10}$ is +5.

Regarding reactivity: Phosphorus in a high oxidation state such as +5 tends to form stable oxides, like $P_4O_{10}$. When it reacts with water, $P_4O_{10}$ forms phosphoric acid ($H_3PO_4$), which indicates that the reactivity involves forming these acidic solutions due to strong electron-withdrawing capacity aligning with periodic trends.

Key Concept: Shielding Effect

c) Electron-electron repulsion in oxygen’s p orbitals

[Solution Description] The first ionization enthalpy of an element depends on its effective nuclear charge and shielding effect. In the second period, oxygen has a lower first ionization enthalpy compared to nitrogen despite having a higher atomic number. This anomaly is due to electron-electron repulsions in the 2p orbital of oxygen, which decreases its ionization energy relative to nitrogen, where such repulsions are less significant.

Key Concept: IUPAC Naming

c) Ununoctium

[Solution Description] According to IUPAC guidelines, the temporary name is formed using Latin-based roots for digits: for 118, “un” for 1, “oct” for 8, and “-ium” at the end. Thus, the name is ununoctium.

Key Concept: Shielding and Penetration Complexities, Advanced Quantum Considerations

a) Lithium (Li)

[Solution Description] The increase in ionization enthalpy from the first to the second stage is primarily influenced by the increase in effective nuclear charge once an electron is removed. When we remove an electron from a neutral atom to form a cation, the remaining electrons experience a greater attraction to the nucleus due to reduced electron-electron repulsion and unchanged nuclear charge. For Li, removing the valence electron leaves a complete inner shell configuration, creating a large jump in ionization enthalpy for the removal of another electron as compared to Be, B, or C where subsequent electrons are still in the same shell.

Key Concept: Advanced Application, Detailed Trend Analysis

c) Addition to higher principal quantum level

[Solution Description] Noble gases possess completely filled valence shells, leading to very stable electronic configurations. Adding an electron requires the addition to a new higher principal quantum level, resulting in unfavorable conditions. The primary contributing factor is the need to input energy to add an electron to a much higher energy level than existing electrons, making the process endothermic and hence having large positive values.

Key Concept: Ionization Enthalpy and Metallic Character

c) Elements with lower ionization enthalpy have greater metallic character.

[Solution Description] Higher ionization enthalpy means it is more difficult to remove an electron, which typically corresponds to non-metallic character. In contrast, lower ionization enthalpy indicates ease of removing an electron, signifying metallic character.

Key Concept: Comparative Analysis, Periodic Trends Analysis

c) The ionic radius of lithium is too small to accommodate such coordination.

[Solution Description] Lithium has a smaller atomic size compared to magnesium, which results in a high charge/radius ratio that limits its ability to expand its valence shell beyond four orbitals. Hence, it cannot accommodate six fluorine atoms as magnesium does.

Key Concept: Comparative Analysis, Theoretical Implications

c) His plotting of atomic volume against atomic weight.

[Solution Description] Lothar Meyer plotted atomic volume against atomic weight to illustrate periodicity. This approach helped in understanding that elements exhibit recurring patterns or trends when arranged by atomic weight. It laid the groundwork for discovering relationships between elemental properties and their atomic structure, leading to the understanding of periodic law.

Key Concept: Triad Analysis, Modern Table Evolution

b) They lacked applicability to many known elements

[Solution Description]

Dobereiner’s Triads were an early attempt to classify elements based on their properties and atomic weights. In each triad, the middle element had an atomic weight approximately equal to the average of the other two. However, this pattern was only applicable to a few sets of elements and could not be generalized across all known elements. Additionally, the discovery of more elements posed challenges that the triads could not address, making them less useful for predicting properties or organizing the table comprehensively.

