Chemistry Ncert Exemplar 2019 Solutions for Class 11 Science Chemistry Chapter 11 - The P Block Elements

Answer:

The melting point of gallium is 30°C and boiling point is 2403°C. Thus, the element exists in liquid state for a wide range of temperature and can be used for measuring high temperature.

Hence, the correct answer is option C.

Answer:

 Lewis acid is defined as a compound which can take electron from a donor compound. AlCl₃ is electron deficient. Al has three electrons in its valence shell so, when it forms a covalent compound  it forms three single bonds with chlorine. It doesn't form an octet by doing so, thus it can take two more electrons by forming a coordinate bond to become an octet and thus it behaves as Lewis acid.

Hence, the correct answer is option A.

Answer:

Boron has electronic configuration: $1s^2 2s^2 2p^1$. It exists in trivalent state and the trichloride on hydrolysis in water form tetrahedral [B(OH)₄]^–species in which the hybridisation of orbitals of central atom, i.e., B is sp³.

Hence, the correct answer is option A.
 

Answer:

While moving down the group acidic strength of oxide decreases and basic strength increases. Hence, B₂O₃ is an acidic oxide.

Hence, the correct answer is option A.

Answer:

The element M in the complex ${\mathrm{MF}}_6^{3-}$ is Boron. Boron (B) belongs to 2nd period and has only 2s and 2p orbitals in the valence shell. So, the maximum covalence possible is 4. Thus, Boron can not form ${\mathrm{MF}}_6^{3-}$.

Hence, the correct answer is option A.

Answer:

Boric acid is a weak acid because it is an electron deficient compound and on dissolving in water, it accepts OH^– ions from water molecule to complete its octet and in turn releases H⁺ ions.

Hence, the correct answer is option C.

Answer:

Catenation is the binding of an element to itself through covalent bonds to form chain or ring molecules.
In Group 14, down the group, the size increases and electronegativity decreases and thereby, tendency to show catenation decreases. The order of catenation is C > > Si > Ge ≈ Sn. Lead does not show catenation.

Hence, the correct answer is option B.

Answer:

The chain length of the polymer can be controlled by adding Me₃SiCl which blocks the ends.



Hence, the correct answer is option C.

Answer:

On moving down the group from B to Tl, a regular decreasing trend in the ionisation energy values is not observed. The order observed is:
B > Al < Ga > In < Tl.
In Ga, there are 10 d electrons in the penultimate shell which screen the nuclear charge less effectively and thus, outer electron is held firmly.  As a result, the ionisation energy of Al is slightly less than Ga. The increase in ionisation energy from In to Tl is due to poor screening effect of 14 f electrons present in the inner shell.

Hence, the correct answer is option D.
 

Answer:

In the structure of diborane, 2 boron atoms and 4 terminal hydrogen atoms lie in the same plane and 2 bridging hydrogen atoms lie above and below this plane, i.e., in the perpendicular plane.


Hence, the correct answer is option B.

Answer:

Answer:

Quartz is a mineral composed of silicon and oxygen atoms and is one of the crystalline forms of silica (SiO₂). It is extensively used as a piezoelectric material.

Hence, the correct answer is option B.

Answer:

The +4 oxidation state of Sn is more stable than +2 oxidation state. Therefore, Sn²⁺ can be easily oxidised to Sn⁴⁺ and hence SnCl₂ acts as a reducing agent.

SnCl₂ + 2Cl⁻ → SnCl₄ + 2e⁻

Hence, the correct answer is option D.

Answer:

Carbon dioxide can be obtained as a solid in the form of dry ice by allowing the liquified CO₂ to expand rapidly. Dry ice is used as a refrigerant for ice-cream and frozen food.

Hence, the correct answer is option C.

Answer:

Cement is a mixture of lime, CaO, an oxide of calcium (present in group 2) with other material such as clay which contains silica, SiO₂, an oxide of silicon (present in group 14) along with the oxides of aluminium (present in group 13), iron (present in group 8) and magnesium (present in group 2).

