InterviewSolution
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InterviewSolution
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A magnetic moment of 1.73 B.M. will be shown by one among the following:A. `[Ni(CN)_(4)]^(2-)`B. `TiCI_(4)`C. `p[CoCI_(6)]^(4-)`D. `[Cu(NH_(3))_(4)]^(2+)` |
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Answer» Correct Answer - D For the compounds of the first series of transition metals, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the spin only 2 formula: `mu = sqrt(n(n + 2))` where n is the number of unpaired electrons and `mu` is the magnetic moment in units of Bohr magneton (BM) `1.73 = sqrt(n(n + 2))` `(1.73)^(2) = n(n +)` `n = 1` Thus a single unpaired electron has a magnetic moment of `1.73` Bohr magnetons (BM). Because `CN^(-)` is a strong field ligand, `[Ni(CN)_(4)]^(2-)` containing `Ni^(2+) (3d^(8))` , is a square planar complex containing zero unpaired electrons: `t_(2g)^(6)d_(z^(2))^(2)` . `TiC1_(4)` containing `Ti^(4+)` `(3d^(0))` also has zero unpaired electrons. Because `C1^(-)` is a weak field ligand, `[CoC1_(6)]^(4-)` containing `Co^(2+)` `(3d^(7))` is a high spin complex with two unpaired electrons. `[Cu(NH_(3))_(4)]^(2+)` containing `Cu^(2+) (3d^(9))` , is a square planar complex due to unsymmetrical electronic arrangement. Thus it has one unpaired electron with both types of ligands: weak field as well as strong field. |
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