Key Concept: Oxide Reaction

c) Phosphoric acid ($H_3PO_4$)

[Solution Description] When phosphorus(V) oxide ($P_4O_{10}$) reacts with water, it forms phosphoric acid ($H_3PO_4$). The reaction can be represented as:

$$P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4$$

Key Concept: Shielding and Ionization Energy, Anomalous Ionization Enthalpy

b) Greater electron-electron repulsion in sulfur

[Solution Description] As we move across a period, ionization energy generally increases due to the increased effective nuclear charge, which holds the electrons more tightly. However, phosphorus has a half-filled 3p sublevel, providing extra stability compared to sulfur’s configuration. Sulfur has one more electron than phosphorus in the 3p subshell, resulting in increased electron-electron repulsion within the same orbit, thus decreasing the energy required to remove an electron. Therefore, even though sulfur is further across the period, its additional electron repulsion lowers its ionization enthalpy compared to phosphorus.

Key Concept: Modification of Law

b) Elements are arranged by increasing atomic numbers

[Solution Description] Mendeleev’s Periodic Law states that the properties of elements are a periodic function of their atomic weights. The Modern Periodic Law, however, modifies this to state that the properties are a periodic function of their atomic numbers, recognizing the significance of atomic number as a more fundamental property.

Key Concept: Naming Elements

d) The IUPAC

[Solution Description] The International Union of Pure and Applied Chemistry (IUPAC) is responsible for providing official names for new elements with atomic numbers greater than 100, ensuring a standardized naming system across the scientific community.

Key Concept: Second Order Thinking, Conceptual Complexity

c) Be < B < Li [Solution Description] For second ionization enthalpy, where $X^+(g) \rightarrow X^{2+}(g) + e^-$, after the first ionization, Li becomes like He and has a significantly higher second ionization enthalpy. Be becomes like Li with a much smaller increase, and B follows similarly. Thus, the trend is: Be < B < Li due to effective nuclear charge post first ionization.

Key Concept: Complex Oxide Behavior, Advanced Reactivity

d) $ZnO$

[Solution Description] An oxide that exhibits both acidic and basic properties is classified as amphoteric. Some oxides, particularly those of transition metals, can display varying behavior depending on reactions with strong acids or bases. Zinc oxide ($ZnO$), typically considered amphoteric, acts differently based on reaction conditions, demonstrating both types of behavior especially due to its position in the periodic table where it transitions between metallic and non-metallic character influences.

Key Concept: Advanced Periodic Trends, Real-world Application

c) Weaker metallic bonding due to larger atomic size

[Solution Description] The melting point of alkali metals decreases as we move down the group due to the increasing atomic size and weakening metallic bonding. As the size increases, the distance between the nucleus and the delocalized electrons also increases, reducing the strength of the metallic bond. This leads to a lower amount of energy required to break these bonds, resulting in a decreased melting point.

Key Concept: Electron Gain Enthalpy

d) Chlorine (Cl)

[Solution Description] Electron gain enthalpy becomes more negative as you move across a period towards the halogens. Chlorine, being one of the halogens, has a highly negative electron gain enthalpy due to its high electronegativity and desire to complete its valence shell.

Key Concept: Complex Electronic Configuration

c) $[Rn]5f^{14}7s^2$

[Solution Description] The element with atomic number 102 is Nobelium (No). For f-block elements, the general filling order follows the pattern $5f^{14}6d^07s^2$. Hence, the electronic configuration for Nobelium is $[Rn]5f^{14}7s^2$.

Key Concept: IUPAC Naming

a) Ununseptium

[Solution Description] According to the IUPAC nomenclature for elements with atomic numbers greater than 100, the element with atomic number 117 is named by taking its numerical roots: un- (1), un- (1), sept- (7) followed by -ium, resulting in the name Ununseptium.

Key Concept: Complex Trend Analysis, Anomalous Behavior

c) Fewer protons in $O^{2-}$ result in less effective nuclear charge, increasing ionic radius.

[Solution Description] The ionic radius trend in isoelectronic species depends on the effective nuclear charge experienced by the electrons. Both $\text{O}^{2-}$ and $\text{F}^{-}$ have the same number of electrons, but different numbers of protons. Oxygen has fewer protons than fluorine, resulting in a lower effective nuclear charge for $\text{O}^{2-}$. This means electrons experience less attraction towards the nucleus, causing the ionic radius to be larger for $\text{O}^{2-}$ compared to $\text{F}^{-}$.