Hence, the correct answer is option B.

Answer:

As we go from Al to Ga, the additional 10 d-electrons offer poor screening effect for the outer electrons. Due to this, effective nuclear charge in Gallium is increased . Hence, atomic radius of Gallium is less than that of aluminium.

Hence, the correct answers are options A and B.

Answer:

In CO₂ molecule carbon atom undergoes sp hybridisation. Two sp hybridised orbitals of carbon atom overlap with two p orbitals of oxygen atoms to make two sigma bonds while other two electrons of carbon atom are involved in pπ– pπ bonding with oxygen atom. This results in its linear shape.

Hence, the correct answers are options A and C.

Answer:

The chain length of the polymer can be controlled by adding Me₃SiCl which blocks the ends of the silicon polymer. 



Hence, the correct answers are options A and B.

Answer:

Fullerenes have smooth structure without having ‘dangling’ bonds and are cage-like molecules. Graphite is thermodynamically most stable allotrope of carbon. It cleaves easily between the layers and, thus, it is very soft and slippery. Therefore, it is used as a dry lubricant in machines running at high temperature, where oil cannot be used as a lubricant

Hence, the correct answers are options B and C.

Answer:

The four terminal hydrogen atoms and the two boron atoms lie in one plane. Above and below this plane, there are two bridging hydrogen atoms. The four terminal B-H bonds are regular two centre-two electron bonds while the two bridge (B-H-B) bonds are different and can be described in terms of three centre–two electron bonds.

Hence, the correct answers are options are B and D.
 

Answer:

Out of all the given structures, O = C = O and ^–O – C ≡ O⁺  are the correct resonating structures of carbon dioxide.

Hence, the correct answers are options B and D.

Answer:

BCl₃ easily accepts a lone pair of electrons from ammonia to form BCl₃⋅NH₃.


AlCl₃ achieves stability by forming a dimer.

Answer:

Boric acid is a weak monobasic acid and acts as a Lewis acid by accepting electrons from a hydroxyl ion when dissolved in water.

The reaction is represented as:

B(OH)₃ + 2H₂O → [B(OH)₄]^- + H₃O⁺ 

Answer:

Boric acid has layer structure in which planar BO₃ units are joined by hydrogen bonds as shown below.

It is a weak monobasic acid and not a protonic acid but acts as a Lewis acid by accepting electrons from a hydroxyl ion in water. [B(OH)₄]⁻ units are present in water.
Boron has sp³ hybridisation in [B(OH)₄]⁻ unit.

Answer:

In trivalent state, the number of electrons around the central atom in a molecule of compounds BCl₃ and AlCl₃ will be only six 
Such electron deficient molecules have tendency to accept a pair of electron to achieve stable electronic configuration and thus, act as Lewis acids. The tendency to behave as Lewis acid decreases with the increase in the size down the group.

Answer:

(i) CCl₄ is a covalent compound while H₂O is a polar compound. Therefore, it is insoluble in water. Alternatively, CCl₄ is insoluble in water because carbon does not have vacant orbitals to accommodate the electrons donated by oxygen atom of water molecules. As a result, there is no interaction between CCl₄ and water molecules and hence CCl₄ is insoluble in water. On the other hand, SiCl₄ has d-orbitals to accommodate the lone pair of electrons donated by oxygen atom of water molecules. As a result, there is a strong interaction between SiCl₄ and water molecules. Consequently, SiCl₄ undergoes hydrolysis by water to form silicic acid.

(ii) The bond dissociation energy decreases rapidly as the atomic size increases. Since the atomic size of carbon is much smaller (77 pm) as compared to that of silicon (118 pm), carbon-carbon bond dissociation energy is much higher (348 kJ mol^-1) than that of silicon-silicon bond (297 kJ mol^-1). Since C−C bonds are much stronger as compared to Si−Si bonds, carbon has a much higher tendency for catenation than silicon.

