Level - I: Solutions (Set-1)
Level - I: Solutions (Set-1)
Level - I: Solutions (Set-1)
Chapter 9
Coordination Compounds
Solutions (Set-1)
SECTION - A
School/Board Exam. Type Questions
Very Short Answer Type Questions :
1. Identify the complex ion and counter ion in K4[Fe(CN)6].
Sol. [Fe(CN)6]4– and K+ respectively.
2. Write coordination number of central metal atom in [Pt(en)2Cl2].
Sol. 6
3. Calculate oxidation number of underlined atom in K3[Fe(C2O4)3].
Sol. +3
4. Calculate EAN of underlined atom in [Cr(NH3)6]Cl3.
Sol. 33
5. Write IUPAC name of Na[Co(CO)4].
Sol. Sodium tetracarbonylcobaltate(–I)
6. Assign the charge (x) on coordination sphere [Ni(DMG)2]x.
Sol. 0 as Ni(II)
7. Why does ammonia readily form complex while ammonium ion does not?
Sol. Ammonium ion (NH4 ) neither has lone pair of electrons nor a vacant orbital.
8. What is the charge (x) present on [Fe(CO)4]x?
Sol. x = O
9. Draw the structure of fac-triaquatrinitrito-N-cobalt(III).
Sol. H2O
O2N H2O
Co
O2N H2O
NO2
10. Calculate magnetic moment in Ni(CO)4.
Sol. Zero, as it has no unpaired electron.
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2 Coordination Compounds Solutions of Assignment (Set-1) (Level-I)
K2[PtCl6]
2K+ + [PtCl4]2–
Mn+2
13. Aqueous solution of potassium ferrocyanide does not give test of iron(II) and it is not poisonous like potassium
cyanide. Why?
Sol. Being a complex salt, it ionizes to 4K+ and [Fe(CN)6]4– ions. Absence of Fe(II) does not give the test of iron.
Absence of free CN– makes it non-poisonous.
14. What is difference between oxygenation and oxidation?
Sol. In oxygenation, O2 ligand is incorporated intact while in oxidation, it loses its identity.
15. Which type of d-electron configuration exhibit both low and high spin in octahedral complexes?
Sol. d4, d5, d6, d7.
16. All the octahedral complexes of Ni2+ are outer orbital complexes, why?
8
3d
+2
Sol. Ni :
Thus, only one 3d-orbital is available if all electrons paired up due to strong field ligand. Therefore, d2sp3
hybridisation is not possible. Only sp3d2 is possible which represents outer orbital complex.
17. NH2 – NH2 although possesses two electron pair for donation but not acts as chelating agent. Why?
Sol. The coordination by NH2 – NH2 leads to a three membered highly unstable strained ring and thus it does not
act as chelating agent.
18. Why complex [Al(NH3)6]3+ does not exist in aqueous solution?
Sol. [Al(NH3)6]3+ undergoes the change into [Al(H2O)6]3+ in aqueous medium due to higher heat of hydration of
aluminium ion on account of its small size.
[Al(NH3)6]3+ + 6H2O [Al(H2O)6]3+ + 6NH3
19. SCN– shows linkage isomerism in coordination compounds. Explain.
Sol. Because it is ambidentate ligand.
20. Explain and arrange the following complexes in increasing order of molar conductivity on the basis of Werner
theory :
(i) [Cr(NH3)3(NO2)3]
(ii) K[Co(NH3)2(NO2)4]
(iii) K2[Cr(NH3)(NO2)5]
Sol. i < ii < iii
Molar conductivity number of ions
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Solutions of Assignment (Set-1) (Level-I) Coordination Compounds 3
21. How would you distinguish between [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4 by chemical test?
Sol. [Co(NH3)5SO4]Br will give yellow precipitate of AgBr with AgNO3, while [Co(NH3)5Br]SO4 will give white
precipitate of BaSO4 with BaCl2.
22. On the basis of VBT, explain geometry, nature of hybridisation and magnetic property of [Co(ox)3]3–.
Geometry – octahedral
24. K2[PtCl6] is a well known compound but K2[NiCl6] does not exist. Why?
Sol. Both Pt and Ni are in (IV) oxidation state and sum of (IE1 + IE2 + IE3 + IE4) is quite high for Ni in comparison
to Pt.
