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coordination compounds Flashcards

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1
Q

moleculare compounds

A

2 or more simple salt combined together in a fixed proportion by their WEIGHT

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2
Q

DOUBLE SALT

A

dissociate into ions completely when dissolved in water
e.g: potash alum. karnalite , mohrs salt

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3
Q

complex compounds

A

it doesn’t dissociate into ions completely when dissolvwd in water

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4
Q

WERNERS THEORY OF CC

A
  1. Metals have two types of links (valences) in coordination compounds: primary and secondary
  2. Negative ions satisfy the primary valances, which are generally ionisable
  3. Secondary valences cannot be ionised. Neutral molecules or negative ions satisfy these needs. For a metal, the secondary valence is equal to the coordination number and is constant
  4. Ions/groups coupled to the metal by secondary connections have distinct spatial configurations that correspond to different coordination numbers
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5
Q

Coordination entity

A

a CMA/i BOUNDED TO A FIXED NO. OF IONS OR MOLECULES

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6
Q

CMA/I
LIGAND

CMA also reffered to as LEWIS ACID

A
  1. the atom which has fixed no. of atoms or molecule bounded to it in a definite geometric arrangements
  2. the ions or molecules bounded to the CMA/i
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7
Q

UNIDENTATE
AND EXAMPLE

A

WHEN ligand is bounded to CMA/i through **single donor atom **

Cl, H2O, NH3

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8
Q

DIDENTATE
AND EXAMPLE

A

WHEN ligand is bounded to CMA/i THROUGH two donor atom

en or C2O42-

ethane - 1,2- diammine

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9
Q

POLYDENTATE AND EXAMPLE

A

WHEN LIGAND is bounded to CMA/i thru more than 2 donor atom

EDTA4-
ethylenediamminetetraacetate
hexadendate

2 N2 and 4O2

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10
Q

CHELATE LIGAND

chelate ring

A

WHEN di or polydendate ligand uses 2 or more donor atoms to bind to the CMA/I simultaneously

if they form ring

chelate ring = cordination no, - 1

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11
Q

DENTICITY

A

No, of ligating grp

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12
Q

AMBIDENDATE LIGAND

LIGETES

A

Ligand which have 2 different donor atom and either one of the 2 ligetes in the complex

e.g: 1. NO2
M<-ON=O or M<-NO2
NITRITO-O AND NITRITO-N
2. SCN
M<-SCN or M<-NCS
THIOCYANATO -S AND THIOCYANATO - N

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13
Q

COORDINATION NO.

A

THE NO. of ligands directly attached to the CMA/i

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14
Q

find the CN of:

  1. [PtCl6]^2-
  2. [Ni(NH3)4]+2
  3. [Fe(C2O4)3]3-
  4. [Co(en)3]3+

C2O42- and en are didendate

CN= denditicity * ligands

A
  1. 6
  2. 4
  3. 6
  4. 6

ONLY SIGMA BONDS

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15
Q

COORDINATION SPHERE

COUNTER ION

A

the CMA/ion and the ligand attached to it are enclosed in a square bracket

the ionisable group written outside the brackett

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16
Q

COORDINATION POLYHEDRON

A

THE spatial arrangement of ligands which are directly bounded to the CMA/i

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17
Q

name the spatial arrangement of the following

  1. [Ni(Co)4]
  2. PtCl4
  3. [Co (NH3)6]

3 most common spatial arrangement

A
  1. tetrahedral
  2. square planar
  3. octahedral
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18
Q

HOMOLEPTIC AND EXAMPLE

A

WHEN THE metal is bounded to only one kind of donor grp
[Co(NH3)6]3+

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19
Q

HETEROLEPTIC AND EXAMPLE

A

COMPLEXES IN WHICH METAL IS bounded to more than one kind of donor grp
[Co(NH3)4Cl2]+

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20
Q

Write the formulas for the following coordination compounds:

(i)Tetraamminediaquacobalt(IlI) chloride
(ii)Potassium tetracyanidonickelate(II)
(iii)Tris(ethanp-1,2-diamine) chromium(III) chloride
(iv)Amminebromidochloridonitrito-N- platinatc(II)
(v)Dichloridobis(ethane-l ,2-diamine) platinum (IV) nitrate
(vi)Iron(III)hexacyanidoferrate(II)

