p-Block Elements

When boric acid reacts with sodium hydroxide, the products formed are sodium tetrahydroxoborate (III), sodium metaborate and water. The reaction is reversible in nature.

The reaction is difficult to proceed in forward direction due to weak monoprotic boric acid. When cis-1, 2-diol is added in the mixture of boric acid and sodium hydroxide, the weak acid gets converted to a comparatively stronger acid due to chelate complex formation. This stronger acid along with NaOH can undergo titration using phenolphthalein as an indicator. $\mathrm{B}(\mathrm{OH})_3+\mathrm{NaOH} \rightleftharpoons \mathrm{Na}\left[\mathrm{B}(\mathrm{OH})_4\right]$

Due to stable nature, it is removed from the solution during reaction. The entire reaction proceed in the same manner, disrupting the reversible chemical reaction and favouring the forward reaction completely.

(A) When $\mathrm{Bi}^{3+}$ is treated with water, it gives $[\mathrm{BiO}]^{+}$ ion and two protons.

$ \mathrm{Bi}^{3+}+\mathrm{H}_2 \mathrm{O} \rightarrow\left[\mathrm{BiO}^{+}+2 \mathrm{H}^{+}\right. $

The hydrolysis is a chemical reaction in which water is used for breaking down chemical bonds. The conversion of $\mathrm{Bi}^{+} \rightarrow[\mathrm{BiO}]^{+}$is carried out with the help of water; hence, it is a hydrolysis reaction.

Hence, option A from column I matches with option Q from column II.

(B) The conversion of $\left[\mathrm{AlO}_2\right]^{-} \rightarrow \mathrm{Al}(\mathrm{OH})_3$ is carried out by acidification. The proton is added to one of the reactant to produce aluminium hydroxide. The acid serve as a source of protons required for the acidification. The chemical reaction for the same is shown below:

$ \left[\mathrm{AlO}_2\right]^{-} \rightarrow \mathrm{Al}(\mathrm{OH})_3 $

Hence, option B from column I matches with option R from column II.

(C) Two molecules of orthosillicate $2 \mathrm{SiO}_4^{4-}$ gives pyrosilicate $\mathrm{Si}_2 \mathrm{O}_7^{6-}$ on strong heating.

$ 2 \mathrm{H}_4 \mathrm{SiO}_4 \xrightarrow{\Delta} \mathrm{H}_6 \mathrm{~S}_2 \mathrm{O}_7 $

Equation can also be written as

$ 2 \mathrm{SiO}_4^{4-} \xrightarrow{\Delta} \mathrm{Si}_2 \mathrm{O}_7^{6-} $

Hence, option C from column I matches with option P in column II.

(D) The conversion of $\left(\mathrm{B}_4 \mathrm{O}_7^{2-}\right)$ into $\left[\mathrm{B}(\mathrm{OH})_3\right]$ can be achieved either by hydrolysis or acidification.

The ( $\mathrm{B}_4 \mathrm{O}_7^{2-}$ ) is an anion in $\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7$.

When $\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7$ is treated with HCl or $\mathrm{H}_2 \mathrm{SO}_4$, it gives $\mathrm{H}_3 \mathrm{BO}_3$.

$ \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \xrightarrow{\mathrm{HCl} \text { or } \mathrm{H}_2 \mathrm{SO}_4} \mathrm{H}_3 \mathrm{BO}_3 \text { or } \mathrm{B}(\mathrm{OH})_3 $

Similarly, hydrolysis of $\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7$ yields $\mathrm{H}_3 \mathrm{BO}_3$.

$ \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \xrightarrow{\mathrm{H}_2 \mathrm{O}} \mathrm{H}_3 \mathrm{BO}_3 $

The $\mathrm{H}_3 \mathrm{BO}_3$ is a chemical formula of boric acid and is written as $\left[\mathrm{B}(\mathrm{OH})_3\right]$.

Hence, option D from column I matches with options Q and R in column II.

