Periodic Properties

To determine which elements have negative electron affinity values, we need to understand the concept of electron affinity. Electron affinity refers to the energy change that occurs when an electron is added to a neutral atom in the gas phase. A negative value indicates that energy is released when an electron is added, making the process exothermic.

Let's analyze each given option:

1. A. Be → Be⁻: Beryllium has a completely filled s-subshell, making it less likely to accept an additional electron. It has a positive electron affinity because adding an electron increases electron-electron repulsion without gaining significant stability.
2. B. N → N⁻: Nitrogen has a half-filled p-subshell, which is stable due to symmetrical electron distribution. Adding an electron to nitrogen introduces repulsion, resulting in a positive electron affinity.
3. C. O → O²⁻: Oxygen generally gains energy when forming a single anion (O⁻). However, adding a second electron (forming O²⁻) involves significant electron-electron repulsion, resulting in a positive electron affinity.
4. D. Na → Na⁻: Sodium has one electron in its outer shell. Losing this electron (to form Na⁺) is favorable, while adding an electron (to form Na⁻) is not, leading to a positive electron affinity.
5. E. Al → Al⁻: Aluminum has a similar situation to sodium where adding an electron is not favorable, resulting in a positive electron affinity.

Thus, the elements where electron affinity values are negative (indicating energy is required to add an electron) are Sodium (Na) and Aluminum (Al), i.e., options D and E.

Hence, the correct answer is: D and E only.

To determine the validity of the given statements about chemical properties and allotropic forms, let's analyze each statement individually.

1. Statement I: and are acidic while and are amphoteric in nature.
   • is known to be an acidic oxide. It reacts with bases to form silicates.
   • also shows acidic properties, reacting with alkalis to form germanates.
   • and are amphoteric, meaning they can react with both acids and bases. For instance, reacts with hydrochloric acid to form and with sodium hydroxide to form sodium stannate.
   • Thus, Statement I is accurate.
2. Statement II: Allotropic forms of carbon are due to the property of catenation and bond formation.
   • Catenation is indeed a property of carbon that allows it to form multiple allotropes, such as diamond, graphite, and fullerenes, through chains or rings.
   • However, bond formation is not typically associated with carbon allotropes. Carbon primarily forms bonds.
   • Therefore, Statement II is incorrect.

Based on the analysis:

• Statement I is true: and are acidic, whereas and are amphoteric.
• Statement II is false: While catenation is a reason for carbon allotropy, the bond formation is not a characteristic property of carbon for its allotropes.

Thus, the most appropriate answer is that Statement I is true but Statement II is false.

To solve the matching problem, we need to understand the chemical reactions between the reactants in List-I and their corresponding products in List-II. Let's analyze each pair:

1. (A) Phenol, Zn/Δ: The reaction of phenol with zinc dust upon heating leads to the removal of the hydroxyl group, yielding benzene as the product. Thus, (A) matches with (III).
2. (B) Phenol, CHCl₃, NaOH, HCl: This is the Reimer-Tiemann reaction, where phenol reacts with chloroform in the presence of a base and then acid, giving salicylaldehyde. Thus, (B) matches with (I).
3. (C) Phenol, CO₂, NaOH, HCl: This is the Kolvbe-Schmitt reaction, where phenol reacts with carbon dioxide in the presence of a base to form salicylic acid. Thus, (C) matches with (II).
4. (D) Phenol, Conc. HNO₃: Nitration of phenol using concentrated nitric acid produces picric acid. Thus, (D) matches with (IV).

Based on the above explanation, the correct matching of List-I with List-II is:

• (A) - (III)
• (B) - (I)
• (C) - (II)
• (D) - (IV)

Therefore, the correct answer is: (A)-(III), (B)-(I), (C)-(II), (D)-(IV).

To match the items from List I with List II:
A. Melting point [K]: Based on the trend, Tl > In > Ga > Al > B. (Matches I).
B. Ionic Radius [M /pm]: Boron has the smallest ionic size, followed by Tl > Al Ga > In (Matches II).
C. [kJ mol ]: Ionization enthalpy follows the trend Tl > In > Al > Ga > B.(Matches III).
D. Atomic Radius [pm]: Atomic radius increases as B > Al > Tl > In > Ga.(Matches IV).
Correct Answer: (3)

To determine which transition metal has the highest third ionization enthalpy, we need to consider the electronic configurations and the stability of the resulting ions upon successive ionizations.

The electronic configuration for each element is as follows:

• Chromium (Cr):
• Manganese (Mn):
• Vanadium (V):
• Iron (Fe):

The third ionization enthalpy refers to the energy required to remove the third electron after removing the first two. When evaluating this, the stability of half-filled d-orbitals plays a significant role.

