p-Block Elements

The melting point of carbon is exceedingly high. The value decreases on moving down the group because the $M-M$ bonds become weaker and thus they are easy to break. So, the correct order of melting point is

Among the given statements (c) is correct while others are incorrect.

• $\mathrm{N}_2$ is colourless, odourless, tasteless and non-toxic gas.

• Ozone is thermodynamically less stable than oxygen because its decomposition into oxygen results in the liberation of heat ( $\Delta H$ is negative) and increase in entropy ( $\Delta S$ is positive). These two effects support each other that results in large negative $\Delta G$ for its conversion into oxygen.

• $\mathrm{Cl}_2$ is greenish yellow gas with pungent and suffocating odour.

Among the given oxo-acids of sulphur, $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7$ contains $\mathrm{S}-\mathrm{O}-\mathrm{S}$ bond.

All the given pairs are correctly matched.

• $\mathrm{Cl}_2$ is used in the preparation of phosgene.

$ \mathrm{CO}(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons \mathrm{COCl}_2(\mathrm{~g}) $

• Iodine pentenoxide ( $\mathrm{I}_2 \mathrm{O}_5$ ) easily oxidises carbon monoxide $(\mathrm{CO})$ to carbon dioxide at room temperature.

$ 5 \mathrm{CO}+\mathrm{I}_2 \mathrm{O}_5 \longrightarrow \mathrm{I}_2+5 \mathrm{CO}_2 $

This reaction can be used to estimation the concentration of CO in a gaseous sample.

• $\mathrm{O}_3$ is used as a germicide, a disinfectant and for sterilising water.

Kernite is the mineral of boron It's formula is $\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_6(\mathrm{OH})_2 \cdot 3 \mathrm{H}_2 \mathrm{O}$ While cryolite is the mineral of aluminium. It's formula is $\mathrm{Na}_3 \mathrm{AlF}_6$.

Inert pair effect is less prominent in Sn than in Pb . Therefore, +2 oxidation state of Sn is less stable than its +4 oxidation state. Hence, $\mathrm{Sn}^{2+}$ can easily lose two electrons to form $\mathrm{Sn}^{4+}$ and thus $\mathrm{Sn}^{2+}$ acts as a reducing agent.

$ \mathrm{Sn}^{2+} \longrightarrow \mathrm{Sn}^{4+}+2 e^{-} $

In $\mathrm{Pb},+2$ oxidation state of Pb is more stable than +4 oxidation state due to inert pair effect. Therefore, $\mathrm{Pb}^{4+}$ can easily gain two electrons to form $\mathrm{Pb}^{2+}$ and hence $\mathrm{Pb}^{4+}$ acts as an oxidising agent.

$ \mathrm{Pb}^{4+}+2 e^{-} \longrightarrow \mathrm{Pb}^{2+} $

Out of $A, B, C$ and $D, C$ is the incorrect order of thermal stability. Stability of hydrides of 15th group decreases from $\mathrm{NH}_3$ to $\mathrm{SbH}_3$. This is because their bond dissociation enthalpy decreases due to increase in size of central atom from N to Sb .

∴ Correct order of thermal stability is

$ \mathrm{NH}_3>\mathrm{PH}_3>\mathrm{AsH}_3>\mathrm{SbH}_3 . $

$\mathrm{SnO}_2+4 \mathrm{HCl} \longrightarrow \mathrm{SnCl}_2+2 \mathrm{H}_2 \mathrm{O}$

$ \mathrm{SnO}_2+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_2 \mathrm{SnO}_3+\mathrm{H}_2 \mathrm{O} $

$\mathrm{SnO}_2$ is amphoteric in nature.

$ \begin{aligned} & \mathrm{SnO}+2 \mathrm{HCl} \longrightarrow \mathrm{SnCl}_2+\mathrm{H}_2 \mathrm{O} \\ & \mathrm{SnO}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_2 \mathrm{SnO}_2+\mathrm{H}_2 \mathrm{O} \end{aligned} $

SnO is also amphoteric in nature.

$\therefore$ The dioxide and monoxide of Sn , both are amphoteric in nature.

$ \begin{aligned} & \mathrm{PbO}_2+4 \mathrm{HCl} \longrightarrow \mathrm{PbCl}_4+2 \mathrm{H}_2 \mathrm{O} \\ & \mathrm{PbO}_2+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_2 \mathrm{PbO}_3+\mathrm{H}_2 \mathrm{O} \end{aligned} $

$\mathrm{PbO}_2$ is amphoteric in nature

$ \begin{aligned} & \mathrm{PbO}+2 \mathrm{HCl} \longrightarrow \mathrm{PbCl}_2+\mathrm{H}_2 \mathrm{O} \\ & \mathrm{H}_2 \mathrm{O}+\mathrm{PbO}+\mathrm{Ca}(\mathrm{OH})_2 \longrightarrow \mathrm{Ca}\left[\mathrm{~Pb}(\mathrm{OH})_3\right]_2 \end{aligned} $

PbO is also amphoteric in nature.

