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)

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

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

CN^– is a strong field ligand. This is due to the fact that:

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)

Coordination compound that would exhibit optical isomerism is:

The atomic numbers of Cr and Fe are respectively 24 and 26. Among the following, the one that is paramagnetic with the spin of the electron is:

1. CrCO₆

2. FeCO₅

3. [Fe(CN)₆]^4-

4. [Cr(NH₃)₆]³⁺