p-Block Elements

(A) Dimethyl silyl chloride undergoes hydration followed by polymerisation owing to the loss of water molecule.

(A) → P, S

$ \begin{aligned} & \text { (B) } 6 \mathrm{XeF}_4+12 \mathrm{H}_2 \mathrm{O} \rightarrow 4 \mathrm{Xe}+2 \mathrm{XeO}_3 +24 \mathrm{HF} \text { (Hydrogen fluoride) }+3 \mathrm{O}_2 \text { (Oxygen) } \\ & \end{aligned} $

Here, xenon undergoes disproportionation reaction (which is a kind of redox reaction) with xenon (in +4 oxidation state) gets oxidized to xenon trioxide $\left(\mathrm{XeO}_3\right)$ and reduced to xenon (Xe) at the same time.

HF reacts with glass as shown :

$ \mathrm{Na}_2 \mathrm{SiO}_3+6 \mathrm{HF} \rightarrow \mathrm{Na}_2 \mathrm{SiF}_6+3 \mathrm{H}_2 \mathrm{O} $

(B) → P, Q, R, T

(C)

$2 \mathrm{Cl}_2+2 \mathrm{H}_2 \mathrm{O} \longrightarrow {4 \mathrm{HCl} \text { (Hydrogen } \text { halide )}+\mathrm{O}_2}$

(C) $\rightarrow$ P, Q

(D) $\mathrm{VCl}_5+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{VOCl}_3+2 \mathrm{HCl}$ (Hydrogen chloride)

(D) $\rightarrow P$

The reactions are as follows:

(A) 3Cu + 8HNO$_3$ (dil.) $\to$ 3Cu(NO$_3$)$_2$ + 4H$_2$O + 2NO

(B) Cu + 4HNO$_3$ (conc.) $\to$ Cu(NO$_3$)$_2$ + 4H$_2$O + 2NO$_2$

(C) 4Zn + 10HNO$_3$ (dil.) $\to$ 4Zn(NO$_3$)$_2$ + 5H$_2$O + N$_2$O

(D) Zn + HNO$_3$ (conc.) $\to$ Zn(NO$_3$)$_2$ + 2H$_2$O + 2NO$_2$

Phosphorus reacts with limited supply of oxygen to form the given product.

P$_4$ + 3O$_2$ $\to$ P$_4$O$_6$

N$_2$ is used to retard the further oxidation.

Tertiary amine will not react due to the bulkiness of trimethyl groups. Tertiary amines instead react with diborane to give simple addition compounds through symmetrical cleavage of diborane.

The bond formation in the given oxides of nitrogen is as follows.

Thus N-N bond exists in N$_2$O, N$_2$O$_3$ and N$_2$O$_4$.

In p-block elements, inert pair effect is observed, due to which lower oxidation state becomes more stable on going down the group. In 14^th group (C, Si, Ge, Sn, Pb) Pb is placed at lower position than Sn, that's why lower oxidation state is more stable for lead (Pb$^{2+}$) and Sn$^{+4}$ is more stable than Pb$^{+4}$. So, Pb$^{+4}$ compounds are stronger oxidising agents than Sn$^{+4}$ compounds. Lower oxidation for the group 14^th elements are more stable for the heavier members.

Due to greater solubility and nature to be prone to microbial action, nitrates are less abundant in earth's crust. Nitrates are soluble. In nitrates, nitrogen has maximum oxidation state and cannot be further oxidised.

Both NH$_3$ and PH$_3$ have one lone pair of electron on it. The lone pair in case of ammonia occupy sp$^3$ orbital is directional.

Hence, the lone pair in case of ammonia is easily available and can be easily donated. But for PH$_3$ the lone pair of electrons occupy spherical s-orbital which is less directional and thus its lone pair is not easily available neither it can be donated easily.

Also the lone pair of electrons in sp$^3$ orbital has lesser s-character and less attracted by nucleus and available for easy donation than in PH$_3$ with pure s orbital, the lone pair in which lone pair are more strongly attracted by nucleus. As we go down a group, the p-character in bond pair increases and the s-character in lone pair increases.

Thus ammonia is a better base than PH$_3$.

White phosphorus on reaction with NaOH gives PH$_3$ as one of the products.

P$_4$ + 3NaOH + 3H$_2$O $\to$ 3NaH$_2$PO$_2$ + PH$_3$

The oxidation state of P in P$_4$, NaH$_2$PO$_2$ and PH$_3$ is 0, +1 and $-$3 respectively. It is an example of disproportionation reaction.

