NCERT Exemplar Class 11 Chemistry solutions Chapter 11 P-Block Elements

NCERT Exemplar Class 11 Chemistry Solutions Chapter The p-block Elements: MCQ (Type 1)

Question:1

The element which exists in a liquid state for a wide range of temperature and
can be used for measuring high temperature is
(i) B
(ii) Al
(iii) Ga
(iv) In
Answer:

The answer is the option (iii) Ga
Galium has different structural properties and can remain in the liquid state for a wide range of temperature since it is melting point is 30°C and boiling point is 224°C.

Question:2

Which of the following is a Lewis acid?
$(i)AlC{l}_3$
$(ii)MgC{l}_2$
$(iii)CaC{l}_2$
$(iv)BaC{l}_2$
Answer:

The answer is the option $(i)AlC{l}_3$
AlCl₃ is electron-deficient and ready to accept electrons and thus acts as a Lewis acid. It is an electron acceptor and a covalent compound.

Question:3

The geometry of a complex species can be understood from the knowledge of type of hybridisation of orbitals of central atom. The hybridisation of orbitals of central atom in $[Be(OH{)}_4{]}^{-}$ and the geometry of the complex are respectively
(i)$s{p}^3$ , tetrahedral
(ii)$s{p}^3$, square planar
(iii) $s{p}^3{d}^2$, octahedral
(iv) $ds{p}^2$ , square planar
Answer:

The answer is the option (i)$s{p}^3$ , tetrahedral
Boron in the excited state has 2s orbitals which are not paired and one electron which is in the p-orbital. In hybridization, the $s{p}^3$ hybridization is formed, and tetrahedral geometry is observed.
Electronic configuration of boron; $1s^{2}2s^{2}2p{x}^{1}2p^{0}y2p^{{}^{\circ }}z$

Question:4

Which of the following oxides is acidic in nature?
$(i)B_2{O}_3$
$(ii)A{l}_2{O}_3$
$(iii)G{a}_2{O}_3$
$(iv)I{n}_2{O}_3$
Answer:

The answer is the option $(i)B_2{O}_3$ because it is acidic and forms metal borates on reaction with basic oxides.

Question:5

The exhibition of highest co-ordination number depends on the availability of vacant orbitals in the central atom. Which of the following elements is not likely to act as central atom in $M{F}_6^{3-}$?
(i) B
(ii) Al
(iii) Ga
(iv) In
Answer:

The answer is the option (i) B
The element M in the complex ion $M{F}_6^{3-}$ has 6 as the coordination number and the B has no d orbitals. Therefore the maximum coordination number it can show is 4 and cannot form the $M{F}_6^{3-}$.

Question:6

Boric acid is an acid because its molecule
(i) contains replaceable $H^{+}$ ion
(ii) gives up a proton
(iii) accepts $O{H}^{-}$ from water releasing proton
(iv) combines with proton from water molecule
Answer:

The answer is the option (iii) accepts $O{H}^{-}$from water releasing proton.
Boron has a small atomic size and only 6 electrons in the valence shell, $B(OH{)}_3$ releases a proton, when it accepts electrons from $O{H}^{-}$ ion of$H_{2}O$.

Question:7

Catenation i.e., linking of similar atoms depends on size and electronic configuration of atoms. The tendency of catenation in Group 14 elements follows the order:
$(i)C>Si>Ge>Sn$
$(ii)C>>Si>Ge\approx Sn$
$(iii)Si>C>Sn>Ge$
$(iv)Ge>Sn>Si>C$

Answer:

The answer is the option $(ii)C>>Si>Ge\approx Sn$
The atomic size decreases when there is movement along down the group and the bond energy, as well as the catenation property, also decreases. Therefore carbon shows maximum catenation in group 14.

Question:8

Silicon has a strong tendency to form polymers like silicones. The chain length of silicone polymer can be controlled by adding
$(i)MeSiC{l}_3$
$(ii)M{e}_{2}SiC{l}_2$
$(iii)M{e}_{3}SiCl$
$(iv)M{e}_{4}Si$
Answer:

The answer is the option $(iii)M{e}_{3}SiCl$
By adding $(C{H}_3{)}_{3}SiCl$ the length of the chain can be controlled, and the ends are blocked.