For manganese (Mn), removing the first two electrons results in a configuration of , which is a stable half-filled configuration.

This stability makes it significantly more challenging to remove a third electron from manganese than from the other elements.

Therefore, the element with the highest third ionization enthalpy is: Mn (Manganese).

Statement I: Number of d and f electrons increases down the group and due to poor shielding of d and f electrons, stability of lower oxidation states increases down the group.

Statement II: The atomic size of aluminium is greater than that of gallium. Thus, Statement I is correct, but Statement II is incorrect.

1. To solve the given question, we must match elements in List-I with their corresponding properties in List-II.
2. First, examine each option in List-I:
   • (A) Cl, S: Chlorine (Cl) and Sulfur (S) are non-metals. Both are known for their high electron gain enthalpies, especially chlorine, which often has the highest negative electron gain enthalpy in its group.
   • (B) Ge, As: Germanium (Ge) and Arsenic (As) are metalloids, showcasing properties of both metals and non-metals.
   • (C) Fr, Ra: Francium (Fr) and Radium (Ra) are metals with large atomic sizes. They are found in the alkali metal and alkaline earth metal groups, respectively.
   • (D) F, O: Fluorine (F) and Oxygen (O) are known for their high electronegativity values, with fluorine being the most electronegative element.
3. Now, map each element pair to the correct property in List-II:
   • (A) Cl, S aligns with (IV) Elements with highest negative electron gain enthalpy.
   • (B) Ge, As fits with (III) Elements which show properties of both metals and non-metals.
   • (C) Fr, Ra matches with (II) Elements with largest atomic size.
   • (D) F, O corresponds to (I) Elements with highest electronegativity.
4. The correct matching is: A-IV, B-III, C-II, D-I.
5. This matches with the given option:

A-IV, B-III, C-II, D-I

Let's analyze each statement about group 15 elements to determine their correctness:

1. Statement (A): "Dinitrogen is a diatomic gas which acts like an inert gas at room temperature."
   Nitrogen typically exists as a diatomic molecule ( ). It is relatively inert under normal conditions due to the strong triple bond between the nitrogen atoms, making it stable and unreactive. Thus, Statement (A) is correct.
2. Statement (B): "The common oxidation states of these elements are −3, +3 and +5."
   The group 15 elements indeed commonly show these oxidation states. For example, nitrogen can exhibit −3 (in ammonia), +3 (in nitrous acid), and +5 (in nitric acid). Statement (B) is correct.
3. Statement (C): "Nitrogen has unique ability to form p -p multiple bonds."
   Nitrogen is capable of forming multiple bonds like -bonds due to its small size and ability to undergo p -p overlap. This is correct, as seen in compounds like nitrogen gas ( ). Statement (C) is correct.
4. Statement (D): "The stability of +5 oxidation states increases down the group."
   The stability of the +5 oxidation state actually decreases down the group. Elements like bismuth prefer lower oxidation states due to the inert pair effect. Statement (D) is incorrect.
5. Statement (E): "Nitrogen shows a maximum covalency of 6."
   Nitrogen, with its electron configuration of 1s² 2s² 2p³, can form a maximum of 4 covalent bonds (in compounds like ammonium ion, ). Statement (E) is incorrect.

Thus, the incorrect statements are: (D) and (E).

Therefore, the correct answer is (D) and (E) only.

The anomalous behavior of oxygen can be attributed to its small size and high electronegativity. Let us understand why:

1. Small Size:
   • Oxygen is one of the smaller atoms due to its position in the periodic table (it is a second-period element).
   • The smaller atomic size results in a stronger attraction between the nucleus and the valence electrons, increasing its ability to attract electrons towards itself.
2. High Electronegativity:
   • Oxygen is the second most electronegative element after fluorine. This high electronegativity leads to the formation of highly polar covalent bonds.
   • The high electronegativity is due to the strong pull from the nucleus due to the small atomic radius, making oxygen excel at attracting shared electrons in a chemical bond.

These characteristics distinguish oxygen from other elements in its group, leading to its unique or anomalous behavior. As a result, oxygen is able to form multiple types of oxides and has distinctive reactivity patterns compared to its group counterparts like sulfur, selenium, etc.

Thus, the correct answer is: Small size and high electronegativity.

Options Analysis:

• Large size and high electronegativity: Oxygen does not have a large size; hence this option is incorrect.
• Small size and low electronegativity: While the size is correct, oxygen exhibits high electronegativity, not low; thus, this option is incorrect.
• Large size and low electronegativity: Both characteristics are incorrect for oxygen, making this option incorrect.

