Question

The pair(s) of complexes wherein both exhibit tetrahedral geometry is(are)
(Note: py = pyridine
Given: Atomic numbers of Fe, Co, Ni and Cu are 26, 27, 28 and 29, respectively)

• A. $\left[FeC{l}_4{]}^{-}\text{and}\right[Fe(CO{)}_4{]}^{2-}$
• B. $\left[Co\right(CO{)}_4{]}^{-}\text{and}[CoC{l}_4{]}^{2-}$
• C. $\left[Ni\right(CO{)}_4{]}^{-}\text{and}\left[Ni\right(CN{)}_4{]}^{2-}$
• D. $\left[Cu\right(py{)}_4{]}^{+}\text{and}\left[Cu\right(CN{)}_4{]}^{3-}$

Answer 1

Answer: (A), (B), (D)

$\left[Cu\right(py{)}_4{]}^{+}\text{and}\left[Cu\right(CN{)}_4{]}^{3-}$

$\left(A\right)[FeC{l}_4{]}^{-}$
$F{e}^{+3}=3d^5,C{l}^{-}$ weak field ligand $s{p}^3$ Hybridisation and Tetrahedral Geometry.

$\left[Fe\right(CO{)}_4{]}^{2-}$
$F{e}^{-2}=3d^{8}4s^2$
$CO$ strong field ligand electron pairing occurs
$F{e}^{-2}=3d^{10}$ hence $s{p}^3$ Hybridisation and Tetrahedral Geometry.

$\left(B\right)\left[Co\right(CO{)}_4{]}^{-}$
$C{o}^{-1}=3d^{8}4s^2$ due to $CO$ strong field ligand electron pairing occurs
Hence $=3d^{10}$ $s{p}^3$ Hybridisation and Tetrahedral Geometry.
$[CoC{l}_4{]}^{-2}$
$C{o}^{+2}=3d^{7}C{l}^{-}$ weak field ligand $s{p}^3$ Hybridisation and Tetrahedral Geometry.

$\left(D\right)\left[Cu\right(py{)}_4{]}^{+1}C{u}^{+1}=3d^{10},s{p}^3\left[Cu\right(CN{)}_4{]}^{-3}C{u}^{+1}=3d^{10},s{p}^3$
Thus the correct Answer are option A,B,D .