p-Block Elements

CCl₄ cannot get hydrolysed as it does not have dorbitals and cannot extend its covalency above four.

The reaction which has slow reactivity requires catalyst.

Among halogens reactivity order is

F₂ > Cl₂ > Br₂ > I₂

So, H₂ + I₂ $ \to $ 2HI, this reaction requires a catalyst.

H₃PO₂ is good reducing agent due to presence of two P $-$ H bonds.

(a) and (b) are properties of silicone.

(d) is the uses of silicone.

Those materials which produce electricity under mechanical stress are called piezoelectric material.

Quartz is an example of piezoelectric material.

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

Here BCl₃ is electron deficient compound as B has 6 electrons. That is why it accept electron pair from NH₃ to form an adduct.

BCl₃ does not form dimer like Al, Ga or In, because its electron deficiency is complemented by the formation of co-ordinate bond between lone pair of electron of chlorine and empty unhybridized P-orbital of boron forming P$\pi $ $-$ P$\pi $ bonding.

$ \therefore $   N₂O₃   and   N₂O₄   has   N $-$ N   bond.

Thermal decomposition reaction of the given compounds are . . .

(NH₄)₂SO₄ $\buildrel \Delta \over \longrightarrow $ 2 NH₃(g) + H₂SO₄

Ba (N₃)₂ $\buildrel \Delta \over \longrightarrow $ Ba + 3N₂(g)

NH₄NO₂ $\buildrel \Delta \over \longrightarrow $ N₂(g) + 2H₂O

(NH₄)₂Cr₂O₇ $\buildrel \Delta \over \longrightarrow $ N₂(g) + 4H₂O + Cr₂O₃

So, here you can see only (NH₄)₂SO₄ does not give N₂ by thermal decomposition it gives NH₃ while other give N₂ gas

Among the given Metal Al is the correct answer.



Al₂O₃ is used as an absorbed in chromatography.

$\therefore\,\,\,$ Number of   P $-$ O   Bonds  =  12

LiAlH₄ + SiCl₄  $\buildrel \, \over \longrightarrow $  LiCl + AlCl₃ + SiH₄

$\therefore\,\,\,$ Number of Bond pairs = 5

Number of $\pi $ bonds = 3

Number of lone pairs = 0

KO₂ is called potassium superoxide. Here O$_2^ - $ is the superoxide ion.

Oxidation state of K is +1 and let oxidation state of oxygent atom is = x

$\therefore\,\,\,$ 1 + 2(x) = 0

$ \Rightarrow $$\,\,\,$ x = $-$ ${1 \over 2}$

The anion is Cl⁻ . Chloride ion reacts with AgNO₃ to give a white precipitate.

AgCl is insoluble in dil. nitric acid.

XeF₆ on hydrolysis with water can produces 3 compounds XeOF₄, XeO₂F₂ and XeO₃.

Here XeOF₄ and XeO₂F₂ are produced on partial hydrolysis and XeO₃ is produced on complete hydrolysis.

XeF₆  +  H₂O  $\buildrel {Partial} \over \longrightarrow $  XeOF₄  +  2HF

XeF₆   +   2H₂O  $\buildrel {Partial} \over \longrightarrow $  XeO₂F₂   +  4HF

XeF₆   +   3H₂O  $\buildrel {Complete} \over \longrightarrow $  XeO₃  +  6HF

So, the compound X and Y are XeOF₄(+6)   or   XeO₂F₂(+6).

Hybridization of carbon in graphite is sp².

So,   %   of p - in graphite = ${2 \over 3} \times 100$ = 67%

Hybridization of carbon in diamond is sp³ .

So,   %   of p - in diamond = ${3 \over 4} \times 100$ = 75%

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 $300^\circ C$

XeF₆ on hydrolysis with water can produces 3 compounds XeOF₄, XeO₂F₂ and XeO₃.

Here XeOF₄ and XeO₂F₂ are produced on partial hydrolysis and XeO₃ is produced on complete hydrolysis.

XeF₆  +  H₂O  $\buildrel {Partial} \over \longrightarrow $  XeOF₄  +  2HF

XeF₆   +   2H₂O  $\buildrel {Partial} \over \longrightarrow $  XeO₂F₂   +  4HF

XeF₆   +   3H₂O  $\buildrel {Complete} \over \longrightarrow $  XeO₃  +  6HF

So, the compound X can be XeOF₄   or   XeO₂F₂

Reaction with silica (SiO₂) of XeF₆ :

When ration of XeF₆ and SiO₂ is 2 : 1 then produce XeOF₄.

2XeF₆   +   SiO₂   $ \to $   2XeF₄   +   SiF₄

When ratio of XeF₆ and SiO₂ is 1 : 1 , then produce XeO₂F₂.

XeF₆   +   SiO₂   $ \to $   XeO₂F₂   +   SiF₄

So, the compound X can be XeOF₄   or  XeO₂F₂.

Here in option only XeOF₄ is given so this will be right answer.