25. Ni(CO)4 possesses tetrahedral geometry, while [Pt(NH3)4]2+ is square planar. Why?
Sol. Ni(CO)4 possesses sp3 hybridisation and then tetrahedral, whereas [Pt(NH3)4]2+ possesses dsp2 hybridisation,
thus square planar.
Sol. [CuI4]2– decomposed as [CuI4]2– 2CuI + 3I2. The instability may be explained as it is complex of oxidising
agent (Cu2+) and reducing agent (I–). Also, I– is poor electron donor than Cl– and steric effect may also play
some role.
28. What are essential requirements for regarding a compound as an organometallic? Give some examples.
Sol. In organometallic compounds, carbon forms bond with atom (metal/non-metal) which is less electronegative
than carbon. For e.g., B(CH3)3, SiCl3(CH3) etc.
29. CO and N2 are isoelectronic, but CO forms a number of complexes while N2 forms very few. Explain.
Sol. N2 is weaker -donor being very symmetrical and weaker -acceptor than CO, so N2 complexes are not much
stable.
Sol. [Cu(en)2]2+ has two rings in the structure. On the other hand, [Fe(EDTA)]– has five rings in structure.
31. Identify (A) and (B) in the given sequence of reaction. Also write their IUPAC names and calculate spin only
magnetic moment of B.
– –
Fe3 (aq)
SCN (excess)
A
F (excess)
B
(blood red) (colourless)
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4 Coordination Compounds Solutions of Assignment (Set-1) (Level-I)
F–
3
Sol. Fe SCN Fe(SCN)3 [FeF6 ]3–
– (excess )
A = Trithiocyanoferrum(III)
B = Hexafluoroferrate(III) ion
F–, being weak, [FeF6]3– shows sp3d2 hybridisation with five unpaired electrons.
5(5 2) 35 5.92 BM
32. A compound Co(en)2 (NO2)2Cl exists in different isomeric forms. If it does not show optical activity, reacts
with AgNO3 but not with ethane-1,2-diammine, identify its structure.
Sol. It is trans isomer. It reacts with AgNO3, so Cl atom is ionizable. It does not react with ‘en’, so two NO2 groups
are not adjacent.
NO2
en Co en Cl
NO2
Sol. A = K2[Ni(CN)4]
B = K2[NiCl4]
A is square planar with dsp2 configuration. Due to absence of unpaired electron, it is diamagnetic (zero magnetic
moment).
2(2 2) 8 BM
(i) [Co(NH3)6]3+
(ii) [Rh(NH3)6]3+
(iii) [Ir(NH3)6]3+
Co, Rh and Ir belongs to 3d, 4d and 5d transition series respectively and CFSE increases by 30% between
two adjacent members down the group.
35. A solution containing 0.319 gram of complex CrCl3 6H2O was passed through cation exchanger and the
solution given out was neutralized by 28.5 ml of 0.125 M NaOH. What is the correct formula of complex?
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Solutions of Assignment (Set-1) (Level-I) Coordination Compounds 5
Sol. The Cl atoms outside the coordination sphere will be ionized to produce acid HCl.
Thus, meq. of Cl– ions outside = meq. of HCl formed
= meq. of NaOH used
= 28.5 × 0.125
= 3.56
Molar mass of CrCl3 6H2O = 266.5
0.319 103
millimoles of complex = 1.20
266.5
3.56
1 millimole complex gives = millimole Cl– = 3
1.20
37. Draw the structures of [Co(NH3)6]3+, [Ni(CN)4]2– and Ni(CO)4 and write hybridisation of each.
Sol. (i) [Co(NH3)6]3+ octahedral, d2sp3
+
NH3
H3N NH3
Co
H3N NH3
NH3
2–
NC CN
Ni
NC CN
Ni CO
OC
CO
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6 Coordination Compounds Solutions of Assignment (Set-1) (Level-I)
O
CH2 NH2 NH2 CH2 CH2 NH2 O C
Sol.
Pt ; Pt
C O O C C O NH2 CH2
O cis-form O O trans-form
SECTION - B
Model Test Paper
Very Short Answer Type Questions :
1. Identify the cation and anion in the complex [Pt(py)4][PtCl4].
Sol. Cation = [Pt(py)4]2+, Anion = [PtCl4]2–.