Question:9

Ionisation enthalpy $(\mathrm{\Delta }i{H}_{1}kJmo{l}^{-1})$ for the elements of Group 13 follows the order.
$(i)B>Al>Ga>In>Tl$
$(ii)B<Al<Ga<In<Tl$
$(iii)B<Al>Ga<In>Tl$
$(iv)B>Al<Ga>In<Tl$
Answer:

The answer is the option l$(iv)B>Al<Ga>In<Tl$
The ionization energy decreases on moving down a group, and the atomic size increases. Till Ga, the ionization energy increases slightly but soon starts decreasing due to the shielding effect of d-electrons and increases once the shielding effect is less. Thus the nuclear energy increase when the ionization energy rises.

Question:10

In the structure of diborane
(i) All hydrogen atoms lie in one plane and boron atoms lie in a plane perpendicular to this plane.
(ii) 2 boron atoms and 4 terminal hydrogen atoms lie in the same plane and 2 bridging hydrogen atoms lie in the perpendicular plane.
(iii) 4 bridging hydrogen atoms and boron atoms lie in one plane and two terminal hydrogen atoms lie in a plane perpendicular to this plane.
(iv) All the atoms are in the same plane.

Answer:

The answer is the option (ii) 2 boron atoms and 4 terminal hydrogen atoms lie in the same plane and 2 bridging hydrogen atoms lie in the perpendicular plane.

Question:11

A compound X, of boron reacts with $N{H}_3$ on heating to give another compound Y which is called inorganic benzene. The compound X can be prepared by treating $B{F}_3$ with Lithium aluminium hydride. The compounds X and Y are represented by the formulas.
$(i)B_2{H}_6,B_3{N}_3{H}_6$
$(ii)B_2{O}_3,B_3{N}_3{H}_6$
$(iii)B{F}_3,B_3{N}_3{H}_6$
$(iv)B_3{N}_3{H}_6,B_2{H}_6$
Answer:

The answer is the option $(i)B_2{H}_6,B_3{N}_3{H}_6$

Question:12

Quartz is extensively used as a piezoelectric material, it contains ___________.
(i) Pb
(ii) Si
(iii) Ti
(iv) Sn

Answer:

The answer is the option (ii) Si because silica is the element that crystallizes to form Quartz.

Question:13

The most commonly used reducing agent is
$(i)AlC{l}_3$
$(ii)PbC{l}_2$
$(iii)SnC{l}_4$
$(iv)SnC{l}_2$

Answer:

The answer is the option $(iv)SnC{l}_2$
Sn has the oxidation state of +4 and is more stable than the other oxidation states. Thus it is easily oxidized to $S{n}^{4+}$ and becomes a reducing agent.
$SnC{l}_2+2Cl\to SnC{l}_4+2e^{-}$

Question:14

Dry ice is
(i) Solid $N{H}_3$
(ii) Solid $S{O}_2$
(iii) Solid $C{O}_2$
(iv) Solid $N_2$

Answer:

The answer is the option (iii) Solid $C{O}_2$
Dry ice is solid $C{O}_2$.

Question:15

Cement, the important building material is a mixture of oxides of several elements. Besides calcium, iron and sulphur, oxides of elements of which of the group (s) are present in the mixture?
(i) group 2
(ii) groups 2, 13 and 14
(iii) groups 2 and 13
(iv) groups 2 and 14
Answer:

The answer is the option (ii) Groups 2,13 and 14
Elements of group 2- calcium and group 13- aluminium and group 14- silicon, constitute Cement.

NCERT Exemplar Class 11 Chemistry Solutions Chapter The p-block Elements: MCQ (Type 2)

Question:16

The reason for the small radius of Ga compared to Al is _______.
(i) poor screening effect of d and f orbitals
(ii) increase in nuclear charge
(iii) presence of higher orbitals
(iv) higher atomic number
Answer:

The answer is the option (i) poor screening effect of d and f orbitals and (ii) increase in nuclear charge.
The poor screening effect of the electrons and additional 10d electrons increase charge in Ga, and thus the radius of Ga is less than Al.

Question:17

The linear shape of $C{O}_2$ is due to _________.
(i) $s{p}^3$hybridisation of carbon
(ii) sp hybridisation of carbon
(iii) $p\pi -p\pi$ bonding between carbon and oxygen
(iv) $s{p}^2$ hybridisation of carbon
Answer:

The answer is the option (ii) sp hybridization of carbon and (iii) $p\pi -p\pi$ bonding between carbon and oxygen
$C{O}_2$ has a structure which is linear, and the bonding of $p\pi -p\pi$structure leads is the reason.

