question_answer

Which of the following facts about the complex \[[Cr{{(N{{H}_{3}})}_{6}}]C{{l}_{3}}\] is wrong?

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 \[AgCl\].