p-Block Elements

It has two 3-centre-2-electron bonds and four 2-centre-2-electron bonds.

Hence, all B–H bonds are not equivalent.

It is an electron deficient compound as the octet of boron is incomplete.

Hence, it can behave as a Lewis acid.

It can be synthesized from both BF₃ and NaBH₄

$2B{F_3} + 6NaH\buildrel {450K} \over \longrightarrow {B_2}{H_6} + 6NaF$

$2NaB{H_4} + {I_2} \to {B_2}{H_6} + 2NaI + {H_2}$

It is a non-planar molecule.

Hence, only Statements (C) and (D) are correct.

The stability of trihalides decreases down the group due to weakening of N – X bond and inability of N to accommodate large sized halogen atoms (Cl, Br, I) around it.

The outer electronic configuration is $n s^2 n p^1$, i.e. the element belongs to group 13 (boron family). As the element in crystalline form is unreactive towards oxygen, it might be boron which does not react with air in crystalline form. In amorphous form it reacts to form $\mathrm{B}_2 \mathrm{O}_3$. Boron trioxide which is acidic in nature. Down the group the nature of oxide changes from acidic to amphoteric ( $\mathrm{Al}_2 \mathrm{O}_3$ and gallium oxide) to basic (indium and thallium oxide). Thus, given element is boron.

The element that does not show catenation is Pb (lead). Catenation is the property/tendency of element to link with one another through covalent bond forming long chains and rings. Down the group, this tendency decreases in order $\mathrm{C} \gg \mathrm{Si}>\mathrm{Ge} \approx \mathrm{Sn}$.

$ \text { ∴ Only } \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8 \text { has peroxy linkage. } $

The balanced reaction is

$ 3 \mathrm{Cl}_2+8 \mathrm{NH}_3 \longrightarrow 6 \mathrm{NH}_4 \mathrm{Cl}+\mathrm{N}_2 $

If 3 moles of chlorine is reacting with 8 moles of ammonia.

$\therefore 1$ mole of chlorine will react with $8 / 3$ moles of ammonia.

Enthalpy of vaporisation $\propto$ van der Waals' forces and van der Waals' forces $\propto$ Molar mass

↑ is molar mass, $\uparrow$ will be enthalpy of vaporisation.

∴ The correct order of enthalpy of vaporisation of noble gases is

(A) is false because $\left[\mathrm{B}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ does not exist, while $\left[\mathrm{B}(\mathrm{OH})_4\right]$ is $s p^3$ hybridised and have tetrahedral complex.

The boron atom is electron deficient. It has 6 electrons instead of 8 electrons in its valence shell. Hence, it acts as a Lewis acid by readily accepting the lone pair from oxygen from a $\mathrm{H}_2 \mathrm{O}$ molecule or $\mathrm{OH}^{-}$ion. Hence, ( R ) is true.

$\mathrm{N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}$

Temperature range for this reaction is $1600-2200 \mathrm{~K}$.

Reaction is endothermic when temperature is high more and more heat will be absorbed and reaction will shift in forward reaction.

So, the correct option is (b), i.e. 2000 K .

(A) is true because both rhombic and monoclinic sulphur have $\mathrm{S}_8$ molecules.

Rhombic $S$ is octahedral in form and monoclinic S is crystalline in nature. As rhombic is melted in a dish and then cooled, monoclinic S is obtained. Therefore, $(R)$ is false.

Aluminium produces sodium tetrahydroxoaluminate $(P)$ and hydrogen $(Q)$ as it reacts with excess of sodium hydroxide solution.

Carbon $(C)$ shows catenation to maximum extent because of its small size and tendency to form strong covalent bonds. The least abundant element on earth is germanium ( Ge ). $\therefore X$ is carbon $(\mathrm{C})$ and $Y$ is germanium (Ge).

Group 16 elements are also called chalcogen. They are so called because most of the copper ores have copper in the form of oxides and sulphides.

$ \text { For reaction, } $

$ \text { Hence, shape of } \mathrm{BrF}_5 \text { is square pyramidal. } $

Hydrolysis of $\mathrm{XeF}_4$ will give $\mathrm{XeO}_3$ and Xe along with HF and $\mathrm{O}_2$. It is a type of disproportionation reaction in which $\mathrm{XeF}_4$ undergoes oxidation as well as reduction.

$ \begin{aligned} & 6 \mathrm{XeF}_4+12 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{Xe}+2 \mathrm{XeO}_3+24 \mathrm{HF}+3 \mathrm{O}_2 \\ & \therefore \quad a=6 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\, b=12 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\, p=4 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\, q=2 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,r=24 \\ & \text { and } \quad s=3 \end{aligned} $

Phosphorus acid gives two $\mathrm{H}^{+}$ ions thus, it is dibasic.

$ \mathrm{H}_3 \mathrm{PO}_3 \longrightarrow 2 \mathrm{H}^{+}+\mathrm{HPO}_3^{2-} $

Phosphoric acid gives three $\mathrm{H}^{+}$ions. Thus, it is tribasic.

