p-Block Elements

• Both nitrogen and oxygen do not contain d-orbitals so it cannot form $\mathrm{d} \pi-\mathrm{p} \pi$ bond.

• Phosphorus and Arsenic can form $\mathrm{d} \pi-\mathrm{d} \pi$ bond with transition metals. Since both have vacant d -orbitals by which it can interact with transition metals and can involve in $\mathrm{d} \pi-\mathrm{d} \pi$ interaction.

• Phosphorous, Arsenic and Antimony show catenation property.

(1) $C$ can form $p \pi-p \pi$ multiple bond with itself.

It is observed when it forms $\mathrm{C}=\mathrm{C}$ and $\mathrm{C} \equiv \mathrm{C}$

(2) $\mathrm{ECl}_3(\mathrm{E}=\mathrm{B}$ and Al$)$

$\mathrm{BCl}_3$ does not form dimer.

$\mathrm{AlCl}_3$ can form dimer in $\mathrm{Al}_2 \mathrm{Cl}_6$.

(3) Catenation property of group $14 \mathrm{C} \gg \mathrm{Si}>\mathrm{Ge} \approx \mathrm{Sn}$.

(4) Oxygen exhibits oxidation state of 0 in $\mathrm{O}_2,-2$ in oxides, -1 in peroxides. So this statement is incorrect.

Statement-I : (False)

$ 2 \mathrm{NaCl}+\stackrel{+4}{\mathrm{MnO}_2}+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Na}_2 \mathrm{SO}_4+\stackrel{+2}{\mathrm{MnSO}} \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2 $

So, reduction of Mn is taking place.

Statement-II : (True)

$ 2 \mathrm{NaI}+\stackrel{+4}{\mathrm{MnO}_2}+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Na}_2 \mathrm{SO}_4+\stackrel{+2}{\mathrm{MnSO}}+\mathrm{I}_2+2 \mathrm{H}_2 \mathrm{O} $

So, Mn has undergone reduction.

Potash alum

$\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3 \cdot 24 \mathrm{H}_2 \mathrm{O}$

No. of $\mathrm{K}^{+}=2$

No. of $\mathrm{SO}_4^{2-}=4$

Ratio $\frac{\mathrm{K}^{+}}{\mathrm{SO}_4^{2-}}=\frac{2}{4}=\frac{1}{2}$

The explanation revolves around the properties and chemical behavior of group 15 elements, specifically nitrogen, arsenic, and antimony.

Statement I: This statement is correct. All elements in group 15, including nitrogen and arsenic, can form hydrides with the formula EH₃. For instance, nitrogen forms ammonia (NH₃), and arsenic forms arsine (AsH₃).

Statement II: This statement is incorrect. Elements in group 15 are capable of forming two types of oxides: E₂O₃ and E₂O₅. Antimony, being a group 15 element, can indeed form antimony pentoxide (Sb₂O₅).

Therefore, Statement I is correct, while Statement II is incorrect.

Let's analyze each statement to identify the incorrect one:

Option A: The acidic strength of $\mathrm{HX}(\mathrm{X}=\mathrm{F}, \mathrm{Cl}, \mathrm{Br}$ and $\mathrm{I})$ follows the order: $\mathrm{HF}>\mathrm{HCl}>\mathrm{HBr}>\mathrm{HI}$.

This statement is incorrect. The correct order of acidic strength for hydrohalic acids is $\mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}$. This is because as we move down the group, the bond strength between hydrogen and the halide decreases, making it easier to ionize in water and making the acid stronger.

Option B: Fluorine exhibits $-1$ oxidation state whereas other halogens exhibit $+1, +3, +5$ and $+7$ oxidation states also.

This statement is correct. Fluorine is the most electronegative element and can only exhibit $-1$ oxidation state. Other halogens can exhibit positive oxidation states due to the availability of d-orbitals.

Option C: The enthalpy of dissociation of $\mathrm{F}_2$ is smaller than that of $\mathrm{Cl}_2$.

This statement is correct. The bond dissociation energy of $\mathrm{F}_2$ is indeed smaller than that of $\mathrm{Cl}_2$ because the F-F bond is weaker due to significant electron-electron repulsion between the non-bonding electrons in the small fluorine molecule.

Option D: Fluorine is a stronger oxidising agent than chlorine.

This statement is correct. Fluorine is the most electronegative element, which makes it the strongest oxidizing agent among the halogens.

Therefore, the incorrect statement is Option A.

