p-Block Elements

Argon is an inert gas; hence, less reactive. It shield metal during welding. Argon is frequently blended with carbon dioxide (CO$_2$), hydrogen (H$_2$), helium (He) or oxygen (O$_2$) to enhance the arc characteristics or facilitate metal transfer in gas metal arc welding.

Hybridisation, H = $\frac{1}{2}$(V + X $-$ C + A) = $\frac{1}{2}$ (8 + 0 $-$ 0 + 0 ) = 4. So, sp$^3$ hybridised and has tetrahedral structure.

No. of lone pair = H $-$ X $-$ D = 4 $-$ 0 $-$ 3 = 1. As lone pair is present the electron pair geometry will be distorted tetrahedral and structure is pyramidal.

H = hybridisation no. (if 2 then sp, if 3 then sp$^2$ etc.)

V = valence electrons of central atom.

X = no. of monovalent atom.

C = cationic charge, A = anionic charge.

D = no. of divalent atom

6XeF$_4$ + 12H$_2$O $\to$ 4Xe + 2XeO$_3$ + 24HF + 3O$_2$

XeF$_6$ + 3H$_2$O $\to$ XeO$_3$ + 6HF

XeF$_4$ and XeF$_6$ are expected to be oxidizing because in both the reactions XeF$_4$ and XeF$_6$ are themselves reduced to xenon (Xe) and oxidises water to oxygen.

When boric acid reacts with sodium hydroxide, the products formed are sodium tetrahydroxoborate (III), sodium metaborate and water. The reaction is reversible in nature.

The reaction is difficult to proceed in forward direction due to weak monoprotic boric acid. When cis-1, 2-diol is added in the mixture of boric acid and sodium hydroxide, the weak acid gets converted to a comparatively stronger acid due to chelate complex formation. This stronger acid along with NaOH can undergo titration using phenolphthalein as an indicator. $\mathrm{B}(\mathrm{OH})_3+\mathrm{NaOH} \rightleftharpoons \mathrm{Na}\left[\mathrm{B}(\mathrm{OH})_4\right]$

Due to stable nature, it is removed from the solution during reaction. The entire reaction proceed in the same manner, disrupting the reversible chemical reaction and favouring the forward reaction completely.

(A) When $\mathrm{Bi}^{3+}$ is treated with water, it gives $[\mathrm{BiO}]^{+}$ ion and two protons.

$ \mathrm{Bi}^{3+}+\mathrm{H}_2 \mathrm{O} \rightarrow\left[\mathrm{BiO}^{+}+2 \mathrm{H}^{+}\right. $

The hydrolysis is a chemical reaction in which water is used for breaking down chemical bonds. The conversion of $\mathrm{Bi}^{+} \rightarrow[\mathrm{BiO}]^{+}$is carried out with the help of water; hence, it is a hydrolysis reaction.

Hence, option A from column I matches with option Q from column II.

(B) The conversion of $\left[\mathrm{AlO}_2\right]^{-} \rightarrow \mathrm{Al}(\mathrm{OH})_3$ is carried out by acidification. The proton is added to one of the reactant to produce aluminium hydroxide. The acid serve as a source of protons required for the acidification. The chemical reaction for the same is shown below:

$ \left[\mathrm{AlO}_2\right]^{-} \rightarrow \mathrm{Al}(\mathrm{OH})_3 $

Hence, option B from column I matches with option R from column II.

(C) Two molecules of orthosillicate $2 \mathrm{SiO}_4^{4-}$ gives pyrosilicate $\mathrm{Si}_2 \mathrm{O}_7^{6-}$ on strong heating.

$ 2 \mathrm{H}_4 \mathrm{SiO}_4 \xrightarrow{\Delta} \mathrm{H}_6 \mathrm{~S}_2 \mathrm{O}_7 $

Equation can also be written as

$ 2 \mathrm{SiO}_4^{4-} \xrightarrow{\Delta} \mathrm{Si}_2 \mathrm{O}_7^{6-} $

Hence, option C from column I matches with option P in column II.

