Question

Which of the following is an outer orbital complex ?

• A. ${\left[Cr{\left(N{H}_3\right)}_6\right]}^{3+}$
• B. ${\left[Ni{\left(N{H}_3\right)}_6\right]}^{2+}$
• C. ${\left[Fe{\left(CN\right)}_6\right]}^{3-}$
• D. ${\left[Mn{\left(CN\right)}_6\right]}^{4-}$

Answer 1

Answer: (B)

(a) In $[Cr(N{H}_3{)}_6{]}^{3+}$, $Cr$ is present as $C{r}^{3+}$
$C{r}^{3+}=[Ar]3d^3$
$[Cr(N{H}_3{)}_6{]}^{3+}=[Ar]$

Since, $(n-1)d$ orbitals are used for hybridisation, it is an inner orbital complex
(b) In $[Ni(N{H}_3{)}_6{]}^{2+}$, $Ni$ is present as $N{i}^{2+}$
$N{i}^{2+}=[Ar]3d^{8}4s^0$
$[Ni(N{H}_3{)}_6{]}^{2+}=[Ar]$

Since, outer d (ie, nd) orbitals are used, it is an outer orbital complex
(c) In $[Fe(CN{)}_6{]}^{3-}$, $Fe$ is present as $F{e}^{3+}$
$F{e}^{3+}=[Ar]3d^5$
$[Fe(CN{)}_6{]}^{3-}=[Ar]$
It is also an inner orbital complex
(d) In $[Mn(CN{)}_6{]}^{4-}$, $Mn$ is present as $M{n}^{2+}$
$M{n}^{2+}=[Ar]3d^{5}4s^0$
$[Mn(CN{)}_6{]}^{4-}=[Ar]$
It is also an inner orbital complex