Answer:

(i) Because of its small size and good π-overlap with other small atoms, carbon forms strong double bonds with two oxygen atoms to give discrete CO₂ molecules.                                         
Silicon atom, on account of large size, does not have good π -overlap with other atoms. It uses its four valence electrons to form four single bonds directed towards the four sides of a tetrahedron (sp³ - hybridisation). Each oxygen is linked with two silicon atoms, i.e. a giant three dimensional structure comes into existence which is very stable. Thus, CO₂ is a gas and SiO₂ is a solid.

(ii) Silicon has 3d - orbitals in the valence shell and thus expands its octet giving sp³d² hybridisation while d-orbitals are not present in the valence shell of carbon. It can undergo sp³ - hybridisation only. Thus, carbon is unable to form CF₆^2- anion.

Answer:

In group 13 and 14, as we move down the group, the tendency of s-electrons of the valence shell to participate in bond formation decreases. This is due to ineffective shielding of s-electrons of the valence shell by the intervening d- and f-electrons. This is called inert pair effect.
Due to this, s-electrons of the valence shell of group 13 and 14 are unable to participate in bonding. Hence, + 1 and + 2 oxidation states, in group 13 and 14 respectively, become more stable with increasing atomic number.

Answer:

Carbon, the first member of group 14 possesses a pronounced ability to form stable p-p multiple bonds with itself and with other first row elements such as nitrogen and oxygen. In CO₂, both the oxygen atoms are linked with carbon atom by double bonds. However, silicon shows its reluctance in forming p-p multiple bonding due to large atomic size. Thus, in SiO₂ oxygen atoms are linked to silicon atom by single covalent bonds giving three dimensional network.

Answer:

If a trivelant atom replaces a few tetrahedral Si atoms in a three dimensional network structure of SiO₂ , then one valence electron of each Si atom will become free. As a result, each substitution of Si atom by a trivalent atom introduces one unit negative charge into the 3D networl structure of SiO₂. Therefore, an overall negative charge is developed on the SiO₂ structure.

Answer:

BCl₃ hydrolyse to give boric acid i.e. B(OH)₃.
BCl₃ + 3H₂O → B(OH)₃ + 3HCl
B(OH)₃ + H₂O → [B(OH)₄]^– + H₃O⁺
B(OH)₃ due to its incomplete octet accepts an electron pair from OH^– to give [B(OH)₄]^–. Since one 2s and three 2p orbitals of boron are involved in hybridisation, therefore, the hybridisation of B in [B(OH)₄]^– is sp³.
Now, Al has vacant d orbitals and therefore, AlCl₃ on hydrolysis forms octahedral [Al(H₂O)₆]³⁺ ion. In this complex ion, one 2s, three 2p two 3d orbitals of Al are involved in hybridisation and therefore, the hybridisation of Al in [Al(H₂O)₆]³⁺ is sp³d². 
 

Answer:

Al is amphoteric in nature and therefore, reacts with acids and alkalies and evolve H₂ gas which burns with a pop sound.
$$\begin{gathered} 2\mathrm{Al} + 6\mathrm{HCl}\to 2{\mathrm{AlCl}}_3+3{\mathrm{H}}_2 \\ 2\mathrm{Al} +2\mathrm{NaOH} +6{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{Na}}^{+}{\left[\mathrm{Al}{\left(\mathrm{OH}\right)}_4\right]}^{-} + 3{\mathrm{H}}_2 \end{gathered}$$
When Al reacts with conc. HNO₃, it leads to the formation of a protective layer of Al₂O₃ on the surface of the metal and prevents the reaction to take place.

Answer:

(i) The ionization enthalpy value of Ga is higher than Al due to the inability of d- electrons, which have a poor screening effect for the outer electrons to compensate for the increase in nuclear charge.

(ii) Boron has a small size and the sum of the three ionisation enthalpies, i.e.,  Δ_i​H₁​+Δ_i​H_2​+Δ_i​H_3​ is very high. Therefore, boron does not form B³⁺ ion and forces it form covalent compounds.

