A complex compound is formed by ligands nitrate and chloride. It gives two moles of AgCl precipitate with $AgN{O}_3$. What will its formula be? 

1. $\left[Co{\left(N{H}_3\right)}_{5}N{O}_3\right]C{l}_2$

2. $\left[Co{\left(N{H}_3\right)}_{5}Cl\right]N{O}_{3}Cl$

3. $\left[Co{\left(N{H}_3\right)}_{4}C{l}_2\right]N{O}_3$

4. $[Co{\left(N{H}_3\right)}_{4}ClN{O}_3]Cl$

The shape of Fe(CO)₅ is:

Mg is present in:

Number of isomers of $[Pt$ ${\left(N{H}_3\right)}_4]$ $\left[CuC{l}_4\right]$ complex are:

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:

Anticancer agent among the following is:

1.		2.
3.		4.

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 possesses a negative charge.