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.You need to login to perform this action.
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