Solutions-Class-11-Science-Chemistry-Chapter-9-Elements of Group 13, 14 and 15-Maharashtra Board

(d) emerald

(c) boron nitride

(b) NH₃

(a) of inert pair effect

(d) Pb

Carbon

Carbon

Pb and Sn

Carbon.

• Both Ga and Tl are in group 13. Ga in 4th period and Tl in 6th period.
• Both form trivalent ions, Ga³⁺ and Tl³⁺.
• Due to poor shielding effect of inner electrons. Tl has a stable 1+ oxidation state. Hence, Tl in 3+ oxidation state has a tendency to undergo reduction to 1+ stable oxidation state. Thus, Tl³⁺ salts are better oxidising agents.
• Ga has 3+ stable oxidation state. Hence, Ga¹⁺ ions have a tendency to undergo oxidation to stable 3+ oxidation state. Thus, they are better reducing agents.

• Pb has electronic configuration [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p².
• Due to poor shielding of 6s² electrons by inner d and f electrons, it is difficult to remove 6s² electrons (inert pair).
• Thus, due to inert pair effect, the +2 oxidation state is more stable than +4 oxidation state. Hence, PbCl₄ is less stable than PbCl₂.

(A) +4 : CO₂

(B) +2 : CO

(C) −4 : CH₄

• Atomic radius shows irregular trends in the elements of group 13.
• In group 13, on moving down the group, the atomic radii increases from boron to aluminium.
• However, the atomic radius of gallium is unusual due to the presence of 3d electrons, which provide poor shielding, causing valence shell electrons to experience greater nuclear attraction, resulting in a lower radius than that of aluminium.
• However, the atomic radii again increase from gallium to thallium.
• Therefore, the atomic radii of the group 13 elements vary in the following order: B < Al > Ga < In < TI

• Ionization enthalpy shows irregular trends in group 13 elements :
• It decreases from B to Al as expected.
• There is marginal difference in ionization enthalpy from Al to Tl.
• The ionization enthalpy slightly increases for Ga, but decreases from Ga to In due to the presence of 10 d−electrons in Ga's inner electronic configuration, which shield the nuclear charge less effectively than s and p−electrons, resulting in a slightly increased enthalpy.
• The last element Tl, has higher ionization enthalpy (IE1) due to poor shielding effect of 10 d−electrons and 14−f electrons on outer electrons.

• Electron affinity shows an irregular trend. It first increases from B to Al and then decreases.
• Electron affinity of group 13 decreases down the group from Al to Tl. As the atomic size increases down the group, nuclear attraction on the added electron decreases due to shielding effect and hence electron affinity decreases.
• However, due to its small atomic size B has lower electron affinity than Al. 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.

Aluminium in AlCl₃ undergoes sp² hybridisation.

Diborane (B₂H₆)

Silicones are organo silicon polymers containing repeated R₂SiO units, where R is an alkyl group or an aryl group held by

 They have empirical formula R₂SiO. Since, it is similar to that of ketones (R₂CO), these compounds are names as silicones.

Uses of silicones :

• Silicones are used as insulating material for electrical appliances.
• They are used to prepare water proof fabrics.
• They are used as sealent and high temperature lubricants.
• They are used for mixing in paints and enamels to make them resistant to high temperature, sunlight and chemicals.

• The elements of group 15 from N to Bi have general valence shell electronic configuration, ns2np3.
• Valence shell has five electrons and common oxidation states are −3, +3, and +5. The range of oxidation states is from −3 to +5.
• The lower oxidation states are stable for the heavier elements.
• Group 15 elements show a tendency to donate electron pairs in −3 oxidation states. This tendency is maximum for nitrogen.
• Thus N has different oxidation states, Bi has 3+ as the stable oxidation state.

When borax is heated with ethyl alcohol and concentrated H₂SO₄ volatile vapours of triethylborate are formed which burn with green flame. This flame test is used to detect borate radical (BO₃⁻³) in the qualitative analysis.

• Boron has electronic configuration, ₅B : 1s² 2s² 2p¹
• In diborane, boron atoms undergo sp³ hybrid
• Three of such hybrid orbitals are half−filled while the fourth sp³ hybrid orbital remains vacant.
• 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−centered−two−electron (2c−2e) bonds.
• When '1s' orbital of each of the remaining two H atoms simultaneously overlap with the half−filled hybrid orbital of one B atom and the vacant hybrid orbital of the other B atom, it produces two three−centered−two electron bonds (3c−2e) or banana bonds.
• Hydrogen atoms involved in (3c−2e) bonds are the bridging H atoms i.e., H atoms in two B − H − B bonds.
• 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.

(a) The compound obtained is borax.

(b) Borax is prepared from its mineral by boiling it with a solution of sodium carbonate.

On crystallisation it forms Na₂B₄O₇.10H₂O. (It is a white solid).

The lone pair of electrons on the nitrogen atom facilitates the complexation of ammonia with transition metal ions. Thus, ammonia is a good complexing agent as it forms a complex by donating its lone pair of electrons.

This reaction is used for the detection of metal ions such as Cu²⁺ and Ag⁺.

False. The acidic nature of oxides of group 13 decreases down the group. The last two elements form amphoteric oxides SnO₂ and PbO₂.

True.

Basic nature of ammonia is due to the formation of OH⁻ ions.

NH₃(g) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)

The following reactions show the basic nature of ammonia.

(1) Ammonia reacts with acids to form salts.

NH₃ + HCl → NH₄Cl

2NH₃ + H₂SO₄ → (NH₄)₂ SO₄.

(2) Aqueous solution of ammonia precipitates hydroxides of metals from their salt solutions.

ZnSO₄(aq) + 2NH₄OH(aq) → Zn(OH)₂(s) + (NH₄)₂ SO₄(aq)

FeCl₃(aq) + 3NH₄OH(aq) → Fe(OH)₃(s) + 3NH₄Cl(aq)

The aqueous salt solution which Shravani used contains Cl⁻ anions. These Cl⁻ ions give white precipitate with silver nitrate solution.

Cl(aq)  +  AgNO₃(aq) →    AgCl(s)   +   NO₃(aq)

Further, on adding ammonium hydroxide to the white precipitate, a clear solution is obtained as white precipitate of AgCl dissolves in ammonium hydroxide forming a complex.

AgCl(s) + 2NH₄OH →  [Ag(NH₃)₂] Cl(aq) + 2H₂O