Key Concept: Simple Trend

a) Increases

[Solution Description] As you move down a group, the atomic radius increases due to additional energy levels being added, which results in a larger size.

Key Concept: Diagonal Relationship

c) Lithium (Li) and Magnesium (Mg)

[Solution Description] A diagonal relationship occurs between certain pairs of elements that are located diagonally adjacent on the periodic table. This is due to similar polarizing power and other chemical properties. For example, lithium (Li) in group 1 and magnesium (Mg) in group 2 exhibit such a relationship.

Key Concept: Newlands’ Law of Octaves, Evolution of Periodic Table

c) Its significance lay in early recognition of periodic recurrence, leading to further development.

[Solution Description] Newlands’ Law of Octaves observed that every eighth element shared properties when ordered by increasing atomic weights. This concept indicated periodicity in elemental properties, foundational for future tables. However, its limitation was evident as heavier elements could not fit this pattern due to the inconsistencies in atomic weight predictions and unknown elements at that time.

Key Concept: Electron Configuration Analysis, Periodic Trends Explanation

d) Bromine (Br)

[Solution Description]

This configuration corresponds to Bromine (Br), which is found in the 4th period and 17th group of the periodic table. Anomalies in ionization enthalpy trends can occur due to sub-level electronic configurations. Generally, elements in a period show increasing ionization enthalpy from left to right; however, between filled and half-filled orbitals, the increase might not be regular. In this case, Bromine will show higher-than-expected ionization energy due to the high nuclear charge and its smaller atomic size causing greater attraction on the outer electrons.

Key Concept: Advanced Trend Analysis, Real-World Application

d) $Mg^{2+}$

[Solution Description] Among isoelectronic species, the one with the highest nuclear charge tends to have the smallest ionic radius and thus often participates more readily in reactions because of its higher charge density. $Mg^{2+}$ would therefore have the smallest ionic radius owing to its greatest nuclear charge, making it likely to participate more actively in chemical reactions.

Key Concept: Conceptual Application, Real-World Application

d) Close to halogens possessing high electron affinities.

[Solution Description] If element “X” has an exceptionally high negative electron gain enthalpy, similar to fluorine, it is likely positioned near the top of Group 17 (the halogens), where elements typically have high tendencies to gain electrons to achieve a stable noble gas configuration.

Key Concept: Configuration Prediction, Real-World Element Placement

a) $8s^26f^{14}7d^{10}8p^3$, p-block

[Solution Description]

An element in Group 15 typically possesses a $p^3$ electron configuration in its outermost energy level. Assuming discovery within the eighth period means considering the filling order with extended blocks. Given that the eighth period commences with the filling of 8s, followed by 6f, 7d, and eventually 8p, the anticipated group 15 element would have an outer config of $8s^28p^3$. It will be placed in the p-block of the hypothetical eighth period.

Key Concept: Radioactive Elements, Periodic Table Evolution

b) The 5f block, highlighted by early Mendeleev classifications.

[Solution Description] The seventh period involves the filling of 7s, 5f, 6d, and 7p orbitals, featuring many man-made radioactive elements within the actinoid series (5f block). Historically, Mendeleev’s prediction of undiscovered elements laid groundwork for modern periodic theory, despite Newlands’ earlier Law of Octaves failing to predict such heavy elements.

Key Concept: Oxide Nature

d) $$\text{Cl}_2\text{O}_7$$

[Solution Description]

The nature of oxides changes across the periodic table. Generally, metals form basic oxides, and non-metals form acidic oxides.

Across periods, acidity of oxides increases. Among the given options, $Cl_2O_7$ is formed by chlorine, a non-metal known for forming acidic oxides.

Correct answer: $Cl_2O_7$.