(D) The conversion of $\left(\mathrm{B}_4 \mathrm{O}_7^{2-}\right)$ into $\left[\mathrm{B}(\mathrm{OH})_3\right]$ can be achieved either by hydrolysis or acidification.

The ( $\mathrm{B}_4 \mathrm{O}_7^{2-}$ ) is an anion in $\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7$.

When $\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7$ is treated with HCl or $\mathrm{H}_2 \mathrm{SO}_4$, it gives $\mathrm{H}_3 \mathrm{BO}_3$.

$ \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \xrightarrow{\mathrm{HCl} \text { or } \mathrm{H}_2 \mathrm{SO}_4} \mathrm{H}_3 \mathrm{BO}_3 \text { or } \mathrm{B}(\mathrm{OH})_3 $

Similarly, hydrolysis of $\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7$ yields $\mathrm{H}_3 \mathrm{BO}_3$.

$ \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \xrightarrow{\mathrm{H}_2 \mathrm{O}} \mathrm{H}_3 \mathrm{BO}_3 $

The $\mathrm{H}_3 \mathrm{BO}_3$ is a chemical formula of boric acid and is written as $\left[\mathrm{B}(\mathrm{OH})_3\right]$.

Hence, option D from column I matches with options Q and R in column II.

The correct answer is Option A: N: tetrahedral, sp³; B: tetrahedral, sp³

Explanation:

• BF₃ (Boron trifluoride): In its isolated state, BF₃ has a trigonal planar geometry around the boron atom with sp² hybridization.


• NH₃ (Ammonia): In its isolated state, NH₃ has a trigonal pyramidal geometry around the nitrogen atom with sp³ hybridization.

However, when BF₃ and NH₃ form a 1:1 complex, the lone pair of electrons on nitrogen donates to the empty p-orbital of boron. This leads to the formation of a coordinate bond between nitrogen and boron.

In the BF₃-NH₃ complex:

• Nitrogen (N): Now has four bonding pairs of electrons (three with hydrogen and one with boron), resulting in a tetrahedral geometry and sp³ hybridization.


• Boron (B): Also has four bonding pairs of electrons (three with fluorine and one with nitrogen), leading to a tetrahedral geometry and sp³ hybridization.

Key point: The formation of the coordinate bond between nitrogen and boron changes the hybridization and geometry of both atoms in the complex.

The oxidation number of an atom in a molecule or an ion is essentially a theoretical charge that the atom would have if all the bonds to atoms of different elements were completely ionic. To determine the oxidation numbers of sulfur in the given compounds: S₈, S₂F₂, and H₂S, let's analyze them one by one.

1. In S₈: This is an elemental form of sulfur, comprising of sulfur atoms only. In elemental forms, the atoms are not bonded to atoms of a different element, but only to themselves. In such cases, the oxidation number is always 0. Hence, the oxidation number of sulfur in S₈ is 0.

2. In S₂F₂: This is a molecule consisting of sulfur and fluorine. Fluorine is more electronegative than sulfur and virtually always has an oxidation number of -1 in its compounds. Let's denote the oxidation number of sulfur as $x$. Since there are two fluorine atoms, each contributing -1, and the overall molecule is neutral, we get:

$2x + (2 \times -1) = 0$

$2x - 2 = 0$

$2x = 2$

$x = +1$

Therefore, the oxidation number of sulfur in S₂F₂ is +1.

3. In H₂S: In this compound, hydrogen has an oxidation number of +1 (except when bonded to metals where it can be -1). Since there are two hydrogens, their total contribution is +2. Let's denote the oxidation number of sulfur in this compound as $x$. The compound is neutral overall, so:

$2(+1) + x = 0$

$2 + x = 0$

$x = -2$

Therefore, the oxidation number of sulfur in H₂S is -2.

Comparing with the given options, the correct answers are 0, +1, and -2 for S₈, S₂F₂, and H₂S, respectively. Therefore, the correct option is Option A: 0, +1, and -2.

