p-Block Elements

The complete reaction is as follows,

$ \mathrm{P}_4+3 \mathrm{OH}^{-}+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{PH}_3+3 \mathrm{H}_2 \mathrm{PO}_4^{-} $

Here, $x$ is $\mathrm{H}_2 \mathrm{PO}_2^{-}$. The conjugate acid of $x$ is hypophosphorus acid.

$ \mathrm{H}_3 \mathrm{PO}_2 \rightleftharpoons \mathrm{H}^{+}+\mathrm{H}_2 \mathrm{PO}_2^{-} $

Statement given in Eqs. (i) and (ii) are correct, while (iii) is incorrect. It's correct form is

Boric acid is monobasic acid in water.

Disproportionation is a type of redox reaction in which an element undergoes both oxidation and reduction to form two different compounds.

The disproportionation reaction for orthophosphorus acid is:

$ 4 \mathrm{H}_3 \mathrm{PO}_3 \xrightarrow{\Delta} \mathrm{H}_3 \mathrm{PO}_4 + \mathrm{PH}_3 $

In this reaction:

Orthophosphorus acid ($\mathrm{H}_3 \mathrm{PO}_3$) is initially at a +3 oxidation state.

It gets oxidized to phosphoric acid ($\mathrm{H}_3 \mathrm{PO}_4$), where phosphorus is at a +5 oxidation state.

Simultaneously, it is reduced to phosphine ($\mathrm{PH}_3$), where phosphorus is at a -3 oxidation state.

Let's look at the reaction step by step.

Aluminium carbide, typically represented as $\mathrm{Al_4C_3},$ reacts with water to produce aluminium hydroxide and methane. The balanced reaction with H₂O is:

$\mathrm{Al_4C_3 + 12\,H_2O \rightarrow 4\,Al(OH)_3 + 3\,CH_4}.$

When deuterium oxide (D₂O) is used instead of water, every hydrogen atom is replaced by deuterium. Hence, aluminium hydroxide becomes aluminium deuteroxide, $\mathrm{Al(OD)_3},$ and methane becomes fully deuterated methane, $\mathrm{CD_4}.$

The modified reaction becomes:

$\mathrm{Al_4C_3 + 12\,D_2O \rightarrow 4\,Al(OD)_3 + 3\,CD_4}.$

So, the product $X'$ in the reaction is $\mathrm{CD_4}.$

Thus, the correct answer is Option D: $\mathrm{CD_4}.$

When chlorine is reacted with hot concentrated NaOH solution, the products are sodium chloride $(\mathrm{NaCl})$, sodium chlorate $\left(\mathrm{NaClO}_3\right)$ and water $\left(\mathrm{H}_2 \mathrm{O}\right)$.

$ \begin{aligned} 3 \mathrm{Cl}_2+6 \mathrm{NaOH} \longrightarrow & 5 \mathrm{NaCl}+\mathrm{NaClO}_3 + 3 \mathrm{H}_2 \mathrm{O} \end{aligned} $

During the water-gas shift reaction, carbon monoxide (CO) + H₂ reacts with steam (H₂O) to produce carbon dioxide (CO₂) and hydrogen gas (H₂). The chemical equation for the reaction is :

Therefore, option B is correct, which states that carbon monoxide is oxidized to carbon dioxide. This reaction is typically catalyzed by metal catalysts such as iron, nickel, or copper, which promote the reaction rate by providing a surface for the reactants to adsorb and react.

Option A is not correct as water is not evaporated during the reaction, but rather reacts with carbon monoxide to produce hydrogen gas and carbon dioxide.

Option C is not correct as carbon is not directly involved in the reaction.

Option D is also not correct as carbon dioxide is not reduced to carbon monoxide during the water-gas shift reaction.

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 - $\mathrm{SO}_{2}(\mathrm{s}), \mathrm{NH}_{3}(\mathrm{s})$.

• Hell-Volhard-Zelinsky reaction involves halogenation and is carried out using halogens like bromine along with red phosphorus and water. So, A matches with III.


• Iodoform reaction involves the reaction of iodine and sodium hydroxide. So, B matches with I.


• Etard reaction involves the oxidation of aromatic aldehydes and is carried out using chromyl chloride ($\mathrm{CrO_2Cl_2}$) and carbon disulfide ($\mathrm{CS_2}$), followed by water. So, C matches with II.


• Gatterman-Koch reaction involves the formation of aromatic aldehydes using carbon monoxide, hydrochloric acid, and anhydrous aluminum chloride. So, D matches with IV.

