Q: How can you explain higher stability of BCl3 as compared to TlCl3?

B shows +3 oxidation state and can form stable BCl3. Thallium shows +1 and +3 oxidation state. Due to inert pair effect, +1 oxidation state is more stable than +3 oxidation state. Hence, BCl3 is stable but TlCl3 not.

Q: Why does boron triflouride behave as a Lewis acid ?

B atom in BF3 has 6 valence electrons. To complete octet, it requires 2 electrons. It readily accepts an electron pair from Lewis bases such as fluoride ion, ethoxide ion and behaves as Lewis acid.BF3+F−→BF4−

Q: Consider the compounds BCl3 and CCl4. How will they behave with water? Justify.

In BCl3, B atom is electron deficient and accepts an electron pair from water. It is hydrolyzed to boric acid.BCl3+3H2O→H3BO3+3HClOn the other hand, CCl4 is an electron precise molecule. C has completed its octet and cannot expand its valency shell due to absence of d orbitals. Hence, it can neither donate nor accept electrons. Hence, CCl4 is not hydrolyzed.

Q: Describe the shapes of BF3 and BH4−. Assign the hybridisation of boron in these species.

The central B atom in BF3 undergoes sp2 hybridization which results in trigonal planar geometry with three bond pairs and 0 lone pairs of electrons.The central B atom in BH4− undergoes sp3 hybridization which results in tetrahedral geometry with 4 bond pairs and 0 lone pairs of electrons.

Question 1

What is the state of hybridisation of carbon in (a) CO3−2​ (b) diamond (c) graphite?

Accepted Answer

(a) In carbonate ion, the carbon atom is sp2 hybridized and forms pi bond.(b) In diamond, carbon atom is sp3 hybridized and has tetrahedral geometry.(c) In graphite, carbon atom is sp2 hybridized.

Question 2

How can you explain higher stability of BCl3​ as compared to TlCl3​?

Accepted Answer

B shows +3 oxidation state and can form stable BCl3​. Thallium shows +1 and +3 oxidation state. Due to inert pair effect, +1 oxidation state is more stable than +3 oxidation state. Hence,  BCl3​ is stable but TlCl3​ not.

Question 3

What happens when,(a) borax is heated strongly,(b) boric acid is added to water,(c) aluium is treated with dilute NaOH,(d) BF3​ is reacted with ammonia?

Accepted Answer

(a) When borax is heated strongly, it loses water and swells into the white mass, which on further heating melts to form a transparent glassy solid called borax glass and borax bead.Na2​B4​O7​.10H2​Oheat​Na2​B4​O7​+10H2​ONa2​B4​O7​heat​sodium meta borate2NaBO2​+B2​O3​​(b) When boric acid is added to water, it accepts electrons from –OH ion. Boric acid is sparingly soluble in cold water however fairly soluble in hot water. B(OH)3​+2H2​O→[B(OH)4​]−+H3​O+(c) Al reacts with dilute NaOH to form sodium tetrahydroxoaluminate(III). Hydrogen gas is liberated in the process. 2Al+2NaOH+6H2​O→Sodium tetrahydroxoaluminate(III)2Na+[Al(OH)4​]−​+3H2​(d) BF3​ (a Lewis acid) reacts with NH3​ (a Lewis base) to form an adduct. This results in a complete octet around B in BF3​.F3​B+:NH3​→F3​B←:NH3​

Question 4

A certain salt X, gives the following results.(i) Its aqueous solution is alkaline to litmus.(ii) It swells up to a glassy material Y on strong heating.(iii) When conc. H2​SO4​ is added to a hot solution of X, white crystal of an acid Z separates out.Write equations for all the above reactions and identify X,Y and Z.

Accepted Answer

The compounds X, Y and Z are borax, sodium metaborate + boric anhydride and boric acid respectively.When borax is heated, it first swells and then forms a transparent glass like bead of sodium meta borate and boric anhydride.BoraxNa2​B4​O7​​heat​sodium metaborate2NaBO2​​+Boric anhydrideB2​O3​​+10H2​OAqueous solution of borax is alkaline due to formation of strong base NaOH. Hence, it turns red litmus blue.Na2​B4​O7​+7H2​O→4H3​BO3​+2NaOHBorax reacts with sulphuric acid to form boric acid and sodium sulphate. BoraxNa2​B4​O7​​ +H2​SO4​+5H2​O→Boric acid4H3​BO3​​ +Sodium sulphateNa2​SO4​​

Question 5

Why does boron triflouride behave as a Lewis acid ?

