Short & long Answer Questions: Coordination Compounds - 2 - JEE PDF Download

Q1. Using IUPAC norms write the formula for the following: Tetrahydroxozincate(II)

Ans. [Zn(OH₄)^2-

Q2. Using IUPAC norms write the formula for the following: Hexaamminecobalt(III)

sulphate

Ans. [Co(NH₃)₆]₂(SO₄)₃

Q3. Using IUPAC norms write the formula for the following: Pentaamminenitrito-cobalt(III)

Ans. [Co(ONO) (NH₃)₅]²⁺

Q4. Using IUPAC norms write the systematic name of the following: [Co(NH₃)₆]Cl₃

Ans. Hexaamminecobalt(III) chloride

Q5. Using IUPAC norms write the systematic name of the following:

[Pt(NH₃)₂Cl(NH₂CH₃)]Cl

Ans. Diamminechlorido(methylamine) platinum(II) chloride

Q6. Using IUPAC norms write the systematic name of the following: c[Cr(C₂O₄)₃]³⁺
Ans. Tris(ethane-1, 2-diammine) cobalt(III) ion 

Q7. What is Spectro chemical series? Explain the difference between a weak field ligand and a strong field ligand.

Ans. A Spectro chemical series is the arrangement of common ligands in the increasing

order of their crystal-field splitting energy (CFSE) values.

I- < BR - < S₂- < SCN- < Cl-<N₃,<F-<OH-<C₂O₄₂-~H₂O-~NCS-~H-<CN-<NH₃<en~ SO₃₂-<NO₂-<phen<CO
Q8. [Cr(NH₃)₆]³⁺is paramagnetic while [Ni(CN)₄]²⁻is diamagnetic. Explain why?

Ans. Cr is in the +3 oxidation state i.e., d3 configuration. Also, NH₃ is a weak field ligand

that does not cause the pairing of the electrons in the 3d orbital. Cr³⁺:

Therefore, it undergoes d²sp³ hybridization and the electrons in the 3d orbitals remain

unpaired. Hence, it is paramagnetic in nature.

In [Ni(CN)₄]²⁻, Ni exists in the +2 oxidation state i.e., d⁸configuration. Ni²⁺:

CN⁻ is a strong field ligand. It causes the pairing of the 3d orbital electrons. Then, Ni²⁺undergoes dsp² hybridization.

Q.9 A solution of [Ni(H₂O)₆ ] ²⁺ is GREEN BUT A SOLUTION OF [Ni(CN)₄ ] ^2- colourless. Explain.

Ans. In [Ni(H₂O)₆]²⁺is a weak field ligand. Therefore, there are unpaired electrons

in Ni²⁺. In this complex, the d electrons from the lower energy level can be excited to the

higher energy level i.e., the possibility of d−d transition is present. Hence, Ni(H₂O)₆]²⁺is n coloured.
 In [Ni(CN)₄]^2-, the electrons are all paired as CN^- is a strong field ligand. Therefore, d-d transition is not possible in [Ni(CN)₄]²⁻. Hence, it is colourless. As there are no unpaired electrons, it is diamagnetic.
Q.10  What is meant by stability of a coordination compound in solution? State the factors which govern stability of complexes.
Ans. The stability of a complex in a solution refers to the degree of association between the two species involved in a state of equilibrium. Stability can be expressed quantitatively in terms of stability constant or formation constant.
M + 3L ↔ ML₃ 

Stability constant

For this reaction, the greater the value of the stability constant, the greater is the

proportion of ML₃ in the solution.
LONG ANSWER TYPE QUESTIONS

Q1. Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:

i. [Fe(CN)₆]⁴⁻

ii. [FeF₆]³⁻

iii. [Co(C₂O₄)₃]³⁻

iv. [CoF₆]³⁻

Ans. i. [Fe(CN)₆]⁴⁻ In the above coordination complex, iron exists in the +II oxidation

state. Fe²⁺: Electronic configuration is 3d₆ Orbitals of Fe²⁺ion:

As CN⁻is a strong field ligand, it causes the pairing of the unpaired 3d electrons. Since

there are six ligands around the central metal ion, the most feasible hybridization is d²sp³. d²sp³ hybridized orbitals of Fe²⁺ are:

 
6 electron pairs from CN⁻ions occupy the six hybrid d₂sp³ orbitals.Then,

Hence, the geometry of the complex is octahedral and the complex is diamagnetic (as there are no unpaired electrons).

ii. [FeF₆]³⁻ In this complex, the oxidation state of Fe is +3.

Orbitals of Fe⁺³ ion:

There are 6 F− ions. Thus, it will undergo d²sp³ or sp³d² hybridization. As F⁻is a weak

field ligand, it does not cause the pairing of the electrons in the 3d orbital. Hence, the

most feasible hybridization is sp³d².sp³d² hybridized orbitals of Fe are:

Hence, the geometry of the complex is found to be octahedral.

iii. [Co(C₂O₄)₃]³⁻ Cobalt exists in the +3 oxidation state in the given complex. Orbitals of

Co³⁺ion: Oxalate is a weak field ligand. Therefore, it cannot cause the pairing of the

3d orbital electrons. As there are 6 ligands, hybridization has to be either sp³d² or d²sp³ hybridization. sp³d² hybridization of Co³⁺ :

The 6 electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand)

occupy these sp³d² orbitals.

Hence, the geometry of the complex is found to be octahedral.

iv. [CoF₆]³⁻ Cobalt exists in the +3 oxidation state.
Orbitals of Co³⁺ion:

Again, fluoride ion is a weak field ligand. It cannot cause the pairing of the 3d

electrons. As a result, the Co³⁺ion will undergo sp³d² hybridization.sp³d² hybridized

orbitals of Co³⁺ion are:

Hence, the geometry of the complex is octahedral and paramagnetic.