Important Questions for CBSE Class 11 Chemistry Chapter 11 - The p-Block Elements

Very Short Answer Questions (1 Marks)

1. How many groups are there in p-block?
Ans: In the periodic table, there are six groups of p-block elements, numbered 13 to 18.

2. What is the 'inert pair effect’?

Ans: The inert pair effect is described as the non-participation of two s-electrons in bonding due to the enormous energy required to unpair them.

3. How does metallic and non-metallic character vary in a group?

Ans: Only the p-block of the periodic table contains nonmetals and metals. The non-metallic nature of elements reduces as the group progresses. In fact, the most metallic element in each p-block group is the heaviest.

4. Why do third-period elements expand their covalence above four?

Ans. Due to the presence of D-orbital in the elements belonging to the p-group and third period. These elements can expand their octet and have a covalence greater than four.

5. Why do heavier elements form π -bonds?

Ans: Because the combined effect of size and availability of d-orbitals significantly determines the capacity of their components to create ${{\pi }}$–bonds, the heavier elements of p-block elements from ${{\pi }}$–bonds.

6. Where do metalloids and non - metals exist?

Ans: Both nonmetals and metalloids are only found in the p-block of the periodic table.

7. Give the chemical formula of inorganic benzene.

Ans: ${{\text{B}}_{\text{3}}}{{\text{N}}_{\text{3}}}{{\text{H}}_{\text{6}}}$ -Borazine

8. Give two examples of electron-deficient molecules.

Ans: ${\text{B}}{{\text{F}}_{\text{3}}}{{\text{B}}_{\text{2}}}{{\text{H}}_{\text{6}}}$.

9. Arrange the following halides of boron in the increasing order of acidic character:

${\text{B}}{{\text{F}}_3},{\text{BC}}{{\text{l}}_3},{\text{BB}}{{\text{r}}_3},{\text{B}}{{\text{I}}_3}.$

Ans: \[{\text{B}}{{\text{F}}_3} < {\text{BC}}{{\text{l}}_3} < {\text{BB}}{{\text{r}}_3} < {\text{B}}{{\text{I}}_3}\]

10. Why is boric acid considered as a weak acid?

Ans: Boric acid does not have the ability to release ions on its own. So, ${{\text{H}}^{\text{ + }}}$ ion is not released by boric acid. It takes ${\text{O}}{{\text{H}}^ - }$ ion from water molecule to complete its octet and then releases ${{\text{H}}^{\text{ + }}}$ ion from itself.

11. Why is boron metalloid?

Ans: Because boron has properties that are similar to both metals and nonmetals, it is classified as a metalloid. When Boron interacts with extremely electro-positive metals like ${\text{Na}}$,${\text{K}}$, it functions as a non-metal. When Boon interacts with halogens like fluorine, it becomes a metal (to produce ${\text{B}}{{\text{F}}_{\text{3}}}$). Hence, boron is a metalloid.

12. Why do boron have an unusual high melting point?

Ans: Boron has a very high melting point due to its extremely strong crystalline lattice.

13. Why does ${\text{B}}{{\text{F}}_{\text{3}}}$ act as Lewis acids?

Ans: Boron's valence shell has just six electrons in its halides. As a result, to complete its octet, it has to accept a pair of electrons from other molecules having excess electrons. Therefore, it is classified as Lewis acid because it behaves as an electron acceptor.

14. What is the electronic configuration of Group-14 elements?

Ans: The electronic configuration is ${\text{n}}{{\text{s}}^{\text{2}}}{\text{n}}{{\text{p}}^{\text{2}}}$.

15. Name the metalloid found in Group 14 element?

Ans: Germanium is a metalloid that belongs to group 14 of the periodic table.

16. Which of the following reacts with water and aqueous solution becomes acidic: ${\text{SiC}}{{\text{l}}_{\text{4}}}$ or \[{\text{CC}}{{\text{l}}_4}\]?

Ans: ${\text{SiC}}{{\text{l}}_{\text{4}}}$.

17. Why ${\text{CC}}{{\text{l}}_{\text{4}}}$ behaves as an electron-precise molecule?

Ans: The number of electrons surrounding the core atom in ${\text{CC}}{{\text{l}}_{\text{4}}}$ is eight, making it an electron-precise molecule.

18. Why is lead unaffected by water?

Ans: Water has little effect on lead, owing to the formation of a protective oxide covering.

19. What is the common name of a recently developed allotrope of carbon i.e. ${{\text{C}}_{{\text{60}}}}$

molecules?

Ans: Fullerene.

20. How are fullerenes obtained?

Ans: Fullerenes are created by heating graphite in the presence of inert gases such as helium or argon in an electric arc.

21. Diamond is the hardest substance known. Why?

Ans. Diamond is the hardest substance on the earth because it is extremely difficult to break strong covalent bonds.

22. What is water gas?

Ans: Water gas, often known as synthesis gas, is a mixture of ${\text{CO &  }}{{\text{H}}_{\text{2}}}$ .

