A) The complex is paramagnetic
B) The complex is an outer orbital complex
C) The complex gives white precipitate with silver nitrate
D) The complex involves \[{{d}^{2}}s{{p}^{3}}\] hybridisation and is octahedral in shape
Correct Answer: B
Solution :
In complex \[[Cr{{(N{{H}_{3}})}_{6}}]\,C{{l}_{3}},\,\,Cr\] is present as \[C{{r}^{3+}}\] ion. Since, \[N{{H}_{3}}\] is a weak field ligand, so it cannot pair up the unpaired electrons \[_{24}Cr=[Ar]\,3{{d}^{5}}\,\,4{{s}^{1}}\] \[C{{r}^{3+}}=[Ar]\,3{{d}^{3}}\,\,4{{s}^{0}}\] Due to \[{{d}^{2}}s{{p}^{3}}\]-hybridisation, it is octahedral in shape and due to presence of 3 unpaired electrons, it is paramagnetic in nature. Since, (n - 1)d/ (3d) orbital takes part in hybridisation, it is an inner orbital complex. Due to presence of \[C{{l}^{-}}\] ions, it reacts with \[AgN{{O}_{3}}\] to give white precipitate of \[AgCl\].You need to login to perform this action.
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