$\therefore$ The dioxide and monoxide of Pb both are amphoteric in nature.

In Group 15 elements, nitrogen does not form pentahalides. This is because nitrogen has the electronic configuration of $1s^2, 2s^2, 2p^3$, indicating it only possesses $s$ and $p$ orbitals. Nitrogen lacks $d$ orbitals, which are required to expand its covalency beyond four. Consequently, it cannot form pentahalides.

Nitrogen can attain a +5 oxidation state, achieving a stable electronic configuration of $1s^2$.

Thus, although both the assertion that nitrogen does not form pentahalides and the reasoning that nitrogen can exhibit a +5 oxidation state are true, the reason does not logically explain the assertion.

In Deacon's method, hydrogen chloride ( HCl ) is passed into a blast furnace and burnt with air to produce chlorine and heat.

$ 4 \mathrm{HCl}+\mathrm{O}_2 \xrightarrow[723 \mathrm{~K}]{\mathrm{CuCl}_2} 2 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2 $

The nitrate of potassium metal of heating liberates $\mathrm{KNO}_2$ and $\mathrm{O}_2$, the reaction involved is as follows:

$ 2 \mathrm{KNO}_3(s) \xrightarrow{\Delta} 2 \mathrm{KNO}_2(s)+\mathrm{O}_2(g) $

So, nitrate of potassium does not librate $\mathrm{NO}_2$ gas on heating.

Let's analyze each statement one by one:

$\textbf{(I)}\quad \text{"The stability of +1 oxidation state follows the order } \mathrm{Tl} > \mathrm{In} > \mathrm{Ga} \text{."}$

In group 13, the inert pair effect increases as we go down the group. This makes the +1 oxidation state particularly stable in thallium (Tl), moderately stable in indium (In), and least stable in gallium (Ga).

Thus, statement (I) is correct.

$\textbf{(II)}\quad \text{"Boron has the lowest melting point and boiling point as it is a non‐metal."}$

Even though boron is a non‐metal (or more correctly a metalloid), it is unique in having very high melting and boiling points. For example, boron has a melting point over 2000°C, contrary to the idea that non‐metals have low melting and boiling points.

Therefore, statement (II) is incorrect.

$\textbf{(III)}\quad \text{"Boron shows a maximum covalency of 4 in its compounds."}$

Being a second-period element, boron does not have d-orbitals to expand its octet. Hence, it can accommodate at most four electron pairs around it. This is seen in species like the tetrahydroborate ion, $\mathrm{BH_4^-},$ where boron forms four bonds.

So, statement (III) is correct.

$\textbf{(IV)}\quad \text{"The order of atomic radius is } \mathrm{Ga} > \mathrm{Al} > \mathrm{In} \text{."}$

Generally, atomic radii increase down a group. For group 13, the expected order is:

$\mathrm{Al} < \mathrm{Ga} < \mathrm{In}.$

The given order is not consistent with the trend (in fact, indium should be larger than gallium), so statement (IV) is incorrect.

$\textbf{(V)}\quad \text{"Aluminium is passive to concentrated nitric acid."}$

Aluminium forms a thin, adherent oxide layer that protects it from reacting with concentrated nitric acid.

Thus, statement (V) is correct.

Summing up, the correct statements are (I), (III), and (V).

The answer is Option A: I, III and V only.

First, let us identify what the species $Q$ is in the given reaction:

$ \mathrm{P}_4 + 3\,\mathrm{NaOH} + 3\,\mathrm{H}_2\mathrm{O} \;\longrightarrow\; Q \;+\; 3\,\mathrm{NaH}_2\mathrm{PO}_2. $

This is a classic reaction in which white phosphorus $\mathrm{P}_4$ reacts with a base (e.g., $\mathrm{NaOH}$) and water to form phosphine ($\mathrm{PH}_3$) and sodium hypophosphite ($\mathrm{NaH}_2\mathrm{PO}_2$):

$ \mathrm{P}_4 + 3\,\mathrm{NaOH} + 3\,\mathrm{H}_2\mathrm{O} \;\longrightarrow\; \mathrm{PH}_3 \;+\; 3\,\mathrm{NaH}_2\mathrm{PO}_2. $

Hence, $Q = \mathrm{PH}_3$ (phosphine).

Checking Each Option

We want to find the reaction that does NOT produce phosphine $(\mathrm{PH}_3)$.

Option A: $\mathrm{Ca}_3 \mathrm{P}_2 + \mathrm{H}_2 \mathrm{O}$

Calcium phosphide and water react to form phosphine ($\mathrm{PH}_3$) and calcium hydroxide:

$ \mathrm{Ca}_3\mathrm{P}_2 + 6\,\mathrm{H}_2\mathrm{O} \;\longrightarrow\; 3\,\mathrm{Ca}(\mathrm{OH})_2 + 2\,\mathrm{PH}_3. $

Phosphine is produced. So $Q$ is formed here.