So, the correct option is :

Option B

A-III, B-I, C-II, D-IV

First, recall that electron-gain enthalpy $\Delta_{\mathrm{eg}}H$ is often quoted as a (negative) numerical value. A “more negative” electron-gain enthalpy means that an atom gains an electron more favorably.

1. Known Trends in the Periodic Table

Halogens (Group 17):

Although one might expect the electron-gain enthalpy to become steadily less negative as we go down the group (due to increasing size and shielding), there is an anomaly at the top:

$ \text{Most negative: } \mathrm{Cl} > \mathrm{F} > \mathrm{Br} > \mathrm{I} > \mathrm{At} \quad(\text{in terms of magnitude of negativity}). $

Numerically (since these values are negative), that ordering translates to:

$ \Delta_{\mathrm{eg}}H(\mathrm{Cl}) < \Delta_{\mathrm{eg}}H(\mathrm{F}) < \Delta_{\mathrm{eg}}H(\mathrm{Br}) < \Delta_{\mathrm{eg}}H(\mathrm{I}) < \Delta_{\mathrm{eg}}H(\mathrm{At}). $

The smaller (more negative) number on a number line corresponds to a larger (more exothermic) electron-gain enthalpy.

Chalcogens (Group 16):

A similar pattern holds, with an anomaly near oxygen. Going down from S to Se to Te to Po, the electron-gain enthalpy becomes less negative:

$ \text{Most negative: } \mathrm{S} > \mathrm{Se} > \mathrm{Te} > \mathrm{Po} \quad(\text{in magnitude}). $

Numerically (all negative values), that means:

$ \Delta_{\mathrm{eg}}H(\mathrm{S}) < \Delta_{\mathrm{eg}}H(\mathrm{Se}) < \Delta_{\mathrm{eg}}H(\mathrm{Te}) < \Delta_{\mathrm{eg}}H(\mathrm{Po}). $

2. Checking Each Option

We interpret each statement as a comparison of numerical values (remembering they are negative).

(A) $\,\Delta_{\mathrm{eg}}H(\mathrm{I}) < \Delta_{\mathrm{eg}}H(\mathrm{At})$

Iodine’s electron-gain enthalpy is more negative than astatine’s.

Numerically, $\Delta_{\mathrm{eg}}H(\mathrm{I})$ is indeed a “smaller” (i.e. more negative) number than $\Delta_{\mathrm{eg}}H(\mathrm{At})$.

So this statement is correct.

(B) $\,\Delta_{\mathrm{eg}}H(\mathrm{Te}) < \Delta_{\mathrm{eg}}H(\mathrm{Po})$

Tellurium (Te) is above polonium (Po) in Group 16, so Te typically has a more negative electron-gain enthalpy than Po.

Numerically, that means $\Delta_{\mathrm{eg}}H(\mathrm{Te})$ is smaller (more negative) than $\Delta_{\mathrm{eg}}H(\mathrm{Po})$.

This statement is correct.

(C) $\,\Delta_{\mathrm{eg}}H(\mathrm{Cl}) < \Delta_{\mathrm{eg}}H(\mathrm{F})$

Among halogens, $\mathrm{Cl}$ actually has the most negative $\Delta_{\mathrm{eg}}H$. Fluorine’s value, while also negative, is a bit less so (because of small-orbital repulsion in F).

Numerically, that means $\Delta_{\mathrm{eg}}H(\mathrm{Cl})$ is indeed a smaller (more negative) number than $\Delta_{\mathrm{eg}}H(\mathrm{F})$.

So this statement is correct.

(D) $\,\Delta_{\mathrm{eg}}H(\mathrm{Se}) < \Delta_{\mathrm{eg}}H(\mathrm{S})$

In Group 16, sulfur (S) has a more negative electron-gain enthalpy than selenium (Se).

Numerically, that implies

$ \Delta_{\mathrm{eg}}H(\mathrm{S}) < \Delta_{\mathrm{eg}}H(\mathrm{Se}), $

not the other way around.

The statement says $\Delta_{\mathrm{eg}}H(\mathrm{Se}) < \Delta_{\mathrm{eg}}H(\mathrm{S})$, which would mean Se is more negative than S—but that is not correct.

Hence, (D) is the incorrect statement.

Statement I is true because Chlorine (Cl) is a highly reactive halogen and has a strong tendency to form compounds with other elements, including oxygen. When chlorine reacts with oxygen, it can form various oxides such as chlorine monoxide (ClO) and chlorine dioxide (ClO₂). These compounds can be explosive in certain conditions.