Anion $\quad-$ Observation on treatment with dil. $\mathrm{H}_2 \mathrm{SO}_4$

(A) Carbonate $\left(\mathrm{CO}_3^{2-}\right)\quad$ Brisk effervescence of colourless and odourless gas $\left(\mathrm{CO}_2\right)$

(B) Sulphide $\left(\mathrm{S}^{2-}\right)\quad$ Evolution of colourless gas with rotten egg like smell $\left(\mathrm{H}_2 \mathrm{~S}\right)$

(C) Sulphite $\left(\mathrm{SO}_3^{2-}\right)\quad$ Gas with pungent smell $\left(\mathrm{SO}_2\right)$

(D) Nitrite $\left(\mathrm{NO}_2^{-}\right)\quad$ Brown fumes $\left(\mathrm{NO}_2\right)$

Correct Match : A – II, B-III, C-IV, D-I

The elements in Group 16 of the periodic table, also known as the chalcogens, include oxygen (O), sulfur (S), selenium (Se), tellurium (Te), and polonium (Po). Generally, these elements are known to commonly exhibit a -2 oxidation state because they have six electrons in their outermost shell and can gain two electrons to complete their octet, thereby achieving a stable electronic configuration similar to the nearest noble gas.

Oxygen, being the most electronegative element in this group, predominantly shows a -2 oxidation state in most of its compounds, such as water ($\mathrm{H_2O}$) and oxides like $\mathrm{SO_2}$.

Selenium and Tellurium also typically exhibit the -2 oxidation state. For instance, in compounds like $\mathrm{H_2Se}$ (hydrogen selenide) and $\mathrm{H_2Te}$ (hydrogen telluride), these elements are in the -2 state.

Polonium (Po), on the other hand, is different from the lighter chalcogens. Being a metalloid with more metallic character, it does not favor the -2 oxidation state nearly as strongly as the other elements in its group. Polonium most commonly exhibits +2 and +4 oxidation states in its compounds, like $\mathrm{PoO_2}$ and $\mathrm{PoCl_2}$. Due to the relativistic effects and its position in the periodic table, the -2 oxidation state is unstable and rare in polonium compounds.

Given this, the correct answer is Option D: Po. Polonium does not commonly exhibit the -2 oxidation state like the other Group 16 elements.

The correct order of group number of ions is $ \underset{(\mathrm{B})}{\mathrm{Cu}^{2+}}<\underset{(\mathrm{A})}{\mathrm{Al}^{3+}}<\underset{(\mathrm{D})}{\mathrm{Co}^{2+}}<\underset{(\mathrm{C})}{\mathrm{Ba}^{2+}}<\underset{(\mathrm{E})}{\mathrm{Mg}^{2+}} $

$\therefore$ The correct order is B, A, D, C, E

$\mathrm{LaH_{2.87}}$ $\to$ non-stoichiometric

$\to$ Metallic/Interstitial hydride.

Assertion is true because He has low solubility in blood. (NCERT)

Coke : It is used as reducing agent in carbon reduction methods. (in metallurgical process)

Diamond : It is a allotrope of carbon in which each carbon is $\mathrm{sp}^{3}$ hybridised.

Fullerene : It contains pentagonal & hexagonal rings (cage like structure)

Graphite : It is soft solid because graphite layers are bonded with weak Vander Wall attractions.

$\mathrm{F}^{-1}, \mathrm{~N}^{-3}, \mathrm{Na}^{+} ~\& ~\mathrm{O}^{-2}$

all ions are isoelectronic containing $10 e^{-}$

$\mathrm{Z}_{\text {eff }} \rightarrow \mathrm{Na}^{+}>\mathrm{F}^{-}>\mathrm{O}^{-2}>\mathrm{N}^{-3}$

order of radius $\rightarrow \mathrm{N}^{-3}>\mathrm{O}^{-2}>\mathrm{F}^{-}>\mathrm{Na}^{+}$

$\rightarrow$ Nitrogen to achieve Noble gas configuration it gain $3 e^{-}$, & form $\mathrm{N}^{-3}$

$\bullet$ Borax : Na₂B₄O₇ . 10H₂O = Na₂[B₄O₅(OH)₄] . 8H₂O

$\bullet$ Kernite : Na₂B₄O₇ . 4H₂O

$\bullet$ Orthoboric acid : H₃BO₃ = B(OH)₃

$\bullet$ Borax bead : NaBO₂ (Sodium metaborate)

4NH₃ + 5O₂ (from air) $\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{\mathrm{500\,K,\,9\,bar}}^{\mathrm{Pt}}} $ 4NO + 6H₂O

2NO(g) + O₂(g) $\rightleftharpoons$ 2NO₂(g)

3NO₂(g) + H₂O(l) $\to$ 2HNO₃(aq) + NO(g)

This is industrial method of preparation of nitric acid.

Na₂B₄O₇ $\buildrel {\Delta} \over \longrightarrow $ B₂O₃ + 2NaBO₂

Product X is B₂O₃

(a) - (iv), (b) - (i), (c) - (ii), (d) - (iii)

Each boron atoms in diborane uses sp³ hybrid orbitals for bonding.

$\bullet$ The boiling points of these hybrids not exactly increases with increase in molar mass.

$\bullet$ H₂O has maximum boiling point due to intermolecular hydrogen bonding.