The compound(s) which react with $\mathrm{NH}_3$ to form Boron nitride $(\mathrm{BN})$ is/are:

(A) Amorphous $\mathrm{B}+\mathrm{NH}_3 \underset{\text { Tempreature }}{\stackrel{\text { Very high }}{\longrightarrow}} \mathrm{BN}+\mathrm{H}_2$

(B) $\mathrm{B}_2 \mathrm{H}_6+\mathrm{NH}_3 \stackrel{\text { High Temp. }}{\longrightarrow}(\mathrm{BN})_{\mathrm{x}}$

(C) $\mathrm{B}_2 \mathrm{O}_3+2 \mathrm{NH}_3 \stackrel{900^{\circ} \mathrm{C}}{\longrightarrow} 2 \mathrm{BN}+3 \mathrm{H}_2 \mathrm{O}$

(D) $\mathrm{HBF}_4+\mathrm{NH}_3 \longrightarrow\left[\mathrm{NH}_4^{\oplus}\right]\left[\mathrm{BF}_4^{\ominus}\right]$

Hence, $(A, B, C)$ are correct

(a) H₃PO₃ gives H₃PO₄ and PH₃ on heating. This reaction is known as disproportionation reaction.

$4{H_3}P{O_3}\buildrel \Delta \over \longrightarrow 3{H_3}P{O_4} + P{H_3}$

(b) P in H₃PO₄ is in its highest oxidation state i.e. + 5 hence, it cannot act as reducing agent. But P in H₃PO₃ is in oxidation state, + 3 hence, it can act as reducing agent.

(c) H₃PO₃ contains two $-$OH groups, hence it is a dibasic acid.


(d) Hydrogen attached to P, does not ionise in water. Therefore, options (a), (b) and (d) are correct.

(a) Order of acid strength different oxyacids of chlorine are :

$\mathop {HClO}\limits_{(Hypochlorous\,acid)}^{( + 1)} < \mathop {HCl{O_3}}\limits_{(Chloric\,acid)}^{( + 5)} < \mathop {HCl{O_4}}\limits_{(Perchloric\,acid)}^{( + 7)} $

Weak acid have strong conjugate base thus hypochlorite ion has strongest conjugate base. Therefore, statement (a) is correct.

(b) Hypochlorite ion is linear and perchlorate ion is tetrahedral and there is no effect of lone pair on hypochlorite ion. Thus, statement (b) is correct.



(c) In the disproportionation reaction, chlorate ion Cl(+5) is oxidised to perchlorate, Cl(+7) and reduce to chloride, Cl($-$1).

$4ClO_3^ - \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} 3ClO_4^ - + C{l^ - }$

While in hypochlorite ion, chlorite ion Cl(+1) is oxidised to chlorate, Cl(+5) and reduced to chloride, Cl($-$1) ion.

$3ClO_{}^ - \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} ClO_3^ - + 2C{l^ - }$

Thus, statement (c) is incorrect.

(d) The hypochlorite ion oxidises the sulphite ion to sulphate ion, because HOCl is the strongest oxidising Cloxyacids,

$Cl{O^ - } + SO_3^{2 - }\buildrel {} \over \longrightarrow SO_4^{2 - } + C{l^ - }$

Thus, statement (d) is correct.

When Zn react with hot conc. H₂SO₄ then SO₂ is released and ZnSO₄ is obtained.



R(SO₂) molecule is V-shaped



Thus, option (b) is correct.

When Zn is react with conc. NaOH then H₂ gas is evolved and Na₂ZnO₂ is obtained.

$Zn + 2NaOH(conc.)\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} \mathop {N{a_2}Zn{O_2}}\limits_{(T)} + \mathop {{H_2}}\limits_{(Q)} \uparrow $

In ground state, H$-$H (Q)

(bond order = 1)

Thus, option (c) is correct.

The oxidation state of Zn in T(Na₂ZnO₂) is +2

Thus, option (a) is incorrect.

$\mathop {ZnS{O_4}}\limits_{(G)} + {H_2}S + N{H_4}OH\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} \mathop {ZnS}\limits_{(Z)} \downarrow + \mathop {2{H_2}O}\limits_{(X)} + \mathop {{{(N{H_4})}_2}S{O_4}}\limits_{(Y)} $

ZnS (Z) compound is dirty white coloured.

