p-Block Elements

$\mathrm{Cr{O_2} + 4NaOH \to \mathop {N{a_2}Cr{O_4}}\limits_{(A)} + 2NaCl + 2{H_2}O}$

$\mathrm{N{a_2}Cr{O_4} + 2{H_2}{O_2} + 2HCl \to \mathop {Cr{O_5}}\limits_{(B)} + \mathop {2NaCl}\limits_{(Mis\sin g\,from\,balanced\,equation)} + 3{H_2}O}$

On moving down the group, bond strength of $\mathrm{M}-\mathrm{H}$ bond decreases, which reduces the thermal stability but increases reducing nature of hydrides, hence A and B are correct statements.

If nitrogen or sulphur is also present in the compound, the sodium fusion extract is first boiled with concentrated nitric acid to decompose cyanide or sulphide of sodium during Lassaigne's test

Nessler's Reagent Reaction :

$\underset{\text { (Nessler's Reagent) }}{2 \mathrm{~K}_2 \mathrm{HgI}_4}+\mathrm{NH}_3+3 \mathrm{KOH} \rightarrow \underset{\substack{\text { (lodine of Millon's base } \\ \text { (Brown precipitate }}}{\mathrm{HgO} . \mathrm{Hg}\left(\mathrm{NH}_2\right) \mathrm{I}}+7 \mathrm{KI}+2 \mathrm{H}_2 \mathrm{O}$

(1) Fluorspar is $\mathrm{CaF}_2$

Statement-I: Oxygen can have oxidation state from -2 to +2 , so statement I is incorrect

Statement- II: On moving down the group stability of +4 oxidation state increases whereas stability of +6 oxidation state decreases down the group, according to inert pair effect.

So both statements are wrong.

Fluorine is the element among the options that does not exhibit a variable oxidation state, unlike its fellow halogens. This is because fluorine is the most electronegative element, and it always forms compounds in the oxidation state of -1 as it has a strong tendency to gain one electron to achieve a stable electronic configuration.

Solid Boron has very strong crystalline lattice so its melting point unusually high in group 13 elements

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.

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