The colourless salt H is either NH$_4$NO$_3$ or NH$_4$NO$_2$. The colourless salt produces NH$_3$ gas (non-inflammable) on boiling with excess of NaOH. On addition of zinc dust (a reducing agent) to this solution, sodium nitrite or sodium nitrate will liberate NH$_3$ gas again.

NH$_4$NO$_3$ + NaOH $\to$ NH$_3$ + NaNO$_3$ + H$_2$O

7NaOH + NaNO$_3$ + 4Zn $\to$ 4Na$_2$ZnO$_2$ + NH$_3$ + 2H$_2$O

NH$_4$NO$_2$ + NaOH $\to$ NaNO$_2$ + NH$_3$ + H$_2$O

3Zn + 5NaOH + NaNO$_2$ $\to$ 3Na$_2$ZnO$_2$ + NH$_3$

Statement‑1:

“Band gap in germanium is small.”

✅ True.

Germanium is a semiconductor with an energy band gap of about 0.66 eV at 300 K, which is much smaller than that of silicon (~1.1 eV).

Statement‑2:

“The energy gap of each germanium atomic energy level is infinitesimally small.”

❌ False.

In a single germanium atom, the energy levels are discrete with finite energy differences between them — not infinitesimally small.

In a solid, when many atoms come together, each discrete atomic level splits into closely spaced energy states due to interactions between atoms, forming continuous bands; the gap between valence and conduction bands (the band gap) is small but finite.

So, the “infinitesimally small” part is incorrect.

✅ Correct Option: C

Statement‑1 is True, Statement‑2 is False.

A molecule is said to be chiral if it cannot be super imposed on its mirror image by any combination of rotations or translation. A chiral molecule exists as two stereoisomers that are mirror images of each other called enantiomers. Chirality is based on molecular symmetry. A chiral molecule cannot have an improper axis of rotation which includes planes of symmetry and inversion center.

Statement- 1 is true as molecules that are not superimposable on their mirror images are a symmetric and hence chiral. Such molecules are known as enantiomers.

Statement- 2: All chiral molecules don’t have to have a stereogenic center. A molecule can be chiral even if it does not have a stereocenter. A molecule’s chirality depends entirely on whether it is a symmetrical. Example-Biphenyls do not have a chiral carbon but they cannot rotate freely due to steric crowding. Hence, they are a symmetrical. Therefore, we can conclude that Statement-2 is false.

P = 5 Value electrons

Steric number = 3 + 1 = 4

Geometry - Tetrahedral hybridisation $-$ sp$^3$

P character $=\frac{3}{4}\times100=75\%$

H$_3$BO$_3$ (orthoboric acid) is a weak Lewis acid.

H$_3$BO$_3$ + H$_2$O $\rightleftharpoons$ B(OH)$^{-}_{4}$ + H$^ \oplus $

It does not donate proton rather it accepts OH$^-$ form water.

So, Statement 1 is True, Statement 2 is False.

Argon is an inert gas; hence, less reactive. It shield metal during welding. Argon is frequently blended with carbon dioxide (CO$_2$), hydrogen (H$_2$), helium (He) or oxygen (O$_2$) to enhance the arc characteristics or facilitate metal transfer in gas metal arc welding.

Hybridisation, H = $\frac{1}{2}$(V + X $-$ C + A) = $\frac{1}{2}$ (8 + 0 $-$ 0 + 0 ) = 4. So, sp$^3$ hybridised and has tetrahedral structure.

No. of lone pair = H $-$ X $-$ D = 4 $-$ 0 $-$ 3 = 1. As lone pair is present the electron pair geometry will be distorted tetrahedral and structure is pyramidal.

H = hybridisation no. (if 2 then sp, if 3 then sp$^2$ etc.)

V = valence electrons of central atom.

X = no. of monovalent atom.

C = cationic charge, A = anionic charge.

D = no. of divalent atom

6XeF$_4$ + 12H$_2$O $\to$ 4Xe + 2XeO$_3$ + 24HF + 3O$_2$

XeF$_6$ + 3H$_2$O $\to$ XeO$_3$ + 6HF

XeF$_4$ and XeF$_6$ are expected to be oxidizing because in both the reactions XeF$_4$ and XeF$_6$ are themselves reduced to xenon (Xe) and oxidises water to oxygen.

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, $HCO_3^-$, $CO_3^{2-}$

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.