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What do you understand by (i) electron deficient (ii) electron precise and (iii) electron rich compounds of hydrogen ? Provide justification with suitable examples. |
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Answer» Solution :Dihydrogen forms molecular compounds with most of the p-block elements. Most familiar examples are `CH_4, NH_3, H_2O` and HF. For convenience hydrogen compounds of non metals have also been CONSIDERED as hydrides. Being covalent, they are volatile compounds. Molecular hydrides are further CLASSIFIED according to the relative numbers of electrons and bonds in their Lewis structure into : (i) electron-deficient, (ii) electron-precise, (iii) electron - RICH hydrides. An electron-deficient hydride, as the name suggests, has too few electrons for writing its conventional Lewis structure. Example : Diborane `(B_2H_6)` In FACT all elements of group 13 will form electron-deficient compounds. They act as Lewis acids i.e., electron acceptors. Electron-precise compounds have the required number of electrons to write their conventional Lewis structures. All elements of group 14 form such compounds (e.g., `CH_4`) which are tetrahedral in geometry. Electron-rich hydrides have EXCESS electrons which are present as lone pairs. Example : Elements of group 15-17 form such compounds. (`NH_3` has 1-lone pair, `H_2O` -2 and HF-3 lone pairs). They will behave as Lewis bases i.e., electron donors. The presence of lone pairs on highly electronegative atoms like N, O and F in hydrides results in hydrogen bond formation between the molecules. This leads to the association of molecules. |
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