To analyze the statements given in the question, we need to understand the concept of electron gain enthalpy within the groups of the periodic table.

Statement I: "Fluorine has the most negative electron gain enthalpy in its group."

Explanation: Electron gain enthalpy refers to the energy change when an electron is added to an isolated gaseous atom. In general, as we move across a period from left to right, electron gain enthalpy becomes more negative. However, this trend doesn't hold perfectly in a group. For Group 17 (halogens), chlorine actually has a more negative electron gain enthalpy than fluorine. This is due to the smaller atomic size of fluorine, resulting in a higher electron-electron repulsion in its compact structure, making chlorine's electron addition more favorable. Therefore, Statement I is false.

Statement II: "Oxygen has the least negative electron gain enthalpy in its group."

Explanation: Oxygen is part of Group 16. Within this group, oxygen does indeed have the least negative electron gain enthalpy. This is because, like with fluorine, the small size of the oxygen atom leads to increased electron-electron repulsion upon gaining an additional electron. Thus, Statement II is true.

Conclusion: Based on the explanations above, the most appropriate choice is: "Statement I is false but Statement II is true."

The question is about finding the element with the highest first ionization enthalpy among the given options. Ionization enthalpy is the energy required to remove an electron from a gaseous atom or ion.

Explanation:

• The given options are Silicon (Si), Aluminium (Al), Nitrogen (N), and Carbon (C).
• Ionization enthalpy generally increases across a period in the periodic table because the nuclear charge increases, making it more difficult to remove an electron.
• Ionization enthalpy decreases down a group as the atomic size increases, the outer electrons are further from the nucleus, and the increase in shielding reduces the effective nuclear charge felt by the valence electrons, making them easier to remove.

Let's analyze each element:

1. Silicon (Si): It is in the same group as Carbon but below it in the periodic table. Its ionization enthalpy is lower than C because it's further down the group.
2. Aluminium (Al): It is in Group 13, after Nitrogen and Carbon in the periodic table. Its first ionization enthalpy is less than that of N and C due to its position.
3. Nitrogen (N): It is in Group 15 and has one more proton than Carbon, making it more efficiently hold on to its electrons than Al or Si. Its ionization enthalpy is quite high due to its small atomic size and half-filled p orbital, making it stable.
4. Carbon (C): It lies before Nitrogen in the periodic table and has fewer protons, thus its ionization enthalpy is lower than that of Nitrogen.

From the above analysis, Nitrogen (N) has the highest first ionization enthalpy among the given elements due to its position in the periodic table, small atomic size, and stable half-filled configuration.

Conclusion: The correct answer is .

The problem asks for the number of valence electrons of an atom B, given that its lowest possible oxidation number in a compound A₂B is -2.

Concept Used:

The oxidation number (or oxidation state) of an atom in a compound represents its degree of oxidation. For a non-metal element, the lowest (most negative) possible oxidation number is related to the number of electrons it needs to gain to achieve a stable electron configuration, typically an octet (8 electrons in its valence shell).

The relationship between the minimum oxidation number and the number of valence electrons ( ) for a main group element is given by the formula:

This formula arises because the lowest oxidation state is achieved when the atom gains enough electrons to complete its octet.

Step-by-Step Solution:

Step 1: Identify the given information.

We are given that the lowest oxidation number of atom B is -2.

Step 2: Apply the formula relating the lowest oxidation number to the number of valence electrons.

Let be the number of electrons in the valence shell of atom B. Using the formula:

Step 3: Substitute the given value and solve for .

Final Computation & Result:

To find the number of valence electrons, we rearrange the equation:

This means that atom B has 6 electrons in its valence shell. Such elements belong to Group 16 of the periodic table (e.g., Oxygen, Sulfur), and their most common negative oxidation state is indeed -2.

The number of electrons in its valence shell is 6.

The given question requires identification of elements among the given list that do not belong to the lanthanoid series. The lanthanoids, also known as the lanthanides, consist of 15 elements ranging from atomic numbers 57 (Lanthanum, La) to 71 (Lutetium, Lu).

Let's evaluate each element in the list:

1. Eu (Europium): Europium is a lanthanoid with an atomic number of 63.
2. Cm (Curium): Curium is an actinoid with an atomic number of 96. It does not belong to the lanthanoid series.
3. Er (Erbium): Erbium is a lanthanoid with an atomic number of 68.
4. Tb (Terbium): Terbium is a lanthanoid with an atomic number of 65.
5. Yb (Ytterbium): Ytterbium is a lanthanoid with an atomic number of 70.
6. Lu (Lutetium): Lutetium is the last element of the lanthanoid series with an atomic number of 71.

