• +91 9971497814
  • info@interviewmaterial.com

Chapter 9- Coordination Compounds Interview Questions Answers

Question 1 :

Write the formulas for the followingcoordination compounds:

Answer 1 :

(i) Tetraamminediaquacobalt(III) chloride

(ii) Potassium tetracyanonickelate(II)

(iii) Tris(ethane−1,2−diamine) chromium(III)chloride

(iv) Amminebromidochloridonitrito-N-platinate(II)

(v) Dichloridobis(ethane−1,2−diamine)platinum(IV)nitrate

(vi) Iron(III) hexacyanoferrate(II)


Answer

Question 2 :

Write the IUPAC names of the followingcoordination compounds:

Answer 2 :

(i) [Co(NH3)6]Cl3

(ii) [Co(NH3)5Cl]Cl2

(iii) K3[Fe(CN)6]

(iv) K3[Fe(C2O4)3]

(v) K2[PdCl4]

(vi) [Pt(NH3)2Cl(NH2CH3)]Cl


Answer

(i) Hexaamminecobalt(III) chloride

(ii) Pentaamminechloridocobalt(III) chloride

(iii) Potassium hexacyanoferrate(III)

(iv) Potassium trioxalatoferrate(III)

(v) Potassium tetrachloridopalladate(II)

(vi) Diamminechlorido(methylamine)platinum(II)chloride

Question 3 : Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers:

Answer 3 :

1.      K[Cr(H2O)2(C2O4)2

2.      [Co(en)3]Cl3

3.      [Co(NH3)5(NO2)](NO3)2

4.      [Pt(NH3)(H2O)Cl2]


Answer

1 Bothgeometrical (cis-, trans-) isomers for canexist. Also, optical isomers for cis-isomer exist.

Trans-isomer is opticallyinactive. On the other hand, cis-isomer is optically active.

                                       

Two optical isomers for  exist. 

Two optical isomers are possible for thisstructure.

   3 

A pair of optical isomers:

It can also show linkage isomerism.

and 

It can also show ionization isomerism.

  4  Geometrical (cis-trans-) isomers of  can exist.

Question 4 :

Giveevidence that [Co(NH3)5Cl]SO4 and[Co(NH3)5SO4]Clare ionization isomers.

Answer 4 :

Ni isin the +2 oxidation state i.e., in d8 configuration.

Thereare 4 CN ions. Thus, it can either have atetrahedral geometry or square planar geometry. Since CN ionis a strong field ligand, it causes the pairing of unpaired 3d electrons.

Itnow undergoes dsp2 hybridization. Since all electrons arepaired, it is diamagnetic.

Incase of [NiCl4]2−, Cl ionis a weak field ligand. Therefore, it does not lead to the pairing of unpaired3d electrons. Therefore, it undergoes sp3 hybridization.

Since there are 2unpaired electrons in this case, it is paramagnetic in nature.

Question 5 :

[NiCl4]2− isparamagnetic while [Ni(CO)4] is diamagneticthough both are tetrahedral. Why?

Answer 5 :

Thoughboth [NiCl4]2− and [Ni(CO)4]are tetrahedral, their magnetic characters are different. This is due to adifference in the nature of ligands. Cl is a weak fieldligand and it does not cause the pairing of unpaired 3d electrons.Hence, [NiCl4]2− is paramagnetic.

InNi(CO)4, Ni is in the zero oxidation state i.e., ithas a configuration of 3d8 4s2.

                     

ButCO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons.Also, it causes the 4s electrons to shift to the 3d orbital, thereby givingrise to sp3 hybridization.Since no unpaired electrons are present in this case, [Ni(CO)4]is diamagnetic.

Question 6 :

[Fe(H2O)6]3+ isstrongly paramagnetic whereas [Fe(CN)6]3− isweakly paramagnetic. Explain.

Answer 6 : In both and , Fe exists in the +3 oxidation state i.e.,

 in d5 configuration.

SinceCN is a strong field ligand, it causes thepairing of unpaired electrons. Therefore, there is only one unpaired electronleft in the d-orbital.

Therefore

Onthe other hand, H2O is a weak field ligand. Therefore, itcannot cause the pairing of electrons. This means that the number of unpairedelectrons is 5.

Therefore,

Thus, it is evident that  is strongly paramagnetic, while is weakly paramagnetic

Question 7 :

Explain[Co(NH3)6]3+ isan inner orbital complex whereas [Ni(NH3)6]2+ isan outer orbital complex.

Answer 7 :

Question 8 :

Predictthe number of unpaired electrons in the square planar [Pt(CN)4]2− ion.

Answer 8 :

Inthis complex, Pt is in the +2 state. It forms a square planar structure. Thismeans that it undergoes dsp2 hybridization.Now, the electronic configuration of Pd(+2) is 5d8.

CN being a strong field ligand causes thepairing of unpaired electrons. Hence, there are no unpaired electrons in

Question 9 :

The hexaquomanganese(II) ion contains five unpaired electrons, while the hexacyanoioncontains only one unpaired electron.

Answer 9 :

 Explain using Crystal Field Theory.

Answer

Question 10 :

Calculatethe overall complex dissociation equilibrium constant for the Cu(NH3)42+ ion, given that β4 forthis complex is 2.1 × 1013.

Answer 10 :

β=2.1 × 1013

Theoverall complex dissociation equilibrium constant is the reciprocal of theoverall stability constant, β4.


Selected

 

Chapter 9- Coordination Compounds Contributors

krishan

Share your email for latest updates

Name:
Email:

Our partners