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Choose the correct answer in each of the followingand explain with reason(i)NaCl, KCl, MgCl_(2), CaCl _(2)- most ionic (ii) Ba to Ba^(2+)-, Be to Be^(2+), Cs to Cs^(+), Li^(+)-maximum ionization energy (iii)AlCl_(3), All_(3), MgI_(2), NaI - most covalent (iv) RbF, CsF, NaF, KF - highest lattice energy (v)Li^(-), Be^(-), B^(-), C^(-) - least stable species (vi)CIO_(3) ,XeF_(4), SF_(4), I_(3)^(-) - maximum number of lone pairs of electrons on central atom |
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Answer» Solution :(i) KCl - on the basis of Fajan's rule that larger the size of the cation greater is the ionic bonding . Here , the cation ` K^(+)`has largestsize (anion `Cl^(-)` is same in all cases ) (ii) ` Be to Be^(2+)` has MAXIMUM ionization energy becausesmaller the atom , more is the energy needed to remove the electron, i.e., ionization energy , Also , removal of two electrons requires more energy . (iii) `AII_(3)` on the basis of Fajan's rules (IV) `Be^(-)`- This is becuae`Li^(-) = 1s^(2) 2x^(2), Be^(-) =1s^(2) 2s^(2)SP^(1)` In `Li^(-)`, addition of electron has taken place in 2s orbital . In ` Be^(-)` , addition of electron has taken place in 2p orbital leaving its 2s orbital COMPLETELY filled `EA_(1)` for Beis more positivethen ` EA_(1) for Li` . Thus , ` Be^(-)` is LEAST stable (vi) ` I_(3)^(-)`- No of lone pairs of electrons are ` I_(3)^(-) = 3 ,XeF_(4) = 2, SF_(4) = 1, CIO_(3)^(-) = 1 ` |
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