question_answer

Which one of the following is an outer orbital complex and exhibits paramagnetic behaviour?

A) \[{{\text{ }\!\![\!\!\text{ Ni(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{2+}}}\]              

B)        \[{{\text{ }\!\![\!\!\text{ Zn(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{2+}}}\]

C) \[{{\text{ }\!\![\!\!\text{ Cr(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{3+}}}\]              

D)        \[{{\text{ }\!\![\!\!\text{ Co(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{3+}}}\]

Correct Answer: A

Solution :

Outer orbital complex utilizes \[nd\] orbitals for bonding and exhibit paramagnetic behaviour, only if there present unpaired electrons.                              In \[{{[Ni{{(N{{H}_{3}})}_{6}}]}^{2+}}:\] \[N{{i}^{2+}}=[Ar]3{{d}^{8}}4{{s}^{o}}\]                 \[{{[Ni{{(N{{H}_{3}})}_{6}}]}^{2+}}=\] So, this is an outer orbital complex having paramagnetic character. In\[{{[Zn{{(N{{H}_{3}})}_{6}}]}^{2+}}:\]\[Z{{n}^{2+}}=[Ar]\,3{{d}^{10}}\] \[{{[Zn{{(N{{H}_{3}})}_{6}}]}^{2+}}=\] Thus, it is also an outer orbital complex but it is diamagnetic as all the electrons are paired. In\[{{[Cr{{(N{{H}_{3}})}_{6}}]}^{3+}},\]\[C{{r}^{3+}}=[Ar]\,3{{d}^{3}}\] \[{{[Cr{{(N{{H}_{3}})}_{6}}]}^{3+}}=\] Because of the involvement of (n - 1) d, i. e., 3d orbital-in hybridization, it is an inner orbital complex. Its nature is paramagnetic because of the presence of three unpaired electrons. In \[{{[CO{{(N{{H}_{3}})}_{6}}]}^{3+}}:\]\[C{{o}^{3+}}=[Ar]\,3{{d}^{6}}\] As all the electrons are paired, it is a diamagnetic complex.