Thus, option (d) is correct.

Sn can exist in +2 or +4 oxidation state. So, Q in the given reactions can be SnCl₂ or SnCl₄. Since X is monoanion having trigonal pyramidal geometry (sp³ with one lone pair as per VSEPR), so Q is SnCl₂, and the first reaction is

The second reaction is Stephen’s reaction, where nitrile is converted to aldehyde via formation of iminium salt.


There is a coordinate bond in the Cl₂Sn.N(CH₃)₃ in between nitrogen and Sn metal.

Z is oxidised product and oxidation state of Sn is +4 in Z compound. Structure of SnCl₄ (Z) is

Option (A) : Correct. The basicity of oxides usually increases on descending a group. Therefore, Bi₂O5 is more basic than N₂O₅.

Option (B) : Correct. Covalent nature of a molecule depends on the electronegativity difference between bonded atoms.

Option (C) : Correct. Boiling point of NH₃ is more than that of PH₃ due to hydrogen bonding.

Option (D) : Incorrect. P$-$P single bond is stronger than N$-$N single bond. This is due to the fact that N is small in size, due to smaller size of atoms lone pair of repulsion will be more.

(A) $N{H_4}N{O_3}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{below\,300^\circ C}} {N_2}O + 2{H_2}O$

${N_2}O\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{above\,600^\circ C}} {N_2} + {O_2}$

(B) ${(N{H_4})_2}C{r_2}{O_7}\buildrel \Delta \over \longrightarrow {N_2} + C{r_2}{O_3} + 4{H_2}O$

(C) $Ba{({N_3})_2}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{180^\circ C}^\Delta } Ba + 3{N_2}$

(D) $M{g_3}{N_2}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{above\,700^\circ C}} 3Mg + {N_2}$

Hence, only ${(N{H_4})_2}C{r_2}{O_7}$ and $Ba{({N_3})_2}$ can provide N₂ gas on heating below $300^\circ C$

Option (A): Correct.

The aluminium compounds are unusual because they have dimeric structures, and appear to have three-centre bonds involving $s p^3$ hybrid orbitals on $\mathrm{Al}$ and $\mathrm{C}$ in $\mathrm{Al}-\mathrm{C}-\mathrm{Al}$ bridges.


Option (B): Correct.

In diborane $\left(\mathrm{BH}_3\right)$ there are 12 valency electrons, three from each $B$ atom and six from the $\mathrm{H}$ atoms. An $s p^3$ hybrid orbital from each boron atom overlaps with the $1 s$ orbital of the hydrogen. This gives a delocalised molecular orbital covering all three nuclei, containing one pair of electrons and making up one of the bridges. This is a three-centre two-electron bond $(3 c-2 e)$.


Option (D): Group 13 elements have only three valency electrons. When these are used to form three covalent bonds, the atom has a share in only six electrons. The compounds are therefore electron deficient. In $\mathrm{AlCl}_3$, effective $\pi$ overlap takes place between $p$ orbitals of $\mathrm{Al}$ and $\mathrm{Cl}$ due to their comparable size while in $\mathrm{BCl}_3$, the $\pi$ overlap is not effective as $p$ orbital of boron is smaller than that of $p$ orbital of chlorine, hence, the acidity of $\mathrm{BCl}_3$ is greater than that of $\mathrm{AlCl}_3$.

Halogens exist as diatomic molecule of different colours:

$ \begin{array}{|c|c|c|} \hline \text { S.No. } & \text { Halogen } & \text { Colour } \\ \hline 1 . & \text { Fluorine } & \text { Pale yellow } \\ \hline 2 . & \text { Chlorine } & \text { Yellow green } \\ \hline 3 . & \text { Bromine } & \text { Red brown } \\ \hline 4 . & \text { Iodine } & \text { Purple } \\ \hline \end{array} $

(i) The molecular orbital energy level diagram explains the appearance of colour by halogens. The MOT for halogens is represented.