Key Concept: Triad Conceptual Challenge, Triad Prediction

b) 22

[Solution Description] For a valid Dobereiner’s Triad, the atomic weight of the middle element must be the average of the atomic weights of the other two elements. Hence,

$$x = \frac{\text{Atomic weight of first element} + \text{Atomic weight of third element}}{2} = \frac{14 + 30}{2} = 22$$

Therefore, the approximate value of x should be 22.

Key Concept: Historical Fact

c) Dmitri Mendeleev

[Solution Description] Dmitri Mendeleev is credited with developing the first widely recognized Periodic Table in 1869, organizing elements by increasing atomic weight and grouping them by similar chemical properties.

Key Concept: Historical Impact, Synthesis of Concepts

c) It prompted the use of atomic numbers

[Solution Description] The Triad system’s failure to provide a comprehensive classification for many elements emphasized the need for a more reliable method. This pushed chemists towards developing the periodic table, which considers both atomic number and electron configuration, thus providing a more accurate framework for element classification beyond just atomic mass.

Key Concept: Group Behavior

b) It has one valence electron and can form a negative ion

[Solution Description]

Hydrogen exhibits dual characteristics due to its electronic configuration:

– Like Group 1 (alkali metals), it has one electron in its outer shell that it can lose to form a positive ion ($H^+$).

– Like Group 17 (halogens), it can gain one electron to achieve a filled outer shell, forming a hydride ion ($H^-$).

This enables hydrogen to participate in reactions characteristic of both groups.

Key Concept: Triad Prediction, Triad Conceptual Understanding

c) 88.71 u

[Solution Description] The atomic weight of an element in a triad can be estimated by taking the average of the other two elements’ atomic weights. Thus, for the triad:

$$\text{Atomic weight of X} = \frac{40.08 + 137.33}{2}$$

Calculating this gives:

$$\text{Atomic weight of X} = \frac{177.41}{2} = 88.705$$

Rounding to two decimal places, we get 88.71 u.

Key Concept: Isoelectronic Species

a) $N^{3-}$

[Solution Description] Among these isoelectronic species, $N^{3-}$, $O^{2-}$, $F^-$, and Ne all have the same number of electrons but different nuclear charges. $N^{3-}$ has the smallest nuclear charge, leading to less pull on the electrons and thus the largest ionic radius.

Key Concept: Trend Explanation

b) Because electronegativity increases

[Solution Description] As you move from left to right across a period, electronegativity increases, which means atoms hold onto their electrons more tightly. This reduces their ability to lose electrons easily, thereby decreasing metallic character.

Key Concept: Comparative Trend Analysis, Real-World Application

d) Electron gain enthalpy becomes more negative and ionization enthalpy increases, promoting non-metallic reactivity.

[Solution Description] As we move across the second period, both electron gain enthalpy and ionization enthalpy generally become more negative and higher, respectively. For electron gain enthalpy, the increasing nuclear charge makes it energetically favorable to add an electron, except for anomalies like fluorine where high repulsion affects the trend. Ionization enthalpy increases because more energy is required to remove an electron as nuclear attraction strengthens. These trends indicate that elements towards the right are more reactive non-metals as they readily gain electrons, while those on the left are more reactive metals due to easier loss of electrons.

Key Concept: Detailed Trend Explanation, Multi-step Solutions

c) $F^-$ is larger because it gains an extra electron

[Solution Description] Lithium ion $Li^+$ has lost one electron compared to its atom and thus becomes significantly smaller due to reduced electron-electron repulsion and increased effective nuclear charge per electron. On the other hand, fluorine gains an electron to form $F^-$, increasing electron-electron repulsion and slightly expanding its radius. Despite both being in the second period, the lack of additional shells means the comparison primarily involves the effects of gaining or losing electrons.

Key Concept: Periodic Table Position

c) Chlorine

[Solution Description] Elements on the extreme right of a period are non-metals with high electronegativity and negative electron gain enthalpy. These characteristics are typically seen in noble gases or halogens.