Accepted Answer

B atom in  BF3​ has 6 valence electrons. To complete octet, it requires 2 electrons. It readily accepts an electron pair from Lewis bases such as fluoride ion, ethoxide ion and behaves as Lewis acid.BF3​+F−→BF4−​

Question 6

Consider the compounds BCl3​ and CCl4​. How will they behave with water? Justify.

Accepted Answer

In  BCl3​, B atom is electron deficient and accepts an electron pair from water. It is hydrolyzed to boric acid.BCl3​+3H2​O→H3​BO3​+3HClOn the other hand, CCl4​ is an electron precise molecule. C has completed its octet and cannot expand its valency shell due to absence of d orbitals. Hence, it can neither donate nor accept electrons. Hence,  CCl4​ is not hydrolyzed.

Question 7

Is boric acid a protic acid? Explain.

Accepted Answer

Boric acid does not ionize in water to give proton. Hence, it is not a 
protic acid. It is a Lewis acid as it accepts electrons from hydroxide ions.B(OH)3​(aq)+2H2​O→[B(OH4​)]−(aq)+H3​O+(aq)therefore, boric acid is not protic acid.

Question 8

Describe the shapes of BF3​ and BH4−​. Assign the hybridisation of boron in these species.

Accepted Answer

The central B atom in BF3​ undergoes sp2 hybridization which results in trigonal planar geometry with three bond pairs and 0 lone pairs of electrons.The central B atom in BH4−​ undergoes sp3 hybridization which results in tetrahedral geometry with 4 bond pairs and 0 lone pairs of electrons.

Question 9

Write reactions to justify amphoteric nature of aluminium.

Accepted Answer

Aluminium is amphoteric in nature as it reacts with acids as well as alkalies.  2Al+6HCl→2Al3++6Cl−+3H2​2Al+2NaOH+6H2​O→2Na[Al(OH)4​]−+3H2​

Question 10

What are electron deficient compounds ? Are BCl3​ and SiCl4​ electron deficient species? Explain.

Accepted Answer

In electron deficient compounds, the central atom either does not have eight electrons in the valence shell or it has eight electrons but can expand its valence beyond 4 due to presence of vacant d orbitals.BCl3​ has 6 valence electrons on B atom and accepts pair of electrons from ammonia to complete its octet. Hence, it is an electron deficient compound. In SiCl4​, the central Si atom has 8 electrons but it can expand its valency beyond 4 due to presence of vacant d orbitals. Hence, it is also an electron deficient compound.

Question 11

Write the resonance structures of CO3−2​ and HCO3​−.

Accepted Answer

The resonance structures of carbonate ion and bicarbonate ion along with resonance hybrid are shown below.

Question 12

Explain the difference in properties of diamond and graphite on the basis of their structures.

Accepted Answer

The difference between diamond and carbon is given below. DiamondGraphite  It has crystalline natureIt has layered structure  Each C is sp3 hybridzed and forms 4 covalent bonds with neighboring C atoms. Each C atom is  sp2  hybridized and forms 3 sigma bonds with 3 other C atoms. Fourth electron forms pi bond. The geometry is tetrahedral.The geometry is planar  C-C bond length is 154 pmC-C bond length is 141.5 pm  It has rigid covalent boning which is difficult to break.It is soft. Its layers can be separated easily.  It is an electrical insulator.It is good conductor of electricity.

Question 13

Rationalise the given statements and give chemical reactions.(a) Lead(II) chloride reacts with Cl2​ to give PbCl4​.(b) Lead(IV) chloride is highly unstable towards heat.(c) Lead is known not to form an iodide, PbI4​.

