Classification Of Elements And Periodicity In Properties

Correct answer is (d) B

From Sc to Mn ionization energy is less than that of Mg.
For 3d series :

For 2nd Group

So , the correct option is (C) : (A) is false but (R) is true

The correct answer is (D) :
According to Henry Moseley
So

Correct answer is (b) He < Xe < Kr < Ne

Electron Gain Enthalpy Trend
Electron gain enthalpy generally becomes less negative (less exothermic) down a group due to increasing atomic size. However, there can be exceptions due to factors like electron-electron repulsion and shielding effects. Fluorine has a smaller atomic size and therefore greater electron density and experiences inter electronic repulsions more than chlorine and hence its magnitude is smaller than that of Cl.
Analyzing the Options
(1) Cl > F (Correct, due to small size and inter electronic repulsions of F) (2) S > Se (Incorrect, electron gain enthalpy becomes less negative down the group; therefore, S should have a more negative electron gain enthalpy than Se. This is an exception). (3) At > I (Correct) (4) Po > Te (Correct)

Correct order is F>N>O>C>Be>B>Li.

The correct option is (B): K > Ca > Be

On moving from top to bottom metallic character increases while from left to right metallic decreases.

The question is based on Moseley's Law, which describes the relationship between the frequency of characteristic X-rays and the atomic number . Moseley's Law: or equivalently,
Here: - is the frequency of the emitted X-rays. - is the atomic number of the element.

1. Graph Interpretation: The graph provided shows a linear relationship between and . For this to hold true: - From Moseley's law, . - Therefore, .
2. Physical Basis: The characteristic X-rays are emitted due to the transition of electrons from higher energy levels to the inner shells (K or L). The energy difference between these levels is proportional to , leading to the proportionality of .
3. Verification of : For (or ), the graph becomes linear because: where is a proportionality constant. This matches the given graph.
Conclusion: The value of that satisfies the given graph is .

Chlorine has the most negative (–349 kJ/mole) whereas Neon has the most positive (116 kJ/mole)

Step 1: Understanding the coagulation process
Blood exists as a negatively charged sol. When ferric chloride (FeCl ) is introduced, the Fe ions interact with the negatively charged blood sol, causing it to coagulate. This occurs due to the electrostatic attraction between the positively charged Fe ions and the negatively charged blood particles.
Step 2: The role of
The Fe ions neutralize the negative charge on the blood particles, leading to their aggregation and coagulation. This explains why Fe ions are responsible for facilitating the coagulation of blood.

The metallic character of elements increases as you move down a group and decreases as you move across a period.
In a group, metallic character increases as the atomic size increases, resulting in a weaker attraction between the valence electrons and the nucleus.
In a period, metallic character decreases as the effective nuclear charge increases, making it more difficult to lose electrons.
Thus, the metallic character decreases from K to Be across the period, and increases from Be to Ca down the group.
Therefore, the correct order of metallic character is:

Assertion (A): Sodium is about 30 times more abundant than potassium in the oceans. This is correct because sodium ions are more soluble in water compared to potassium ions, and sodium salts are more readily dissolved and transported into water bodies.

Reason (R): Potassium is larger in size than sodium due to its position below sodium in Group 1 of the periodic table. This is also correct.

However, the larger size of potassium does not explain the higher abundance of sodium. The abundance is influenced by the higher solubility of sodium salts rather than atomic size. Hence, the reason does not justify the assertion.

Conclusion: Both the assertion and reason are correct, but the reason is not the correct explanation for the assertion. The correct answer is option (1).

• Acidic Strength is inversely proportional to the effect and directly proportional to the effect.
• The effect order of halogens is: F > Cl > Br.

Compound Analysis:

1. (A) CH₃−COOH : The methyl group exhibits a effect, which decreases acidic strength.
2. (B) F−C(F)−F−COOH: Contains three fluorine atoms with a strong effect, significantly increasing acidic strength.
3. (C) Cl−CH₂−COOH : The chlorine atom exhibits a effect, increasing acidic strength but less than fluorine.
4. (D) F−CH₂−COOH : Fluorine exhibits a stronger effect compared to chlorine, making this compound more acidic than (C).
5. (E) Br−CH₂−COOH : Bromine has the weakest effect among the halogens, so its acidic strength is less than (C) and (D).

Acidic Strength Order:

(B) > (D) > (C) > (E) > (A)

(A) Incorrect:

The reduction potential for the M³⁺/M²⁺ couple for manganese is greater than that for iron:

Therefore, this statement is incorrect.

(B) Correct:

Higher oxidation states of first-row d-block elements are stabilized by oxide ions (O²⁻) due to the formation of strong metal-oxygen bonds. This statement is correct.

(C) Correct:

Chromium in the Cr²⁺ oxidation state can reduce H⁺ to H₂ in aqueous solution:

The reduction potential confirms this. This statement is correct.

(D) Incorrect:

V²⁺ has three unpaired electrons, resulting in a magnetic moment of approximately 3.87 BM, which is not within the range of 4.4-5.2 BM. This statement is incorrect.

Conclusion:

• (A) Incorrect
• (B) Correct
• (C) Correct
• (D) Incorrect

• Reactivity is influenced by the electron density on the benzene ring:

.
• Dimethylamine ( ) and methoxy are strong electron-donating groups, en hancing reactivity.
• Alkyl groups (CH₃) show moderate activating effects, while electron-withdrawing groups like trifluoromethyl (CF₃) reduce reactivity.

