Question

How many in each of the following complexes obey the rule of $18$ (EAN rule)?

(a)$\left[Cu\right(N{H}_3{)}_4{]}^{2+},\left[Cu\right(en{)}_3{]}^{2+},\left[Cu\right(CN{)}_4{]}^{3-}$

(b)$\left[Fe\right(CN{)}_6{]}^{4-},\left[Fe\right(CN{)}_6{]}^{3-},Fe(CO{)}_6$

(c)$\left[Ni\right(N{H}_3{)}_6{]}^{2+},\left[Ni\right(CN{)}_4{]}^{2-},\left[Ni\right(CO{)}_4]$

(d)$\left[Cr\right(N{H}_3{)}_6{]}^{3+},\left[Cr\right(CO{)}_6]$

(e)$\left[Co\right(N{H}_3{)}_6{]}^{3+},[CoC{l}_4{]}^{2-}$

• A. 2,2,1,1,2
• B. 1,2,1,1,1
• C. 1,2,1,1,2
• D. 1,1,2,2,1

Answer 1

The correct option is B 1,2,1,1,1

To obey the rule of $18$ (EAN), the sum of the number of electrons in
the metal valence shell $\left[\right(n-1\left)dsp\right]$ and those donated by the ligands
should be $18$, filling the low-energy metal orbitals.
$Cu\left[Ar\right]3d^{10}4s^1$ $Ni\left[Ar\right]3d^{8}4s^2$
$Cr\left(Ar\right)3d^{5}4s^2$ $Co\left[Ar\right]3d^{7}4s^2$
$Fe\left(Ar\right)3d^{6}4s^2$
$C{u}^{+}\left[Ar\right]3d^{10}$ $N{i}^{2+}\left[Ar\right]3d^8$
$F{e}^{2+}\left[Ar\right]3d^6$ $C{r}^{3+}\left[Ar\right]3d^3$
$F{e}^{3+}\left[Ar\right]3d^5$ $C{o}^{3+}\left[Ar\right]3d^6$
(a)$\left[Cu\right(N{H}_3{)}_4{]}^{2+}$ $9+8=17e^{-}$
$\left[Cu\right(en{)}_3{]}^{2+}$ $9+12=21e^{-}$
$\left[Cu\right(CN{)}_4{]}^{3-}$ $10+8=18e^{-}$
Thus, $\left[Cu\right(CN{)}_4{]}^{3-}$
(b)$\left[Fe\right(CN{)}_6{]}^{4-}$ $6+12=18e^{-}$
$\left[Fe\right(CN{)}_6{]}^{3-}$ $5+12=17e^{-}$
$\left[Fe\right(CO{)}_5]$ $8+10=18e^{-}$
Thus, $\left[Fe\right(CN{)}_6{]}^{4-}$ and $\left[Fe\right(CO{)}_5]$
(c)$\left[Ni\right(N{H}_3{)}_6{]}^{2+}$ $8+12=20e^{-}$
$\left[Ni\right(CN{)}_4{]}^{2-}$ $8+8=16e^{-}$
$\left[Ni\right(CO{)}_4]$ $10+8=18e^{-}$
Thus, $\left[Ni\right(CO{)}_4]$
(d)$\left[Cr\right(N{H}_3{)}_6{]}^{3+}$ $3+12=15e^{-}$
$\left[Cr\right(CO{)}_6]$ $6+12=18e^{-}$
Thus, $\left[Cr\right(CO{)}_6]$
(e)$\left[Co\right(N{H}_3{)}_6{]}^{8+}$ $6+12=18e^{-}$
$[CoC{l}_4{]}^{2-}$ $7+8=15e^{-}$
Thus, $\left[Co\right(N{H}_3{)}_6{]}^{3+}$