Question:18

$M{e}_{3}SiCl$ is used during polymerisation of organic silicones because
(i) the chain length of organic silicone polymers can be controlled by adding $M{e}_{3}SiCl$
(ii) $M{e}_{3}SiCl$ blocks the end terminal of a silicone polymer
(iii) $M{e}_{3}SiCl$ improves the quality and yield of the polymer
(iv) $M{e}_{3}SiCl$ acts as a catalyst during polymerisation
Answer:

The answer is the option (i) The chain length of organic silicone polymers can be controlled by adding Me₃ and (ii) $M{e}_{3}SiCl$ blocks the end terminal of a silicone polymer.
The length of the chain can be in control by increasing the amount of $M{e}_{3}SiCl$, which helps in blocking of the polymer ends.

Question:19

Which of the following statements are correct?
(i) Fullerenes have dangling bonds
(ii) Fullerenes are cage-like molecules
(iii) Graphite is a thermodynamically most stable allotrope of carbon
(iv) Graphite is slippery and hard and therefore used as a dry lubricant in Machines

Answer:

The answer is the option (ii)Fullerenes are cage-like molecules and (iii) Graphite is a thermodynamically most stable allotrope of carbon.
The cage-like molecules are known as Fullerenes and are known to be the most stable carbon allotropes.

Question:20

Which of the following statements are correct. Answer based on the given figure,
(i) The two bridged hydrogen atoms and the two boron atoms lie in one plane;
(ii) Out of six B–H bonds two bonds can be described in terms of 3 centre 2-electron bonds.
(iii) Out of six B-H bonds, four B-H bonds can be described in terms of 3 centres 2 electron bonds;
(iv) The four-terminal B-H bonds are two centre-two electron regular bonds.

Answer:

The answer is the option: -
(i) The two bridged hydrogen atoms and the two boron atoms lie in one plane.
(ii) Out of six B – H, bonds two bonds can be described in terms of 3 centre 2-electron bonds.
(iv) The four-terminal B – H bonds are two centre-two electron regular bonds.
The two boron atoms have a $s{p}^3$-hybrid state. Only three atoms have one electron each while the fourth is empty. They lie in one plane and two bridging hydrogen atoms.

Question:21

Identify the correct resonance structures of carbon dioxide from the ones given below :
(i) O – C ≡ O
(ii) O = C = O
(iii) ^–O ≡ C – O⁺
(iv) ^–O – C ≡ O⁺

Answer:

The answer is the option (ii) O = C = O and (iv) ^–O – C ≡ O⁺

NCERT Exemplar Class 11 Chemistry Solutions Chapter The p-block Elements: Short Answer Type

Question:22

Draw the structures of $BC{l}_3.N{H}_3$ and $AlC{l}_3$ (dimer)

Answer:

$BC{l}_3$ is the covalent compound which has 6 electrons in the valence shell. Thus it is an electron acceptor and needs 2 electrons to become an octet. It is a Lewis acid and can react with $N{H}_3$ as it donates electrons easily and is a Lewis Base. They combine as shown

In $AlC{l}_3$ , Al has 6 electrons in the valance shell and functions similarly as B and thus combines with Chlorine to form an octet.

Question:23

Explain the nature of boric acid as a Lewis acid in water.
Answer:

Boric acid is a weak monobasic acid that acts as a Lewis acid as it accepts electrons from a hydroxyl ion. Boric acid accepts $O{H}^{-}$ and then the formation of hydroxyl ion.
$B(OH{)}_3+2H_{2}O\to [B(OH{)}_4{]}^{-}+H_3{O}^{+}$

Question:24

Draw the structure of boric acid showing hydrogen bonding. Which species is present in water? What is the hybridisation of boron in this species?

Answer:

$B(OH{)}_3+2H_{2}O\to [B(OH{)}_4{]}^{-}+H_3{O}^{+}$
The$H_{3}B{O}_4$ forms hexagonal rings through hydrogen bonds and is a weak acid. The hybridization is $s{p}^3$. Acting as an electron-deficient, it is a Lewis acid.

Question:25

Explain why the following compounds behave as Lewis acids?
$(i)BC{l}_3$
$(ii)AlC{l}_3$
Answer:

$(i)BC{l}_3$ – It is an electron-deficient compound due to the presence of 6 electrons in its outermost orbital along with a vacant p orbital. Therefore, it behaves as Lewis acid and accepts a lone pair of electrons.
$(ii)AlC{l}_3$ – It forms a covalent bond with chlorine through formation of three single bonds of chlorine. There are three electrons in the valence shell of Aluminium and thus it acts as an electron-deficient compound, therefore behaving as a Lewis acid.

