question_answer

                Which one of the following is an inner orbital complex as well as diamagnetic in behaviour?                                                                                                                 (Atomic no. : Zn = 30, Cr = 24, Co = 27, Ni = 28)

A)                 \[{{[Zn\,{{(N{{H}_{3}})}_{6}}]}^{2+}}\]

B)                 \[{{[Cr{{(N{{H}_{3}})}_{6}}]}^{3+}}\]

C)                 \[{{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}}\]

D)                 \[{{[Ni\,{{(N{{H}_{3}})}_{6}}]}^{2+}}\]

Correct Answer: C

Solution :

                In [Co(NH3)6]3+, oxidation state of Co = + 3 and its co-ordination number is six.                 So           \[_{27}Co=1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}}3{{p}^{6}},3{{d}^{7}},4{{s}^{2}}\]                 \[C{{o}^{3+}}=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{6}}\]                                 In \[{{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}}\] shows inner orbital complex as well as diamagnetic in behaviour (due to absence of upaired electron).                 \[{{[Zn{{(N{{H}_{3}})}_{6}}]}^{3+}}\] \[\to \] \[s{{p}^{3}}{{d}^{2}}\] hybridisation   (outer) and diamagnetic.                 \[{{[Cr{{(N{{H}_{3}})}_{6}}]}^{2+}}\to \,{{d}^{2}}s{{p}^{3}}\] hybridisation   (inner) and paramagnetic.