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p-Block Elements - Group 13 Elements (Boron Family)

Grade 11ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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The Group 13 elements (Boron family) include Boron (BB), Aluminum (AlAl), Gallium (GaGa), Indium (InIn), and Thallium (TlTl). Their general valence shell electronic configuration is ns2np1ns^2 np^1.

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Atomic Radii Anomaly: The atomic radius of GaGa (135135 pm) is unexpectedly smaller than that of AlAl (143143 pm). This is due to the poor shielding effect of the intervening 10d10d electrons in GaGa, which increases the effective nuclear charge.

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Inert Pair Effect: The stability of the +1+1 oxidation state increases down the group (Al<Ga<In<TlAl < Ga < In < Tl), while the +3+3 state becomes less stable. This is due to the reluctance of the s2s^2 electron pair to participate in bond formation.

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Lewis Acid Character: Group 13 halides (like BF3BF_3) are electron-deficient (66 electrons in the valence shell). They act as strong Lewis acids by accepting electron pairs to complete their octet.

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Diborane (B2H6B_2H_6) Structure: It is an electron-deficient molecule containing four terminal Bβˆ’HB-H bonds (22-centre-22-electron) and two bridging Bβˆ’Hβˆ’BB-H-B bonds (33-centre-22-electron), also known as 'banana bonds'.

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Anomalous properties of Boron: Due to its small size and absence of dd-orbitals, Boron differs from other members. It forms covalent compounds and has a maximum covalency of 44, whereas other members can expand their coordination number to 66 using vacant dd-orbitals (e.g., [AlF6]3βˆ’[AlF_6]^{3-}).

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Amphoteric Nature: Aluminum oxide (Al2O3Al_2O_3) and Aluminum hydroxide (Al(OH)3Al(OH)_3) are amphoteric, reacting with both acids and bases.

πŸ“Formulae

Na2B4O7⋅10H2O→ΔNa2B4O7→Δ2NaBO2+B2O3Na_2B_4O_7 \cdot 10H_2O \xrightarrow{\Delta} Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2 + B_2O_3

B2H6+3O2β†’B2O3+3H2O(Ξ”Hc∘=βˆ’2035Β kJΒ molβˆ’1)B_2H_6 + 3O_2 \rightarrow B_2O_3 + 3H_2O \quad (\Delta H_c^{\circ} = -2035 \text{ kJ mol}^{-1})

3LiBH4+4BF3β†’2B2H6+3LiBF43LiBH_4 + 4BF_3 \rightarrow 2B_2H_6 + 3LiBF_4

B2H6+6NH3β†’[BH2(NH3)2]+[BH4]βˆ’β†’Ξ”B3N3H6Β (InorganicΒ Benzene)B_2H_6 + 6NH_3 \rightarrow [BH_2(NH_3)_2]^+[BH_4]^- \xrightarrow{\Delta} B_3N_3H_6 \text{ (Inorganic Benzene)}

H3BO3+3H2Oβ‡Œ[B(OH)4]βˆ’+H3O+H_3BO_3 + 3H_2O \rightleftharpoons [B(OH)_4]^- + H_3O^+

πŸ’‘Examples

Problem 1:

Explain why the Lewis acid strength of Boron halides follows the order: BF3<BCl3<BBr3<BI3BF_3 < BCl_3 < BBr_3 < BI_3.

Solution:

The order is BF3<BCl3<BBr3<BI3BF_3 < BCl_3 < BBr_3 < BI_3.

Explanation:

This is due to pΟ€βˆ’pΟ€p\pi - p\pi back-bonding. In BF3BF_3, the small size of FF and BB allows effective overlap between the filled 2p2p orbital of FF and the vacant 2p2p orbital of BB. This reduces the electron deficiency of Boron. As the size of the halogen increases (Cl,Br,ICl, Br, I), the pp-orbital size increases (3p,4p,5p3p, 4p, 5p), making back-bonding less effective, thus increasing Lewis acidity.

Problem 2:

What happens when Boric acid (H3BO3H_3BO_3) is heated?

Solution:

H3BO3β†’370KHBO2β†’410KH2B4O7β†’redΒ hotB2O3H_3BO_3 \xrightarrow{370K} HBO_2 \xrightarrow{410K} H_2B_4O_7 \xrightarrow{\text{red hot}} B_2O_3

Explanation:

On heating, Orthoboric acid (H3BO3H_3BO_3) first loses water to form Metaboric acid (HBO2HBO_2). Further heating yields Tetraboric acid (H2B4O7H_2B_4O_7), and finally, at high temperatures, it decomposes into Boric anhydride (B2O3B_2O_3).

Problem 3:

Why is AlCl3AlCl_3 usually found as a dimer Al2Cl6Al_2Cl_6?

Solution:

2AlCl3β†’Al2Cl62AlCl_3 \rightarrow Al_2Cl_6

Explanation:

In AlCl3AlCl_3, the Aluminum atom is electron-deficient (only 66 electrons). To complete its octet, it forms a dimer where two chlorine atoms act as bridges by donating their lone pairs to the vacant pp-orbitals of the neighboring Aluminum atoms through coordinate covalent bonding.

Group 13 Elements (Boron Family) - Revision Notes & Key Formulas | ICSE Class 11 Chemistry