When carbon dioxide (CO₂) is dissolved in water, it undergoes a series of chemical reactions that result in various species being present in the solution. The reactions can be described as follows:

1. Carbon dioxide reacts with water to form carbonic acid:

$CO_2 + H_2O \rightleftharpoons H_2CO_3$

2. Carbonic acid can dissociate into a bicarbonate ion and a proton:

$H_2CO_3 \rightleftharpoons HCO_3^- + H^+$

3. Bicarbonate ion can further dissociate into a carbonate ion and a proton:

$HCO_3^- \rightleftharpoons CO_3^{2-} + H^+$

Based on these reactions, the species present in the solution when CO₂ is dissolved in water include CO₂, $H_2CO_3$, $HCO_3^-$, and $CO_3^{2-}$. Thus, the correct option is:

Option A: CO₂, H₂CO₃, $HCO_3^-$, $CO_3^{2-}$

When phosphine gas (PH₃) is mixed with chlorine gas (Cl₂), the reaction between them depends on the conditions and the proportions of the reactants. Let's consider the options :

1. Option A : PCl₃ and HCl are formed and the mixture warms up.

   
   
   • When phosphine gas reacts with chlorine gas, one of the possible products is phosphorus trichloride (PCl₃) along with hydrogen chloride (HCl). This reaction is exothermic, meaning it releases heat and causes the mixture to warm up.



2. Option B : PCl₅ and HCl are formed and the mixture cools down.

   
   
   • Phosphorus pentachloride (PCl₅) can be formed if an excess of chlorine is used, but this reaction is also exothermic. The formation of PCl₅ is less likely with PH₃ as it requires a higher proportion of chlorine and controlled conditions.



3. Option C : PH₃·Cl₂ is formed with warming up.

   
   
   • There is no known stable compound like PH₃·Cl₂ formed directly from the reaction of phosphine and chlorine gas. This option does not represent a typical chemical reaction product.



4. Option D : The mixture only cools down.

   
   
   • Chemical reactions between phosphine and chlorine are exothermic, so this statement is incorrect.

Based on this analysis, the most likely and correct reaction is described in Option A : PCl₃ and HCl are formed, and the mixture warms up. This reaction is feasible and aligns with the known chemistry of phosphine and chlorine gases.

The correct answer is Option A: N: tetrahedral, sp³; B: tetrahedral, sp³

• BF₃ (Boron trifluoride): In its isolated state, BF₃ has a trigonal planar geometry around the boron atom with sp² hybridization.


• NH₃ (Ammonia): In its isolated state, NH₃ has a trigonal pyramidal geometry around the nitrogen atom with sp³ hybridization.

However, when BF₃ and NH₃ form a 1:1 complex, the lone pair of electrons on nitrogen donates to the empty p-orbital of boron. This leads to the formation of a coordinate bond between nitrogen and boron.

• Nitrogen (N): Now has four bonding pairs of electrons (three with hydrogen and one with boron), resulting in a tetrahedral geometry and sp³ hybridization.


• Boron (B): Also has four bonding pairs of electrons (three with fluorine and one with nitrogen), leading to a tetrahedral geometry and sp³ hybridization.

The oxidation number of an atom in a molecule or an ion is essentially a theoretical charge that the atom would have if all the bonds to atoms of different elements were completely ionic. To determine the oxidation numbers of sulfur in the given compounds: S₈, S₂F₂, and H₂S, let's analyze them one by one.

1. In S₈: This is an elemental form of sulfur, comprising of sulfur atoms only. In elemental forms, the atoms are not bonded to atoms of a different element, but only to themselves. In such cases, the oxidation number is always 0. Hence, the oxidation number of sulfur in S₈ is 0.

2. In S₂F₂: This is a molecule consisting of sulfur and fluorine. Fluorine is more electronegative than sulfur and virtually always has an oxidation number of -1 in its compounds. Let's denote the oxidation number of sulfur as $x$. Since there are two fluorine atoms, each contributing -1, and the overall molecule is neutral, we get:

$2x + (2 \times -1) = 0$

$2x - 2 = 0$

$2x = 2$

$x = +1$

Therefore, the oxidation number of sulfur in S₂F₂ is +1.

3. In H₂S: In this compound, hydrogen has an oxidation number of +1 (except when bonded to metals where it can be -1). Since there are two hydrogens, their total contribution is +2. Let's denote the oxidation number of sulfur in this compound as $x$. The compound is neutral overall, so:

$2(+1) + x = 0$

$2 + x = 0$

$x = -2$

Therefore, the oxidation number of sulfur in H₂S is -2.

Comparing with the given options, the correct answers are 0, +1, and -2 for S₈, S₂F₂, and H₂S, respectively. Therefore, the correct option is Option A: 0, +1, and -2.