p-Block Elements

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.

$Cs{l_3}\,\,$ dissociates as

$Cs{l_3} \to C{s^ + } + I_3^ - $

Acidic strength increases as the oxidation number of central atom increases.

Hence acidic strength order is

$\eqalign{ & \left( { + 7} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( { + 5} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( { + 3} \right)\,\,\,\,\,\,\,\,\,\,\,\left( { + 3} \right) \cr & HCl{O_4}\, > \,HCl{O_3}\, > \,HCl{O_2}\, > \,HClO \cr} $

Nitric oxide is paramagnetic in the gaseous state because of the presence of one unpaired electron in its outermost shell.

The electronic configuration of $NO$ is

$\sigma _{1s}^2\,\sigma _{1s}^{ * 2}\,\sigma _{2s}^2\,\sigma _{2s}^{ \circ 2}\,\sigma _{{2_{{p_z}}}}^2\,\pi _{{2_{{p_x}}}}^2$

$ = \pi _{{2_{Py}}}^2\,\pi _{{2_{Px}}}^{ * 1}$

The reaction involved is

P₄ + 8SOCl₂ $\to$ 4PCl₃ + 4SO₂ + 2S₂Cl₂

The reaction involved is

(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.

Option A: O₃ molecule is bent

✅ True

• The structure of ozone (O₃) is angular (bent) with a bond angle of about 117°.

• It has resonance structures and sp²-hybridized central oxygen.

Option B: Ozone is violet-black in solid state

✅ True

• In solid state, ozone is violet-black.

• In liquid state, ozone is dark blue.

• In gas state, it is pale blue.

Option C: Ozone is diamagnetic gas

✅ True

• Although O₂ is paramagnetic, ozone (O₃) has all paired electrons, hence diamagnetic.

Option D: ONCl and ONO⁻ are isoelectronic

❌ False

Let’s check the total number of electrons:

• ONCl:

  O (8) + N (7) + Cl (17) = 32 electrons

• ONO⁻:

  O (8×2 = 16) + N (7) + 1 (extra for negative charge) = 24 electrons

→ They are not isoelectronic.

✅ Correct answer: Option D is the wrong statement.

The reactions involved are

Cl₂ + 2NaOH (cold, dil.) $\to$ $\mathop {NaOCl}\limits_{(P)} $ + NaCl + H₂O

3Cl₂ + 6NaOH (hot, conc.) $\to$ $\mathop {NaCl{O_3}}\limits_{(Q)} $ + 5NaCl + 3H₂O

where NaOCl and NaClO₃ are salts of hypochlorous acid (HOCl) and chloric acid (HClO₃), respectively.

The reactions involved are

$S{O_2} + C{l_2}\buildrel {Charcoal} \over \longrightarrow \mathop {S{O_2}C{l_2}}\limits_{(R)} $

$\mathop {10S{O_2}C{l_2}}\limits_{(R)} + {P_4} \to \mathop {4PC{l_5}}\limits_{(S)} + 10S{O_2}$

$\mathop {PC{l_5}}\limits_{(S)} + 4{H_2}O \to \mathop {{H_3}P{O_4}}\limits_{(T)} + 5HCl$

The reactions involved are as follows:

(P) $Pb{O_2} + {H_2}S{O_4}\buildrel {Warm} \over \longrightarrow PbS{O_4} + {H_2}O + {1 \over 2}{O_2}$

(Q) $2N{a_2}{S_2}{O_3} + 2{H_2}O + C{l_2} \to 2NaCl + 2NaHS{O_4} + 2S$

(R) ${N_2}{H_4} + 2{I_2} \to {N_2} + 4HI$

(S) $Xe{F_2} + 2NO \to Xe + 2NOF$

The reaction involved is 4HNO₃ $\to$ 4NO₂ + 2H₂O + O₂. The yellow-brown color appears due to the formation of NO₂.

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.

All the members form volatile halides of the type $A{X_3}.$ All halides are pyramidal in shape. The bond angle decreases on moving down the group due to decrease in bond pair-bond pair repulsion.

$\eqalign{ & NC{L_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,PC{l_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,AsC{l^3} \cr & {107^ \circ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{94^ \circ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{92^ \circ } \cr} $

According to the reaction

P₄ + NaOH + 3H₂O $\to$ PH₃ + 3NaH₂PO₂

oxidation state of phosphorus in P₄ is zero while in PH₃ it is $-$3 and in NaH₂PO₂, it is +1. This shows that this is a type of disproportionation reaction because there is an increase as well as decrease in the oxidation state of phosphorus.

Bleaching powder is CaOCl₂ which contains HOCl as an oxoacid and the anhydride of which is Cl₂O.

Consider the titration reaction

CaOCl₂ + 2KI $\to$ I₂ + Ca(OH)₂ + KCl

According to this reaction, 25 mL of CaOCl₂ reacts with 30 mL of 0.50 M KI

I₂ + 2Na₂S₂O₃ $\to$ Na₂S₄O₆ + 2NaI

Given that 48 mL of 0.25 N Na₂S₂O₃ was used to reach the end point. So, the number of moles of I₂ produced = 48 $\times$ 0.25/2 = 6.

According to the reaction,

Number of millimoles of bleaching powder = Number of moles of I₂ = (1/2) $\times$ Number of moles of Na₂S₂O₃ = 6

So, the molarity of CaOCl₂ is

${{Number\,of\,moles\,of\,bleaching\,powder} \over {Volume\,of\,solution}} = {{6\,mmol} \over {25\,mL}} = 0.24M$

In HNO₃, the oxidation state of N is +5 while in NO, it is +2. In N₂, it is zero and in NH₄Cl, the oxidation state of nitrogen is $-$3. So, the decreasing order of oxidation state of nitrogen is

HNO₃ > NO > N₂ > NH₄Cl