The P Block Elements

In graphite, each carbon is bonded to three others in a planar structure, indicating sp\textsuperscript{2} hybridization.
In diamond, each carbon is tetrahedrally bonded to four other carbons, representing sp\textsuperscript{3} hybridization.
In C60} (buckminsterfullerene), carbon atoms also form a structure using sp\textsuperscript{2} hybrid orbitals.

Halogens (X ) can act as oxidizing agents. Their oxidizing power is related to their tendency to gain electrons (i.e., their standard reduction potentials). The general reaction of a halogen X acting as an oxidizing agent is: X + 2e 2X The standard reduction potentials ( ) for halogens are: F (g) + 2e 2F (aq) \quad Cl (g) + 2e 2Cl (aq) \quad Br (l) + 2e 2Br (aq) \quad (or +1.09 V) I (s) + 2e 2I (aq) \quad A higher (more positive) standard reduction potential indicates a stronger tendency to be reduced, and thus a stronger oxidizing agent. Based on these values, the decreasing order of oxidizing power is: F >Cl >Br >I . When halogens react with water, they can oxidize water or disproportionate. \begin{itemize} \item F : Fluorine is the strongest oxidizing agent. It oxidizes water vigorously: (Fluorine oxidizes O from -2 in H O to 0 in O ). \item Cl : Chlorine reacts with water to form HCl and HOCl (hypochlorous acid). This is a disproportionation reaction for Cl (0 to -1 in HCl and +1 in HOCl), but in this context, Cl can also oxidize other substances in water. Its oxidizing power is less than F . \item Br : Bromine reacts with water similarly to chlorine, but to a lesser extent. \item I : Iodine has limited solubility and reactivity with water. The reaction is very slight. Iodine is the weakest oxidizing agent among the common halogens. In fact, I can be oxidized to I by stronger oxidizing agents like Cl or Br . (equilibrium lies far to the left). \end{itemize} The question asks for "oxidizing power of halogens with water". This refers to their general strength as oxidizing agents, which is best compared using their standard reduction potentials. The specific reactions with water illustrate this trend. Fluorine's ability to oxidize water to O highlights its supreme oxidizing power. The other halogens' reactions with water are more complex (disproportionation), but their overall oxidizing strength follows the trend. So, the correct decreasing order of oxidizing power is F >Cl >Br >I . This matches option (d).

I. Boron trihalides do not form dimeric structures, unlike AlCl3}.
II. Boron does not commonly exhibit +1 as a stable oxidation state.
III. Boron being a group 13 element has a maximum covalency of four due to its small size and absence of d-orbitals.
IV. Boron does not form BF6}\textsuperscript{−} ion because it cannot expand its octet (no available d-orbitals).

A conjugate base is formed when an acid loses a proton (H ). The ammonium ion (NH ) is a weak acid that donates a proton to form NH : Hence, NH is the conjugate base of NH .

Interstitial hydrides are formed by transition metals, especially those from the d-block, where small hydrogen atoms occupy the voids (interstices) in the metal lattice. These are generally formed by metals like Ti, V, Cr, etc. Chromium (Cr), being a first-row transition metal, has the right atomic size and lattice structure to allow hydrogen atoms to occupy interstitial positions and form interstitial hydrides.

Boron has a high melting point due to its strong covalent structure.
It is a metalloid with poor electrical conductivity. The isotope B-10 is used as a neutron absorber in nuclear reactors. Hence only statements I and IV are correct.

PbI\textsubscript{4} does not exist because Pb in +4 oxidation state is unstable in presence of I\textsuperscript{–}, which is a good reducing agent. Pb\textsuperscript{4+} gets reduced to Pb\textsuperscript{2+} and iodine gets oxidized.

Statement I is incorrect: Boron does exhibit allotropy (e.g., crystalline and amorphous forms).
Statement II is correct: Crystalline boron is extremely hard and appears black in color.

Cryolite (Na AlF ): Used in the electrolytic extraction of aluminum. Not directly associated with water molecules in its primary application.
Bauxite (Al O .xH O): An ore of aluminum, it is a hydrated oxide and thus contains water molecules.
Kernite (Na B O .4H O): A hydrated borate mineral containing water molecules.
Borax (Na B O .10H O): A hydrated borate mineral containing water molecules.