Key Concept: Anomalous Properties

d) Oxygen

[Solution Description]

In the second period, oxygen shows an anomaly in its first ionization enthalpy. Normally, ionization enthalpy increases across a period; however, due to oxygen’s electronic configuration ($1s^2 2s^2 2p^4$), it requires less energy to remove an electron than nitrogen ($1s^2 2s^2 2p^3$), despite being to the right of nitrogen on the periodic table.

This is because removing one paired electron in oxygen reduces electron-electron repulsion significantly compared to breaking a half-filled stable p-orbital set in nitrogen.

Key Concept: Historical Naming

b) Nobelium

[Solution Description] Element 102 was historically called Nobelium after Alfred Nobel. This name was used before IUPAC’s standardized naming conventions were established.

Key Concept: Exception Identification, Diagonal Relationship

b) Lithium and Magnesium

[Solution Description]

Lithium and magnesium exhibit a diagonal relationship, leading them to exhibit similarities in properties like solubility patterns and formation of complex compounds.

Key Concept: Element Placement

a) Because they have similar properties

[Solution Description] In Mendeleev’s table, elements were grouped based on similar chemical properties rather than strictly by increasing atomic weight. Cobalt and Nickel shared similar properties, justifying their placement in the same group.

Key Concept: Comparative Analysis, Electronegativity and Chemical Properties

c) Sulfur (S)

[Solution Description]

Chlorine is highly electronegative, so it tends to form compounds with elements having moderate electronegativity values that allow for shared bonding. Sulfur has higher electronegativity than both silicon and phosphorus, making it more compatible with chlorine for sharing electrons and forming polar covalent bonds. Therefore, sulfur is more likely to form a stable compound with chlorine due to its relatively high electronegativity compared to silicon and phosphorus.

Key Concept: Valence Shell Expansion

d) Sulfur can expand its valence shell

[Solution Description] Sulfur can expand its valence shell due to the presence of available d-orbitals, which allows it to accommodate more than 8 electrons. Oxygen lacks such d-orbitals in its valence shell, restricting it to the octet configuration and preventing the formation of $OF_6$.

Key Concept: Periodic Trends in Ionization Enthalpy

d) Nitrogen (N)

[Solution Description]

First ionization enthalpy generally increases across a period from left to right due to increasing nuclear charge and decreasing atomic radius. Between Beryllium (Be), Boron (B), Carbon (C), and Nitrogen (N), nitrogen has the highest first ionization enthalpy due to its half-filled stable p orbital configuration.

Key Concept: Quantum Level Influence, Shielding vs. Nuclear Charge

c) Sodium will have higher second ionization enthalpy due to penetration into a noble gas core

[Solution Description]

The second ionization enthalpy is the energy needed to remove an electron from a singly charged cation in its gaseous state. For sodium, the electronic configuration after losing one electron becomes $[Ne]$ $3s^0$. Removing another electron is challenging as it involves breaking into the stable noble gas core, significantly raising the second ionization enthalpy.

Magnesium loses its first electron transforming into $[Ne]$ $3s^1$. The second ionization removes an electron from this partially filled s orbital, not penetrating the neon core yet.

Hence, the second ionization enthalpy of sodium is much greater than that of magnesium because removing an electron from a filled shell (neon core) requires substantially more energy than removing from a partially filled or unfilled subshell.

Key Concept: Octaves Limitations

b) Calcium

[Solution Description] Newlands’ Law of Octaves states that every eighth element has similar properties, much like musical octaves. However, this pattern was observed to hold true only up until the element Calcium in Mendeleev’s table due to the lack of noble gases at that time and increasing complexity of heavier elements.

Key Concept: Periodic Trends and Oxidation, Anomalous Behavior

d) Nitrogen

[Solution Description] In the second period, nitrogen often displays unique behavior due to its small size and high electronegativity. It can have several oxidation states ranging from -3 in ammonia ($NH_3$) to +5 in nitric acid ($HNO_3$). This variability is anomalous compared to other group members.

Considering the options:

– Lithium (Li) commonly has an oxidation state of +1.

– Beryllium (Be) typically shows +2.

