InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Among the following complexes : `K_(3)[Fe(CN)_(6)],[Co(NH_(3))_(6)]Cl_(3)` , `Na_(3)[Co ( o x )_(3)],[Ni(H_(2)O)_(6)]Cl_(2)`, `K_(2)[Pt(CN)_(4)]` and `[Zn(H_(2)O)_(6)(NO_(3))_(2)]` The diamagnetic are .A. (a) K, L, M, NB. (b) L, M, O, PC. (c) K, M, O, PD. (d) L, M, N, O |
|
Answer» Correct Answer - B `[K]: K_3[Fe(CN)_6]:` `Fe^(3+)` has `3d^5` configuration, `CN^(-)` is strong field ligand and thus four electrons are paired leaving one unpaired electron with `d^2sp^3` hybridization and thus paramagnetic. `[L]: [Co(NH_3)_6]Cl_3:Co^(3+)` has configuration, `NH_3` is strong field ligand thus all six electrons are paired with `d^2sp^3` hybridization and therefore diamagnetic. `[M]: Na_3[Co(o x)_3]: Co^(3+)` has `3d^6` configuration, `C_2O_4^(2-)` is strong field ligand and thus all the six electrons are paired with `d^2sp^3` hybridisation and therefore diamagnetic. `[N]: [Ni(H_2O)_6]Cl_2: Ni^(2+)` has `3d^8` configuration with two unpaired electrons. `H_2O` is weak field ligand and thus `sp^3d^2` hybridisation and paramagnetic. `[O]: K_2[Pt(CN)_4]: Pt^2` has `3d^8` configuration, `CN^(-)` is strong field ligand and thus all the eight electrons are paired with `dsp^2` hybridisation and therefore diamagnetic. `[P]: [Zn(H_2O)_6](NO_3)_2: Zn^(2+)` has `3d^(10)` configuration and thus all paired showing `sp^3d^2` hybridisation and so diamagnetic. |
|