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Arrange the following a.CaH_(2),BeH_(2)and TiH_(2) in order of increasing electrical conductance. b. LiH,NaH and CsH in order of increasing ionic character. c. H-H,D-D and F-F in order of increasing bond dissociaton enthalpy. d. NaH,MgH_(2) and H_(2)O in order of increasing reducing property. |
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Answer» Solution :a. The order of increasing ELECTRIC conductance is `BeH_(2)ltCH_(2)ltTiH_(2)` Skince `CaH_(2)` is an ionic hydride, it conducts electricity in fused state, `BeH_(2)` being covalent hydride does not conduct electricity at all. `TiH_(2)`, a metallic hydride conducts electricity at room temperature. B. The order of increasing ionic character is `LiHltNaHltCsH` Down the group `(DARR)`, from `Li` to `Cs`, electronegativity decreases adn and hence ionic character increases. c. Order of increasing bond dissociation enthalpy: `F-FltH-HltD-D` Due to greater nucleus mass, the bond pair in `D-D` bond is attracted more strongly than in `H-H` bond. THUS, the bond dissociation enthalpy of `D-D` is higher than `H-H` bond. However, due to repulsion between lone pairs of `F` and the bond, `F-F` bond dissociation enthalpy is the minimum, and hence the order. d. Order of increasing reducing property: `H_(2)OltMgH_(2)ltNaH` `NaH` is ionic hydride whereas `H_(2)O` adn `MgH_(2)` are covalent hydrides. Ionic hydrides are STRONGER reducing agent as compared to covalent hydrides. Hence `NaH` is stronger reducing agent as compared to `H_(2)O` and `MgH_(2)`. Out of `H_(2)O` and `MgH_(2)`, hence `H_(2)O` is weaker reducing agent as compared to `MgH_(2)`. |
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