Question:26

Give reasons for the following:
(i) $CC{l}_4$ is immiscible in water, whereas $SiC{l}_4$ is easily hydrolyzed.
(ii) Carbon has a strong tendency for catenation compared to silicon.

Answer:

(i) $CC{l}_4$ is a compound that is unable to form H bonds with water as it is a polar compound. It is insoluble, and carbon does not have d orbital and cannot take in any extra electrons from the oxygen molecule. If the central atom cannot take in lone pair of electrons, then it cannot be hydrolyzed. $SiC{l}_4$ can undergo hydrolysis as it can accept a lone pair of electrons.

(ii) As the atomic size increases, the bond dissociation energy decreases, and since carbon has a size much smaller than silicon, then the dissociation energy of carbon is higher. Thus the carbon bonds are stronger and higher tendency for catenation than silicon.

Question:27

Explain the following :
(i) $C{O}_2$ is a gas whereas $Si{O}_2$ is a solid.
(ii) Silicon forms $Si{F}_6^{2-}$ ion whereas corresponding fluoro compound of carbon is not known.

Answer:

(i) In $C{O}_2$­ , sp hybridization occurs, and the carbon atoms overlap to form two sigma bonds while $p\pi -p\pi$ bonding occurs with oxygen. This brings out the linear shape and no dipole energy.
Silicon dioxide is covalent and forms a tetrahedral structure, and each corner is covalent and shares with another tetrahedron.

(ii) The 3d orbitals in silicon are all in the valence shell, and thus the octet expands to give a $s{p}^3{d}^2$ hybridization. But carbon does not have d-orbitals present in the valence shell. It can only acquire $s{p}^3$ hybridisation. Thus, carbon is unable to form $C{F}_6^{2-}$ anion.

Question:28

The +1 oxidation state in group 13 and +2 oxidation state in group 14 becomes more and more stable with increasing atomic number. Explain.

Answer:

The group 13 and 14, on going down a group, the energy to form bonds decreases. This is given the weak shielding of the s-electrons and the interference by the d-electrons. The inert pair effect arises, and thus the s-electrons, especially of groups 13 and 14, don’t indulge in bonding and the oxidation states of group 13 and 14 become stable due to an increase in atomic number.

Question:29

Carbon and silicon both belong to the group 14, but despite the stoichiometric similarity, the dioxides, (i.e., carbon dioxide and silicon dioxide), differ in their structures. Comment.
Answer:

Carbon in group 14 has the ability to form stable p-p bonds with itself and other elements of the first row.
With $C{O}_2$ ­ the link formed between the atoms is a double bond.

The silicon does not form p-p bonds easily and instead link by single covalent bonds due to its large nuclear size.

Question:30

If a trivalent atom replaces a few silicon atoms in a three-dimensional network of silicon dioxide, what would be the type of charge on an overall structure?
Answer:

The Si atoms are in 3-dimensional structure in $SiO{�}_2$, and the tetrahedral atoms are responsible for this. These atoms are replaced by trivalent atoms, and the valence electron is freely available, and thus there is one negative charge in the structure. This makes the entire compound negatively charged.

Question:31

When BCl₃ is treated with water, it hydrolyses and forms $[B[OH{]}_{}4{]}^{-}$ only whereas$AlC{l}_3$ in acidified aqueous solution forms $[Al(H_{2}O{)}_6{]}^{3+}$ ion. Explain what is the hybridisation of boron and aluminium in these species?

Answer:

$BC{l}_3+3H_{2}O\to B(OH{)}_3+3HCl$
$B(OH{)}_3+H_{2}O\to [B(OH{)}_4{]}^{-}+H^{+}$
$B(OH{)}_3$ in order to complete its octet, accepts an electron pair as $(O{H}^{-})$ to give $[B(OH{)}_4{]}^{-}$. Boron here has one 2s orbital and three 2p orbitals. Thus, hybridization of B is $s{p}^3$ in $[B(OH{)}_4{]}^{-}$
$AlC{l}_3+6H_{2}O\to [Al(H_{2}O{)}_6{]}^{3+}+3C{l}^{-}$
The 6 H2O molecules get attached with Al i.e. they donate 6 electron pairs to the 3s,3p and 3d orbital of $A{l}^{3+}$ ion. Therefore, the hybridization of the Al atom in $[Al(H2O{)}_6{]}^{3+}$+ is $s{p}^3{d}^2$

Question:32

Aluminium dissolves in mineral acids and aqueous alkalies and thus shows amphoteric character. A piece of aluminium foil is treated with dilute hydrochloric acid or dilute sodium hydroxide solution in a test tube and on bringing a burning matchstick near the mouth of the test tube, a pop sound indicates the evolution of hydrogen gas. The same activity when performed with concentrated nitric acid, reaction does not proceed. Explain the reason.

