Question

Which of the following is an outer orbital complex ?

• A. $ \left[Cr\left(NH_{3}\right)_{6}\right]^{3+} $
• B. $ \left[Ni\left(NH_{3}\right)_{6}\right]^{2+} $
• C. $ \left[Fe\left(CN\right)_{6}\right]^{3-} $
• D. $ \left[Mn\left(CN\right)_{6}\right]^{4-} $

Answer 1

The Correct Option is B

(a) In $[Cr(NH_{3})_{6}]^{3+}$, $Cr$ is present as $Cr^{3+}$
$Cr^{3+}=[Ar]3d^{3}$
$[Cr(NH_{3})_{6}]^{3+}=[Ar]$

Since, $(n-1)d$ orbitals are used for hybridisation, it is an inner orbital complex
(b) In $[Ni(NH_{3})_{6}]^{2+}$, $Ni$ is present as $Ni^{2+}$
$Ni^{2+}=[Ar] 3d^{8} 4s^{0}$
$[Ni(NH_{3})_{6}]^{2+}=[Ar]$

Since, outer d (ie, nd) orbitals are used, it is an outer orbital complex
(c) In $[Fe(CN)_{6}]^{3-}$, $ Fe$ is present as $Fe^{3+}$
$Fe^{3+}=[Ar] 3d^{5}$
$[Fe(CN)_{6}]^{3-}=[Ar]$

It is also an inner orbital complex
(d) In $[Mn(CN)_{6}]^{4-}$, $Mn$ is present as $Mn^{2+}$
$Mn^{2+}=[Ar] 3d^{5} \, 4s^{0}$
$[Mn(CN)_{6}]^{4-}=[Ar]$

It is also an inner orbital complex