Oleum, i.e. $\mathrm{H}_2 \mathrm{S}_2 \mathrm{O}_7$ can be prepared by adsorption of $\mathrm{SO}_3$ in conc. $\mathrm{H}_2 \mathrm{SO}_4$ as follows :

$\mathrm{SO}_3+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \underset{\text { Oleum }}{\mathrm{H}_2 \mathrm{S}_2 \mathrm{O}_7}$

It's structure is as follows :

Thus, it has no $\mathrm{O}-\mathrm{O}$ bond and has $2-\mathrm{OH}$ groups. Thus, option (a) is the correct.

Neil Bartlett prepared a red compound, $\mathrm{O}_2^{+}[\mathrm{PtF}_6]^{-}$ by reacting platinum hexafluoride $(\mathrm{PtF}_6)$ with oxygen.

He found that ionisation energy of $\mathrm{Xe}(1170 \mathrm{~kJ} \mathrm{~mol}^{-1})$ is nearly equal to that of molecular oxygen $(1175 \mathrm{~kJ} \mathrm{~mol}^{-1})$ available. Therefore, he was able to form compound of xenon.

$\mathrm{Xe}+\mathrm{PtF}_6 \longrightarrow \mathrm{Xe}^{+}\left[\mathrm{PtF}_6\right]$, On the other hand $\mathrm{He}$ and $\mathrm{Ne}$ has very high value of ionisation energy.

Therefore, only Xe can participate in chemical reactions and form compounds.

Hence, option (a) is the correct.

As the size of the halogen atom increases from $\mathrm{F}$ to $\mathrm{I}, \mathrm{H}-X$ bond length in halogen acids also increases from $\mathrm{H}-\mathrm{F}$ to $\mathrm{HI}$.

$(\mathrm{HF}<\mathrm{HCl}<\mathrm{HBr}<\mathrm{HI})$. The increase in $\mathrm{H} X$ bond length decreases the strength of $\mathrm{H} X$ bond from $\mathrm{HF}$ to $\mathrm{HI}(\mathrm{HF}>\mathrm{HCl}>\mathrm{HBr}>\mathrm{HI})$.

Due to successive decrease in the strength of $\mathrm{H} X$ bonds, bond dissociation enthalpy decreases from $\mathrm{HF}$ to $\mathrm{HI}$.

Hypophosphoric acid $\left(\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_6\right)$ does not contain $\mathrm{P}-\mathrm{O}-\mathrm{P}$ bond whereas, isohypophosphoric acid $\left(\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_6\right)$, diphosphorus acid $\left(\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_5\right)$ and diphosphoric acid $\left(\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7\right)$ all contain $\mathrm{P}-\mathrm{O}-\mathrm{P}$ bonds. The structures of the following oxyacids of phosphorus is shown below:

$\mathop {8N{H_3}}\limits_{(Excess)} + 3C{l_2}\buildrel {} \over \longrightarrow 6NH_4Cl+N_2$

$N{H_3} + \mathop {3C{l_2}}\limits_{(Excess)} \buildrel {} \over \longrightarrow NCl_3+3HCl$

Hence,

$\begin{aligned} & X=6 \mathrm{NH}_4 \mathrm{Cl}+\mathrm{N}_2 \\ & Y=\mathrm{NCl}_3+3 \mathrm{HCl} \end{aligned}$

The products of the reaction of copper with $\mathrm{HNO}_3$ depends upon the concentration of $\mathrm{HNO}_3$ used.

Copper metal reacts with dilute $\mathrm{HNO}_3$ to form nitrogen (II) oxide (NO).

$3 \mathrm{Cu}+8 \mathrm{HNO}_3 \text { (dil.) } \longrightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}+4 \mathrm{H}_2 \mathrm{O}$

Copper metal reacts with conc. $\mathrm{HNO}_3$ to form nitrogen (IV) oxide or nitrogen dioxide $(\mathrm{NO}_2)$

$\mathrm{Cu}+4 \mathrm{HNO}_3 \text { (conc.) } \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O}$

A halate will be formed from halogen and the greenish yellow gas is $\mathrm{Cl}_2$.

The halate which is used in fireworks and safety matches is $\mathrm{KClO}_3$. The reaction is as follow

$3 \mathrm{Cl}_2+6 \mathrm{KOH} \longrightarrow \mathrm{KClO}_3+5 \mathrm{HCl}+3 \mathrm{H}_2 \mathrm{O}$

Assertion is true but reason is false. $\mathrm{N}_2$ is less reactive than $\mathrm{P}_4$ due to high value of bond dissociation energy, which is because of the presence of triple bond between two $\mathrm{N}$-atoms of $\mathrm{N}_2$ molecule.

Both assertion and reason is correct and reason is the correct explanation of the assertion. $\mathrm{NaCl}$ reacts with concentrated $\mathrm{H}_2 \mathrm{SO}_4$ to give colourless fumes with pungent smell. This pungent smell is due to formation of $\mathrm{HCl}$.

$\mathrm{NaCl}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{Na}_2 \mathrm{SO}_4+2 \mathrm{HCl}$

But on adding $\mathrm{MnO}_2$, the fumes become greenish yellow due to formation of chlorine gas.