MPBSE Solutions For Class 11 Chemistry Some P Block Elements

P-Block Elements Class 11 Notes

Class 11 Chemistry Some P Block Elements Long Questions And Answers

Question 1. Unlike diamond, graphite is a good conductor of electricity —explain.
Answer:

In a diamond crystal, each carbon atom is sp³-hybridized, meaning that all electrons in the valence shell of the carbon atom engage in the formation of sigma bonds with four adjacent carbon atoms.

• Consequently, there are no free electrons remaining in the lattice, rendering the diamond incapable of conducting electricity.
• Conversely, each carbon atom in graphite, being sp²-hybridized, utilizes three of its valence electrons to establish three σ-bonds with three adjacent carbon atoms.
• The fourth electron in the 2pz orbital, oriented perpendicularly to the carbon layer, establishes an n-bond through lateral overlap with the 2pz orbital of any of the three neighboring carbon atoms.
• The electrons of the π-bonds are delocalized throughout the entire crystal via resonance. Graphite exhibits excellent electrical conductivity owing to the existence of mobile electrons.

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Question 2. Diamond is extremely hard but graphite is soft and slippery —explain with reason.
Answer:

In diamond crystal, each sp³-hybridized carbon atom is tetrahedrally bonded to four other carbon atoms via covalent connections.

• A multitude of tetrahedral units interconnected to create a three-dimensional extensive array of molecules with robust bonds extending in all directions. Due to its three-dimensional network of robust covalent connections, diamond exhibits exceptional hardness.
• Conversely, in graphite crystals, each sp²-hybridized carbon atom is bonded to three other carbon atoms, creating a network of planar hexagons.
• These two-dimensional layers are arranged in parallel planes, positioned horizontally above one another at a distance of 3.35 Å.
• Due to the weak van der Waals forces binding these layers, one layer can effortlessly glide over another with minimal pressure applied. This elucidates the reasons behind graphite’s softness and lubricity.

P-Block Elements Class 11 Notes

Question 3. Carbon monoxide possesses both oxidising and reducing properties—why?
Answer:

The oxidation slate of carbon in carbon monoxide is +2. It stands in between its highest oxidation state (+4) and lowest oxidation state (-4). As a result, in chemical reactions, the carbon atom in CO may increase its oxidation number from +2 to +4 or it may decrease its oxidation number from +2 to -4. When the oxidation number increases, CO is oxidised and plays the role of a reducing agent. For example, at high temperatures, CO reduces black CuO to red metallic Cu.

On the other hand, in case of a decrease in the oxidation number of carbon, carbon monoxide undergoes reduction, i.e., it acts as an oxidising agent. For instance, in the following reaction, carbon monoxide oxidises hydrogen-producing water and itself gets reduced to methane

Question 4. How will you convert a mixture of CO and CO₂ completely into

1. CO₂ and
2. CO?

Answer:

When a mixture of CO₂ and CO is passed over strongly heated CuO and kept in a combustion tube, CO₂ remains unchanged while CO gets oxidised to CO₂. As a consequence, only CO₂ comes out of the combustion tube. In this way, the mixture is completely converted into CO₂

On the other hand, when a mixture containing CO and CO₂ is passed over white-hot coke, CO remains unchanged but CO₂ gets reduced to CO. In this way, the mixture is completely converted into CO.

Mpbse Class 11 Chemistry P-Block Solutions

Question 5. Mention 3 similarities like B and Si.
Answer:

Three similarities between boron and silicon are

1. Both boron and silicon react with caustic soda to form H₂ gas.

2B + 6NaOH → + 3H₂

Si + 2NaOH + H₂O →  Na₂SiO₃ + 2H₂

2. Halides of both boron and silicon undergo hydrolysis to form weak acids

BCl₃ + 3H₂O→H₃BO₃+ 3HCl

SiCl₄ + 4H₂O→ H₄SiO₄ + 4HCl

3. Halides of both boron and silicon form complex compounds

BF₃ +HF →  HBF₄; SiF₄ + 2HF → H₂SiF₆

Question 6. When boron trichloride reacts with water, it only B3 forms [B(OH)₄]^– whereas aluminium trichloride forms [AI(H₂O)₄]³⁺ in acidified aqueous solution. State the hybridisation of boron and aluminium in these species and explain your answer
Answer:

Boron trichloride (BCl₃ ) undergoes hydrolysis to form orthoboric acid [B(OH)₃] at first As the atomic size of B is small and its electronegativity is high, B(OH)₃ polarises HO molecule by accepting an OH^– ion thereby forming [B(OH)₄]^– and releasing a proton.

BCl₃ + 3H₂O→B(OH)₃ + 3HCl

B(OH)₃ + H₂O→[B(OH)₄]^–+ H⁺

There are no vacant d -orbitals in B as it lies in the second period and has only one s – and three p – orbitals. So, B can have four pairs of electrons in its valence shell, i.e., its maximum coordination number is 4.

