If 1/3rd of the total chlorine of the compound is precipitated by adding AgNO₃ to its aqueous solution, then the most likely structure of CrCl₃.6H₂O among the following is:

1.	CrCl3.6H2O	2.	[Cr(H2O)3Cl3](H2O)3
3.	[CrCl2(H2O)4]Cl.2H2O	4.	[CrCl.(H2O)5]Cl2.H2O

The correct statement among the following is: 

The complex that does not have a tetrahedral shape is:

1. ${\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}$             

2. $\left[Ni{\left(CO\right)}_4\right]$

3. ${\left[NiC{l}_4\right]}^{2-}$                     

4. ${\left[Zn{\left(N{H}_3\right)}_4\right]}^{2+}$

The complex that contains only one unpaired electron is:

1. [CoF₆]^3-

2. [Co(C₂O₄)₃]^3-

3. [Fe(CN)₆]^3-

4. [MnCl₆]^3-

Total geometrical isomers possible for [Pd²⁺(NH₂-CH(CH₃)-CO₂^-)₂]⁰ are -

1. 2

2. 3

3. 4

4. 6

Isomerism exhibited by [Co(EDTA)]^- is/are:
 

1. Optical & geometrical isomerism.

2. Geometrical isomerism.

3. Optical isomerism.

4. No isomerism is exhibited by [Co(EDTA)]^-

The total number of possible co-ordination isomers for the following compounds are:

$\left[\mathrm{Co}{\left(\mathrm{en}\right)}_3\right]\left[Cr{\left(C_2{O}_4\right)}_3\right]$
$\left[Cu{\left(N{H}_3\right)}_4\right]\left[CuC{l}_4\right]$
$\left[Ni{\left(en\right)}_3\right]\left[Co{\left(N{O}_2\right)}_6\right]$

1. 4, 4, 4

2. 2, 2, 2

3. 2, 2, 4

4. 4, 2, 4

Which of the following complex ions is Bronsted-Lowry acid?

1. [Cu(NH₃)₄]²⁺

2. [FeCl₄]^-

3. [Fe(H₂O)₆]³⁺

4. [Zn(OH)₄]^2-

The expected number of unpaired electrons for the complex ion [Cr(NH₃)₆]³⁺ are: 

1. 2

2. 3

3. 4

4. 5

The correct order of energies of d-orbitals of metal ions in a square planar complex is:

1. ${\mathrm{d}}_{\mathrm{xy}}={\mathrm{d}}_{\mathrm{yz}}={\mathrm{d}}_{\mathrm{zx}}>{\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}={\mathrm{d}}_{{\mathrm{z}}^2}$

2. ${\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}={\mathrm{d}}_{{\mathrm{z}}^2}>{\mathrm{d}}_{\mathrm{xy}}={\mathrm{d}}_{\mathrm{yz}}={\mathrm{d}}_{\mathrm{zx}}$

3. ${\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}>{\mathrm{d}}_{{\mathrm{z}}^2}>{\mathrm{d}}_{\mathrm{xy}}>{\mathrm{d}}_{\mathrm{yz}}={\mathrm{d}}_{\mathrm{zx}}$

4. ${\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}>{\mathrm{d}}_{\mathrm{xy}}>{\mathrm{d}}_{{\mathrm{z}}^2}>{\mathrm{d}}_{\mathrm{zx}}={\mathrm{d}}_{\mathrm{yz}}$