P Block Elements

The correct answer is (B): NO₂

Explanation:

The oxide of nitrogen which contains odd electrons is NO₂

The correct answer is (A): Ce(At. No. 58)

Explanation:

Ce → [Xe]4f¹ 5d¹ 6s²
Ce⁺⁴ → [xe] 4f° 5d° 6s°
Cerium in +4 oxidation state acquires inert gas configuration.

The correct answer is 2

and have 2 lone pairs.

The correct answer is (D) :
+
Acid Conjugate base
Conjugate base of acid is always a stronger nucleophile.

The correct answer is (A) : 3s², 3p⁴
Element E is Selenium
The element which is just above ‘E’ in periodic table is sulphur, its electronic configuration is 1s², 2s², 2p⁶, 3s², 3p⁴.

Element above

The correct answer is (D) : Statement I false, but Statement II is true.
Statement I is false, as E₂O₅ is more acidic than E₂O₃
Statement II is correct.
As +ve oxidation state increases, EN of element increases hence acidic character increases. Down the group, non-metallic character decreases, acidic character decreases.
Acidic character : E2O5 > E2O3 Down the group, acidic character of E2O3 decreases

The question involves the borax bead test, a qualitative analysis technique used to identify certain metal ions based on the color of the bead formed when a metal compound is fused with borax in a flame. Let's go through the process to identify why the blue-colored bead is formed when borax is heated with cobalt oxide (CoO).

When borax is heated, it loses water and forms sodium metaborate and boric oxide :

The boric oxide then reacts with the cobalt oxide present on the platinum loop to form cobalt metaborate . This compound is responsible for the distinctive blue color of the bead:

Now, let's analyze why the correct answer is :

1. Formation of the compound: During the fusion in the flame, the formed cobalt metaborate imparts a blue color to the glassy bead as it is integrated into the structure of the glass matrix.
2. Ionic influence: The presence of cobalt ion specifically imparts the deep blue coloration which is typical in cobalt-containing glass and glazes.
3. Elimination of other options:
   • : Boric oxide itself does not contribute to any specific coloration; it acts as a basic constituent enabling the formation of metaborate structures.
   • and : These structures do not typically form or exist in the bead under the heated conditions from borax bead preparation.

Thus, the blue-colored bead is indeed due to the formation of cobalt metaborate, .

Borazine, also known as "inorganic benzene," has the molecular formula and is a six-membered cyclic compound. It has the following characteristics:
- It is indeed a cyclic compound, so statement (1) is correct.
- Borazine exhibits electronic delocalization similar to benzene due to the alternating single and double bonds between boron and nitrogen, making statement (2) correct.
- Borazine can react with water to form boric acid and ammonia, so statement (3) is also correct.
- The term "banana bonds" typically refers to bonds in compounds like diborane, where boron atoms form unconventional bonds with hydrogen. Borazine does not contain banana bonds; instead, it has regular covalent bonds. Therefore, statement (4) is incorrect.

Thus, the incorrect statement is (4).

Beryllium has a less negative reduction potential compared to other alkaline earth metals (AEM). This is because of:
• High hydration energy: Due to the small size of the Be2+ ion, the hydration energy is very large, which stabilizes the ion in aqueous solutions.
• Large atomization enthalpy: Beryllium has strong metallic bonding due to its small atomic size, which increases its atomization enthalpy. These factors reduce the tendency of beryllium to get reduced, leading to a less negative
reduction potential.
Since the reason (R) explains why beryllium has a less negative reduction potential, both A and R are correct, and R is the correct explanation of A.

Assertion A is incorrect. Alkali metals and their salts typically impart distinct colors to an oxidizing flame, not a reducing flame. When alkali metal salts are heated in a flame, they excite the metal ions, which then emit characteristic colors as they return to their ground state. These colors are observed in an oxidizing flame (such as that produced by a Bunsen burner with sufficient oxygen), not in a reducing flame, which lacks the necessary oxidizing conditions for such reactions to occur.

Reason R is correct. Flame tests are indeed a common and reliable method for identifying alkali metals and other metal ions based on the characteristic colors they emit when heated. For example, lithium produces a red flame, sodium a bright yellow flame, and potassium a lilac flame. This principle is widely used in qualitative analysis.

Correct answer is (d) A-II, B-I, C-IV, D-III

To correctly match List I with List II, let's analyze each item:
A. Chlorophyll: - Chlorophyll contains magnesium ( ) as its central metal ion. - Match: A – III ( ).
B. Soda ash: - Soda ash is chemically sodium carbonate ( ). - Match: B – I ( ).
C. Dentistry, Ornamental work: - Calcium sulfate ( ) is used in dentistry and ornamental work, especially in plaster form. - Match: C – II ( ).
D. Used in white washing: - Slaked lime ( ) is commonly used in white washing. - Match: D – IV
Conclusion: The correct matches are:

reduces to and itself gets oxidised to

The correct option is (c ):

Bonds in Each Compound:

• (Pyrophosphoric Acid): Contains one P–O–P bond as it has a linear structure connecting two phosphate groups.
• (Cyclotriphosphoric Acid): Contains three P–O–P bonds as it forms a cyclic structure with alternating P and O atoms.
• (Phosphorus Pentoxide): Contains six P–O–P bonds in its tetrahedral structure, where four phosphorus atoms are connected by oxygen bridges.

Summary:

• : 1 P–O–P bond
• : 3 P–O–P bonds
• : 6 P–O–P bonds

• Peroxodisulphuric acid: Structure: Number of -bonds = 4 (two in each group)
• Pyrosulphuric acid: Structure: Number of -bonds = 4 (two in each group).
Total -bonds = .

The structure of pyrophosphoric acid (H₄P₂O₇) consists of an O=P-O-P=O backbone with terminal hydroxyl groups (OH).

Step 1: Counting Sigma ( ) Bonds:

• Each single bond (P-O and O-H) is a sigma bond.
• Each double bond (P=O) contains 1 sigma bond.

Total sigma bonds: 12 (8 single P-O and O-H bonds + 4 sigma components from P=O bonds).

Step 2: Counting Pi ( ) Bonds:

• Each double bond (P=O) contains 1 pi bond.

Total pi bonds: 2 (from P=O bonds).

Step 3: Ratio of Sigma to Pi Bonds:

Conclusion:

The ratio of sigma to pi bonds in pyrophosphoric acid is 6:1.

Calculation of Radial Nodes:

The number of radial nodes for an electron in an orbital is given by:

Radial nodes = n - l - 1,

where:

• n: Principal quantum number
• l: Azimuthal quantum number

Calculations for the given orbitals:

• 5s: radial nodes
• 6s: radial nodes
• 7s: radial nodes
• 6p: radial nodes
• 4p: radial node

Conclusion:

Only the 6s orbital has five radial nodes. Therefore, the correct answer is 6s.