$ \mathrm{H}_3 \mathrm{PO}_4 \longrightarrow 3 \mathrm{H}^{+}+\mathrm{PO}_4^{3-} $

The order of back-bonding in boron halides is

$ \mathrm{BF}_3>\mathrm{BCl}_3>\mathrm{BBr}_3>\mathrm{BI}_3 $

due to the strength of overlapping which is in order

$ 2 p-2 p>2 p-3 p>2 p-4 p>2 p-5 p $

Hence, stronger is the back-bonding, lesser the availability of empty $p$-orbital and thus, weaker the tendency to act as a Lewis acid.

So, the correct order of Lewis acidic character of boron trihalides is

$ \mathrm{BI}_3>\mathrm{BBr}_3>\mathrm{BCl}_3>\mathrm{BF}_3 . $

Oxides of Sn (i.e. SnO and $\mathrm{SnO}_2$ ) and Pb , (i.e. $\mathrm{PbO}_2$ ) are amphoteric in nature However, oxide of silicon ( $\mathrm{SiO}_2$ ) is acidic in nature.

The test for nitrate ion is called "brown ring test". It can be performed by adding $\mathrm{FeSO}_4$ to a solution of nitrate followed by careful addition of conc. $\mathrm{H}_2 \mathrm{SO}_4$ along the sides of test tube. In this reaction, NO (nitric oxide) is formed in situ which then reacts with the iron (II) complex to give brown ring along the sides of test tube.

Reactions involved

$ \begin{aligned} & 2 \mathrm{NO}_3^{-}+6 \mathrm{H}^{+}+6 \mathrm{Fe}^{2+} \longrightarrow 6 \mathrm{Fe}^{3+}+2 \mathrm{NO}+4 \mathrm{H}_2 \mathrm{O} \end{aligned} $

The colour of $\mathrm{F}_2, \mathrm{Cl}_2, \mathrm{Br}_2$ and $\mathrm{I}_2$ molecules is yellow, greenish yellow, red and violet respectively. They absorb different quanta of radiations in visible region which results in the excitation of outer electrons to higher energy levels.

∴ The correct match is- A-III, B-IV, C-I, D-II

The oxidation state of central atom, Xe increases from +2 to +4 to +6 in $\mathrm{XeF}_2, \mathrm{XeF}_4$ and $\mathrm{XeF}_6$ respectively. Thus, its tendency to accept electrons also increases and thus, its oxidising power increases.

Also, $\mathrm{XeF}_6$ can give more fluorine than $\mathrm{XeF}_4$ and $\mathrm{XeF}_2$. So, the fluorinating agent order is $\mathrm{XeF}_6>\mathrm{XeF}_4>\mathrm{XeF}_2$. Hence, the correct decreasing order of the given ' $\mathrm{Xe}^{\prime}$ compound to act as both fluorinating and oxidising agent is $\mathrm{XeF}_6>\mathrm{XeF}_4>\mathrm{XeF}_2$.

When aluminium chloride is dissolved in excess of water, hydrated aluminium ion and chloride ions are formed.

$ \mathrm{AlCl}_3+\underset{\text { (Excess) }}{6 \mathrm{H}_2 \mathrm{O}} \longrightarrow\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} $

$ (a q)+3 \mathrm{Cl}^{-}(a q) $

(A) The $\mathrm{N}_2 \mathrm{O}$ produced by heating $\mathrm{NH}_4 \mathrm{NO}_3$ is a gas.

$ \mathrm{NH}_4 \mathrm{NO}_3 \xrightarrow{\text { Heat }} \mathrm{N}_2 \mathrm{O}+2 \mathrm{H}_2 \mathrm{O} $

It is diamagnetic as it does not have unpaired electron.

(B) No gaseous product is formed $2 \mathrm{NaNO}_2+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow 2 \mathrm{HNO}_2+\mathrm{Na}_2 \mathrm{SO}_4$

$ \begin{aligned} &\text { }\\ &\begin{aligned}(C)\,\, 2 \mathrm{~Pb}\left(\mathrm{NO}_3\right)_2 \xrightarrow{673 \mathrm{~K}} & 4 \mathrm{NO}_2 +2 \mathrm{PbO}+\mathrm{O}_2 \end{aligned} \end{aligned} $

Nitric oxide and oxygen are gases produced in this reaction.

$\mathrm{O}_2$ is paramagnetic because it has two unpaired electrons.

$\mathrm{NO}_2$ is paramagnetic as it has unpaired electron at nitrogen.

$2 \mathrm{Se}_2 \mathrm{Cl}_2 \longrightarrow \mathrm{SeCl}_4+3 \mathrm{Se}$

In a disproportionation reaction, the element is oxidised and reduced simultaneously.

The reactions in which a more reactive element displaces another element from its compounds are called displacement reactions.

$ \begin{aligned} & 2 \mathrm{Al}(s)+2 \mathrm{NaOH}(l)+6 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow 2 \mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_4\right](s)+3 \mathrm{H}_2(g) \end{aligned} $

The reaction of aluminium with sodium hydroxide gives a salt and hydrogen gas. In this reaction, aluminium displaces sodium from its compound, therefore, it is a displacement reaction.