– Boron (B) usually is +3.

– Nitrogen (N), however, varies widely from -3 to +5.

This wide range of possible oxidation states due to nitrogen’s ability to form multiple bonds is considered anomalous behavior.

Key Concept: Periodic Table Exceptions, Ionization Enthalpy Trend

d) It has a completely filled valence shell

[Solution Description] Beryllium has a filled 2s orbital, resulting in a stable electronic configuration. This stability makes it more difficult to remove an electron, hence its higher ionization enthalpy. Other s-block elements do not have this filled subshell stability, leading to lower ionization enthalpies.

Key Concept: Valence Shell Configuration

b) $1s^2$

[Solution Description] Helium is placed in the p-block due to its filled $1s^2$ configuration. Although it doesn’t have any p-electrons, its fully filled s-orbital gives it stability, aligning it with the characteristics of noble gases.

Key Concept: Periodic Trends Explanation, Transition Element Trends

c) Increases slightly; d-orbitals contribute to gradual rise

[Solution Description] Ionization enthalpy generally increases from left to right across a period due to an increase in nuclear charge, which attracts the electrons more strongly. However, transition metals show anomalies because of their d-orbitals. As the number of electrons increases in d-orbitals, they provide some shielding effect, but the increased effective nuclear charge predominates, thus gradually increasing the ionization enthalpy across the transition series.

Key Concept: Comparative Analysis, Real-World Application

d) Uranium (f-block)

[Solution Description]

To determine which element is less likely to form an ionic bond, consider their blocks: Sodium (s-block), Chlorine (p-block), Iron (d-block), Uranium (f-block). Generally, f-block elements like Uranium have complex bonding behaviors and often do not form simple ionic bonds due to their large size and involvement of inner f-electrons; instead, they show covalent character as well.

Key Concept: Complex Configuration Analysis, Real-World Application

c) p-block, Group 17; Reactive halogen

[Solution Description] The given configuration is $[Xe]4f^{14}5d^{10}6s^26p^5$, which indicates that the outermost electron is in a p-orbital, specifically the 6p orbital. This places the element in the p-block. Counting from Group 13 to 18, there are 5 electrons in the p-shell, making it part of Group 17. Being one electron short of having a filled valence shell (like noble gases), elements in Group 17 are highly reactive non-metals known as halogens, typically gaining an electron during reactions.

Key Concept: Ionization Enthalpy Trends

d) Nitrogen

[Solution Description] Ionization enthalpy refers to the energy required to remove an electron from a gaseous atom or ion. Across a period, ionization enthalpy generally increases due to increased effective nuclear charge and decreased atomic radii. Among the given elements, Nitrogen (N) has the highest ionization enthalpy because it has a more stable half-filled p orbital configuration compared to the others.

Key Concept: Covalency and Bonding, Stability and Reactivity

c) Carbon

[Solution Description] Carbon is well known for forming stable multiple bonds such as double ($C=C$) and triple ($C≡C$) bonds. This is due to its small size, which allows effective orbital overlap, resulting in strong $\pi$-bonding. The stability of these bonds arises because carbon can participate in extensive $\pi$ conjugation.

Key Concept: Second Order Thinking, Critical Analysis

c) Increased electron repulsion

[Solution Description] The chloride ion ($Cl^{-}$) forms when an extra electron is added to a neutral chlorine atom. This additional electron increases electron-electron repulsion within the ion. Additionally, the effective nuclear charge experienced per electron decreases as there are more electrons but the same number of protons. Hence, the increased repulsion and decreased effective nuclear charge cause the ionic radius to expand compared to the atomic radius.

Key Concept: Real-World Application, Transition Metal Reactivity

b) Lower reactivity than s-block elements

[Solution Description]

Transition metals are characterized by their ability to form complex ions with different oxidation states and moderate reactivity. This is due to the relatively low energy difference between their s and d orbitals, allowing them to lose varying numbers of electrons to form multiple oxidation states.