Number of P - OH bonds = 4

Let the oxidation number of P = x

$ \therefore $   In H₄ P₂ O₇

for 7 oxygen = 7 $ \times $ ($-$ 2) = $-$ 14

for 4 Hydrogen = 4 $ \times $ (+1) = 4

For 2 Phosphorus = 2x

$ \therefore $   2x + 4 $-$ 14 = 0

$ \Rightarrow $   x = + 5

Peroxidisulphuric Acid (H₂ S₂ O₈) :

Number of S = O, bonds = 4

Number of S $-$ OH, bond = 2

Pyrosulphuric Acid (H₂ S₂ O₇) :

Number of S = O, bonds = 4

Number of S $-$ OH, bonds = 2

(i) Potassium chlorate $\left(\mathrm{KClO}_3\right)$ is decomposed in presence of $\mathrm{MnO}_2$ as catalyst to form potassium chloride and oxygen gas.

$2 \mathrm{KClO}_3(s) \underset{\mathrm{MnO}_2}{\stackrel{\Delta}{\longrightarrow}} 2 \mathrm{KCl}+\underset{(w)}{3 \mathrm{O}_2(g)}$

The gas $(w)$ is oxygen.

(ii) Reaction of white phosphorous with excess of gas $w$ (i.e., $\mathrm{O}_2$ ) gives $\mathrm{P}_4 \mathrm{O}_{10}$ (a dimer of phosphorous pentaoxide).

$ \underset{\text{White}}{\mathrm{P}_4(s)}+\underset{\text{(excess)}}{5 \mathrm{O}_2(g)} \rightarrow \underset{(X)}{\mathrm{P}_4 \mathrm{O}_{10}(s)}$

phosphorous oxygen

The compound $(X)$ is $\mathrm{P}_4 \mathrm{O}_{10}$.

Reaction of $\mathrm{P}_4 \mathrm{O}_{10}$ with pure nitric acid $\left(\mathrm{HNO}_3\right)$ gives dinitrogen pentaoxide $\left(\mathrm{N}_2 \mathrm{O}_5\right)$ and metaphosphoric acid $\left(\mathrm{HPO}_3\right)$

$ \mathrm{P}_4 \mathrm{O}_{10}+4 \mathrm{HNO}_3 \rightarrow \underset{Z}{4 \mathrm{HPO}_3}+\underset{Y}{2 \mathrm{~N}_2 \mathrm{O}_5} $

The compound $\mathrm{Y}$ is dinitrogen pentaoxide $\left(\mathrm{N}_2 \mathrm{O}_5\right)$ and compound $\mathrm{Z}$ is metaphosphoric $\operatorname{acid}\left(\mathrm{HPO}_3\right)$.

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^{-} $

NaF is composed of Na⁺ and F^–.

Na⁺ : 1s²2s²2p⁶

F^- : 1s²2s²2p⁶

Hence do not match with the configuration given in the question.

In diamond, each C - atom is covalently bonded with four other carbon atom. So it utilizes its four unpaired electrons in bond formation. Due to this reason diamond is a bad conductor of electricity.

In graphite, each carbon atom is covalently bonded to three carbon atom. In those bond formation 3 out of 4 valence electrons of each C - atom are used while the fourth electron is free to move in the structure of graphite. Due to this reason graphite is a good conductor of electricity.

In diamond, each carbon is bonded with 4 other carbon so, carbon is sp³ hybridized.

In graphite, each carbon is bonded with 3 other carbon so carbon is sp² hybridized.

So, here both Assertion and Reason are correct but hybridization is not the reason for the Assertion.

S₈ ring has crown shape and bond angle is 107^o

S₆ ring has chair form hexagon ring and bond angle is 102.2^o.

Instantaneous dipole-induced dipole forces or van der Waals’ forces are most responsible in allowing xenon gas to liquify.

(A) Silica gel (SiO₂) is a good dehydrating agent. It is used good dehydrating agent. It is used to absorb moisture.

SiO₂ + 2H₂O $ \to $SiO₂.2H₂O

(B) Silicon is used in transistor as it is a semiconductor.

(c) Silicon is a good resistant of heat, water. So it used as sealant to seal some element.

(d) Silicate including zeolites used in ion-exchange beds in domestic and commercial water purification, softening and other applications.

Fluorine is the most electronegative element and it shows only –1 oxidation state.

Phosphorous acids contain $P$ in $+3$ oxidation state.

Acid	Formula	Oxidation state of Phosphorous
Pyrophosphorous acid	H4P2O5	+3
Pyrophophoric acid	H4P2O7	+5
Orthophosphorous acid	H3PO3	+3
Hypophosphoric acid	H4P2O6	+4

The increasing order of atomic radii is as follows:

Ga < Al < In < Tl

The atomic radius generally increases on moving down a group in the periodic table. As an anomaly, the atomic radius of Ga is less than that of aluminium because of poor shielding of nuclear charge by 10 number of 3d electrons. As a result, the outershell electrons are held more firmly by the nucleus and contraction of radius is observed. This contraction is also called d-block contraction. The size of Tl is similarly affected by 14 number of 4f electrons (lanthanoid contraction) and the atomic radius of Tl is almost similar in size to In.

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.

$ICl$

Order of reactivity of halogens

$C{l_2} > B{r_2} > {I_2}$

But, the interhalogen compounds are generally more reactive than halogens (except ${F_2}$), since the bond between two dissimilar electronegative elements is weaker than the bond between two similar atoms i.e, $X-X$

Nitrogen and oxygen in air do not react to form oxides of nitrogen in atmosphere because the reaction between nitrogen and oxygen requires high temperature.

$Xe.$ As we move down the group, the melting and boiling points show a regular increase due to corresponding increase in the magnitude of their van der waal forces of attraction as the size of the atom increases.

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