Analysis of Complexes and Unpaired Electrons:

1. Option (A) : The complex [Cr(CN)₆]³⁻ contains Cr³⁺ with an electronic configuration of 3d³. Cyanide (CN⁻) is a strong field ligand (SFL), which causes pairing of electrons.
   Number of unpaired electrons = 3.
2. Option (B) : The complex [Fe(H₂O)₆]²⁺ contains Fe²⁺ with an electronic configuration of 3d⁶. Water (H₂O) is a weak field ligand (WFL), which does not cause electron pairing.
   Number of unpaired electrons = 4.
3. Option (C) : The complex [Co(NH₃)₆]³⁺ contains Co³⁺ with an electronic configuration of 3d⁶. Ammonia (NH₃) is a strong field ligand (SFL), which causes pairing of electrons.
   Number of unpaired electrons = 0.
4. Option (D) : The complex [Ni(NH₃)₆]²⁺ contains Ni²⁺ with an electronic configuration of 3d⁸. Ammonia (NH₃) is a strong field ligand (SFL), which causes partial pairing of electrons.
   Number of unpaired electrons = 2.

Final Matching: A-II, B-IV, C-I, D-III

The correct answer is option (1).

Understanding Electron Gain Enthalpy: Electron gain enthalpy is the amount of energy released when an electron is added to a neutral atom in the gaseous state to form a negative ion. Higher negative values indicate that an element releases more energy when gaining an electron, signifying higher electron affinity.

Considering the Elements: We need to arrange Ar, Br, F, and S based on their electron gain enthalpy.

• Argon (Ar): As a noble gas, Argon has a complete octet, making its electron affinity negligible.
• Fluorine (F): Fluorine has a high electron affinity because it is one electron short of a complete octet, and it is known for having the highest electron affinity among halogens.
• Bromine (Br): Bromine, being another halogen, also exhibits high electron affinity but not as much as Fluorine. It has less electron gain enthalpy compared to Fluorine.
• Sulfur (S): Sulfur has a moderate electron affinity. It is less than that of halogens like Bromine or Fluorine.

Ranking based on Electron Gain Enthalpy: Considering the above information, the order of electron gain enthalpy from the most negative (most energy release) to the least is as follows:

• The least electron gain enthalpy is Argon because it is a noble gas.
• After Argon, Sulfur has lower electron gain enthalpy than Bromine.
• Bromine has lower electron gain enthalpy than Fluorine.
• Thus, the order is:

Conclusion: Therefore, the correct sequence of electron gain enthalpy for the given elements is A > D > B > C.

To determine which statements are true about elements A', B', C', and D', we refer to the periodic trends:

1. Elements in the same period (row) are represented as A' and B'. The atomic radius typically decreases across a period from left to right due to the increasing nuclear charge that pulls electrons closer to the nucleus.
2. Elements in the same group (column) are represented as A', C', and D'. The atomic radius generally increases down a group due to the addition of electron shells.

Let's evaluate each option:

1. Option A: Order of atomic radii:
   • This is true. Since A' is left of B' in the same period, . And since D' and C' are below A', .
2. Option B: Order of metallic character:
   • This is true. Metallic character decreases across a period and increases down a group. Hence, and .
3. Option C: Size of the element:
   • This is false. As discussed, the atomic size increases down a group and decreases across a period.
4. Option D: Order of ionic radii:
   • This is true, following the same general trends as atomic radii and taking into account the ionic charge effect.

Thus, the correct answer is: A, B, and D only.

To solve this problem, we need to match the tests in List-I with their correct observations in List-II. Let's analyze each test and its typical outcome:

1. Br₂ water test: This test is used to identify the presence of unsaturated compounds like alkenes and alkynes. When Br₂ reacts with such compounds, the reddish-orange color of bromine disappears. Thus, it matches with (Reddish orange color disappears).
2. Ceric ammonium nitrate test: This test is typically used to detect alcohols. The presence of alcohol changes the color of the ceric ammonium nitrate reagent to red. Thus, it matches with (Red color appears).
3. Ferric chloride test: This test is used for phenols. Phenols react with ferric chloride to produce a variety of colored complexes, often blue, green, violet, or red. Thus, it matches with (Blue, Green, Violet, or Red color appear).
4. 2, 4-DNP test: This is a test for carbonyl compounds (aldehydes and ketones). The test results in the formation of a yellow, orange, or red precipitate. Thus, it matches with (Yellow orange or orange-red precipitate formed).

Based on the analysis above, the correct matching is:

• A-II: Br₂ water test - Reddish orange color disappears
• B-III: Ceric ammonium nitrate test - Red color appears
• C-IV: Ferric chloride test - Blue, Green, Violet, or Red color appear
• D-I: 2, 4-DNP test - Yellow orange or orange-red precipitate formed

The correct answer is: A-II, B-III, C-IV, D-I.

The question asks us to identify the strongest oxidizing agent among the given ions. To determine this, we need to understand the concept of oxidizing agents.

An oxidizing agent is a substance that has the ability to oxidize other substances, i.e., it gains electrons and is reduced itself. The strength of an oxidizing agent is typically determined by its tendency to accept electrons.

Let's evaluate each option:

• : Lutetium has completed f-electron shells (4f¹⁴), which gives it stability and a low tendency to accept electrons.
• : Europium in its +2 state tends to revert to the more stable +3 state (Eu³⁺), making it a moderate oxidizing agent but not very strong.
• : Terbium in the +4 state tends to gain an electron to form the more stable +3 state (Tb³⁺), making it a strong oxidizing agent.
• : Cerium in the +3 state can further oxidize to the +4 state (Ce⁴⁺) under certain conditions, thus acting as a mild oxidizing agent.

The ion is the strongest oxidizing agent among the given options. This is because it tends to effectively accept an electron owing to the stability associated with the conversion from +4 to +3 oxidation state.

In conclusion, the correct answer is .