(ii) It represents molecular orbital energy level diagram for fluorine. Similar molecular energy level diagram exist. Other halogens.

(iii) Antibonding $\pi$ orbitals, i.e., $\pi^*{ }_{2 p x}$ and $\pi^*{ }_{2 p y}$ forms the highest occupied molecular orbital (HOMO) and antibonding sigma* orbitals forms the lowest unoccupied molecular orbital (LUMO) for halogens.

(iv) Absorption of energy of suitable wavelength (or colour) results in transition of electron from HOMO to LUMO. As electron returns back to ground state, i.e, HOMO, it releases energy corresponding to a different wavelength (colour complementary to the colour absorbed). This gives halogens their characteristic colour.

(v) As we move down the group 17, size of halogen atom increases and nuclear force of attraction for the outermost shell electrons decrease. This affects the energy gap between $\mathrm{HOMO}$ and LUMO.

$ \mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2 $

HOMO-LUMO energy gap decreases $\rightarrow$

Wavelength of light emitted decreases $\rightarrow$

(vi) The energy gap between HOMO and LUMO keeps on decreasing as we move down the group. As a result, the energy required for transition of electron from HOMO to LUMO decreases and less energy (or light of lower wavelength is absorbed).

(vii)When electron moves back to HOMO, energy is emitted. It corresponds to the light of complementary colour to the colour of the light absorbed.

Option (A): Correct. The structures of the ions formed are shown in below figure. All these structures are based on a tetrahedron. The $s p^3$ hybrid orbitals used for bonding form only weak $\sigma$ bonds, because the $s$ and $p$ levels differ appreciably in energy. The ions are stabilised by strong $p \pi-d \pi$ bonding between full $2 p$ orbitals on oxygen with empty $d$ orbitals on the halogen atoms.


Option (B): $\mathrm{HClO}_4$ is an extremely strong acid, while $\mathrm{HOCl}$ is a very weak acid. Oxygen is more electronegative than chlorine. The more oxygen atoms that are bonded, the more the electrons will be pulled away from the $\mathrm{O}-\mathrm{H}$ bond, and the more this bond will be weakened. Thus $\mathrm{HClO}_4$ requires the least energy to break the $\mathrm{O}-\mathrm{H}$ bond and form $\mathrm{H}^{+}$.

Option (C): The reaction is $\mathrm{Cl}_2+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{HOCl}+\mathrm{HCl}$

Option (D): $\mathrm{As} ~\mathrm{HClO}_4$ is a stronger acid than $\mathrm{H}_2 \mathrm{O}$, therefore, its conjugate base $\left(\mathrm{ClO}_4^{-}\right)$will be weaker than $\left(\mathrm{OH}^{-}\right)$that of $\mathrm{H}_2 \mathrm{O}$.

$ \mathrm{HClO}_4+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{ClO}_4^{-} $

The reaction of HNO₃ and P₄O₁₀ produces N₂O₅.

4HNO₃ + P₄O₁₀ $\to$ 2N₂O₅ + 4HPO₃

The reaction of HNO₃ with P₄ does not yield N₂O₅.

P₄ + 20 HNO₃ $\to$ 4H₃PO₄ + 20 NO₂ + 4 H₂O

The structure of N₂O₅ has one N$-$O$-$N bond, but no N$-$N bond. It is diamagnetic in nature.

N₂O₅ reacts with sodium metal to produce NO₂ (brown gas)

N₂O₅ + Na $\to$ NaNO₃ + $\mathop {N{O_2} \uparrow }\limits_{Brown\,gas} $

Structure of crystalline form of borax


(a) Borax contains four boron atom; hence, it is tetranuclear.

(b) Only 2 boron atoms lie in the same plane.

(c) Two boron atoms are $s p^3$ hybridised other two borons are $s p^2$ hybridised.

(d) Each of the boron has hydroxyl group attached to it with two hydroxyl groups present as hydroxyl salts.

In all the oxyacids of chlorine : Cl undergoes sp³ hybridization HClO₄ is the strongest acid among the given compounds.