Answer:

Al when dissolved in acids and alkalis and thus is amphoteric and thus gives out H2 gas which can be tested as it burns with a pop sound.
$2Al+6HCl\to 2AlC{l}_3+3H_2$
$2Al+2NaOH+2H_{2}O\to 2NaAl{O}_2+3H_2$
When the nitric acid becomes passive, the reaction stops, and the thin layer of aluminium oxide ceases the reaction.
$2Al+6HN{O}_3\to A{l}_2{O}_3+6N{O}_2+3H_{2}O$

Question:33

Explain the following :
(i) Gallium has higher ionisation enthalpy than aluminium.
(ii) Boron does not exist as $B^{3+}$ ion.
(iii) Aluminium forms$[Al{F}_6{]}^{3-}$ion but boron does not form$[B{F}_6{]}^{3-}$ ion.
(iv)$Pb{X}_2$ is more stable than $Pb{X}_4$
(v) $P{b}^{4+}$ acts as an oxidising agent but $S{n}^{2+}$ acts as a reducing agent.
(vi) Electron gain enthalpy of chlorine is more negative as compared to fluorine.
(vii)$Tl(N{O}_3{)}_3$ acts as an oxidising agent.
(viii) Carbon shows catenation property but lead does not.
(ix) $B{F}_3$ does not hydrolyse.
(x) Why does the element silicon, not form a graphite like structure whereas carbon does.
Answer:

(i) The ionization energy of Al is lower than that of Ga and Ga has less d and f orbital electrons and thus has an increases nuclear charge in order to balance the screening effect.
(ii) The atomic size of Boron is small, and the energy of hydrogen is immensely high. Therefore Boron cannot ionize to form B³⁺ but instead forms covalent compounds.
(iii) The vacant orbitals in Al are the d orbitals which can expand, and the coordination number can rise to 6; thus Al undergoes sp³d² hybridization to form [AlF₆]^3- ion, which is octahedral. But only 4 as the coordination number is possible and thus forms [BF₄]^– not [BF₆]^3-.
(iv) The +4 oxidation state of Pb is less stable than its +2 oxidation state given the inert pair effect. PbX₄ has the oxidation state of Pb is +4 and is less stable than PbX₂, which has the oxidation state of Pb is +2.
(v) Pb has a more dominant inert pair effect than Sn. Pb⁴⁺ can gain 2 electrons to become Pb ²⁺ which is more stable and then Sn²⁺ when it loses 2 electrons. Thus Pb is an oxidizing agent, and Sn is a reducing one.
(vi) F has a small atomic size, and the repulsion between the electrons is quite strong, and any extra electron is not accepted and as the Cl atom where the repulsion is weak. Thus the energy given out by Cl through gaining an electron is much lesser and negative as compared to F.
(vii)The +3 oxidation state of Tl is less stable than its +1 oxidation state due to the inert pair, and it is a strong effect. In Tl(NO₃)₃, Tl’s oxidation state is +3; therefore, it can accept two electrons to form TlNO₃. This will make the oxidation state of Tl is +1. Thus it is an oxidizing agent.
(viii)The strength of element-element bond determines the strength and properties of catenation. This also depends on the atomic size and thus if carbon has a small atomic size in comparison to Pb, then the strength between the carbon bonds is much higher than that of Pb-Pb bonds which explains carbon’s higher tendency for catenation.
(ix)$B{F}_3+3H_{2}O\to H_{3}B{O}_3+3HF(x4)$
$H_{3}B{O}_3+4HF\to H^{+}+B{F}_4^{-}+3H_{2}O(x3)$
$4B{F}_3+3H_{2}O\to H_{3}B{O}_3+3[B{F}_4^{-}]+3H^{+}$
$B{F}_3$ does not completely hydrolyze, and it forms boric and fluoroboric acid instead.
(x) Carbon is $s{p}^2$-hybridised to form graphite, and it forms hexagonal rings and links 3 other atoms. This leaves one p orbital which is unhybridized and forms double bonds instead. Thus this forms a layered structure and rings are fused together. Silicon is not formed out of carbon as it does not undergo the same hybridization or form p-p bonds but has three-dimensional network due to $s{p}^3$-hybridisation.