(iii) Al has vacant 3d-orbitals and can expand its co-ordination number and forms [AlF₆​]³⁻. On the other hand, boron does not have d- orbitals and can not form [BF₆]^3– ion and cannot expand its covalence beyond 4.

(iv) On moving down the group in group 14, +2 oxidation state is more stable than +4 oxidation state for heavier members due to the inert pair effect. Thus, PbX₂ is more stable than PbX₄.

(v) Due to inert pair effect, Pb⁴⁺changes into Pb²⁺ by gaining 2 electrons which is more stable while Sn²⁺ is less stable and gets converted to Sn⁴⁺ by losing 2 electrons. Therefore, Pb⁴⁺ acts as an oxidizing agent while Sn²⁺ acts as a reducing agent.

(vi) The size of fluorine atom is very small and incoming electron feels interelectronic repulsion from the valence shell electrons while the size of chlorine is comparatively bigger and  interelectronic repulsion is very less. Therefore, electron gain enthalpy of F is less negative as compared to Cl.

(vii) Due to the inert pair effect, Tl is more stable in +1 oxidation state than that +3 oxidation state. Therefore, Tl in +3 oxidation state in Tl(NO_3​)₃​ acts as a strong oxidizing agent and itself gets reduced to +1 oxidation state.

(viii) Carbon atoms have the tendency to link with another through covalent bonds to form chains and rings. This property is called catenation. This is because C−C bonds are very strong. Down the group, the size increases, and electronegativity decreases, and thereby, the tendency to show catenation decreases. Therefore, lead does not show catenation.

(ix) BF₃​ does not hydrolyze completely due to strong pπ-pπ backbonding and absence of d orbitals.

(x) In graphite, carbon is sp² hybridized and makes three sigma bonds with three neighbouring carbon atoms. Fourth electron forms a π bond. Carbon has a tendency to form multiple pπ−pπ bonds due to its small size and highest electronegativity in group 14. Silicon, due to its large size and less electronegativity cannot form multiple bonds. Thus, silicon cannot form a graphite-like structure.

Answer:

$$\begin{gathered} \underset{\underset{\mathrm{Borax}}{\left(\mathrm{A}\right)}}{{\mathrm{Na}}_2{\mathrm{B}}_4{\mathrm{O}}_7} + 2\mathrm{HCl} + 5{\mathrm{H}}_2\mathrm{O}\to 2\mathrm{NaCl} + \underset{\underset{\mathrm{Boric} \mathrm{acid}}{\left(\mathrm{X}\right)}}{4{\mathrm{H}}_3{\mathrm{BO}}_3} \\ \underset{\underset{\mathrm{Boric} \mathrm{acid}}{\left(\mathrm{X}\right)}}{{\mathrm{H}}_3{\mathrm{BO}}_3}\stackrel{∆, 370\mathrm{K}}{\to }{\mathrm{HBO}}_2 \underset{>370 \mathrm{K}}{\overset{∆}{\to }}\underset{\underset{\mathrm{Boric} \mathrm{oxide}}{\left(\mathrm{Z}\right)}}{{\mathrm{B}}_2{\mathrm{O}}_3} \end{gathered}$$

Answer:

$$\begin{gathered} 4{\mathrm{BF}}_3+3 {\mathrm{LiAlH}}_4 \stackrel{}{\to } 2{\mathrm{B}}_2{\mathrm{H}}_6+3 \mathrm{LiF}+3{\mathrm{AlF}}_3 \\ \left(\mathrm{X}\right) \\ {\mathrm{B}}_2{\mathrm{H}}_6+6{\mathrm{H}}_2\mathrm{O} \stackrel{}{\to }2{\mathrm{H}}_3{\mathrm{BO}}_3+6{\mathrm{H}}_2 \\ \left(\mathrm{X}\right) \left(\mathrm{Y}\right) \\ {\mathrm{B}}_2{\mathrm{H}}_6+3{\mathrm{O}}_2\stackrel{∆}{\to }{\mathrm{B}}_2{\mathrm{O}}_3+3{\mathrm{H}}_2\mathrm{O} \\ \left(\mathrm{X}\right) \end{gathered}$$

