The coordination compound that will give four isomers is:

1. [Fe(en)₃]Cl₃

2. [Co(en)₂Cl₂]Cl

3. [Fe(PPh₃)₃NH₃ClBr]Cl

4. [Co(PPh₃)₃Cl]Cl₃

The IUPAC name of $\left[Co{\left(N{H}_3\right)}_{3}ClBrN{O}_2\right]$ will be:

1. Triamminebromochloronitrocobaltate (III)
2. Triamminebromochloronitrocobalt (III)
3. Triamminebromonitrochlorocobalt (III)
4. Triamminenitrobromochlorocobalt (III)

$[Cu{\left(N{H}_3\right)}_4{]}^{+2}$ reacts with $HN{O}_3$ in excess of water and gives: 

1. $Cu{\left(OH\right)}_2$

2. $Cu{\left(N{O}_3\right)}_2$

3. $Cu{\left(H_{2}O\right)}^{-2}$

4. None of the above 

In [Cr(NH₃)₆](Br)₃, number of unpaired electrons in Cr is:

Which of the following does not have a metal-carbon bond:

1. ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{MgBr}$

2. $\mathrm{K}\left[\mathrm{Pt}\left({\mathrm{C}}_2{\mathrm{H}}_4\right){\mathrm{Cl}}_3\right]$

3. $\mathrm{Ni}(\mathrm{CO}{)}_4$

4. $\mathrm{Al}{\left({\mathrm{OC}}_2{\mathrm{H}}_5\right)}_3$

Anticancer agent among the following is:

1.		2.
3.		4.

In an octahedral structure, the pair of d orbitals involved in d²sp³ hybridization is:

1. ${\mathrm{d}}_{\mathrm{xz}},{\mathrm{d}}_{\mathrm{x}}^2-{\mathrm{y}}^2$

2. ${\mathrm{d}}_{\mathrm{z}}^2,{\mathrm{d}}_{\mathrm{xz}}$

3. ${\mathrm{d}}_{\mathrm{xy}},{\mathrm{d}}_{\mathrm{yz}}$

4. ${\mathrm{d}}_{\mathrm{x}}^2-{\mathrm{y}}^2,{\mathrm{d}}_{\mathrm{z}}^2$

The strong field nature of CN⁻ as a ligand can be attributed to:

1. It is a pseudohalide.

2. It can accept electrons from metal species.

3. It forms high-spin complexes with metal species.

4. It carries a negative charge.

Considering H₂O as a weak field ligand, the number of unpaired electrons in \([Mn(H_{2}O)_{6}]^{2+}\) will be:
(At. no. of Mn = 25)

Among [Ni(CO)₄], [Ni(CN)₄]²⁻, [ NiCl₄]²⁻ species, the hybridization states of the Ni atom are respectively:
(At. No. of Ni = 28)