13 th group elements (p-block) practice · Mixed difficulty

Option A ($\mathrm{N}{\mathrm{H}_3}$) is correct because boron nitride reacts with caustic alkali to produce ammonia.

Option A ($N{H_3}$ in the ratio $1:2$) is correct because borazole is formed by the reaction of diborane (${B_2}{H_6}$) with ammonia ($N{H_3}$) in a 1:2 ratio.

The correct answer is C (Assertion is true, but Reason is false) because the solubility of aluminum hydroxide decreases at higher temperatures, but this is due to its endothermic dissolution, not a decrease in the solubility product.

Option B is correct because Gallium (Ga) is known to expand upon solidification, a property it shares with water and bismuth.

Option A ($NaB{O_2} + {B_2}{O_3}\,$) is the correct composition of borax glass, which is formed upon heating borax.

Option C (Tl > In > Al > Ga > B) is correct because atomic radii generally increase down a group, but gallium is smaller than aluminum due to the d-block contraction.

The correct answer is A (Both (A) and (R) are true and (R) is the correct explanation of (A)) because back bonding in BF3 reduces its electron deficiency, making it a weaker Lewis acid than BCl3.

The correct answer is C (6) because in diborane (B2H6), the four terminal hydrogen atoms and the two boron atoms lie in one plane.

Option B is correct because Gallium does have a low melting point and its valence shell configuration is 4s²4p¹, but the configuration doesn't directly explain the low melting point.

Option D ($B{F_3}$) is correct because Fluorine has the highest tendency to back bond due to its small size and high electronegativity, leading to maximum $P\pi - P\pi$ back bonding.

Option A is correct because Beryllium's extremely small size and high ionization energy make the formation of B³⁺ highly unfavorable.

Option D (Oxidation state of $Ga$ in the salt $GaAICI_4$ is +3) is correct because in $GaAlCl_4$, the oxidation state of Ga is +1 and Al is +3, forming $Ga^+[AlCl_4]^-$.

Option D ($Al\,<\,Ga\,<\,In <\,TI$) is correct because the inert pair effect becomes more prominent down the group, stabilizing the +1 oxidation state for heavier elements like Thallium.

Option C is correct because Alum is commonly used as a mordant in dyeing processes to fix the dye to the fabric.

Option D ($\mathbf{A}$ is false but $\mathbf{R}$ is true) is correct because $AlCl_3$ does not form a dimer in aqueous solution; it forms hydrated ions, and the high enthalpy of hydration is indeed sufficient to break any potential covalent dimer.

Option D ($B > Al < Ga > In < Tl$) is correct because ionization enthalpy generally decreases down the group, but there are exceptions due to the presence of d- and f-electrons.

Option B (Spreading water over fires) is correct because bentonite slurry dropped from aeroplanes helps to extinguish fires by spreading water.

Option C (Aluminium oxide is highly stable) is correct because aluminium oxide's high stability prevents it from being easily reduced by chemical reactions.

Option D is correct because all listed options (dry powder, NaHCO3 + H2SO4, and foamite extinguishers) are effective fire extinguishing agents.

Option C is correct because the larger size of chlorine atoms prevents them from fitting between boron atoms to form a dimer, unlike the smaller hydrogen atoms in BH3.

Option A ($\mathrm{B}{\mathrm{F}_3}$) is correct because Boron trifluoride exists as a gas at room temperature.

Option A (8) is correct because diborane has 4 $2c-2e$ bonds and 2 $3c-2e$ bonds, so $x=4$ and $y=2$, and $x^2 - y^3 = 4^2 - 2^3 = 16 - 8 = 8$.

Option B (3, sp³) is correct because $BCl_3$ reacts with water to form boric acid, $B(OH)_3$, where boron has an oxidation state of +3 and is sp³ hybridized.

Option D (Anodizing) is correct because anodizing is an electrolytic passivation process used to increase the thickness of the natural oxide layer on the surface of aluminum, enhancing its corrosion resistance.

The correct answer is C (2,6,2,6,3) because it represents the balanced stoichiometric coefficients for the given reaction, ensuring mass conservation.

Option B (${\mathrm{H}_2}{\mathrm{B}_4}{\mathrm{O}_7}$) is correct because boric acid when heated at 150°C gives metaboric acid, which further decomposes to tetraboric acid.

Option D (Al) is correct because powdered aluminium suspended in oil is used as paint.

Option C ($B{\left( {OH} \right)_3}$) is correct because boric acid is a weak Lewis acid, while the others are amphoteric or basic hydroxides.

Option C (Water softening) is correct because alum is used in water softening to remove impurities.

Option A (Al) is correct because aluminum forms dimeric halides like $Al_2Cl_6$.