9.Elements of Group 13 ,14 and 15

Qno.1. Choose correct option.

A. Which of the following is not an allotrope of carbon?

a. bucky ball 

b. diamond

c. graphite 

d. emerald

B. ............ is inorganic graphite

a. borax 

b. diborane

c. boron nitride 

d. colemanite

C. Haber’s process is used for preparation of .........

a. HNO₃ 

b. NH₃ 

c. NH₄CONH₂

d. NH₄OH

D. Thallium shows different oxidation state because .........

a. of inert pair effect

b. it is inner transition element

c. it is metal

d. of its high electronegativity

E. Which of the following shows most prominent inert pair effect?

a. C 

b. Si

c. Ge 

d. Pb

Qno.2. Identify the group 14 element that best fits each of the following description. [1 Mark Each]

A. Non metallic element

Ans:- Carbon(C)

B. Form the most acidic oxide

Ans:- Carbon

C. They prefer +2 oxidation state.

Ans:- Tin(Sn) and lead (pb)

D. Forms strong π bonds.

Ans:- Carbon

Qno.3. Give reasons.

A. Ga³⁺ salts are better reducing agent while Tl¹⁺ salts are better oxidising agent.

[2 Marks]

Ans:-

i. Both gallium (Ga) and thallium (Tl) belong to group 13.

ii. Ga is lighter element compared to thallium Tl. Therefore, its +3 oxidation state is stable. Thus, Ga⁺ loses two electrons and get oxidized to Ga³⁺. Hence, Ga⁺ salts are better reducing agent.

iii. Thallium is a heavy element. Therefore, due to the inert pair effect, Tl forms stable compounds in +1 oxidation state.

Thus, Tl³⁺ salts get easily reduced to Tl¹⁺ by accepting two electrons. Hence, Tl³⁺ salts are better oxidizing agent.

B. PbCl₄ is less stable than PbCl₂.[Marks]

Ans:-

i. Pb has electronic configuration [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p².

ii. Due to poor shielding of 6s² electrons by inner d and f electrons, it is difficult to remove 6s² electrons (inert pair).

iii. Thus, due to inert pair effect, +2 oxidation state is more stable than +4 oxidation state. Hence, PbCl₄ is less stable than PbCl₂.

Qno.4. Give the formula of a compound in which carbon exhibit an oxidation state of

[1 Mark Each]

A. +4 

Ans:- CCl₄

B. +2 

Ans:- CO

C. -4

Ans:- CH₄

Qno.5. Explain the trend of the following in group 13 elements: [2 Marks Each]

A. atomic radii 

Ans:-

a. In group 13, on moving down the group, the atomic radii increases from B to Al.

b. However, there is an anomaly observed in the atomic radius of gallium due to the presence of 3d electrons. These inner 3d electrons offer poor shielding effect and thus, valence shell electrons of Ga experience greater nuclear attraction. As a result, atomic radius of gallium is less than that of aluminium.

c. However, the atomic radii again increases from Ga to Tl.

d. Therefore, the atomic radii of the group 13 elements varies in the following order: B < Al > Ga < In < Tl or B < Ga < Al < In < Tl

B. ionization enthalpy

Ans:-

a. Ionization enthalpies show irregular trend in the group 13 elements.

b. As we move down the group, effective nuclear charge decreases due to addition of new shells in the atom of the elements which leads to increased screening effect. Thus, it becomes easier to remove valence shell electrons and hence, ionization enthalpy decreases from B to Al as expected.

c. However, there is a marginal difference in the ionization enthalpy from Al to Tl.

d. The ionization enthalpy increases slightly for Ga but decreases from Ga to In.

In case of Ga, there are 10 d-electrons in its inner electronic configuration which shield the nuclear charge less effectively than the s and p-electrons and therefore, the outer electron is held fairly strongly by the nucleus. As a result, the ionization enthalpy increases slightly.

e. Number of d electrons and extent of screening effect in indium is same as that in gallium. However, the atomic size increases from Ga to In. Due to this, the first ionization enthalpy of In decreases.

f. The last element Tl has 10 d-electrons and 14 f-electrons in its inner electronic configuration which exert still smaller shielding effect on the outer electrons. Consequently, its first ionization enthalpy increases considerably.

