Question

5. Using HSAB theory, choose the better acid or base in the following pairs and explain your choice: a. CH3NH2 or NH3 in reaction with H b. Which end of SCN will coordinate to Cr3+: Pt2+? c. Boric acid, B(OH)3, acts as an acid in water, but does not do so via ionization of a proton. Rather, it serves as a Lewis acid towards OH-. Explain with the use of a balanced equation.

Answer 1

• a) H⁺ is an extremely Hard Species. In Methylamine, the electron density on N is increased due to the +I effect of -CH_{​​​​3}​​​. So, Methylamine is a Hard Base Compared to Ammonia. Hence, according to the HSAB Theory, Reaction of H⁺ with Methylamine will be more facile.
• b) Due to greater polarisability, the S-end of SCN^{​​​​-} is Soft and N-End is Hard, due to high Electronegativity and Low Polarisability. Now, Cr^{​​​​​3+} is Hard due to greater positive charge, while Pt^{​​​​​2+} is Soft, due to low Positive Charge. Thus, according to HSAB Theory, Cr^{​​​​​3+} will co-ordinate to the N-End of SCN^{​​​​-}, while Pt^{​​​​​2+} will Co-ordinate to the S-End of SCN^{​​​​-}, because Hard-Hard and Soft-Soft Combinations are more stable according to HSAB Theory.
• c) B has a vacant p-orbital in B(OH)₃, B atom in B(OH)₃ is quite electrophilic, making it succeptible to Nucleophilic Attack. B(OH)₃ is a hypervalent compound of B, in which B has a Sextet. The High Electronegativity of the 3 O atoms, make B atom more electrophilic, and the triagonal planar geometry of B(OH)₃ also sterically favours the Nucleophilic Attack. Thus, H_{​​​​​​2}​​​O attacks the vacant p-orbital of Boron to form Tetrahedral B(OH)₄^- Anion, in which B has an Octet. Thus, the Acidity of B(OH)_{3 ^​}​​​​ is Exclusively due to the Abstraction of OH^{​​​​​-}​​​​ from Water, and not due to Proton Donation from O-H bonds.

The Reaction can be expressed as :

B(OH)₃ + 2 H₂O ⇌ B(OH)₄^- + H₃O^​+