Co-ordination Compounds MCQs for NEET — Chemistry Questions with Answers

In complex compounds, a ligand is an ion or molecule that donates a pair of electrons to the central metal atom or ion to form a coordinate covalent bond. Therefore, ligands are electron pair donors.

Common donor atoms in ligands include nitrogen (N) and oxygen (O). These atoms often have lone pairs of electrons that they can donate to the central metal atom or ion in a coordination complex. Therefore, both nitrogen and oxygen are common donor atoms.

Finely divided iron combines with carbon monoxide (CO) to form iron pentacarbonyl, $Fe(CO)_5$. This can be remembered as a specific reaction where iron and CO form a stable coordination complex.

The Ziegler-Natta catalyst is used in the polymerization of alkenes to produce polymers like polyethylene and polypropylene. This catalyst allows for the control of the polymer structure and molecular weight.

The compound $Fe_4[Fe(CN)_6]_3$ dissociates in water to produce 4 $Fe^{3+}$ ions and 3 $[Fe(CN)_6]^{4-}$ ions. The total number of ions in solution is 4 + 3 = 7. Therefore, the correct option is 7 ions. Hence, option o3 is correct.

The compound $[Co(H_2O)_6]Cl_2$ dissociates in water to form one $[Co(H_2O)_6]^{2+}$ ion and two $Cl^-$ ions. Thus, the total number of ions produced is 1 + 2 = 3. Therefore, the correct option is 3 ions. Hence, option o2 is correct.

The correct formula for dichlorido bis (urea) copper (II) is $ [CuCl_2(O=C(NH_2)_2)_2] $. In this complex, the copper ion (Cu^2+) is coordinated to two chloride ions (Cl^-) and two urea molecules. This is indicated by the bis (urea) and dichlorido in the name.

The correct formula for diammine silver (I) chloride is $ [Ag(NH_3)_2]Cl $. In this complex, the silver ion (Ag^+) is coordinated to two ammonia molecules (NH_3), forming the diammine complex, and there is one chloride ion (Cl^-) balancing the charge.

The correct formula for sodium nitroprusside is $ Na_2[Fe(CN)_5NO] $. In this complex, the iron (Fe) is in the +3 oxidation state and is coordinated to five cyanide ions (CN^-) and one nitrosyl group (NO). There are two sodium ions (Na^+) to balance the charge of the complex.

To determine the oxidation state of chromium in $Na_2[CrF_4O]$, we need to consider the charges of the other components in the complex. Sodium (Na) has an oxidation state of +1. Fluoride (F) has an oxidation state of -1, and oxygen (O) usually has an oxidation state of -2. Let the oxidation state of chromium be x. The net charge of the complex ion $[CrF_4O]^{2-}$ should be equal to -2. Therefore, the equation can be set up as follows:

$$2(+1) + x + 4(-1) + (-2) = 0$$

Solving for x, we get:

$$2 + x - 4 - 2 = 0$$

$$x = +4$$

Hence, the oxidation state of chromium in $Na_2[CrF_4O]$ is IV.