The number of unpaired electron (uparrow) in the tetrahedral complex ion [FeCl_4
ID: 511372 • Letter: T
Question
The number of unpaired electron (uparrow) in the tetrahedral complex ion [FeCl_4]^- is: (remember the tetrahedral has the reverse splitting of the octahedral, Cl^- is a weak field ligand, Fe has +3 charge: a) 1 b) 5 c) 3 The name of the complex: [Co(CN)_2(NH_3)(en)Cl]^-2 is: a) amminechlorocyanoethylenediammineCobalt(II) b) amminechlorodicyanoethylenediammineCobaltate(II) c) a) amminechlorodicyanoethylenediammineCobaltate(III) Which of the following statement is TRUE about the octahedral complexes of Ni^+2 (3d^8) a) both strong and weak field complexes are diamagnetic b) the strong-field complex is magnetic and the weak-filed complex is paramagnetic c) both strong and weak complexes are paramagnetic The possible geometries for the ML_4 complex (M is the central metal ion, L is the ligand) are: a) octahedral & tetrahedral b) octahedral & square planar c) tetrahedral & square planar Which of the following metal ions would have the possibility of both high-spin and low-spin configurations in an octahedral crystal field? a. Zn (II) b. Co (III) c. V(II) A ligand is any _____ forming a coordinate bond to a metal cation. a) Lewis acid b) Lewis base c) organic molecule d) any negative ion species The 3d orbitals that have higher energy in an octahedral field a) d_xy and d_xz b) d^2_z and d_y2 c) d^2_z and d^2_x -^2_y d) d^2_x - y^2 and d_xz The orbitals that hybridized in the complex ion Fe(EDTA)^3+ a. One s and the three p's b. One s, three p's, one d c. One s, three p's and two d's d. Three p's and three d's orbitalsExplanation / Answer
Ans 9 - ML4 will form tetrahedral and square planar
Ans 8 - option B is correct as square planar of Ni (II) is dimagnatic and Tetrahedral is Paramagnetic in nature