p-Block Elements

(A) Compound X is used for sterilizing drinking water.

Chlorine gas ($\mathrm{Cl_2}$) is widely used as a disinfectant in water treatment. It kills harmful bacteria and microorganisms, making the water safe for drinking.

(B) Compound Y is $\mathrm{NCl_3}$ which has a pyramidal shape.

This is because the nitrogen atom has one lone pair of electrons. Due to the repulsion between the bond pairs and lone pair, the molecular shape becomes pyramidal (similar to $\mathrm{NH_3}$).

(C) Compound Z is used in the enrichment of ${}^{235}\mathrm{U}$.

Chlorine trifluoride ($\mathrm{ClF_3}$) is a very strong fluorinating agent. In the nuclear industry, it converts uranium into uranium hexafluoride ($\mathrm{UF_6}$). This compound is important in the gaseous diffusion or centrifugation process used to separate ${}^{235}\mathrm{U}$ from ${}^{238}\mathrm{U}$.

$ \mathrm{U(s)} + 3\,\mathrm{ClF_3} \rightarrow \mathrm{UF_6(g)} + 3\,\mathrm{ClF(g)} $

(D) $\mathrm{NH_3}$ is a stronger Lewis base than $\mathrm{NCl_3}$.

In $\mathrm{NH_3}$, nitrogen’s lone pair is more available for donation because hydrogen is less electronegative. In $\mathrm{NCl_3}$, the chlorine atoms pull electron density away from nitrogen, making its lone pair less available. Hence, $\mathrm{NH_3}$ acts as the stronger Lewis base.

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$_3$