Question

Choose the correct statement(s).
I. The geometry of $Ni(CO{)}_4$ and $Ni(PP{h}_3{)}_{2}C{l}_2$ are tetrahedral.
II. $K_2\left[Pt\right(CN{)}_4]$ and $N{a}_3\left[Co\right(C_2{O}_4{)}_3]$ are diamagnetic.
III. All the geometrical isomers of $\left[M\right(AB\left)b_{2}cd\right]$ are optically active.

• A. I only
• B. I and II
• C. III only
• D. All of these

Answer 1

The correct option is B I and II

$Ni(CO{)}_4$:
$CO$ is a neutral ligand. Hence, the charge on Ni is zero.
$\text{Outer electronic configuration of Ni}:3d^{8}4s^2$

It has $s{p}^3$ hybridisation. Thus, it is tetrahedral.

$Ni(PP{h}_3{)}_{2}C{l}_2$:
$PP{h}_3$ is a weak field ligand of pairing of electrons does not occurs.
Here, the $Ni$ is in +2 oxidation state.
$\text{Outer electronic configuration of}N{i}^{+2}:3d^8$
It also has $s{p}^3$ hybridisation and the geometry is tetrahedral.

$K_2\left[Pt\right(CN{)}_4]$
Here, $CN$ is a strong field ligand so pairing occurs.
The oxidation state of $Pt$ is +2.

Hence, it is a diamagnetic.

$N{a}_3\left[Co\right(C_2{O}_4{)}_3]$:
Oxalate is a bidentate ligand and strong field ligand.
Oxidation state of $Co$ is +3
Outer electronic configuration of $Co$ is $3d^6$
Since, oxalate is strong field ligand all electrons are paired so it is a diamagnetic compound.

Not all geometrical isomers of $\left[M\right(AB\left)b_{2}cd\right]$ is optically active.
Thus, correct option is (b).