Bihar Board - Class 12 Chemistry - Chapter 9: Coordination Compounds Short Answer Question
Short Question Answer
Q. 1. Write the name of the coordination isomer of the complex [Co(en)3] [Cr(CN6)].
Answer ⇒ [Co(en)3] [Cr(CN6)]= tris (ethane1,2-diamine) chromium (III) hexacyanocobaltate (III).
Q. 2. How many geometrical isomers are possible in the following coordination entities ?
(i) [Cr(C2O4)3]3- (ii) [Co(NH3)3Cl3]
Answer ⇒ (i) Nil (ii) Two (fac and mer)
Q. 3. Square planar complexes with coordination number four exhibit geometric isomerism whereas tetrahedral complexes do not. Why ?
Answer ⇒ Tetrahedral complexes do not show geometrical isomerism because the relative positions of the ligand attached to the central metal atom/ion are the same witā respect to each other.
Q. 4. Write the IUPAC names of the following coordination compounds :
(a) [Co(NH3)6]Cl3,
(b) [Co(NH3)5Cl]Cl2,
(c) K3[Fe(CN)6],
(d) K3[Fe(C2O4)],
(e) K2[PdCl4],
(f) [Pt(NH3)2Cl(NH2CH3)]Cl
Answer ⇒ (a) Hexa-amino cobalt (III) chloride.
(b) Penta amine chlorido cobalt (III) chloride.
(c) Potassium hexacyano ferrate (III).
(d) Potassium trioxalatoferrate (III).
(e) Potassium tetrachloride palladate (II).
(f) Diammine chlorido (methylamine) platinum (II) chloride.
Q. 5. (a) What are ligands ? Give examples. (b) What are ambident ligands ? What is their importance ?
Answer ⇒ (a) The atoms or molecules or ions which donate a pair of electrons to the central metal atom and thus form a co-ordinate bond with the central metal atom. Common ligands are-NH3.CO ,Cn- , Cl- , I- , H2O etc.
(b) Ligands which can attach themselves (called ligating through two different atoms of the same molecule are called ambident ligands or group, e.g. NO2m SCN-. They introduce linkage isomerism in the complexes.
Q. 6. What will be the correct order for the wavelength of absorption in the visible region for the following:
[Ni- (NO2)6]4- , [Ni (NH3)6]2+, [Ni(H2O)6]2+
Answer ⇒ (i) [Ni- (NO2)6]4- =wavelength of light is 498 nm.
(ii) [Ni (NH3)6]2+= wavelength of light is 475 nm.
(iii) [Ni(H2O)6]2+= wavelength of light is 500 nm.
Q7. Explain the bonding in coordination compounds in terms of Werner’s postulates.
Answer ⇒
( a ) A metal shows two kinds of valencies viz primary valency and secondary valency. Negative ions satisfy primary valencies, and secondary valencies are filled by both neutral ions and negative ions.
( b ) A metal ion has a fixed amount of secondary valencies about the central atom. These valencies also orient themselves in a particular direction in the space provided to the definite geometry of the coordination compound.
( c ) Secondary valencies cannot be ionized, while primary valencies can usually be ionized.
Q.8. FeSO4 solution mixed with (NH4)2SO4 solution in 1 : 1 molar ratio gives the test of Fe2+ ion, but CuSO4 solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu2+ ion. Explain why.
Answer ⇒
FeSO4 solution, when mixed with (NH4)2SO4 in 1 : 1 molar ratio, produces a double salt FeSO4(NH4)2SO4-6H2O . This salt is responsible for giving the Fe2+.
CuSO4 mixed with aqueous ammonia in a ratio of 1 : 4 gives a complex salt. The complex salt does not ionize to give Cu2+, hence failing the test.
Q. 9. A solution of [Ni (H2O)6]2+ is green but a solution of [Ni(CN)4]2- is colorless. Explain.
Answer ⇒ Ni2+ ion in [Ni (H2O)6]2+ is green in color because it has two unpaired electrons in the 3d-orbitals and due to d-d The transition complex is green in color. On the other hand [Ni(CN)4]2- - has no unpaired electrons in the 3d-orbitals [3d10 configuration), hence it is colorless.
