Question

Answer is following by appropriately matching the lists based on the information given in the paragraph. Ligands are broadly classified into two classes classical and non classical ligands depending on thier doner and acceptability. Bonding mechanism which is non classical is called synegic bonding.
In hte following list-I contains some complex and list-II contains the characterestics of complex.
$\begin{array}{cc}\text{List-I}& \text{List-II}\\ \left(Compounds\right)& \\ \left(I\right)\left[Pt\right(N{O}_2{)}_2(en{)}_2{]}^{2+}& \left(P\right)\text{Stable according to EAN-Rule}\\ \left(II\right)\left[Cr\right(\pi -C_6{H}_6\left)\right(NO{)}_2]& \left(Q\right)\text{Both ligands acts as ambidentate}\\ \left(III\right)\left[Ir\right(SCN\left)\right(S{O}_4\left)\right(N{H}_3{)}_4]& \left(R\right)\text{Bond order M - L bond > 1.0}\\ \left(IV\right)\left[V\right(C_2{H}_4\left)\right(CN{)}_5]& \left(S\right)BondorderofLigandsdecreases\\ & \left(T\right)Co-ordinationnumberofcentralmetalissix\end{array}$
Which of the following options has the correct combination considering List-I and List-II?

• A. I, Q
• B. IV, Q
• C. I, R
• D. II, P

Answer 1

The correct option is D II, P

$\left(a\right)\left[Pt\right(N{O}_2{)}_2\left(en{)}_2{]}^{2+}or\right[Pt\left(ONO{)}_2\right(en{)}_2{]}^{2+}$
$N{O}_2$ is an ambidentate ligand that can bind with $ONO$ site as well as with N- atom to $Pt$ atom also but ethelene diamine is not-an ambidentate ligand.
(b) Here $C{N}^{-}$ can act as ambidentate ligand, but $C_6{H}_6$ isn't an ambidentate ligand.
(c) $\left[Pt\right(N{O}_2{)}_2(en{)}_2{]}^{2+}$
Here metal is in $+4$ oxidation state, and ligands don't participate in back bonding. So due to lack of back bonding the bond order between (M $\to$ L) doesn't increase.
(d) EAN of $\left[Cr\right(\pi -C_6{H}_6\left)\right(NO{)}_2]=$ Atomic number of metal atom + number of $e^{-}$ donate by ligands
$=24+6+3\times 2$
$=36$ stable configuration
Therefore, the correct option is D.