2. Give IUPAC name for [Mn3(CO)12].
Sol. Dodecacarbonyltrimanganese(0)
3. What is hybridisation of [Cr(NH3)6]3+?
Sol. d2sp3
4. What is the relationship between CFSE (o) and CFSE (t)?
4
Sol. o t
9
5. Will [NiCl4]2– show geometrical isomerism?
Sol. No, because it is tetrahedral complex.
6. Why [Zn(NH3)4]2+ solution is colourless?
Sol. Because it has 3d10 configuration of Zn+2.
Sol. In d2sp3, (n – 1)d z 2 and (n – 1)d x 2 – y 2 orbitals are involved in hybridisation because they are more directional
toward coordinate axis.
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Solutions of Assignment (Set-1) (Level-I) Coordination Compounds 7
8. Explain why secondary valencies are directional in nature.
Sol. Because, geometry of complex and stereoisomerism due to different arrangements of electron donor groups
are decided by secondary valencies.
Sol. KCl does not precipitize Ag+ in presence of aqueous NH3 due to formation of complex [Ag(NH3)2]Cl.
10. Identify the geometrical shapes of complexes formed by reaction of Ni2+ with Cl–, CN– and H2O respectively
with hybridisation.
Sol. Tetrahedral,
3
square planar and octahedral
3 2
( sp ) ( dsp 2 ) ( sp d )
11. Why ammonia forms the complex [Cu(NH3)4]2+ with copper ions in alkaline solutions but not in acidic solution?
Sol. In acidic solution protons coordinate with ammonia forming NH4 ions and NH3 molecules are not available for
coordination.
SCN– is weak ligand, does not pair up electrons. Co+2 has three unpaired electrons in 3d7 configuration. Hence,
3(3 2) 15 B.M.
13. Identify the oxidation states of Fe in the complex formed by reaction of FeCl3 and K4[Fe(CN)6].
14. A complex Co(en)2 (NO2)2Cl shows optical activity, and it reacts with both AgNO3 and ethane-1,2-diammine.
What is the structure of complex? Explain.
Sol. Optical activity shows that it is cis-isomer. Reaction with AgNO3 proves that Cl atom is ionizable. Reaction
with ‘en’ proves that two –NO2 groups are adjacent to each other. Hence, structure is
en
NO2
Co Cl
NO2
en
15. Among the complexes [Ti(NO3)4], K3[VF6], [CuNC(CH3)4]+BF4– and [Cr(NH3)6]Cl3; which are expected to be
coloured? Explain.
Sol. [Cr(NH3)6]Cl3 and K3[VF6] are coloured because Cr3+ and V3+ have 3d3 and 3d2 configuration respectively and
thus show d-d transition. Ti+4 and Cu+ has 3d0 (empty) and 3d10 (fully filled) configuration, hence colourless.
16. Classify Na2[CrOF4] and [Cr(H2O)6]Cl3 between cationic and anionic complex. Explain.
Sol. In Na2[CrOF4], central atom is involved in anionic part [CrOF4]2–, hence it is anionic complex.
In [Cr(H2O)6]Cl3, central atom is involved in cationic part [Cr(H2O)6]3+, hence it is cationic complex.
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8 Coordination Compounds Solutions of Assignment (Set-1) (Level-I)
17. Are geometrical isomers of [M(NH3)4Cl2] optically active? Explain with structures.
Sol. No, both are optically inactive, because both possess axis of symmetry.
Cl Cl
H3N NH3 H3N Cl
M M
H3N NH3 H3N NH3
Cl NH3
trans-form cis-form
18. Explain –NO2 and –CN are ambidentate ligands with structures.
Sol. An ambidentate ligand has two sites on different species available for coordination, only one is used at a time.
O
In NO2 N and O–N=O
O
(nitrito-N) (nitrito-O)
In CN C N and N C
(cyano) (iso cyano)
19. A metal complex having composition Cr(NH3)4Cl2Br has been isolated in (A) and (B). (A) reacts with AgNO3
forms white precipitate readily soluble in dilute aq. NH3, while (B) gives pale yellow precipitate in concentrated
ammonia solution. Identify (A) and (B) with hybridisation of chromium and calculate spin only magnetic moment.