Statement II is also true because the chemical reactivity of an element can be determined by its ability to react with oxygen and halogens. Elements that react readily with oxygen or halogens are considered to be highly reactive, while elements that do not react easily with these substances are considered to be less reactive. The reactivity of an element can give important information about its behavior in chemical reactions and its potential uses in various industrial and technological applications.

Boric acid is solid because of strong hydrogen bond in it.

Ammonium salts produce haze in atmosphere. Ozone is produced when atmospheric oxygen reacts with oxygen atoms and not chlorine atoms. Polychlorinated biphenyls have number of applications including their use as cleansing solvents.

‘Blue baby’ syndrome occurs due to the presence of excess of nitrate ions and not sulphate ions in water.

$\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ acts as a strong oxidising agent in acidic medium. During this process, oxidation state of $\mathrm{Cr}$ changes from $+6$ to $+3$.

$ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O} $

Oxyacid having P–H bond can reduce AgNO₃ to Ag.

To match the reactions given in List-I with the products in List-II, let's analyze each reaction individually based on its reactants and expected products.

• Reaction (P): $\mathrm{P}_2 \mathrm{O}_3 + 3 \mathrm{H}_2 \mathrm{O} \rightarrow$ involves the hydration of phosphorus trioxide. The product of this reaction is phosphorous acid:

$ \mathrm{P}_2 \mathrm{O}_3 + 3 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}_3 \mathrm{PO}_3 $

Hence, P matches with (2) $\mathrm{H}_3 \mathrm{PO}_3$.

• Reaction (Q): $\mathrm{P}_4 + 3 \mathrm{NaOH} + 3 \mathrm{H}_2 \mathrm{O} \rightarrow$ involves the reaction of phosphorus with a base (NaOH) and water, leading to the production of phosphine and sodium hypophosphite, not exactly reflected in the products listed, but given the context and possible products, we align with phosphorus chemistry principles. Typically, reactions involving phosphorus and strong bases can generate phosphine:

$ \mathrm{P}_4 + \mathrm{NaOH} + 3 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{PH}_3 +\mathrm{NaH}_2 \mathrm{PO}_2 $

Thus, Q matches best with (3) $\mathrm{PH}_3$ given the options, acknowledging the simplicity of the given answer choices.

• Reaction (R): $\mathrm{PCl}_5 + \mathrm{CH}_3 \mathrm{COOH} \rightarrow$ involves phosphorus pentachloride and acetic acid. The key reaction here is that $\mathrm{PCl}_5$ reacts with $\mathrm{CH}_3\mathrm{COOH}$ to replace $\mathrm{OH}$ groups in acetic acid with $\mathrm{Cl}$, creating acetyl chloride $\mathrm{CH}_3\mathrm{COCl}$ and $\mathrm{POCl}_3$:

$ \mathrm{PCl}_5 + \mathrm{CH}_3 \mathrm{COOH} \rightarrow \mathrm{CH}_3\mathrm{COCl} + \mathrm{POCl}_3 $

Therefore, R matches with (4) $\mathrm{POCl}_3$.

• Reaction (S): $\mathrm{H}_3 \mathrm{PO}_2 + 2 \mathrm{H}_2 \mathrm{O} + 4 \mathrm{AgNO}_3 \rightarrow$ involves hypophosphorous acid reacting in an oxidative environment given by silver nitrate, leading to the production of orthophosphoric acid among other products:

$ \mathrm{H}_3 \mathrm{PO}_2 + 2 \mathrm{H}_2 \mathrm{O} + 4 \mathrm{AgNO}_3 \rightarrow \mathrm{H}_3 \mathrm{PO}_4 +\mathrm{Ag}+\mathrm{HNO}_3 $

S, therefore, matches with (5) $\mathrm{H}_3 \mathrm{PO}_4$.

Given these reactions and the best fits for the products listed,

• P matches with 2,


• Q matches with 3,


• R matches with 4,


• S matches with 5.

Thus, the correct option is Option D: $\mathrm{P} \rightarrow 2 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 4 ; \mathrm{S} \rightarrow 5$.

Zeolite is silicate containing two metal ions $\mathrm{Na}^{+}$and $\mathrm{Al}^{3+}$. Therefore, $X$ and $Y$, respectively are $\mathrm{Na}^{+}$and $\mathrm{Al}^{3+}$.

The order of melting point is

$ \mathrm{Si}>\mathrm{Ge}>\mathrm{Pb}>\mathrm{Sn} $

The lead $(\mathrm{Pb})$ has high melting point than tin ( Sn ) because Sn forms a distorted 12 -coordinated structure. Also, Sn atoms are bonded covalently, which are weak in nature. So, Sn has a low melting point.