Question:34

Identify the compounds A, X and Z in the following reactions :
$A+2HCl+5H_{2}O\to 2NaCl+X$
$X\underset{370K}{\overset{\mathrm{\Delta }}{\to }}HB{O}_2\underset{>370K}{\overset{\mathrm{\Delta }}{\to }}Z$
Answer:

In the given reaction A is Borax which reacts with HCl in the presence of water to give Orthoboric acid i.e. X.
When the Orthoboric acid is heated, it gives metaboric and on further heating gives Boron trioxide i.e. the compound Z.

Question:35

Complete the following chemical equations :
$Z+3LiAi{H}_4\to X+3LiF+3Al{F}_3$
$X+6H_{2}O\to Y+6H_2$
$3X+3O_2\stackrel{\mathrm{\Delta }}{\to }{B}_2{O}_3+3H_{2}O$
Answer:

$B{F}_3+3LiAi{H}_4\to 2B_2{H}_6+3LiF+3Al{F}_3$
$B_2{H}_6+6H_{2}O\to 2H_{3}B{O}_3+6H_2$
$3B_2{H}_6+3O_2\to B_2{O}_3+3H_{2}O$

NCERT Exemplar Class 11 Chemistry Solutions Chapter The p-block Elements: Matching Type

Question:36

Match the species given in Column I with the properties mentioned in Column II.

Answer:

(i→e);
$B{F}_4^{-}$: Tetrahedral shape
Due to the $s{p}^3$ hybridization, it has regular geometry with no lone pairs but 4 pairs that have bonded.
(ii → c);
$A{l}_{3}C{l}_3$: Lewis acid
The octet of is incomplete and thus is an electron acceptor making it a Lewis acid.
(iii —> d);
SnO: Can be further oxidized
$S{n}^{2+}$ can lose 2 more electrons to form a +4 oxidation state.
(iv →a, b)
$Pb{O}_2$ : Oxidation state of the central atom is +4, Strong oxidizing agent

Question:37

Match the species given in Column I with properties given in Column II.

Answer:

(i → c);

• BH₃ is not stable, has $s{p}^3$ hybridization and shows banana bonds, and forms diborane $B_2{H}_6$ by 3 centres -2 electron bonds

(ii → d)

• Gallium has a different structure as it can exist in the liquid state for high ranges of temperature and the low melting point and high boiling point make it flexible to be used in fluctuating temperature.

(iii → a);

• Borax is used as a flux for soldering metals and coating as it has a high melting point and can sustain heat.

(iv → e);

• Aluminosilicates also contain zeolites and widely used as a catalyst in petrochemical industries.

(v → b)

• Silica crystallizes to form Quartz.

Question:38

Match the species given in Column I with the hybridisation given in Column II.

Answer:

(i →b)

• Boron in $[B(OH{)}_4{]}^{-}$ is $s{p}^3$ and is the central atom.

(ii→ c);

• Aluminium in $[Al(H_{2}O{)}_6{]}^{3+}$ is $s{p}^3{d}^2$ hybridized and has octahedral geometry.

(iii → b);

• Boron in $B_2{H}_6$ is $s{p}^3$, and the 4 hybrids have the same geometric configuration except one.

(iv→ a);

• Carbon in Buckminsterfullerene $s{p}^2$ is hybridized. Each atom forms a sigma bond with 3 others.

(v →b);

• Silicon in $Si{O}_4^{4-}$ is $s{p}^3$ and has tetrahedral geometry.

(vi→ c)

• Germanium in $[GeC{l}_6{]}^{2-}$ is $s{p}^3{d}^2$ and has octahedral geometry.

NCERT Exemplar Class 11 Chemistry Solutions Chapter The p-block Elements: Assertion and Reason Type

Question:39

Assertion (A): If aluminium atoms replace a few silicon atoms in three dimensional network of silicon dioxide, the overall structure acquires a negative charge.
Reason (R) : Aluminium is trivalent while silicon is tetravalent.
(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) Both A and R are not correct
(iv) A is not correct but R is correct.
Answer:

The answer is the option (i) Both A and R are true, and R is the correct explanation of A.
The aluminium atoms are trivalent while silicon’s tetravalency gives negatively charged ion.

Question:40

Assertion (A): Silicons are water-repelling in nature.
Reason (R): Silicons are organosilicon polymers, which have $(-R_{2}SiO-)$ as repeating unit.
(i) A and R both are correct, and R is the correct explanation of A.
(ii) Both A and R are correct, but R is not the correct explanation of A.
(iii) A and R both are not true.
(iv) A is not true but R is true.
Answer:

The answer is the option (ii) Both A and R are true, but R is not the correct explanation of A.
Silicones are hydrophobic and thus, are organosilicon polymers. So they neither react nor absorb water molecules.