The electronegativity decreases down the group. $\mathrm{C}, \mathrm{Ge}$ and Pb all belong to 14th group. So, the order should be C $>\mathrm{Ge}>\mathrm{Pb}$. But the actual order is $\mathrm{C}> \mathrm{Pb}>\mathrm{Ge}$. This can be explained on the basis of shielding effect.

The valence shell electronic configuration of Ge is $[\mathrm{Ar}] 3 d^{10} 4 s^2 4 p^2$ and that of Pb is $[\mathrm{Xe}] 6 s^2 4 f^{14} 5 d^{10} 6 p^2$. Due to the poor shielding effect of intervening ' $f$ ' electrons in Pb , the $Z_{\text {eff }}$ gets highly increased in Pb . Therefore, the electronegativity of Pb is higher than Ge .

Reaction of ammonia with nitrous acid results in production of $\mathrm{N}_2$ gas.

Thermal decomposition of ammonium dichromate results into $\mathrm{N}_2$ production.

$ \left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7 \longrightarrow \mathrm{Cr}_2 \mathrm{O}_3+4 \mathrm{H}_2 \mathrm{O}+\mathrm{N}_2 $

Treating aqueous solution of ammonium salt with nitrite results into $\mathrm{N}_2$ production.

$ \mathrm{NH}_4^{+}+\mathrm{NO}_2^{-} \longrightarrow \mathrm{N}_2+2 \mathrm{H}_2 \mathrm{O} $

Reaction of ammonia with sodiumhypochlorite is a oxidation reduction reaction in which NaOCl acts as oxidising agent and $\mathrm{NH}_3$ acts as reducing agent. It does not results into $\mathrm{N}_2$ production.

$ 2 \mathrm{NH}_3+\mathrm{NaOCl} \longrightarrow \mathrm{NaCl}+\mathrm{N}_2 \mathrm{H}_4+\mathrm{H}_2 \mathrm{O} $

Gradual increase in boiling points down the group happens as the relative molecular mass of the molecules increases. But the water has anomalously highest boiling point due to strong intermolecular hydrogen bonding in water molecules. Therefore, the order is $\mathrm{H}_2 \mathrm{O}>\mathrm{H}_2 \mathrm{Te}>\mathrm{H}_2 \mathrm{Se}>\mathrm{H}_2 \mathrm{~S}$.

Carnallite is not the mineral of fluorine, whereas all others are mineral of fluorine. Carnallite (also carnalite) is an evaporite mineral, a hydrated potassium magnesium chloride with formula $\mathrm{KMgCl}_3 \cdot 6\left(\mathrm{H}_2 \mathrm{O}\right)$.

The element that can even diffuse through silica glass is He due to very small size.

Here’s the reaction we need to identify P and Q for:

$P + Q \longrightarrow \left[ B(\mathrm{OH})_4 \right]^{-} + \mathrm{H}_3 \mathrm{O}^{+}$

Explanation:

The balanced reaction is:

$\mathrm{B}(\mathrm{OH})_3 + 2 \mathrm{H}_2 \mathrm{O} \longrightarrow \left[ \mathrm{B}(\mathrm{OH})_4 \right]^{-} + \mathrm{H}_3 \mathrm{O}^{+}$

This means that $P = \mathrm{H}_3 \mathrm{BO}_3$ and $Q = 2 \mathrm{H}_2 \mathrm{O}$. In this reaction, $\mathrm{H}_3 \mathrm{BO}_3$ acts as a Lewis acid.

$\mathrm{P}_4$ reacts with NaOH. On heating, it gives $\mathrm{NaH}_2 \mathrm{PO}_2$ and $\mathrm{PH}_3$.

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

This reaction is a disproportionation reaction as oxidation state of P changes from 0 to +1 and $-$3 . Which implies that, it undergoes oxidation and reduction simultaneously.

$\mathrm{S}_8$ reacts with NaOH on heating to given $\mathrm{Na}_2 \mathrm{~S}$ and $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$.

This reaction is a disproportionation reaction as oxidation state of $S$ changes from 0 to $-$2 and +2 . Which implies that, it undergoes oxidation and reduction simultaneously. But, $\mathrm{N}_2$ is an inert gas. So, it does not react with NaOH . Thus, only $\mathrm{P}_4$ and $\mathrm{S}_8$ undergo disproportionation when heated with NaOH solution.

(I) Size of boron atom is very small, so it cannot accommodate 4 large halide atoms and thus, does not form dimer. Also, the electron deficiency of boron is removed by $p \pi-p \pi$ back-bonding. Hence, the boron trihalides do not form dimers.

(II) In boron family, as we move down the group, two oxidation states are shown by them, i.e. +1 and +3 . However, boron shows only +3 oxidation state because of the absence of inert pair effect. Boron does not show +1 oxidation state.

(III) Boron shows a maximum covalency of four because of absent of $d$-orbitals.

(IV) Boron does not form $\mathrm{BF}_6^{6-}$ ion because of absence of empty $d$-orbitals.

Hence, only statements III and IV are correct.