Option B: $\mathrm{H}_3 \mathrm{PO}_3 \xrightarrow{\Delta}$

On heating phosphorous acid, it disproportionates to form phosphoric acid ($\mathrm{H}_3\mathrm{PO}_4$) and phosphine ($\mathrm{PH}_3$):

$ 4\,\mathrm{H}_3\mathrm{PO}_3 \;\xrightarrow{\Delta}\; 3\,\mathrm{H}_3\mathrm{PO}_4 + \mathrm{PH}_3. $

Phosphine is produced. So $Q$ is formed here too.

Option C: $\mathrm{PH}_4 \mathrm{I} + \mathrm{KOH}$

Phosphonium iodide reacts with a strong base (KOH) to release phosphine ($\mathrm{PH}_3$) plus $\mathrm{KI}$ and $\mathrm{H}_2\mathrm{O}$:

$ \mathrm{PH}_4\mathrm{I} + \mathrm{KOH} \;\longrightarrow\; \mathrm{PH}_3 + \mathrm{KI} + \mathrm{H}_2\mathrm{O}. $

Phosphine is produced. So $Q$ is formed in this reaction as well.

Option D: $\mathrm{PCl}_3 + \mathrm{H}_2\mathrm{O}$

Phosphorus trichloride hydrolyzes (usually vigorously) to form phosphorous acid ($\mathrm{H}_3\mathrm{PO}_3$) and hydrochloric acid ($\mathrm{HCl}$):

$ \mathrm{PCl}_3 + 3\,\mathrm{H}_2\mathrm{O} \;\longrightarrow\; \mathrm{H}_3\mathrm{PO}_3 + 3\,\mathrm{HCl}. $

Phosphine ($\mathrm{PH}_3$) is not produced in this reaction.

Conclusion

Since we are looking for the reaction in which phosphine $(Q)$ is not a product, the correct choice is:

$ \boxed{\text{Option D: } \mathrm{PCl}_3 + \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_3\mathrm{PO}_3 + 3\,\mathrm{HCl} \ (\text{no } \mathrm{PH}_3).} $

The formula of peroxydisulphuric acid is $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8$. It's structure is given by

From the structure, we can clearly see that, the oxidation state for both the sulphur atom is +6 and it has two $\mathrm{S}-\mathrm{OH}$ bonds.

Sulphur does not exist as $\mathrm{S}_2$ molecule primarily because of its inability to constitute $p \pi-p \pi$ bonds.

In the case of oxygen, the small size of $O$ atoms and the high electronegativity allow the $2p$ orbitals to overlap effectively, forming a strong $p \pi-p \pi$ double bond in $\mathrm{O}_2$. As a result, molecular oxygen exists as a diatomic molecule. However, sulphur atoms are larger and less electronegative compared to oxygen. This significant size difference makes the overlap between $3p$ orbitals (the corresponding orbitals in sulphur) less effective for forming strong $p \pi-p \pi$ double bonds.

Instead, sulphur stabilizes itself more effectively by catenating to form larger molecular structures like $\mathrm{S}_8$, where each sulphur atom forms single bonds with adjacent sulphur atoms, creating a puckered ring structure. This results in the elemental sulphur commonly found in the form of octa-atomic rings.

So, the correct answer is:

Option A: it is not able to constitute $p \pi-p \pi$ bonds

The term "amphoteric" refers to a substance that can react both as an acid and as a base. In the context of oxides, an amphoteric oxide can react with both acids and bases to form salts and water. Let's analyze the options presented:

Option A: $\mathrm{SnO}_2$ (Tin(IV) oxide)
This oxide is known to be amphoteric. It can react with acids to form stannic salts and with strong bases to form stannates.

Option B: $\mathrm{SiO}_2$ (Silicon dioxide)
This oxide is acidic in nature. It primarily reacts with bases (such as sodium hydroxide) but not with acids.

Option C: $\mathrm{CO}_2$ (Carbon dioxide)

Option D: $\mathrm{CaO}$ (Calcium oxide)
This oxide is basic. It reacts with acids to form calcium salts and water but not with bases.

Among the given options, the correct answer is Option A: $\mathrm{SnO}_2$, which is amphoteric in character.

High concentration of NO₂ can damage plant leaves and retard photosynthesis.

$Na_{2}B_{4}O_{7}+7H_{2}O\longrightarrow 2H_{3}BO_{3}+2Na\left[ B\left( OH\right)_{4} \right] $

Aqueous solution of borax is buffer whose pH $ \simeq $ 9

Other compounds are less basic than this.

$2\text{SO}_{2} +\text{O}_{2} +\text{2H}_{2} \text{O} \longrightarrow \text{2H}_{2} \text{SO}_{4} $

Acid rain occurs due to increased concentration of oxides of sulphur and Nitrogen.

Caesium is used in devising photoelectric cells.

Boron fibres are used in making bullet–proof vest.

Silicones being surrounded by non–polar alkyl groups are water repelling in nature.

Gallium is less toxic and has a very high boiling point, so it is used in high temperature thermometers.