question_answer

A correct statement is

A)  \[{{[Co{{(N{{H}_{3}})}_{6}}]}^{2+}}\] is paramagnetic

B)  \[{{[MnB{{r}_{4}}]}^{2-}}\] is tetrahedral

C)  \[{{[CoB{{r}_{2}}{{(en)}_{2}}]}^{-}}\] exhibits linkage isomerism

D)  \[{{[Ni{{(N{{H}_{3}})}_{6}}]}^{2+}}\] is an inner orbital complex.

Correct Answer: A

Solution :

: [a]  \[{{[Co{{(N{{H}_{3}})}_{6}}]}^{2+}}\]   \[_{27}C{{o}^{2+}}:[Ar]\,\,3{{d}^{7}}\,4{{s}^{0}}\]   \[n=3\](where n = no. of unpaired electrons) Hence, complex is paramagnetic in nature. [b]:  \[{{[MnB{{r}_{4}}]}^{2-}}\]   \[_{25}M{{n}^{2+}}:[Ar]3{{d}^{5}}\]   [c] \[{{[CoB{{r}_{2}}{{(en)}_{2}}]}^{-}}\]exhibit geometrical                                                            isomerism.   [d]      \[{{[Ni{{(N{{H}_{3}})}_{6}}]}^{2+}}\]   \[_{28}N{{i}^{2+}}:\,\,[Ar]\,\,3{{d}^{8}}\]     Both statements [a] and [b] are correct.