Accepted Answer

(a) Due to inert pair effect, Pb(II) is more stable than Pb(IV). Hence, there is no reaction between PbCl2​ and  Cl2​.(b) Due to inert pair effect, Pb(II) is more stable than Pb(IV). Hence,  PbCl4​ is thermally unstable and on heating decomposes to form  PbCl2​.PbCl4​Δ​PbCl2​+Cl2​(c)  PbI4​ does not exist due to strong oxidizing power of Pb(IV) and strong reducing power of I−.

Question 14

Suggest reasons why the BF bond lengths in BF3​ (130 pm) and BF4​− (143 pm) differ.

Accepted Answer

In BF3​ molecule, B−F bond has partial double bond due to presence of pπ−pπ backbonding between B and F, which removes the electron deficienty of the molecule.Due to this, B−F bond length is reduced to 130 pm. When BF3​ changes to BF4−​, the hybridization changes from sp2 to  sp3.The double bond character changes to single bond and the B−F bond length becomes 143 pm.

Question 15

If BCl bond has a dipole moment, explain why BCl3​ molecule has zero dipole moment.

Accepted Answer

B-Cl bond is polar due to electronegativity difference between B and Cl.In BCl3​, the central B atom undergoes sp2 hybridization which results in plane triangular geometry. The molecule has symmetry and the individual bond dipoles cancel each other. Hence, the molecule has zero dipole moment.

Question 16

Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluium trifluoride precipitates out of the resulting solution when gaseous BF3​ is bubbled through. Give reasons.

Accepted Answer

Anhydrous HF is a covalent compound and forms strong intramolecular H bonds. It does not give fluoride ions and cannot dissolve  AlF3​ in it. KF is an ionic compound and contains fluoride ions. It combines with AlF3​ to form the soluble complex. AlF3​+3NaF→Na3​[AlF6​]B has small size and high electronegativity. It has mauch higher tendency for complex formation than Al. Hence, when BF3​ is added to above solution, AlF3​ is precipitated.Na3​[AlF6​]+3BF3​→3NaBF4​+AlF3​

Question 17

Suggest a reason as to why CO is poisonous.

Accepted Answer

CO is very toxic as it can form complex with haemoglobin in the red blood cells. Haemoglobin combines with oxygen to form oxyhaemoglobinHaemoglobin+O2​⇌Oxyhaemoglobin Oxyhamoglobin (formed in lungs) is carried to different cells where it gives its oxygen. CO has stronger affinity for Haemoglobin than oxygen.  Due to this, the oxygen carrying capacity of blood is destroyed.. This results in suffocation and finally death.

Question 18

How is excessive content of CO2​ responsible for global warning ?

Accepted Answer

During combustion reactions,  CO2​ is formed. During photosynthesis, plants utilize  CO2​ and release  O2​. Due to excessive combustion, the concentration of  CO2​ increases beyond certain limit and some  CO2​ remains unutilized which absorbs heat irradiated by earth. Some of the heat is dissipated back in the atmosphere and the remaining is redirected back towards earth surface. This increases the temperature and is known as green house effect which results in global warming.

Question 19

Explain structures of diborane and boric acid.

Accepted Answer

(a) Diborane is an electron deficient compound in which only 12 valence electrons are present. 2 B atoms and 4 terminal H atoms are co-planar whereas two bridging hydrogen are present in perpendicular plane. One bridging hydrogen is above the plane and other bridging H atom is below the plane. Terminal B-H bonds are 2c-2e bonds and bridging B-H-B bonds are 3c -2e bonds.(b) Boric acid has layered structure. Planar BO3​ unit is linked to another unit through H atoms. The H atoms form covalent bond with BO3​ unit and hydrogen bond with other BO3​ unit.

Question 20

What happens when (a) Borax is heated strongly, (b) Boric acid is added to water, (c) Aluminium is treated with dilute NaOH, (d) BF3 is reacted with ammonia ?

Accepted Answer

When boric acid is heated, boron trioxide is obtained.H3​BO3​Heat373K​Metaboric acidHBO2​​+H2​O4HBO2​Heat−H2​O​H2​B4​O7​Heat​Boron trioxide2B2​O3​​+H2​O

Chapter 11 : The p-Block Elements