(a) It does not undergo self ionisation in water but accepts an electron pair from water, so it behaves as weak monobasic acid

H₃BO₃ + H₂O $\rightleftharpoons$ B(OH)$_4^ - $ + H⁺

Hence, (a) is incorrect.

(b) When treated with 1, 2-dihydroxy or polyhydroxy compounds, they form chelate (ring complex) which effectively remove [B(OH)₄]^$-$ species from solution and thereby produce maximum number of H₃O⁺ or H⁺ ions, i.e., results in increased acidity.

(c) Baric acid crystallises in a layer structure in which planar triangular BO$_3^{3 - }$ ions are bonded together through hydrogen bonds.

(d) In water the pK_a value of H₃BO₃ is 9.25

H₃BO₃ + H₂O $\rightleftharpoons$ B(OH)$_4^ - $ + H⁺ ; pK_a = 9.25

So, it is weak electrolyte in water.

As all electrons are paired, ozone is diamagnetic in nature. The structure is bent or V-shaped. The structure of ozone is resonance hybrid of the two structures with a delocalised p-orbital which covers all three atoms. Because of this, the two O-O bond lengths are equal.

Diamond is hard and graphite is soft. Graphite is a good conductor of electricity as it has one free electron which is responsible for the conduction.

Diamond has higher thermal conductivity than graphite because the structure of diamond is precise and thus the transfer of heat is faster in it.

In case of graphite, the C-C bond has a double bond character so its bond order becomes higher than that of diamond which has only single C-C bonds.

Only HF does not react with AgNO₃. All others react with AgNO₃ to give precipitate of AgX which is soluble in Na₂S₂O₃

$HX + AgN{O_3} \to AgX \downarrow + HN{O_3}(X = Cl,Br,I)$

$AgX + N{a_2}{S_2}{O_3} \to \mathop {N{a_3}[Ag{{({S_2}{O_3})}_2}]}\limits_{Soluble} \, + NaX$

Tertiary amine will not react due to the bulkiness of trimethyl groups. Tertiary amines instead react with diborane to give simple addition compounds through symmetrical cleavage of diborane.

The bond formation in the given oxides of nitrogen is as follows.

Thus N-N bond exists in N$_2$O, N$_2$O$_3$ and N$_2$O$_4$.

The colourless salt H is either NH$_4$NO$_3$ or NH$_4$NO$_2$. The colourless salt produces NH$_3$ gas (non-inflammable) on boiling with excess of NaOH. On addition of zinc dust (a reducing agent) to this solution, sodium nitrite or sodium nitrate will liberate NH$_3$ gas again.

NH$_4$NO$_3$ + NaOH $\to$ NH$_3$ + NaNO$_3$ + H$_2$O

7NaOH + NaNO$_3$ + 4Zn $\to$ 4Na$_2$ZnO$_2$ + NH$_3$ + 2H$_2$O

NH$_4$NO$_2$ + NaOH $\to$ NaNO$_2$ + NH$_3$ + H$_2$O

3Zn + 5NaOH + NaNO$_2$ $\to$ 3Na$_2$ZnO$_2$ + NH$_3$

When carbon dioxide (CO₂) is dissolved in water, it undergoes a series of chemical reactions that result in various species being present in the solution. The reactions can be described as follows:

1. Carbon dioxide reacts with water to form carbonic acid:

$CO_2 + H_2O \rightleftharpoons H_2CO_3$

2. Carbonic acid can dissociate into a bicarbonate ion and a proton:

$H_2CO_3 \rightleftharpoons HCO_3^- + H^+$

3. Bicarbonate ion can further dissociate into a carbonate ion and a proton:

$HCO_3^- \rightleftharpoons CO_3^{2-} + H^+$

Based on these reactions, the species present in the solution when CO₂ is dissolved in water include CO₂, $H_2CO_3$, $HCO_3^-$, and $CO_3^{2-}$. Thus, the correct option is:

Option A: CO₂, H₂CO₃, $HCO_3^-$, $CO_3^{2-}$