AgNO
Sol. [Cr(NH3 )4 Cl2 ]Br
3
[Cr(NH3 )4 Cl2 ] NO3– AgBr
(yellow ppt.)
AgNO
[Cr(NH3 )4 BrCl]Cl
3
[Cr(NH3 )4 BrCl] NO3– AgCl
(white ppt.)
n(n 2) B.M.
3(3 2) 15 B.M.
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Solutions of Assignment (Set-2) (Level-I) Coordination Compounds 9
Solutions (Set-2)
Objective Type Questions
(Classification of Ligands and Nomenclature of Coordination Compounds)
1. The total number of electrons donated by ligands to platinum ion in [Pt(en)2Cl2] is
(1) 8 (2) 10 (3) 12 (4) 14
Sol. Answer (3)
In [Pt(en)2Cl2], there are 2 monodentate and 2 bidentate ligands.
Hence, electrons donated by ligands = 2 2 + 2 4
=4+8
= 12
Primary valency = 2
Cu is bonded to 4, NH3 ligands.
Secondary valency = 4
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10 Coordination Compounds Solutions of Assignment (Set-2) (Level-I)
12. Aqueous solution of CoCl3 6NH3 upon addition with AgNO3 produces 3 moles white precipitate. Primary and
secondary valency of metal in this complex is
(1) 3, 6 (2) 2, 6 (3) 3, 3 (4) 6, 4
Sol. Answer (1)
3 moles white ppt. 3 moles of AgCl
3Cl– are outside the coordination sphere
Primary valency = 3
6 NH3 are co-ordinated with Co, in the coordination sphere Secondary valency = 6.
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Solutions of Assignment (Set-2) (Level-I) Coordination Compounds 11
14. Which of the following octahedral complexes do not show geometrical isomerism?
(1) [Co(NH3)3Cl3] (2) [PtCl2(NH3)4] (3) [Pt(NH3)2Cl2] (4) [Co(en)3]3+
Sol. Answer (4)
[Co(en)3]3+ will not show geometrical isomerism, because only one structure is possible. (en is bidentate ligand).
16. Which one of the following complexes will have six isomers?
(1) [Co(en)NH3Cl2]Cl (2) [Cr(H2O)4Cl2]Cl (3) [Co(ox)3]3– (4) [Co(en)2Br2]Cl
Sol. Answer (4)
[Co(en)2Br2]Cl will show 6 isomers.
[Cr(en)2Br2] Br
en Cr en
Br
en en
Br Br
Cr Cr
Br Br
en en
3
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12 Coordination Compounds Solutions of Assignment (Set-2) (Level-I)
n(n + 2) 1(1 + 2) 3 BM
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Solutions of Assignment (Set-2) (Level-I) Coordination Compounds 13
Sol. Answer (4)
25. The spin only magnetic moment of [MnBr4]2– is 5.9 B.M. Geometry of the complex ion is
(1) Tetrahedral (2) Octahedral (3) Square planar (4) Pentagonal pyramidal
Sol. Answer (1)
[MnBr4]2– Mn2+ 3d5
Given that M = 5.9 BM
Let the number of unpaired electrons be n
n(n + 2) 5.9
n=5
sp3 hybridization Tetrahedral
26. In the formation of octahedral complex, ligands approach towards _______ and _______ orbital of central metal.
(1) d xy , d 2 2
x –y
(2) dx2 – y 2 ,dz2 (3) dxy, dyz (4) d z 2 , d xz
29. According to crystal field theory, five d-orbitals of an octahedral complex split to give
(1) Two orbitals with lower energy and three orbitals with higher energy
(2) Three orbitals with lower energy and two orbitals with higher energy
(3) One orbital with lower energy and four orbitals with higher energy
(4) Four orbitals with lower energy and one orbital with higher energy
Sol. Answer (2)
d-orbital splits in 2 energy levels i.e. t2g and eg. eg is higher (2), t2g is lower (3).
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14 Coordination Compounds Solutions of Assignment (Set-2) (Level-I)
Trans platin ⇒ Pt
NH3 Cl
It is not an organometallic complex.
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