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Explain the following : (i) Gallium has higherionisation enthalpy than aluminium. (ii) Borondoes notexistsas B^(3+) ion. (iii)Aluminium forms [AlF_(6)]^(3-) ion butboron does notform [BF_(6)]^(3-) ion. (iv) PbX_(2) is more stablethan PbX_(4). (v) Pb^(4+)acts as an oxidisingagent but Sn^(2+)acts as a reducingagent. (vi) Electron gain enthalpy of chlorineis more negative as compared to fluorine. (vii) Tl(NO_(3))_(3) acts as an oxidising agent. (viii) Carbonshows catenationproperty but leaddoes not. (ix) BF_(3) does nothydrolysecompletely (modified). (x) Why does the elemnet silicon, not forma graphitelike structurewhereas carbondoes. |
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Answer» Solution :(i) Due toineffectiveshieldingof valence electronsby the inventing 3d-electrons, the effectivenucleuschargeon Gais slightly higher thanthat onAl and hence the`DeltaH_(i)` of galliumis slightly higher than theAl. (ii) N/A . (iii) Al hasvacantof d-orbitalsand hence can expand its coodination numberfrom 4 to 6 and hence for octahedral `[AlF_(6)]^(3-)`ion in whichAl undergoes `sp^(3) d^(2)`hybridization . in contrast,B does not have d-orbital. Therefore. It canhave a maximumcoordinationnumber of 6.Therefore, B forms `[BF_(4)]^(-)`(in which B is `sp^(3)` hybridized) but not `[BF_(6)]^(3-)`. (iv) Due to inert effect,+2 oxidationstate of Pb is more stablethan its +4 oxidation state.Consequently `PbX_(2)` in whichthe oxidationstate of Pbis +2 is more stablethan `PbX_(4)` in whichthe oxidation state is `+4`. (v) Inert pair effectis less prominentin Sn than it Pb.Therefore +2 oxidation of Sn is less stable the `+4` oxidation state. In otherwords, `Sn^(2+)` can easily lose two electrons to form `Sn^(4+)` and hence`Sn^(2+)` acts acts a reducing agent. `Sn^(2+) rarr Sn^(4+) +2e^(-)` In contrast, the inert pair effectis more prominant in Pb than in Sn. Therefore `+2` oxidation state of Pb in more stablethan its+4 oxidationstate. In other words , `Pb^(4+)` can easily accept two electronsto form `Pb^(2+)` and hence`Pb^(4+)`acts as an oxidisingagent. `Pb^(4+) + 2e^(-) rarr Pb^(2+)` (vi) Due to small size, theelectron-electronrepulsionsin the relatively compact `2p`-subshell of F are quite strong and hence the incomingelectron is not accepted with the same ease as in case of bigger Cl atom where repulsionsare comparatively weak. Thus, electron gain enthalpy of chlorineis more negative as comparedto that of fluorine. (vii) Due tostrong inert paireffect, the +3 oxidation state of Tl is less stable than its +1 oxidation state. Since in `Tl(NO_(3))_(3)`,oxidation state of `Tl`is `+3`, therefore, it caneasily gain twoelectronsto form `TlNO_(3)` in which inoxidationstate of `Tl` is `+1`.Consequently, `Tl(NO_(3))_(3)` acts as an oxidisingagent. (VIIII) Property of catenationdepends upon the strengthof element-elementbond which, in turn,dependsupon the sizeof the element . Since the atomic size of carbonis MUCH smaller than that of lead, therefore, carbon -carbonbond strength is much higher than that of lead-lead bond. Due to stronger C-Cthan `Pb-Pb`bonds,carbon has a muchhigher tendency for catenation than lead. (ix) UNLIKE other boron halids, `B_(3)` does not hydrolysecompletely, Instead, it hydrolysesincompletely to form boric acidand fluoroboric acid. This is because the HF FIRST formed reacts with `H_(3)BO_(3)`. `{:([BF_(3)+3H_(2)OrarrH_(3)BO_(3)+3HF]xx4),([H_(3)BO_(3)+4HFrarrH^(+)+[BF_(4)]^(-)+3H_(2)O]xx3):}/(4BF_(3)+3H_(2)OrarrH_(3)BO_(3)+3[BF_(4)]^(-)+3H^(+))` (x) N/A |
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