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InterviewSolution
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A metal complex having composition Cr(NH_(3))_(4) Cl_(2)Br has been isolated in two forms (A) and (B). The form (A) reacts with AgNO_(3) to give a white precipitate readily soluble in dilute aqueous ammonia, whereas (B) gives a pale yellow precipitate soluble in concentrated ammonia. Write the formula of(A) and (B) and state the hybridization of chromium in each Calculate their magnetic moments (spin--only value) |
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Answer» SOLUTION : `Cr(NH_(3)_(4),Cl_(2)Br` have two forms A and B. `[Cr(NH_(3))_(4)BrCl]Cl, (d^(2)sp^(3)), [Cr(NH_(3))_(4)CI_(2)]`Br.`d^(2)sp^(3))`, 3.87 BM, 3.87 BMForm .A. gives white ppt. wth `AgNO_(3)` hence it must have chloride ion in form of noncomplex ion i.e. outside the complex sphere as< `Cr(NH_(3))_(4)CIBr]Cl` `UNDERSET(From A)[Cr(NH_(3))_(4)CIBr]Cl+AğNO_(3)tounderset(White ppt)(AgCI)downarrow(+)[Cr(NH_(3))_(4)CIBr]+NO_(3)` These precipitates of AgCl are soluble in `NH_(4) OH` due to formation of complex salt `underset(whiteppt)(AgCI)+2NH_(4)OH tounderset(complex salt)([Ag(NH_(4))_(2))CI]+2H_(2)O` Similarly, form B gives pale yellow precipitate of AgBr which are sparingly soluble in `NH_(4)OH` Form .B. is `[Cr(NH_(3))_(4)CI_(2)]Br` `AgNO_(3)+[Cr(NH_(3))_(4)C1_(2)]Br tounderset(pale yellow ppt) (AgBr)downarrow(+)+[Cr(NH_(3))_(4)Cl_(2)]^(+)+ NO_(3)^(-)` `underset(pale yellow ppt)(AgBr)+2NH_(4)OHto [Ag(NH_(3))_(2)Br]+2H_(2)O` In both complexes, chromium is present as central ion and its oxidation number is +3 So in these `_(24) Cr = 1s^(2), 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(5), 4s^(1)` `Cr6(3+) = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(3)` Number of ligands are six and `Cr^(3+)` shows `d^(2)sp^(3)` hybridization in both complexes A and B Hence in it, number of unpaired electrons are 3. So MAGNETIC moment `(mu) = sqrt(n(n + 2)).= sqrt(3(3+2)) = sqrt(15)` = 3.872 BM` 
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