From this evaluation, we can see that the element Cm (Curium) is the only one that does not belong to the lanthanoid series.

Conclusion: The correct answer is option , as Cm (Curium) is the only element that is not a lanthanoid.

- are isoelectronic (same number of electrons = 10).

- has the largest ionic radius due to its lowest effective nuclear charge ( ).

- has the smallest ionic radius due to its highest effective nuclear charge.
Hence, statements (A), (C), and (D) are correct.

The question asks to identify the primary factor that affects the size of the isoelectronic species F⁻, Ne, and Na⁺.

Concept Used:

The key concepts for determining the size of isoelectronic species are:

1. Isoelectronic Species: These are atoms or ions that have the same number of electrons.
2. Nuclear Charge (Z): This is the total positive charge of the nucleus, which is equal to the number of protons in the atom.
3. Effective Nuclear Charge (Z_eff): This is the net positive charge experienced by an electron in a multi-electron atom. The electrons in the outer shells are shielded from the full nuclear charge by the inner-shell electrons. The effective nuclear charge is approximately the nuclear charge minus the screening constant ( ), .

For an isoelectronic series, the number of electrons is constant. Therefore, the screening effect from the inner electrons is also roughly constant. The size of the species is then primarily determined by the magnitude of the nuclear charge (number of protons). A higher nuclear charge exerts a stronger electrostatic pull on the same number of electrons, causing the electron cloud to contract and resulting in a smaller atomic or ionic radius.

Step-by-Step Solution:

Step 1: Determine the number of protons (Nuclear Charge, Z) and electrons for each species.

• For Fluoride ion (F⁻):
  • The atomic number of Fluorine (F) is Z = 9. So, it has 9 protons.
  • The charge is -1, meaning it has gained one electron.
  • Number of electrons = 9 (protons) + 1 = 10 electrons.
• For Neon atom (Ne):
  • The atomic number of Neon (Ne) is Z = 10. So, it has 10 protons.
  • It is a neutral atom.
  • Number of electrons = 10 (protons) = 10 electrons.
• For Sodium ion (Na⁺):
  • The atomic number of Sodium (Na) is Z = 11. So, it has 11 protons.
  • The charge is +1, meaning it has lost one electron.
  • Number of electrons = 11 (protons) - 1 = 10 electrons.

Since all three species (F⁻, Ne, Na⁺) have 10 electrons, they are isoelectronic.

Step 2: Compare the nuclear charge of the species.

The number of protons (nuclear charge) for each species is:

• F⁻: Z = 9
• Ne: Z = 10
• Na⁺: Z = 11

The nuclear charge increases in the order F⁻ < Ne < Na⁺.

Step 3: Relate nuclear charge to ionic/atomic size.

All three species have the same number of electrons (10) arranged in the same electronic configuration (1s²2s²2p⁶). However, they have different numbers of protons in their nuclei.

The Na⁺ ion, with 11 protons, has the highest nuclear charge. It will exert the strongest electrostatic attraction on its 10 electrons, pulling them closer to the nucleus. This results in the smallest radius.

The F⁻ ion, with only 9 protons, has the lowest nuclear charge. Its nucleus exerts the weakest attraction on the 10 electrons, resulting in the largest radius.

The Ne atom is intermediate with 10 protons and 10 electrons.

Thus, the order of size is: Na⁺ < Ne < F⁻.

Final Result:

For the isoelectronic species F⁻, Ne, and Na⁺, the number of electrons is the same, but the nuclear charge is different. The electrostatic attraction between the nucleus and the electrons increases with the increase in nuclear charge. Therefore, the size of these species is primarily affected by the nuclear charge (number of protons).

To determine the incorrect trend among the given atomic radii, let's review the basic concept of atomic radii trends in the periodic table:

• Atomic radius generally decreases across a period from left to right due to the increase in nuclear charge which pulls the electrons closer to the nucleus, thereby decreasing the size.
• Atomic radius generally increases down a group due to the addition of electron shells, which outweighs the effect of increased nuclear charge, resulting in a larger size.

Now, let's evaluate each option:

1. :
   Bromine ( ) is located to the right of Potassium ( ) in the periodic table, and both elements are in the same period. Thus, will have a smaller atomic radius than . This trend is correct.
2. :
   Magnesium ( ) and Aluminum ( ) are in the same period, with located to the left of . Therefore, should have a smaller atomic radius than due to the increased nuclear charge. Thus, the given trend is incorrect.
3. :
   Rubidium ( ) and Cesium ( ) are in the same group with below . Therefore, should have a larger atomic radius than due to the additional electron shell. This trend is correct.
4. :
   Astatine ( ) is below Bromine and is a halogen, while Cesium ( ) is an alkali metal. Despite their group differences, Cesium is known to have one of the largest atomic radii in its period and is generally larger than Astatine. This trend is correct.