Answer:

The correct match between column I and column II is given as:
(i)→(e), (ii)→(c), (iii)→(d), (iv)→(a),(b)

Answer:

The correct match between column I and column II is given as:
(i)→(c), (ii)→(d), (iii)→(a), (iv)→(e), (v)→(b)

Answer:

The match between column I and column II is given as:
(i)→(b), (ii)→(c), (iii)→(b), (iv)→(a), (v)→(b), (vi)→(c)

Answer:

In aluminosilicates, some of the silicon atoms are replaced by aluminium. As aluminium is trivalent while silicon is tetravalent, therefore, an overall negative charge is acquired on the 3D network structure of silicon dioxide. 

Hence, the correct answer is option (i),

Answer:

Silicones are organosilicon polymers having (–R₂SiO^–) as repeating unit. They are hydrophobic in nature. Silicones neither react nor absorb water molecules and therefore, they are water repelling in nature.

Hence, the correct answer is option (ii).

Answer:

Answer:

(i)  In AlCl₃, the central atom, i.e., Al has only six electrons in its valence shell. Therefore, it is an electron deficient species and acts as a Lewis acid, i.e., electron acceptor.

(ii) In BF₃, due to the vacant 2p-orbital in boron and fully filled 2p orbital of fluorine, back bonding takes place. Since an electron pair from the filled 2p orbital of fluorine is donated to the vacant 2p orbital of boron, it results in the formation of ${\displaystyle p\pi -p\pi }$ bond. But in BCl₃, no effective overlap takes place between the vacant 2p orbital of boron and fully filled 3p orbital of chlorine, thus, the extent of backbonding is very less. Therefore, electron deficiency in B atom of BCl₃ is higher and hence, it acts as a stronger Lewis acid in comparison to BF₃.

(iii) In PbO₂ and SnO₂, both lead and tin are present in +4 oxidation state. But due to stronger inert pair effect in lead, Pb²⁺ ion is more stable than Sn²⁺ ion. In other words, Pb⁴⁺ ion, i.e., PbO₄ is more easily reduced to Pb²⁺ ion than Sn⁴⁺ ion reduced to Sn²⁺ ion. Therefore, PbO₂ acts as a stronger oxidising agent than SnO₂.

(iv) Due to inert pair effect, the stability of lower oxidation state increases down the group. Therefore, +1 oxidation state of Tl is more stable than its +3 state.
 

Answer:

When an aqueous solution of borax is acidified with hydrochloric acid,  boric acid is formed. The reaction involved is as written below:
Na₂B₄O₇ + 2HCl + 5H₂O → 2NaCl + 4H₃BO₃ 
Boric acid is not a protonic acid i.e., it does not give H⁺ ion but accepts a pair of electrons from hydroxyl ion and acts as a Lewis acid. Thus, it is is a weak monobasic acid and is soapy to touch.

Answer:

(i) TlCl is more stable and it is an ionic compound. Inert pair effect is maximum in thallium and therefore, +1 oxidation state is more stable than +3 state.

(ii) AlCl₃ is more stable. Anhydrous state is covalent while aqueous state is ionic in nature. The +3 oxidation state is more stable than +1 oxidation state. Inert pair effect is negligible.

(iii) lnCl₃ is more stable. Anhydrous state is more covalent less ionic in nature. Inert pair effect is considerable but less than in thallium. Therefore, +3 oxidation state is more stable.

Answer:

BCl₃ exists as monomer in spite of the fact that it is an electron deficient compound. It achieves stability by forming pπ−pπ back bonding. Chlorine transfers two electrons from its filled 3p orbital to vacant 2p-orbital of boron. All the three bond lengths are same and the backbonding gives double bond character that is delocalized in the molecule.

AlCl₃ is also an electron deficient compound. Since aluminium has larger size than boron, the back bonding is not possible in AlCl₃. Aluminium metal completes its octet by forming coordinate bond with chlorine atom of the other AlCl₃ molecule. Therefore, coordinate bond forms bridges by chlorine atoms between two Al atoms make a dimer molecule.