C. electron affinity

Ans:-

a. Electron affinity shows irregular trend. It first increases from B to Al and then decreases. The less electron affinity of boron is due to its smaller size. Adding an electron to the 2p orbital in boron leads to a greater repulsion than adding an electron to the larger 3p orbital of aluminium.

b. From Al to Tl, electron affinity decreases. This is because, nuclear charge increases but simultaneously the number of shells in the atoms also increases. As a result, the effective nuclear charge decreases down the group resulting in increased atomic size and thus, it becomes difficult to add an electron to a larger atom. The electron affinity of Ga and In is same.

Note: Electron affinity of group 13 elements:

Qno.6. Answer the following.[1 Mark Each]

A. What is hybridization of Al in AlCl₃ ?

Ans:- Al is sp³ hybridized in AlCl₃.

B. Name a molecule having banana bond.

Ans:- Diborane [B₂H₆].

Qno.7. Draw the structure of the following.[1 Mark Each]

A. Orthophosphoric acid.

Ans:-

Orthophosphoric Acid

Chemical Formula: H₃PO₄

Orthophosphoric acid is a triprotic acid, meaning it can donate three protons (H⁺) in aqueous solution.

B. Resonance structure of nitric acid.

Ans:-

Qno.8. Find out the difference between:

[2 Marks]

A. Diamond and Graphite.

Ans:-

No.	Diamond	Graphite
i	It has a three-dimensional network structure.	It has a two-dimensional hexagonal layered structure.
ii	In diamond, each carbon atom is sp3 hybridized.	In graphite, each carbon atom is sp2 hybridized.
iii	Each carbon atom in diamond is linked to four other carbon atoms.	Each carbon atom in graphite is linked to three other carbon atoms.
iv	Diamond is poor conductor of electricity due to absence of free electrons.	Graphite is good conductor of electricity due to presence of free electrons in its structure.
v	Diamond is the hardest known natural substance.	Graphite is soft and slippery.

B. White phosphorus and Red phosphorus.

Ans:-

No.	White phosphorus	Red phosphorus
i	It consists of discrete tetrahedral P4 molecules.	It consists chains of P4 molecules linked together by covalent bonds.
ii	It is less stable and more reactive.	It is stable and less reactive.
iii	It exhibits chemiluminescence.	It does not exhibit chemiluminescence.
iv	It is poisonous.	It is nonpoisonous.

Qno.9. What are silicones? Where are they used?[3 Marks]

Ans:-

i.  

a. Silicones are organosilicon polymers having R₂SiO (where, R = CH₃ or C₆H₅ group) as a repeating unit held together by  

    |    |       

        —Si — O —Si—  linkage.    

      |               |

b. Since the empirical formula R₂SiO (where R = CH₃ or C₆H₅ group) is similar to that of ketones (R₂CO), these compounds are named as silicones.

ii. Applications: They are used as  

 a. insulating material for electrical appliances.  

 b. water proofing of fabrics.  

 c. sealant.  

 d. high temperature lubricants.  

 e. for mixing in paints and enamels to make them resistant to high temperature, sunlight and chemicals.

Qno.10. Explain the trend in oxidation state of elements from nitrogen to bismuth.[2 Marks]

Ans:-

i. Group 15 elements have five valence electrons (ns² np³). Common oxidation states are –3, +3 and +5. The range of oxidation state is from –3 to +5.

ii. Group 15 elements exhibit positive oxidation states such as +3 and +5. Due to inert pair effect, the stability of +5 oxidation state decreases and +3 oxidation state increases on moving down the group.

iii. Group 15 elements show tendency to donate electron pairs in –3 oxidation state. This tendency is maximum for nitrogen.

iv. The group 15 elements achieve +5 oxidation state only through covalent bonding. 

e.g. NH₃, PH₃, AsH₃, SbH₃, and BiH₃ contain 3 covalent bonds. PCl₅ and PF₅ contain 5 covalent bonds.

Qno.11. Give the test that is used to detect borate radical is qualitative analysis.[2 Marks]

Ans:-

i.Borax when heated with ethyl alcohol and concentrated H₂SO₄ produces volatile vapours of triethyl borate, which burn with green edged flame.

a.

Na₂B₄O₇ + H₂SO₄ + 5H₂O ⟶ (Δ) Na₂SO₄ + Borax        (conc.)            Sodium sulphate

4H₃BO₃  

Orthoboric acid          

b.

H₃BO₃ + 3C₂H₅OH ⟶ B(OC₂H₅)₃ + 3H₂O  

Orthoboric acid+Ethyl alcohol →Triethyl borate

ii.The above reaction is used as a test for the detection and removal of borate radical (BO₃³⁻) in qualitative analysis.

