Short & long Answer Question(Part-2) - Coordination Compounds Class 12 Notes | EduRev

Chemistry Class 12

Created by: Mohit Rajpoot

Class 12 : Short & long Answer Question(Part-2) - Coordination Compounds Class 12 Notes | EduRev

The document Short & long Answer Question(Part-2) - Coordination Compounds Class 12 Notes | EduRev is a part of the Class 12 Course Chemistry Class 12.
All you need of Class 12 at this link: Class 12

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, NHis a weak field ligand

that does not cause the pairing of the electrons in the 3d orbital. Cr3+:
Short & long Answer Question(Part-2) - Coordination Compounds Class 12 Notes | EduRev

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+:
Short & long Answer Question(Part-2) - Coordination Compounds Class 12 Notes | EduRev

CN is a strong field ligand. It causes the pairing of the 3d orbital electrons. Then, Ni2+undergoes dsp2 hybridization.
Short & long Answer Question(Part-2) - Coordination Compounds Class 12 Notes | EduRev
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 Short & long Answer Question(Part-2) - Coordination Compounds Class 12 Notes | EduRev

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:
Short & long Answer Question(Part-2) - Coordination Compounds Class 12 Notes | EduRev

As CNis 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:

 Short & long Answer Question(Part-2) - Coordination Compounds Class 12 Notes | EduRev
6 electron pairs from CNions occupy the six hybrid d2sporbitals.Then,
Short & long Answer Question(Part-2) - Coordination Compounds Class 12 Notes | EduRev
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:
Short & long Answer Question(Part-2) - Coordination Compounds Class 12 Notes | EduRev

There are 6 F− ions. Thus, it will undergo d2sp3 or sp3d2 hybridization. As Fis 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:
Short & long Answer Question(Part-2) - Coordination Compounds Class 12 Notes | EduRev
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+ :
Short & long Answer Question(Part-2) - Coordination Compounds Class 12 Notes | EduRev

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

occupy these sp3d2 orbitals.
Short & long Answer Question(Part-2) - Coordination Compounds Class 12 Notes | EduRev
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:
Short & long Answer Question(Part-2) - Coordination Compounds Class 12 Notes | EduRev

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:

Short & long Answer Question(Part-2) - Coordination Compounds Class 12 Notes | EduRev
Hence, the geometry of the complex is octahedral and paramagnetic.

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