A) \[{{I}^{-}}\] is a weaker base than \[{{F}^{-}}\]
B) \[H{{O}^{-}}\] is a stronger base than \[{{H}_{2}}{{N}^{-}}\]
C) \[HON{{H}_{2}}\] is a weaker base than \[N{{H}_{3}}\]
D) \[{{F}_{3}}{{C}^{-}}\] is a stronger base than \[C{{l}_{3}}{{C}^{-}}\]
Correct Answer: C
Solution :
| [b] [a] For elements in the same periodic group, the larger the basic site atom the more spread out (delocalized) is the charge and the weaker is the base. Thus \[{{l}^{-}}\] is a weaker base than \[{{F}^{-}}\]. Alternatively, HI is stronger acid than HF, the conjugate base follows the reverse order. |
| [b] \[{{H}_{2}}O\] is a stronger acid than \[N{{H}_{3}},\] thus the conjugate base \[O{{H}^{-}}\] is weaker base than \[NH_{2}^{-}\] |
| [c] Electron - attracting inductive effect of \[-\,OH\] decreases the electron density on N and thus decreases the basicity of \[-NH_{2}^{{}}.\] |
| [d] In \[C{{l}_{3}}{{C}^{-}},\] the unshared electron pair of the carbon in the p-orbital undergoes extended \[p-d\pi \] bonding into an emtpy d-orbital of each of the three Cl atoms. |
| \[\left[ \underset{\bullet \bullet }{\overset{\bullet \bullet }{\mathop{\mathbf{:}\text{Cl}}}}\,-\underset{\begin{smallmatrix} | \\ \underset{\bullet \bullet }{\overset{{}}{\mathop{\mathbf{:}\text{Cl}}}}\,\mathbf{:} \end{smallmatrix}}{\mathop{C}}\,=\underset{\bullet \bullet }{\overset{\bullet \bullet }{\mathop{\text{Cl}}}}\,\mathbf{:}{{\leftrightarrow }^{-}}\underset{\bullet \bullet }{\overset{\bullet \bullet }{\mathop{\mathbf{:}\text{Cl}}}}\,=\underset{\begin{smallmatrix} | \\ \underset{\bullet \bullet }{\overset{{}}{\mathop{\mathbf{:}\text{Cl}}}}\,\mathbf{:} \end{smallmatrix}}{\mathop{C}}\,{{-}^{-}}\underset{\bullet \bullet }{\overset{\bullet \bullet }{\mathop{\text{Cl}}}}\,\mathbf{:}\leftrightarrow \underset{\bullet \bullet }{\overset{\bullet \bullet }{\mathop{\mathbf{:}\text{Cl}}}}\,-\underset{\begin{smallmatrix} | \\ \underset{\bullet \bullet }{\overset{{}}{\mathop{\mathbf{:}\text{Cl}}}}\,\mathbf{:}- \end{smallmatrix}}{\mathop{C}}\,-\underset{\bullet \bullet }{\overset{\bullet \bullet }{\mathop{\text{Cl}}}}\,\mathbf{:} \right]\] |
| Resonance stabilization makes a more dominating role than the inductive effect, making \[C{{l}_{3}}{{C}^{-}}\] a weaker base than \[{{F}_{3}}{{C}^{-}}.\] |
| In F, there is no d orbitals, hence, no stabilization occurs in \[{{F}_{3}}{{C}^{-}}.\] |
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