Qno.12. Explain structure and bonding of diborane.[ 3 Marks]

Ans:-

i. Electronic configuration of boron is 1s² 2s² 2p¹. Thus, it has only three valence electrons.

ii. In diborane, each boron atom is sp³ hybridized. Three of such hybrid orbitals are half filled while the fourth sp³ hybrid orbital is vacant.

iii. The two half-filled sp³ hybrid orbitals of each B atom overlap with 1s orbitals of two terminal H atoms and form four B-H covalent bonds. These bonds are also known as two-centred-two-electron (2c-2e) bonds.

iv. When ‘1s’ orbital of each of the remaining two H atoms simultaneously overlap with half-filled hybrid orbital of one B atom and the vacant hybrid orbital of the other B atom, it produces two three-centred-two-electron bonds (3c-2e) or banana bonds.

v. Hydrogen atoms involved in (3c-2e) bonds are the bridging H atoms i.e., H atoms in two B-H-B bonds.

vi. In diborane, two B atoms and four terminal H atoms lie in one plane, while the two bridging H atoms lie symmetrically above and below this plane.

Qno.13. A compound is prepared from the mineral colemanite by boiling it with a solution of sodium carbonate. It is white crystalline solid and used for inorganic qualitative analysis. [1 Mark Each]

a. Name the compound produced.

Ans:- Borax

b. Write the reaction that explains its formation.

Ans:-

Borax is obtained from its mineral colemanite by boiling it with a solution of sodium carbonate.

Ca₂B₆O₁₁ + 2Na₂CO₃ (boil)→ Na₂B₄O₇ + 2NaBO₂ + 2CaCO₃↓

Colemanite + Sodium carbonate → Borax + Sodium metaborate + Calcium carbonate

Qno.14. Ammonia is a good complexing agent. Explain.[2 Marks]

Ans:-

i. The lone pair of electrons on Nitrogen atom facilities complexation of ammonia with transition metal ions.Thus,ammonia is a good complexing agent as it forms complex by donating it's lone pair of electrons.

a.Cu²⁺(aq) + 4NH₃(aq) → [Cu(NH₃)₄]²⁺(aq)  

 (Blue)         (Deep blue)             

b.AgCl(s) + 2NH₃(aq) → [Ag(NH₃)₂]Cl(aq)  

 (White                              (Colourless)   ii. This reaction is used for the detection of metal ions such as Cu²⁺ and Ag⁺.

Qno.15. State true or false. Correct the false statement.[1 Mark Each]

A. The acidic nature of oxides of group 13 increases down the group.

Ans:-

False

The acidic nature of oxides of group 13 decreases down the group.It changes from acidic through amphoteric to basic.

B. The tendency for catenation is much higher for C than for Si.

Ans:- True

Qno.16. Match the pairs from column A and B.[½ Mark Each]

A                              B

BCl₃          Angular molecule

SiO₂           linear covalent molecule

CO₂            Tetrahedral molecule

                   Planar trigonal molecule

Ans:- i-d,ii-c,iii-b

Qno.17. Give the reactions supporting basic nature of ammonia.[2 Marks]

Ans:-

In the following reactions ammonia reacts with acids to form the corresponding ammonium salts which indicates basic nature of ammonia.

i. NH₃ + HCl → NH₄Cl  

Ammonia + Hydrochloric acid →  Ammonium chloride

ii. 2NH₃ + H₂SO₄ → (NH₄)₂SO₄  

Ammonia + Sulphuric acid → Ammonium sulphate

Qno.18. Shravani was performing inorganic qualitative analysis of a salt. To an aqueous solution of that salt, she added silver nitrate. When a white precipitate was formed. On adding ammonium hydroxide to this, she obtained a clear solution. Comment on her observations and write the chemical reactions involved.

Ans:-

i. When silver nitrate (AgNO₃) is added to an aqueous solution of salt sodium chloride (NaCl), a white precipitate of silver chloride (AgCl) is formed.

NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)

Sodium chloride + Silver nitrate   →      Silver chloride +  Sodium nitrate

   (White ppt) 

ii. On adding ammonium hydroxide (NH₄OH) to this, the white precipitate of silver chloride gets dissolved and thus, a clear solution is obtained.

AgCl(s) + 2NH₄OH(aq) →

Ag(NH₃)₂Cl(aq) + 2H₂O

White ppt + Ammonium hydroxide→ 

Diamine silver chloride (Colourless)