Let's analyze each statement regarding Group 13 elements:
Statement 1: Nature of aqueous solution of borax is alkaline.
Borax (sodium tetraborate decahydrate, ) dissolves in water and undergoes hydrolysis. The hydrolysis products are boric acid ( , a weak acid) and sodium hydroxide ( , a strong base).

Since a strong base ( ) and a weak acid ( ) are formed, the solution will be basic (alkaline). This statement is correct.
Statement 2: Orthoboric acid is a weak tribasic acid.
Orthoboric acid is (or ). It is a weak acid. However, it is not a Brønsted acid (proton donor) but rather a Lewis acid, accepting a hydroxide ion from water, thereby releasing a proton from water:

Since one molecule of orthoboric acid generates only one ion per molecule in aqueous solution, it is a monobasic acid. Therefore, the statement that it is a tribasic acid is incorrect.
Statement 3: Metaboric acid on heating gives an acidic oxide.
Metaboric acid ( ) is an intermediate formed during the heating of orthoboric acid. On further strong heating, metaboric acid undergoes dehydration to form boron trioxide ( ).

Boron trioxide ( ) is an acidic oxide, meaning it reacts with bases to form borates. This statement is correct.
Statement 4: LiBH acts as a reducing agent.
Lithium borohydride ( ) is a well-known and powerful reducing agent commonly used in organic synthesis. It can reduce various functional groups, including aldehydes, ketones, esters, and carboxylic acids. This statement is correct.
Conclusion:
Based on the analysis, the incorrect statement is Statement 2.
The final answer is .

Step 1: Identify Gas X from the Reaction
The reaction between Ammonium chloride ( ) and Calcium hydroxide ( ) is a standard laboratory method for the preparation of ammonia gas.
The balanced chemical equation is:
From the reaction, the gas X produced is Ammonia ( ). Step 2: Evaluate Statement: "X is ammonia"
As identified in Step 1, gas X is indeed ammonia.
So, this statement is Correct.
Step 3: Evaluate Statement: "Shape of X is trigonal pyramidal"
For Ammonia ( ):
The central atom is Nitrogen (N).
Nitrogen has 5 valence electrons. It forms 3 single bonds with 3 Hydrogen atoms. This leaves 5 - 3 = 2 electrons, which form 1 lone pair. According to VSEPR (Valence Shell Electron Pair Repulsion) theory, the presence of three bond pairs and one lone pair around the central atom results in a trigonal pyramidal molecular geometry.
So, this statement is Correct.
Step 4: Evaluate Statement: "X is an electron rich hydride"
Hydrides are compounds formed between hydrogen and other elements. Electron-rich hydrides are those in which the central atom possesses lone pair(s) of electrons in addition to forming covalent bonds with hydrogen atoms. Elements of group 15, 16, and 17 form electron-rich hydrides.
In , the central Nitrogen atom has one lone pair of electrons.
Therefore, ammonia is an electron-rich hydride.
So, this statement is Correct.
Step 5: Evaluate Statement: "Hybridisation of central atom in X is sp "
For Ammonia ( ):
The central Nitrogen atom has 3 bond pairs (with H) and 1 lone pair. The steric number (number of bond pairs + number of lone pairs) = 3 + 1 = 4.
A steric number of 4 corresponds to hybridization. Thus, the hybridization of the central nitrogen atom in is }, not .
So, this statement is Incorrect.
Step 6: Identify the Incorrect Statement
The question asks for the incorrect statement about X. Based on our analysis, the statement "Hybridisation of central atom in X is sp " is incorrect. Step 7: Analyze the Options
\begin{itemize} \item Option (1): X is ammonia. (Correct statement, so not the answer) \item Option (2): Shape of X is trigonal pyramidal. (Correct statement, so not the answer) \item Option (3): X is an electron rich hydride. (Correct statement, so not the answer) \item Option (4): Hybridisation of central atom in X is sp . (Incorrect statement, this is the answer) \end{itemize}