In hydrides of group 16 elements, boiling point increases down the group because of increasing molecular mass but water ( $\mathrm{H}_2 \mathrm{O}$ ) has exceptionally high boiling point due to intermolecular hydrogen bonding and $\mathrm{H}_2 \mathrm{~S}$ has lowest boiling point due to least molecular mass. Therefore, $X$ is $\mathrm{H}_2 \mathrm{~S}$ and $Y$ is $\mathrm{H}_2 \mathrm{O}$.

Sodium nitrite with hydrochloric acid given two oxides of nitrogen i.e. NO and $\mathrm{NO}_2$ alongwith water and sodium chloride.

The reaction involved is as follows.

Iodine monochloride ( ICl ) reacts with pure water to from HCl and iodic acid ( HOI ).

The reaction involved is a follows

$ \mathrm{ICl}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{HOI}+\mathrm{HCl} $

Therefore, statement (c) is incorrect.

Aluminium reacts with dilute hydrochloric acid at room temperature. The metal dissolve in HCl , yielding aluminium chloride and colourless hydrogen gas.

$ 2 \mathrm{Al}(s)+6 \mathrm{HCl}(a q) \longrightarrow 2 \mathrm{AlCl}_3(a q)+3 \mathrm{H}_2(g) $

When aluminium reacts with aqueous alkali it produces hydrogen gas.

$ \begin{aligned} \mathrm{Al}(s)+\mathrm{NaOH}(a q) & +\mathrm{H}_2 \mathrm{O}(a q) \longrightarrow \mathrm{NaAlO}_2(a q)+\mathrm{H}_2(g) \end{aligned} $

Statement given in option (c) is incorrect.

Due to inert pair effect, the lower elements in group 14 are stablised in $(+2)$ oxidation state than $(+4)$.

Hence, $\mathrm{Pb}^{2+}$ is more stable than $\mathrm{Pb}^{4+}$.

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

Hence, $\mathrm{Pb}^{4+}$ is good oxidising agent. Since, it is unstable.

Statement given in option (b) is incorrect, because 4 is hypophosphorus acid.

$\mathrm{H}_3 \mathrm{PO}_2$ is called hypophosphorus acid, conical has only one ionisable $(\mathrm{H})$ ions, hence its basicity is 1 and is a reducing agent.

$ \begin{aligned} \mathrm{H}_3 \mathrm{PO}_2 & \\ & 3 \times 1+x+2(-2)=0 \\ \Rightarrow & 3+x-4=0 \end{aligned} $

or $[x=1] \rightarrow$ oxidation state

Statement given in A, C and D are correct. While that in B is incorrect.

(A) $\mathrm{OF}_2$ is thermodynamically stable at 298 K because of the higher electronegativity of fluorine then oxygen.

(B) Order of stability of oxide of halogen $=\mathrm{I}>\mathrm{Cl}>\mathrm{Br}>\mathrm{F}$.

(C) Iodine pentoxide $\left[\mathrm{I}_2 \mathrm{O}_5\right]$ easily oxidises corbon monoxide to carbon dioxide at room temperature.

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

(D) $\mathrm{Cl}_2 \mathrm{O}$ is used as a bleaching agent for paper pulp and textiles.

Shape of $\mathrm{XeF}_6=\mathrm{bp}=6, \mathrm{lp}=1$

Shape = Distorted octahedral

Shape of $\mathrm{XeOF}_4=\mathrm{bp}=5, \mathrm{lp}=1$

Shape $=$ Square pyramidal

Among the given statements (A) and $(C)$ are correct while $(B)$ is an incorrect statement. $\mathrm{MgH}_2$ is not an interstitial hydride. Instead, it has significant covalent character and has polymeric structure.

Among the given statements C and D are incorrect.

(C) In case of aluminium, the sum of first three ionisation energies decreases considerably and hence, its value can never be positive. Its $E^{\circ}{ }_M{ }^{3+} / M(\mathrm{~V})$ value is -1.66 (i.e. in negative).

(D) In group 13, the +1 oxidation state becomes more and more stable as we move down the group i.e.

$\mathrm{Al}<\mathrm{Ga}<\mathrm{In}<\mathrm{Tl}$. This is because of the inert pair effect. Thus, in $\mathrm{Tl}+1$ oxidation state is more predominant. Whereas the +3 oxidation state is strongly oxidising in nature.