The correct splitting diagram of d orbitals in an octahedral crystal field is:

1.		2.
3.		4.	None of the above

Optical isomerism is exhibited by among the following is/are:

$1.$ ${\mathrm{K}}_2\left[\mathrm{Cr}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right]$
$2.$ ${\mathrm{K}}_3\left[\mathrm{Cr}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right]$
$3.$ ${\mathrm{K}}_3\left[\mathrm{Cr}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_2\right]$
$4.$ ${\mathrm{K}}_2\left[\mathrm{Cr}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_4\right]$

The correct representation of tetrahydroxidozincate (II) is:
1. \([Zn(OH)_4]^{2-}\)
2. \([Zn(OH)_4]^{-}\)
3. \([Zn(OH)_4]^{3-}\)
4. \([Zn(OH)_4]\)

A solution of [Ni(H₂O)₆]²⁺ is green, but a solution of [Ni(CN)₄]^2– is colorless because:

[Cr(NH₃)₆]³⁺ is paramagnetic ,while ${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$ is diamagnetic because:

The crystal field splitting energy is:

How many optical isomers of [Pt(NH₃)(Br)(Cl)(py)] can exist?

1. 3

2. 2

3. 1

4. 0

A metal ion present in vitamin B₁₂ is:

The hypothetical complex
chlorodiaquatriamminecobalt(III) chloride can be
represented as:

1. $\left[\mathrm{CoCl}{\left({\mathrm{NH}}_3\right)}_3{\left({\mathrm{H}}_2\mathrm{O}\right)}_2\right]{\mathrm{Cl}}_2$

2. $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_3\left({\mathrm{H}}_2\mathrm{O}\right){\mathrm{Cl}}_3\right]$

3. $\left[\mathrm{Co}{\left({\mathrm{NH}}_2\right)}_3{\left({\mathrm{H}}_2\mathrm{O}\right)}_2\mathrm{Cl}\right]$

4. $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_3{\left({\mathrm{H}}_2\mathrm{O}\right)}_3{\mathrm{Cl}}_3\right]$