23. Silicon dioxide is treated with hydrogen fluoride. Explain?

Ans: When silicon dioxide is treated with hydrogen fluoride; silicon tetrafluoride is formed as a product. Generally, the \[{\text{Si - O}}\] bond is strong and it does not allow the reaction of halogens or acids with silicon dioxide. But, this reaction takes place at high temperature, where \[{\text{HF}}\]attacks silicon and oxygen of silicon is substituted by the high electronegative fluorine forming \[{\text{Si}}{{\text{F}}_4}\] as the main product.

\[{\text{Si}}{{\text{O}}_2} + 4{\text{HF}} \to {\text{Si}}{{\text{F}}_4} + 2{{\text{H}}_2}{\text{O}}\]

24. What are silicones?

Ans: Simple silicones are made up of \[{( - {\text{Si}} - {\text{O}} - )_{\text{n}}}\] chains with alkyl or phenyl groups occupying the remaining bonding positions. They are naturally hydrophobic.

25. What is dry ice?

Ans: Dry ice is made up of solid ${\text{C}}{{\text{O}}_{\text{2}}}$. It is extremely cold at temperature     \[{\text{ - 78}}{\text{.}}{{\text{5}}^{\text{o}}}{\text{C}}\] and used as a refrigerant or cooling agent.

26. What are silicates?

Ans: Silicates have a structural unit\[{\text{SiO}}_4^{4 - }\]  in which a silicon atom is tetrahedrally connected to four oxygen atoms. 

27. Write the resonance structures of carbon dioxide.

Ans: Carbon dioxide shows resonance structures which are as follows:

28. What is silica gel used as?

Ans: Silica gel is a drying agent that also serves as a support for chromatographic materials and catalysts.

Short Answer Questions (2 Marks)

1. Why are the elements of group 13 called p-block elements?

Ans: Because the last electron is present in the p –orbital $\left( {{\text{n}}{{\text{p}}^{\text{1}}}} \right)$. Group 13 elements are referred to as p-block elements. The configurations of the valence shells are as follows:

$\left( {{\text{n}}{{\text{p}}^{\text{1}}}} \right)$${\text{B}}\left( {{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{1}}}} \right){\text{, Al}}\left( {{\text{3}}{{\text{s}}^{\text{2}}}{\text{,3}}{{\text{p}}^{\text{1}}}} \right){\text{, Ga}}\left( {{\text{4\;}}{{\text{s}}^{\text{2}}}{\text{,4}}{{\text{p}}^{\text{1}}}} \right){\text{, In}}\left( {{\text{5}}{{\text{s}}^{\text{2}}}{\text{5}}{{\text{p}}^{\text{1}}}} \right){\text{, Tl}}\left( {{\text{6}}{{\text{s}}^{\text{2}}}{\text{6}}{{\text{p}}^{\text{1}}}} \right)$

2. The elements ${\text{B, Al, Ga, In \&  Tl}}$ are placed in the same group of the periodic table. Give a reason.

Ans: Because they have the same amount of electrons in their valence shells, the elements, ${\text{B, Al, Ga, In \&  Tl}}$ are classified in the same group of the periodic table. 

3. Aluminum forms ${\left[ {{\text{Al}}{{\text{F}}_6}} \right]^{3 - }}$ whereas ${\left[ {{\text{B}}{{\text{F}}_6}} \right]^{3 - }}$ is not formed why?

Ans: Aluminium can enlarge its octet to create bonds with six fluoride ions due to the presence of unoccupied d-orbitals, whereas boron does not contain d- orbital due to which it is not able to expand its octet.

4. The atomic radius of ${\text{Ca}}$ is less than that of ${\text{Al}}$. Why?

Ans: This is due to differences in the electrical configuration's inner core. Because of the increased nuclear charge in gallium, the presence of additional electrons only provides a weak screening effect for the outside electrons. 

5. \[{\text{C}}\] and ${\text{S}}$ are always tetravalent but \[{\text{Ge, Sn &  Pb}}\] show divalency. Why?

Ans: As we continue down the list of p-block components, the inert pair becomes more noticeable. The inert pair effect causes divalency in \[{\text{Ge, Sn &  Pb}}\] .

6. Some halides of group 14 elements form complexes of the type \[{\left[ {{\text{M}}{{\text{x}}_6}} \right]^{2 - }}.\] Give a reason.

Ans: Halides like \[{\left[ {{\text{Si}}{{\text{F}}_6}} \right]^{2 - }}\] and ${\left[ {{\text{SnC}}{{\text{l}}_6}} \right]^{2 - }}$ are formed by elements that possess empty d-orbitals. Because of the availability of unoccupied d-orbitals, the central atom can increase its coordination number from 4 to 6.

7. \[{\left[ {{\text{Si}}{{\text{F}}_6}} \right]^{2 - }}\] is lead known whereas \[{\left[ {{\text{SiC}}{{\text{l}}_6}} \right]^{2 - }}\] not. Give a reason.