Compared to main group metals, transition metals generally have:

– Intermediate ionization enthalpies.

– The ability to form colored compounds due to d-d transitions.

– Moderate reactivity, especially when comparing with highly reactive alkali metals.

Among the given choices, one should look at a property characteristic of transition metals versus main group elements. Given the context, the presence of intermediate ionization enthalpies allows for the variability in oxidation states common to transition elements.

Key Concept: Historical Impact, Octaves and Modern Table

b) It is a precursor to the modern understanding

[Solution Description] Newlands’ comparison of the periodic recurrence of elemental properties to musical octaves laid foundational groundwork for recognizing patterns within the periodic chart. Though initially considered humorous rather than scientific, this analogy highlighted the concept of periodic repetition. Its influence set the stage for later work by Mendeleev, who refined these ideas into the modern periodic law based on atomic numbers, thereby validating the relevance of recurring chemical properties albeit in a different framework.

Key Concept: Electronic Configuration

c) Copper

[Solution Description] The electronic configuration $3d^{10}4s^1$ corresponds to Copper (Cu). Copper has an atomic number of 29 and its configuration reflects the filling of the 3d subshell followed by one electron in the 4s orbital.

Key Concept: Radioactivity

c) Np

[Solution Description] All actinoids are radioactive. Among the given options, Neptunium (Np) belongs to the actinoid series and is well-known for its radioactivity. This property is characteristic of all actinoid elements due to their unstable nuclei, leading to radioactive decay.

Key Concept: Diagonal Relationships, Periodicity of Valence or Oxidation States

c) Aluminum

[Solution Description] The diagonal relationship is typically found between elements that have similar sizes and charges. Lithium (Li) shows a diagonal relationship with magnesium (Mg). Additionally, lithium can form compounds such as $LiAlH_4$, where it exhibits an oxidation state of +1. However, aluminum is known for regularly exhibiting a +3 oxidation state in most of its compounds.

Key Concept: Anomalous Behavior

c) Small atomic size and high electronegativity

[Solution Description] Fluorine exhibits anomalous behavior due to its small atomic size and high electronegativity, which causes it to form hydrogen bonds and have different chemical properties compared to other halogens.

Key Concept: Mendeleev’s Anomalies, Quantum Numbers

b) Using principal quantum number for period alignment.

[Solution Description] In terms of quantum numbers, iodine and tellurium are positioned based on electron configuration, which relates to their properties. Tellurium is placed before iodine due to its lower atomic number despite having a higher atomic weight, aligning with the principal quantum number n that determines period position.

Key Concept: Real-world Applications, Comparative Analysis

b) Copper; Catalysts like Cu offer variable oxidation states improving efficiency.

[Solution Description] The Deacon process involves the oxidation of hydrogen chloride gas to produce chlorine using oxygen from the air. Copper (Cu) is commonly used as a catalyst in this process. This showcases the superior catalytic abilities of d-Block elements compared to s-Block elements, as they provide variable oxidation states and facilitate electron transfer reactions more efficiently.

Key Concept: Periodic Table Anomalies, Real-World Application

a) Fluorine

[Solution Description] The boiling point of halogens generally increases with increasing atomic number due to stronger Van der Waals forces as the size of the molecule increases. However, fluorine ($F_2$) deviates from this trend because its molecules are very small, leading to unusually weak Van der Waals forces compared to chlorine ($Cl_2$). Thus, despite being lighter, it has a surprisingly low boiling point.

Key Concept: Reactivity Explanation, Nature of Oxides

c) ${Al}_2{O}_3$

[Solution Description]

Amphoteric oxides can react both as acids and bases. For example, aluminum oxide (${Al}_2{O}_3$) displays this property owing to its intermediate electronegativity and positioning in the periodic table where it can form bonds with both $H^+$ and $OH^{-}$
. Other group 13 and early transition metal oxides display such characteristics, particularly when moving diagonally in the periodic table from metal to non-metal sections—balancing metallic and non-metallic properties.