Q. 10. [Fe(CN)6]4- and [Fe(H2O)6]2+ are of different colors in dilute solutions. Why?
Answer ⇒ [Fe(CN)6]4- has no unpaired electrons due to d2sp3 hybridisation. All electrons in 3d orbitals are paired, whereas in [Fe(H2O)6]2+, there are four unpaired electrons in 3d orbitals and the hybridization involved in sp3d2 . In this complex d-d transitions are possible and therefore, it shows different colors.
Q. 11. Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex.
Answer ⇒ In the presence of NH3 the 3d electrons pair up leaving two d-orbitals empty to be involved in d2sp3 hybridization forming inner orbital complex in [Co(NH3)6]3+. In [Ni(NH3)6]2+ , Ni is in +2 oxidation state and has d8 configuration, the hybridization involved is sp3 d2 forming an outer orbital complex.
Q. 12. Predict the number of unpaired electrons in the square planar [Pt(CN)4]2- ion.
Answer ⇒ Since the complex [Pt(CN)4]2- -is square planar. Therefore the hybridization is dsp2. Hence the unpaired electrons in 5d orbitals pair up to make one d-orbital empty for dsp2 hybridization. Thus there is no unpaired electron.
Q. 13. Out of tetracarbonylnickel (O) and tetraammine copper (II) sulfate which is paramagnetic and why ?
Answer ⇒ Tetraamine copper (II) sulfate is paramagnetic because it has one unpaired 3d electron, wheres tetracarbonyl nickel (O) is diamagnetic as it has all paired electrons
Q. 14. Why only transition metals are known to from complexes ?
Answer ⇒ Transition metal ions have empty d-orbitals into which electron pairs can be donated by ligands containing electrons, e.g.,CH2=CH2 , C5H5 , C6H6 etc.
Q. 15. Do we call metal carbonyls as organometallics ? Why or Why not?
Answer ⇒ Yes metal carbonyls are called organometallics because C atom of CO is linked to the metal atom. If fact, the metal-carbon bonds have both and character.
Q. 16. Which type of ligands form chelastes ?
Answer ⇒ Oxidation state of Ni is 0.
Q. 17. What are ambidentate ligands? Give an example.
Answer ⇒ Unidentate ligands containing more than one coordinating atom are called ambidentate ligands. For
example - NO2- can coordinate through nitrogen or oxygen.
Q. 18. What is a bidentate ligand ?
Answer ⇒ A ligand which contains two donor atoms so that it can form two coordinate bonds with the central metal atom. e.g., ethylene, diammine NH2-CH2-CH2-NH2.
Q. 19. What is crystal field splitting energy ?
Answer ⇒ The difference of energy between the two sets of d-orbitals after splitting on the approach of the ligands is called crystal field splitting energy.
Q.20. What is meant by unidentate, bidentate and ambidentate ligands? Give two examples for each.
Answer ⇒
( i ) Unidentate ligands: These are ligands with one donor site. For example, Cl- , NH3
( ii ) Ambidentate ligands: These are ligands that fasten themselves to the central metal ion/atom via two different atoms.
Example NO2- or ONO- , CN- or NC-
( iii ) Bidentate: These are ligands with two donor sites.
For example, Ethane-1 , 2-diamine , Oxalate ion (C2O42-)
Q.21. What is the spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Answer ⇒ A series of common ligands in ascending order of their crystal-field splitting energy (CFSE) is termed as the Spectrochemical series.
Strong field ligands have larger values of CFSE. Whereas weak field ligands have smaller values of CFSE.
Q.22. What is the oxidation number of cobalt in K[Co(CO)4] ?
Answer ⇒ K[Co(CO)4]= K+ [Co(CO)4]-
We know,
x+0=-1 [ Where x is the oxidation number.] x= -1.
Q.23. What is a spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Answer ⇒ A spectrochemical set of ligands is the organization of their crystal field splitting energy (CFSE) values in the ascending order. The ligands on the R.H.S are strong ligands of fields whereas L.H.S are weak ligands of field. In d'orbitals, strong field ligands are more divided than weak field ligands. The series is given below:
I- < Br- < S2- < SCN- < Cl- < F- < OH- < C2O42- <O2- <H2O <
NCS- < NH3 < en < NO2- < CN- < CO
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