p-Block Elements

Boron is indeed a very hard substance, which does indicate high lattice energy.

This is because the lattice energy is a measure of the strength of the forces between the ions in an ionic solid; the greater the lattice energy, the stronger the forces and the harder and more stable the compound.

This makes Statement I correct.

Statement II is also correct.

Boron, as the first member of group 13 elements, does indeed have the highest melting and boiling points among its group members (Al, Ga, In, Tl).

This can be attributed to the strong covalent bonding in boron due to its small size, which requires a high amount of energy to break.

Therefore, Both statement I and Statement II are correct is the correct answer.

Statement I : SbCl₅ is more covalent than SbCl₃ In general, the covalent character of a compound is directly proportional to the polarization, and the polarization depends upon the size of the cation and the charge on the cation.

If the charge on the cation increases or if the size of the cation decreases, the covalent character increases.

In the case of SbCl₅ and SbCl₃, the antimony (Sb) cation in SbCl₅ has a higher charge due to the greater number of chlorine (Cl) atoms, thus increasing the polarization and making SbCl₅ more covalent than SbCl₃.

Statement II : The higher oxides of halogens also tend to be more stable than the lower ones.

This statement is also correct.

Halogens can form a variety of oxides, and generally, the stability of these oxides increases with increasing oxidation state.

This is because the halogens, being highly electronegative, stabilize the high positive oxidation state.

For example, in the case of chlorine, Cl₂O₇ (where chlorine is in the +7 oxidation state) is more stable than Cl₂O (where chlorine is in the +1 oxidation state).

Polar molecular solids are formed by molecules that have polar covalent bonds and possess a net dipole moment.

A net dipole moment arises when there is an asymmetric distribution of charge in the molecule, resulting in a separation of positive and negative charges.

Out of the given options, only option C consists of two polar molecular solids - $\mathrm{SO}_2(\mathrm{s})$ and $\mathrm{NH}_3(\mathrm{s})$. $\mathrm{SO}_2$ has polar covalent bonds due to the difference in electronegativity between sulfur and oxygen, and the molecule has a net dipole moment due to its bent molecular geometry.

Similarly, $\mathrm{NH}_3$ has polar covalent bonds due to the difference in electronegativity between nitrogen and hydrogen, and the molecule has a net dipole moment due to its trigonal pyramidal molecular geometry.

Therefore, the correct answer is - $\mathrm{SO}_{2}(\mathrm{s}), \mathrm{NH}_{3}(\mathrm{s})$.

cannot undergo disproportionation reaction as chlorine is present in it's highest oxidation state.

Option A : Cl forms bond with both Al and Ga in GaAlCl4 - This is incorrect.

In GaAlCl4, chloride ions (Cl-) form bonds with Al to create the AlCl4- ion.

There is an ionic bond between the Ga+ cation and the AlCl4- anion.

Option B : The oxidation state of Ga in the salt GaAlCl4 is +3 - This is incorrect.

The oxidation state of Ga in GaAlCl4 is actually +1.

Option C : Ga is coordinated with Cl in GaAlCl4 - This is incorrect.

Ga is not coordinated with Cl.

The Ga+ ion forms an ionic bond with the AlCl4- ion.

Option D : Ga is more electronegative than Al and is present as a cationic part of the salt GaAlCl4 - This is correct.

According to the Pauling scale, aluminum (Al) has an electronegativity of 1.61 and gallium (Ga) has an electronegativity of 1.81, which makes Ga more electronegative than Al, and Ga exists as a cationic part (Ga+) in the compound.

Both statements need to be evaluated to determine which (if any) are true: Statement (I) : $\mathrm{SiO}_2$ and $\mathrm{GeO}_2$ are acidic while $\mathrm{SnO}$ and $\mathrm{PbO}$ are amphoteric in nature.

This statement is generally true.

The oxides of silicon ($\mathrm{SiO}_2$) and germanium ($\mathrm{GeO}_2$) are indeed acidic.

These elements belong to group 14 of the periodic table and as you move down the group, the acidic character of the oxides decreases while the basic character increases.

Tin oxide ($\mathrm{SnO}$) and lead oxide ($\mathrm{PbO}$) are lower in the group and they are amphoteric, which means they show both acidic and basic properties depending on the reacting substance.

Statement (II) : Allotropic forms of carbon are due to the property of catenation and $\mathrm{p} \pi-\mathrm{d} \pi$ bond formation.

This statement is false.

Catenation is indeed the property that carbon has to form long chains and rings with other carbon atoms, which is responsible for the vast number of organic compounds.

However, in the context of pure carbon allotropes—such as diamond, graphite, and fullerene—the different physical structures and properties are mainly due to the $\mathrm{sp}^3$, $\mathrm{sp}^2$, and $\mathrm{sp}$ hybridizations of carbon, not to $\mathrm{p} \pi-\mathrm{d} \pi$ bonding.

Carbon does not use d-orbitals in forming the allotropes.

The $\mathrm{p} \pi-\mathrm{d} \pi$ bonding is typically discussed in the context of transition metal complexes, not allotropic forms of carbon.

Thus, the correct answer is Option D : Statement I is true but Statement II is false.

The reducing strength of a substance is generally associated with its ability to donate electrons to other substances.

In the context of hydrides of Group 15 elements, such as $\mathrm{SbH}_3$, $\mathrm{NH}_3$, $\mathrm{BiH}_3$, and $\mathrm{PH}_3$, the reducing strength can be considered based on their chemical reactivity, stability, and tendency to donate electrons.

In general, the stability of hydrides decreases as we move down the group in the periodic table, which means that heavier hydrides are less stable and hence act as stronger reducing agents.

This trend is primarily due to the increase in the size of the central atom as we move down the group, leading to weaker bonds between the central atom and hydrogen in the hydride.

As the bond strength decreases, the hydrides can more readily release hydrogen in the form of protons (H+) or hydride ions (H−), making them stronger reducing agents.

The order of reducing strength for Group 15 hydrides is typically as follows:

Among the options given, $\mathrm{BiH}_3$ (bismuthine) is the strongest reducing agent.

It is the least stable of the listed hydrides, due to the large size and low electronegativity of bismuth compared to antimony, phosphorus, and nitrogen.

This makes $\mathrm{BiH}_3$ more prone to releasing electrons and acting as a reducing agent.

In contrast, $\mathrm{NH}_3$ (ammonia) is the most stable among the given hydrides and, therefore, the weakest reducing agent.

Its bond with hydrogen is comparatively stronger, and nitrogen's higher electronegativity makes the molecule more resistant to donating electrons.

Thus, the correct answer is Option C, $\mathrm{BiH}_3$.

Solid Boron has very strong crystalline lattice so its melting point unusually high in group 13 elements

Statement-I: Oxygen can have oxidation state from -2 to +2 , so statement I is incorrect Statement- II: On moving down the group stability of +4 oxidation state increases whereas stability of +6 oxidation state decreases down the group, according to inert pair effect.

So both statements are wrong.

In trivalent state most of the compounds being covalent are hydrolysed in water.

Trichlorides on hydrolysis in water form tetrahedral

species, the hybridisation state of element

is

. In case of aluminium, acidified aqueous solution forms octahedral

ion.