Key Concept: Nomenclature Derivation

d) Ununquadium

[Solution Description] According to IUPAC nomenclature rules, the name of an element with atomic number 114 is derived using numerical roots: un (1), and quad (4). Therefore, its systematic name is Ununquadium.

Key Concept: Electron Gain Enthalpy, Periodic Law Application

d) Fluorine

[Solution Description] Fluorine is located in Group 17 and has a high tendency to gain electrons due to its small size and high electronegativity. These factors contribute to it having the most negative electron gain enthalpy compared to other halogens or group members.

Key Concept: Scale Comparison, Non-metallic and Metallic Properties

c) Pauling scale, due to qualitative measurement based on bond energies.

[Solution Description]

The Pauling scale is the most commonly used electronegativity scale and assigns high values to non-metals such as oxygen because it measures the tendency of an element to attract bonding electrons. Non-metals, particularly those with smaller atomic radii, have higher electronegativities due to their increased ability to attract shared electrons.

Key Concept: Reactivity and Nature, Anomalous Behavior

c) High electronegativity

[Solution Description] Beryllium has a small atomic size and high charge/radius ratio, which leads to high electronegativity compared to other alkaline earth metals. As a result, it prefers to form covalent bonds rather than giving up its electrons completely to form ionic bonds. This behavior is anomalous when compared to the typical ionic character observed in s-block elements.

Key Concept: Reactivity Trends

d) Fluorine

[Solution Description] The reactivity of halogens decreases as we move down the group in the periodic table. This trend is due to the increase in atomic size and decrease in electronegativity, making it harder for outer electrons to participate in reactions. Fluorine, being at the top of the group with the smallest atomic radius and highest electronegativity, is the most reactive.

Key Concept: Mendeleev’s Arrangement

d) Iodine

[Solution Description] Mendeleev placed iodine (atomic weight 126.90) after tellurium (atomic weight 127.60) because of their chemical properties, even though iodine has a lower atomic weight.

Key Concept: Periodic Trends: Chemical Reactivity

b) It increases from left to right.

[Solution Description] Ionization enthalpy tends to increase across a period from left to right. This is due to the increasing nuclear charge, which holds the valence electrons more tightly, thereby requiring more energy to remove an electron. Hence, option b) is correct.

Key Concept: Historical Naming

b) USA and Soviet Union

[Solution Description]

The naming controversy of element 106 involved the USA and the Soviet Union, as both had claims to its discovery and suggested different names.

Key Concept: Variable Valency, Diagonal Relationships

a) Lithium (Li) and Magnesium (Mg)

[Solution Description] In the periodic table, diagonal relationships occur between certain pairs of elements where they share similarities in properties despite being in different groups. For example, lithium in Group 1 shows similarity with magnesium in Group 2 because they have comparable electronegativities and ionic potentials. Similarly, sulfur from Group 16 shows some chemical resemblance to phosphorus in Group 15 in terms of oxidation states and reactivity.

Key Concept: Transition Metal Alloys, Real-World Application

c) Nickel

[Solution Description]

Nickel is extensively used in superalloys for jet engines because it provides excellent resistance to thermal creep deformation, stability at high temperatures, and oxidation resistance. These properties are crucial for performance in extreme conditions.

Key Concept: Periodic Trends

c) Atomic radius decreases due to increased effective nuclear charge.

[Solution Description] As we move across a period from left to right, the atomic radius generally decreases. This is because electrons are added to the same energy level while the nuclear charge increases, pulling the electron cloud closer to the nucleus.

Key Concept: Oxidation State Calculation

b) +6

[Solution Description]

In $SF_6$, fluorine is more electronegative than sulfur and typically has an oxidation state of -1. Since there are six fluorine atoms, their total contribution is $6 \times (-1) = -6$. For the compound to be neutral, the oxidation state of sulfur must balance this out, resulting in:

$x + 6(-1) = 0$

Solving for x, the oxidation state of sulfur:

$x – 6 = 0$

$x = +6$

Therefore, the oxidation state of sulfur in $SF_6$ is +6.