The correct answer is the option where the stated trend is wrong: .

Given elements are: Indium (In), Thallium (Tl), Aluminum (Al), Lead (Pb), Tin (Sn), and Germanium (Ge).

We need to determine the most stable oxidation states of the elements with the highest and lowest first ionization enthalpies from this group.

Position in the periodic table:

• Aluminum (Al): Group 13, Period 3
• Germanium (Ge): Group 14, Period 4
• Tin (Sn): Group 14, Period 5
• Indium (In): Group 13, Period 5
• Lead (Pb): Group 14, Period 6
• Thallium (Tl): Group 13, Period 6

Ionization enthalpy trends:
- Ionization enthalpy generally increases across a period and decreases down a group.
- Among these elements, **Aluminum (Al)** has the highest first ionization enthalpy because it is the lightest and topmost element in its group.
- Thallium (Tl) has the lowest first ionization enthalpy due to its position in Period 6, where the inert pair effect becomes significant.

Oxidation states:

Aluminum (Al), with the highest ionization enthalpy, typically shows a +3 oxidation state. This is the most stable oxidation state for Al, as it readily loses all three valence electrons.

Thallium (Tl), with the lowest ionization enthalpy, shows both +3 and +1 oxidation states. However, due to the **inert pair effect**, the +1 oxidation state is more stable than +3.

This trend is also seen in other heavier p-block elements like Pb and Sn, where the +2 oxidation state becomes more stable compared to the +4 state because of the inert pair effect.

Conclusion:
- The element with the highest first ionization enthalpy is **Aluminum (Al)**, which has a most stable oxidation state of **+3**.
- The element with the lowest first ionization enthalpy is **Thallium (Tl)**, which has a most stable oxidation state of **+1**.

Final Answer:
The final answer is .

- For the reaction at equilibrium, the concentrations of the reactants ( and ) decrease, and the concentration of the product ( ) increases over time.
- As the system reaches equilibrium, the concentration of the reactants will stabilize and the product concentration will also stabilize. Thus, the graph where the concentrations of and decrease and the concentration of increases correctly represents the attainment of equilibrium.

Final Answer: Option (3).

The question involves determining the order of melting points for pairs of elements (Mn, Fe), (Tc, Ru), and (Re, Os). We have four options to consider. Let's analyze each pair:

1. Mn (Manganese) and Fe (Iron):
   Manganese has a lower melting point compared to Iron. Therefore, Mn < Fe.
2. Tc (Technetium) and Ru (Ruthenium):
   Technetium, being in the same column group earlier than Ruthenium, has a lower melting point. Thus, Tc < Ru.
3. Re (Rhenium) and Os (Osmium):
   Rhenium has a lower melting point compared to Osmium, so Re < Os.

Based on the analysis, the correct order of the melting points is Mn < Fe, Tc < Ru and Re < Os.

This corresponds to the correct option given: Mn < Fe, Tc < Ru and Re < Os

The ionic character of a compound depends on the difference in ionisation enthalpy and electron gain enthalpy. The larger the difference, the greater the ionic character. A compound with a more negative electron gain enthalpy will result in a stronger ionic bond. Given that the electron gain enthalpy values for elements A, B, C, and D are as follows:
- has
- has
- has
- has
The order of ionic character is given by the electron gain enthalpy, where has the highest ionic character and has the least. Thus, the correct order is (3): .

Concept: Atomic and ionic radii depend on several factors such as:

• Nuclear charge
• Number of shells
• Shielding effect
• Charge on the ion

Electronegativity is not the sole deciding factor for atomic size.

Analysis of Statement-I:

Consider neutral atoms and :

• and lie in the same period (Period 3).
• Across a period, atomic radius decreases due to increasing nuclear charge.

Hence, the given order is incorrect because of the trend of atomic radius across a period.

Now consider the ions:

• Cations are smaller than their parent atoms, so .
• Higher the positive charge, smaller the ionic radius.

The correct ionic order should be:

Thus, the overall order should be , and Statement-I is incorrect.

Analysis of Statement-II:

The statement says: and

However, the statement is incorrect because:

• Atomic size primarily depends on nuclear charge and shielding effect.
• Electronegativity is a consequence of atomic size, not the controlling factor.
• The word "always" makes the statement absolutely false.

Thus, Statement-II is also incorrect.

Final Conclusion:

Both Statement-I and Statement-II are incorrect.