Oxygen family Questions in English

(C) The ozone $(O_3)$ molecule is angular in shape and is a resonance hybrid of two contributing structures. Due to resonance, the bond order in ozone is $1.5$, which is intermediate between a single bond $(1)$ and a double bond $(2)$. In molecular oxygen $(O_2)$, the bond order is $2$. Since bond length is inversely proportional to bond order, the $O-O$ bond length in $O_3$ is greater than the $O=O$ bond length in $O_2$. Therefore, statement $C$ is incorrect.

(C) $H_2S_2O_8$ (Peroxodisulphuric acid or Marshall's acid) contains a peroxide linkage ($-O-O-$ bond). The structure is as follows:
$HO-S(=O)_2-O-O-S(=O)_2-OH$

(D) $H_{2}SO_{4}$ saturated with $SO_{3}$ is called $Oleum$ or $Sulphan$.
The chemical reaction is: $H_{2}SO_{4} + SO_{3} \longrightarrow H_{2}S_{2}O_{7}$.

(C) At room temperature and pressure, $P_4O_{10}$ exists as a white solid.
$SO_2$ is a colourless gas.
$NO_2$ is a brown gas.
However, $SO_3$ exists as a colourless, crystalline transparent solid at room temperature, not a gas.
Therefore, the statement that $SO_3$ is a colourless gas is not true.

(B) The reaction of sodium sulfide with sulfur is given by: $Na_{2}S + nS \rightarrow Na_{2}S_{(n+1)}$, where $(X) = Na_{2}S_{(n+1)}$.
The reaction of sodium sulfite with sulfur is given by: $Na_{2}SO_{3} + S \rightarrow Na_{2}S_{2}O_{3}$, where $(Y) = Na_{2}S_{2}O_{3}$.

(D) Statement $I$ is false. While the oxidation state of oxygen in $OF_2$ is $+2$, in $SO_2$, oxygen is in a $-2$ oxidation state. However, the covalency of oxygen is restricted to two due to the absence of $d$-orbitals in its valence shell, meaning it cannot exceed a covalency of two.
Statement $II$ is true; the small size, high electronegativity, and absence of $d$-orbitals account for the anomalous properties of oxygen compared to other group $16$ elements.
Thus, Statement $I$ is false and Statement $II$ is true.

(D) $1$. The reducing property of hydrides of Group $16$ elements depends on the bond dissociation enthalpy of the $E-H$ bond.
$2$. As we move down the group from $O$ to $Te$, the atomic size increases, which leads to an increase in the bond length and a decrease in the bond dissociation enthalpy.
$3$. $A$ weaker $E-H$ bond is more easily broken, making the hydride a stronger reducing agent.
$4$. Since the $H-Te$ bond is the weakest among the given hydrides, $H_2Te$ has the highest reducing property.

(D) $1$. Acidic strength of hydrides of group $16$ elements depends on the bond dissociation enthalpy of the $H-E$ bond.
$2$. As we move down the group from $O$ to $Te$, the atomic size increases, which leads to an increase in the bond length and a decrease in the $H-E$ bond dissociation enthalpy.
$3$. $A$ weaker $H-E$ bond is easier to break, making it easier to release $H^+$ ions.
$4$. Therefore, the acidic strength increases in the order: $H_2O < H_2S < H_2Se < H_2Te$.
$5$. Thus, $H_2Te$ is the most acidic compound.

(A) $1$. The acidity of hydrides of group $16$ elements increases down the group.
$2$. As we move down the group from $S$ to $Te$, the atomic size increases.
$3$. The increase in atomic size leads to a decrease in the bond dissociation enthalpy of the $H-E$ bond (where $E = S, Se, Te$).
$4$. $A$ weaker $H-E$ bond makes it easier for the molecule to release $H^+$ ions in an aqueous solution.
$5$. Therefore, the acidity increases in the order: $H_2S < H_2Se < H_2Te$.

(C) $1$. Ozone $(O_3)$ acts as a strong oxidizing agent because it readily decomposes to give nascent oxygen $(O_3 \rightarrow O_2 + [O])$.
$2$. In option $(C)$, $H_2O_2$ is oxidized to $O_2$ by $O_3$, while $O_3$ itself is reduced to $O_2$. This reaction demonstrates the reducing nature of ozone.
$3$. In options $(A)$, $(B)$, and $(D)$, ozone acts as an oxidizing agent by oxidizing $NO$ to $NO_2$, $I^-$ to $I_2$, and $PbS$ to $PbSO_4$ respectively.