For this reason, B(OH)₃ accepts one OH” ion to form [B(OH)₄]- in which B-atom is sp³ -hybridised. On the other hand, AlCl₃ undergoes hydrolysis in acidic medium to form [Al(H₂O)g]³⁺

AlCl₃ + 6H₂O →[Al(H₂O)₆]₃+ + 3Cl(aq)^–

In an acidic medium, the concentration of OH^– ions is lower than H+ ions. Thus, Al³⁺ ions coordinate with H₂O molecules and not with OH^– ions. Due to the availability of vacant d -d-orbitals in Al³⁺ ions, it can expand its coordination number from 4 to 6. Thus, it can form the complex ion, [Al(H₂O)₆]³⁺ in which hybridisation state of Al is sp³d²

Chemical Properties Of P-Block Elements Class 11

Question 7. Give reasons for the following:

1. Diamond-tipped tools are used for drilling and cutting purposes.
2. Graphite is used as a lubricant.
3. Silicones are water-repelling in nature.
4. CO gets absorbed by ammoniacal cuprous chloride to form a complex but CO₂ does not

Answer:

1. Diamond has the highest thermal conductivity among all known substances. Because of its high thermal conductivity, diamond-tipped tools do not overheat. Diamond is extremely hard as well. For these reasons, diamond-tipped tools are extensively used for drilling and cutting purposes.

2. Graphite has a layered structure in which any two successive layers are held together by weak van der Waals forces of attraction and thus one layer can smoothly slide over the other. For this reason, graphite acts as a lubricant.

3. Silicones are water-repelling in nature because they are surrounded by non-polar alkyl groups.

4. Carbon monoxide acts as a Lewis base due to the presence of a lone pair of electrons on carbon :C=0: and forms a soluble complex with ammoniacal cuprous chloride (CuCl).

CuCl + NH₃ + CO→ [Cu(CO)NH₃]⁺Cl^–

On the other hand, carbon dioxide  does not have a lone pair ofelectrons on carbon and cannot form complexes.

Question 8. What happens when

1. A mixture of sand and sodium carbonate is melted on heating. 0 At 200°C and under high pressure, carbon monoxide is passed through caustic soda solution and the product is heated to 300°C.
2. At high temperatures, metallic calcium is made to react with carbon and the product obtained is treated with water.
3. Potassium ferrocyanide is heated in the presence of concentrated H₂SO₄ and the gas thus obtained is passed over finely divided nickel powder at 50°C.
4. Silicon is heated with methyl chloride at high temperatures in the presence of copper.
5. SiO₂ is treated with HF.

Answer:

1. When a mixture of sand and sodium carbonate is melted by tremendous heat, sodium silicate and carbon dioxide are obtained.

SiO₂ + Na₂ CO₃ →  Na₂SiO₃ + CO₂↑

2. When CO is passed through NaOH solution at 200°C under high pressure, sodium formate is produced.

NaOH + CO→ HCOONa

When sodium formate is heated at 300°C, sodium oxalate and hydrogen gas are formed.

3. Metallic calcium reacts with carbon at high temperatures to form calcium carbide. Calcium carbide on treatment with water produces acetylene and calcium hydroxide

4. When potassium ferrocyanide is heated with concentrated sulphuric acid, potassium sulphate, ammonium sulphate, ferrous sulphate and carbon monoxide are formed. When the resulting CO gas is passed over finely divided nickel at 50°C, nickel tetracarbonyl (a coordinated complex) is formed

Ni + 4CO →(50°C) Ni(CO)₄

5. When silicon is heated with methyl chloride at high temperature in the presence of copper, a mixture of mono-, di- and trimethylchlorosilane along with a small amount of tetramethylsilane is obtained

SiO₂ reacts with HF to form silicon tetrafluoride. The initially formed SiF₄ dissolves in HF to form hydrofluorosilicic acid.

SiO₂ + 4HF→SiF₄ + 2H₂O ; SiF₄ + 2HF → H₂SiF₆

Electronic Configuration Of P-Block Elements

Question 9. Starting from boric acid how can you prepare—

1. Boric anhydride
2. Boron trichloride
3. Boron trifluoride
4. Meta and tetraboric acid
5. Boron hydride
6. Ethyl borate

Answer:

1. Boric anhydride is prepared by heating boric acid at high temperatures

2. Boric acid is first converted to boric anhydride, which on heating with Cl₂ forms boron trichloride.

B₂O₃ + 3C + 3Cl₂→2BCl₃ + 3CO

3. Boric acid, on heating with CaF₂ and cone. H2S04 forms boron trifluoride.

CaF₂ + H₂SO₄→CaSO₄ + H₂F₂

2H₃BO₃ + 3H₂F₂ → 2BF₃ + 6H₂O

4. Orthoboric acid on heating at 100°C and 160°C forms metaboric acid and tetraboric acid respectively.

Metaboric Acid:

Tetraboric Acid:

5. Boric acid is first converted to boric anhydride which on heating with Mg-powder forms magnesium boride. On reacting dilute HCl with magnesium boride, boron hydrides are formed.

Ethyl borate is formed by heating a mixture of boric acid and ethanol in the presence of concentrated H₂SO₄.

3C₂H₅OH + H₃BO₃ → (C₂H₅)₃BO₃ + 3H₂O

Electronic Configuration Of P-Block Elements

Question 10. Consider the compounds, BCI₃ and CCl₄. How will they behave with water? Justify.
Answer:

BCl₃ is an electron-deficient molecule because the core boron atom possesses only six electrons in its valence shell. Consequently, it can take a pair of electrons provided by water and undergo hydrolysis to produce boric acid (H₃BO₃) and HCl.

The central atom of the molecule, designated as atom B, possesses six valence electrons. Shell

Conversely, the carbon atom in CCl₄ possesses 8 electrons in its valence shell and lacks unoccupied d-orbitals to expand its octet. CCl₄ cannot take a pair of electrons from H2O; therefore, it does not undergo hydrolysis.