Answer:

BF₃ exists as a monomer due to pπ−pπ back bonding. Fluorine having a fully filled 2p orbital transfers two electrons to vacant 2p-orbital of boron. The delocalisation reduces the deficiency of electrons on boron thereby increasing the stability of BF₃ molecule.
Due to absence of lone pair of electrons on H the back bonding does not occur in BH₃. The electron deficiency of boron remains and BH₃ does not exist. To reduce electron deficiency, BH₃ dimerises to form B₂H₆ i.e., diborane. 

The four terminal hydrogen atoms and the two boron atoms lie in one plane. Above and below this plane, there are two bridging hydrogen atoms. The four terminal B-H bonds are regular two centre-two electron bonds while the two bridge (B-H-B) bonds are different and can be described in terms of three centre–two electron bonds.

Answer:

(i) Silicones are a group of organosilicon polymers, which have (R₂SiO) as a repeating unit. They are prepared by the hydrolysis of alkyl or aryl derivatives of SiCl₄, like RSiCl₃,R₂SiCl₂ and R₃SiCl and polymerisation of alkyl or aryl hydroxy derivatives obtained by hydrolysis.
Uses of silicones:
1. They are used as sealant, greases, electrical insulators and for water proofing of fabrics.
2. They are used in surgical and cosmetic plants.

(ii) Boron forms a number of covalent hydrides known as boranes with general formulae B_n H_{n + 4}and B_n H_{n + 6}. B₂H₆ and B₄H₁₀ are the representative compounds of the two series respectively.
Preparation of Diborane (B₂H₆ ):
(a) It is prepared by treating boron trifluoride with LiAlH₄ in diethyl ether
4BF₃+3LiAlH₄→2B₂H₆+3LiF+3AlF₃
(b) On industrial scale it is prepared by the reaction of BF₃ with sodium hydride
2BF₃+6NaH$\stackrel{450\mathrm{K}}{\to }$B₂H₆+6NaF

Answer:

(A) is diborane i.e., B₂H₆ as it reacts with NMe₃ to give an adduct (B).

$$\begin{gathered} {\mathrm{B}}_2{\mathrm{H}}_6 +2{\mathrm{NMe}}_3\to 2{\mathrm{BH}}_3.{\mathrm{NMe}}_3 \\ \left(\mathrm{A}\right) \left(\mathrm{B}\right) \end{gathered}$$

(B) on hydrolysis gives boric acid i.e., H₃BO₃ (C).

$$\begin{gathered} {\mathrm{BH}}_3.{\mathrm{NMe}}_3+3{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_3{\mathrm{BO}}_3 +{\mathrm{NMe}}_3 +3{\mathrm{H}}_2 \\ \left(\mathrm{B}\right) \left(\mathrm{C}\right) \end{gathered}$$
 

Answer:

Non-metallic element of group 13 is boron. It is black solid and very hard in nature and is used to make bullet proof vests. It has a high melting point of about 2453 K. It exists in many allotropic forms. It forms a trifluoride, BF₃ which acts as a Lewis acid as it is an electron deficient compound. Since it is a Lewis acid, it forms an adduct with NH₃ as given below:
BF₃ + NH₃→ H₃N→BF₃
                        Adduct

NH₃ donates an electron pair which is accepted by boron to complete its octet.
Maximum covalency of boron is four as its valence shell contains only four orbitals (one 2s and three 2p) and due to non-availability of d orbitals.

Answer:

The teravalent element is carbon which forms two oxides namely carbon monoxide (CO) and carbon dioxide (CO₂).
When air is passed over heated element at 1273 K, producer gas is obtained. The chemical equation is represented as:
2C + O₂ + N₂→ 2CO + N₂
                           Producer gas
CO is a powerful reducing agent and reduces ferric oxide to iron.
Fe₂O₃ + CO → 2FeO + CO₂
FeO + CO → Fe + CO₂