NCERT Exemplar Class 11 Chemistry Solutions Chapter The p-block Elements: Long Answer Type

Question:41

Describe the general trends in the following properties of the elements in Groups 13 and 14.
(i) Atomic size
(ii) Ionisation enthalpy
(iii) Metallic character
(iv) Oxidation states
(v) Nature of halides

Answer:

(i) Group 13 has a greater atomic size as when going down a group, the atomic size increases, so an extra shell with electrons is added, and the poor shielding effect is observed.

Group14- These have a covalent radius, and a small radius is observed, and the presence of the shielding effect is high as there are completely filled d and f orbitals.
(ii) The ionization enthalpy is unaffected by the group, and it decreases with increase in size. Group 13 has erratic ionization enthalpy as the nuclear charge is complemented by the screening effect.
Group 14 experiences a high ionization enthalpy, and it subsequently decreases and due to the poor shielding effect.
(iii) All elements in group 13 are non-metallic except Boron, and the shielding effect is responsible for the low metallic character.
Group 14 has Sn and Pb as metals and the metallic character increases on going down the group, so the carbon is classified as a non-metal and all others are metalloids
(iv) Group 13 has prevalent +3 oxidation states due to the lone pairs but slowly change to +1 as the group order persist
Group14 has + 4, and +2 oxidation states and the enthalpies are high, and they are covalent in nature.
(v)Group 13 forms trihalides and some electron acceptors act as Lewis Acids.
Group 14 forms Halides by accepting electrons or sharing, making them covalent compounds.

Question:42

Account for the following observations:
(i) $AlC{l}_3$ is a Lewis acid
(ii) Though fluorine is more electronegative than chlorine yet $B{F}_3$ is a weaker Lewis acid than $BC{l}_3$
(iii) $Pb{O}_2$ is a stronger oxidising agent than $Sn{O}_2$
(iv) The +1 oxidation state of thallium is more stable than its +3 state.
Answer:

(i) To complete the octet, it accepts electrons and thus acts as a Lewis Acid.
(ii) Boron’s vacant p orbital is not utilized and thus forms back bonding which is strong in F. The increasing atomic size decreases the strength of this bond, and thus Boron is completed using this bond
(iii)The inert pair effect is responsible, which increases in a group and Pb experiences this effect strongly and is an oxidizing agent.
(iv)The inert pair effect causes the strengthening of the oxidation state.

Question:43

When aqueous solution of borax is acidified with hydrochloric acid, a white crystalline solid is formed which is soapy to touch. Is this solid acidic or basic in nature? Explain.
Answer:

HCl acidifies the borax to give boric acid, which monobasic and weak. It needs to accept electrons in order to complete its octet and thus acts as a Lewis Acid.
$N{a}_2{B}_4{O}_7+HCl+H_{2}O\to 2NaCl+4H_{3}B{O}_3$
$B(OH{)}_3+H_{2}O\to [B(OH{)}_4{]}^{-}+H^{+}$

Question:44

Three pairs of compounds are given below. Identify that compound in each of the pairs which has group 13 element in more stable oxidation state.
Give reason for your choice. State the nature of bonding also.
$(i)TlC{l}_3,TlCl$
$(ii)AlC{l}_3,AlCl$
$(iii)InC{l}_3,InCl$
Answer:

i)
TlCl is more stable as the inert pair effect strengthens the oxidation state.
ii)
$AlC{l}_3$ has no inert pair effect but instead is a covalent compound and thus accept electrons to become stable, acting as a Lewis acid.
iii)
Due to inert pair effect, Indium exists in both +1 and +3 oxidation states out of which +3 oxidation state is more stable is more stable than +1 oxidation state. In other words, InCl₃ is more stable than InCl.

Question:45

$BC{l}_3$ exists as monomer whereas $AlC{l}_3$ is dimerised through halogen bridging. Give reason. Explain the structure of the dimer of $AlC{l}_3$ also.
Answer:

$BC{l}_3$ has 6 electrons in its valence shell and is in trivalent state. It requires two additional electrons to complete the octet. $BC{l}_3$ act as a Lewis acid and $N{H}_3$ can donate its electron. Therefore, $BC{l}_3$ readily accepts the lone pair of an electron from $N{H}_3$ and forms $BC{l}_3.N{H}_3.$

$BC{l}_3+N{H}_3\to BC{l}_3.N{H}_3$

$AlC{l}_3$ makes a stable molecule by forming a Dimer. Aluminium’s valence shell has 6 electrons and it requires 2 additional electrons for completing the octet. While chlorine has 3 lone pairs of electrons. Therefore, Aluminium takes up 1 lone pair and forms Dimer with chlorine.