Ans: This is possibly due to:

(i) The size restriction, six big chloride ions cannot fit around the ${\text{S}}{{\text{i}}^{{\text{4 + }}}}$ .

(ii) The interaction between the lone pair of chloride ions and the ${\text{S}}{{\text{i}}^{{\text{4 + }}}}$ is weak.

 8.  ${\text{Pb}}{{\text{I}}_{\text{4}}}$ does not exist. Why?

Ans:  The bonding made between${\text{Pb  -  I}}$ during the reaction does not release enough energy to unpair the electrons in the \[6\;{{\text{s}}^2}\] orbital and excite one of them to a higher orbital, resulting in four unpaired electrons surrounding the lead atom. Hence,  ${\text{Pb}}{{\text{I}}_{\text{4}}}$ does not exist.

9. Why is carbon different from other members of the group?

Ans: Carbon is distinguished from the rest of its group elements by its smaller size, higher electronegativity, higher ionisation enthalpy, and absence of d-orbitals.

10. Why does the covalency of carbon not expand beyond four?

Ans: Because only the s and p orbitals are available for bonding in carbon, it can only hold four pairs of electrons. Due to the absence of d-orbitals in carbon, this limits its maximum covalence to four, whereas other members can grow their covalence.

11. Why does carbon show different allotropic forms?

Ans: Allotropes are different structural forms of the same element that show variety in their physical and chemical characteristics. For example, graphite and diamond are two allotropes of carbon present in nature. Carbon can hold on several allotropic forms due to its catenation and ${{p\pi }} - {{p\pi }}$ bond-forming properties. 

12. Silicon has no allotropic form analogous to graphite. Why?

Ans: With the lower size of carbon atoms in graphite structure, they easily form ${{p\pi }} - {{p\pi }}$ bonds which are absent in the case of silicon. Due to the enormity of size and little or no ${\text{p\pi }} - {\text{p\pi }}$ bonding capability in silicon, Silicon has no graphite structure.

13. Why does graphite conduct electricity?

Ans: \[{\text{s}}{{\text{p}}^{\text{2}}}\] hybridization occurs when graphite forms a hexagonal ring. The electrons are scattered throughout the sheet. Because electrons are mobile, graphite transmits electricity across the sheet.

14. Graphite is used as a lubricant. Give reason.

Ans: Graphite has \[{\text{s}}{{\text{p}}^{\text{2}}}\] hybridized carbon with a layer structure, and the two neighbouring layers can readily roll over one other due to the large gap and weak interlayer interactions. Hence, graphite is used as a lubricant.

15. How are silicones manufactured?

Ans: Chlorosilanes are hydrolyzed and eliminate hydrochloric acid (the unwanted byproduct) to produce cross-linked silicones –

where ${\text{R}}$ represents the alkyl group; either methyl or phenyl group.

16. Why does ${\text{C}}{{\text{O}}_{\text{2}}}$ have a linear shape with no dipole moment

Ans: The carbon atom in a ${\text{C}}{{\text{O}}_{\text{2}}}$ molecule undergoes \[{\text{sp}}\] hybridization. Two ${\text{sp}}$ hybridised orbitals of carbon atoms overlap with two p-orbitals of oxygen atoms to form two sigma bonds, while the rest two electrons of carbon atoms bond with oxygen atom in ${{p\pi }} - {{p\pi }}$ bonding. This gives it a linear shape with zero resultant dipole moment [since both c-o bonds are the same length \[\left( {115{\text{ pm}}} \right)\] ].

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Short Answer Questions (3 Marks)

1. Why does carbon not form ionic compounds?

Ans: The electronic configuration of carbon is ${\text{1}}{{\text{s}}^{\text{2}}}\,{\text{2}}{{\text{s}}^{\text{2}}}\,{\text{2p}}{{\text{x}}^{\text{1}}}\,{\text{2p}}{{\text{y}}^{\text{1}}}$, from the electronic configuration, we can infer that four valence electrons are present in the carbon atom. It should either lose four electrons or receive four electrons to supply an associate degree of ionic compound. As it needs loads of energy to do so. Ionic compounds don't seem to be shaped by carbon. It builds valency compounds by sharing electrons and finishing its octet.

2. Why do the heavier elements do not form ${{p\pi }} - {{p\pi }}$ multiple bonds as carbon does?

Ans: Carbon is that the sole component which will create ${{p\pi }} - {{p\pi }}$ multiple bonds with itself and with different atoms because of its tiny size and high electronegative, however, heavier elements cannot form ${{p\pi }} - {{p\pi }}$ bonds as a result of their atomic orbitals area unit is too large and diffuse to permit for effective overlapping.

3. Why is ${\text{CO}}$ considered poisonous?

Ans: ${\text{CO}}$ has the ability to form a complex with haemoglobin that is approximately 300 times more stable than the oxygen-haemoglobin complex. This inhibits haemoglobin in red blood cells from transporting oxygen throughout the body, ultimately leading to death. Hence, ${\text{CO}}$ is considered poisonous for the human body.

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