Question:46

Boron fluoride exists as $B{F}_3$ but boron hydride doesn’t exist as $B{H}_3$. Give reason. In which form does it exist? Explain its structure.

Answer:

The lone pair of electrons in Boron fluoride form a pπ- pπ bonds and accept electrons to reduce the deficiency of them. Thereby reducing the ability of Boron to accept electrons and not becoming a Lewis Acid but instead increasing stability.


In $B{H}_3$ hydrogen has a lone pair. It therefore, cannot fulfil the deficiency of boron to dimerise to form $B_2{H}_6$ which is in the shape of a banana.

Question:47

(i) What are silicones? State the uses of silicones.
(ii) What are boranes? Give chemical equation for the preparation of diborane.

Answer:

(i) Silicones are organosilicon polymers and have a repeating unit. The alkyl groups form silicones and the methyl chloride in the presence of the copper as a catalyst with silicon form $M{e}_{4}Si$. The straight-chain polymers are formed by condensation and polymerization.
Uses: - They are used in industries as greasing agents and sealants as well as insulators. They are also expensively employed in the cosmetic industry for surgical implants.
(ii) The binary forms of boron and hydrogen form alkanes and the covalent hybrids are called diborane.
$4B{F}_3+3LiAl{H}_4\to 2B_2{H}_6+3LiF+3Al{F}_3$

Question:48

A compound (A) of boron reacts with NMe₃ to give an adduct (B) which on hydrolysis gives a compound (C) and hydrogen gas. Compound (C) is an acid. Identify the compounds A, B and C. Give the reactions involved

Answer:

Compound A reacts with Boron and gives [B], and thus A is a Lewis acid as it accepts electrons. [B] reacts with [C] and liberates hydrogen and thus [A] is$B_2{H}_6$. B is thus $2B{H}_{3}NM{e}_3$ and C is boric acid.
A = $B_2{H}_6$ (DIBORANE)
B = $2B{H}_{3}NM{e}_3$ (ADDUCT)
C = $2B_3{N}_3{H}_6$ (INORGANIC BORAZINE)
$B_2{H}_6+2NM{e}_3\to 2B{H}_{3}NM{e}_3$
$3B_2{H}_6+6N{H}_3\to 3[B{H}_3(N{H}_3{)}_2]+[B{H}_4{]}^{+}\to 2B_3{N}_3{H}_6+12H_2$

Question:49

A non-metallic element of group 13, used in making bullet proof vests is extremely hard solid of black colour. It can exist in many allotropic forms and has unusually high melting point. Its trifluoride acts as Lewis acid towards ammonia. The element exhibits maximum covalency of four. Identify the element and write the reaction of its trifluoride with ammonia. Explain why does the trifluoride act as a Lewis acid.
Answer:

The element being discussed is Boron. It has an extremely high melting point and is black in colour along with being extremely hard.
Reaction of Boron Trifluoride with ammonia: - $B{F}_3+N{H}_3\to B{F}_3\leftarrow N{H}_3$
Monomeric trihalides are deficient in the electrons, thus they act as strong Lewis acid. In order to complete the octet of boron, trifluoride of boron reacts conveniently with Lewis bases like $N{H}_3$. Due to the absence of d electrons in boron, it is not able to exceed the above mentioned covalence.

Question:50

A tetravalent element forms monoxide and dioxide with oxygen. When air is passed over heated element (1273 K), producer gas is obtained. Monoxide of the element is a powerful reducing agent and reduces ferric oxide to iron. Identify the element and write formulas of its monoxide and dioxide. Write chemical equations for the formation of producer gas and reduction of ferric oxide with the monoxide.
Answer:

The element being discussed is Carbon which is tetravalent. It can thus produce carbon monoxide and dioxide on reaction with oxygen.

1. $2C+O_2\to 2CO$ (As a result of the direct oxidation of carbon)

2. $HCOOH\to H_{2}O+CO$ (Dehydration of formic acid with concentrated $H_{2}S{O}_4$ at 373k)

3. $C+H_{2}O\to CO+H_2$ (The mixture of CO and $H_2$ is known as water gas)

4. $2C+O_2+4N_2\to 2CO+4N_2$ (The mixture of CO and$N_2$ is known as the producer gas)

Reduction Of Ferric Oxide With Monoxide

$F{e}_2{O}_3+3CO\to \to 2Fe+3C{O}_2$