A
  1. [Co(NH3)4(H20)]Cl2
  2. K2[Ni(CN)4]
  3. [Cr(en)3]Cl3
  4. [Pt(NH3)BrClNO2]-
  5. [Pt Cl2(en)2] (NO3)2
  6. Fe4[Fe(CN)6]3
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21
Q

Write IUPAC names of following co-ordination compounds :
(a) [CO(NH3)6]Cl3
(b) [CO(NH3)Cl]Cl2
(C) K3[Fe(CN)6]
(d) [K3[Fe(C2O4)3]
(e) K2[PdCl4]
(f) [Pt(NH3)2ClNH2CH3]Cl

A

(a) hexaamminecobalt (III) chloride
(b) pentaamminechloridocobalt (III) chloride
(c) potassium hexacyanoferrate (III)
(d) potassium trioxalatoferrate (III)
(e) potassium tetrachloridoplatinum (II)
(f) diamminechlorido (methylamine) platinum(II) chloride.

22
Q

Using IUPAC norms, write the formulae for the following :
(a) tetrahydroxozincate(II)
(b) hexaammineplatinum (TV)
(c) potassiumtetrachloridopalladate(II)
(d) tetrabromidocuprate (II)
(e) hexaaminecobalt(III) sulphate
(f) potassiumtetracyanonicklate (II)
(g) potassiumtrioxalatochromate(III)
(h) pentaamminenitrito-O-cobalt(III)
(i) diamminedichloridoplatinum(II)
(j) pentaamminenitrito-N-cobalt (III).

A

(a) [Zn(OH)4]2-
(b) [Pt(NH3)6]4+
(c) K2[PdCl4]
(d) [Cu(Br)4]2-
(e) [CO(NH3)6]2 (SO4)3
(f) K2[Ni(CN)4]
(g) K3 [Cr(OX)3]
(h) [CO(NH3)5ONO]2+
(i) [Pt(NH3)2Cl2]
(j) [CO(NH3)5NO2]2+.

23
Q

Using IUPAC norms write the systematic names of the following:
(i) [Co(NH3)6]CI3,
(ii)[Pt(NH3)2CI (NH2CH3)] Cl
(iii) [Ti(H20)6]3+
(iv) [Co(NH3)4Cl(N02)]CI
(v)|Mn(H20)6]2+
(vi)[NiCl4]2-
(vii)[Ni(NH3)6]CI2
(viii)[Co(en)3]3+
(ix) [Ni(CO)4]

A

(i) Hexaammine cobalt (III) chloride.
(ii) Diammine chlorido (methylamine) platinum (II) chloride.
(iii) Hexaaquatitanium (III) ion.
(iv) Tetraammine chlorido nitrito-N-cobalt (IV) chloride.
(v)Hexaaquamanganese (II) ion.
(vi)Tetrachloridonickelate (II) ion.
(vii)Hexaammine nickel (II) chloride.
(viii)Tris (ethane -1,2-diamine) cobalt (III) ion.
(ix) Tetra carbonyl nickel (0).

24
Q

Linkage isomerism And EXAMPLE

A
  • ARISES IN coordination compound containing AMBIDENTATE LIGAND

[Co(NH3)5 (NO2)]Cl2
obtained in red form when nitrite ligand is bound thru OXYGEN
obtained in yellow form when nitrite ligand is bound thru NITROGEN

25
Q

COORDINATION isomerism and EXAMPLE

A
  • ARISES due to interchange of cationic and anionic entities of different metal ions

[Co(NH3)6] [Co(CN)6 ]
ITS COORDINATION ISOMERISM: [Cr(NH3)6] [Co(CN)6]

oxi no. is same: co3+ and cr3+

26
Q

IONISATION ISOMERISM AND EXAMPLE

A
  • arises when counter ion is itself a potential ligand
  • and displace the ligand which can then become counter ion

[Co(NH3)5(SO4)]Br and its ionisation isomerism:
[Co (NH3)5 Br]SO4

27
Q

SOLVATE ISOMERISM AND EXAMPLE

AKA HYDRATE ISOMERISM

A
  • water is involved as a solvent
  • similar to IONISATION isomerism
  • it differ by whether or not a solvent molecule is directly bonded to the metal ion or merely present as free solvent molecules in the crystal lattice

[Cr(H20)6]Cl3 - VIOLET
and its solvate isomer[Cr (H2O)5Cl]Cl2. H2O- grey green

28
Q

limitation of werners theory

A

It could not explain
* the inability of all elements to form coordination compounds.
* the directional properties of bonds in various coordination compounds.
* the colour, the magnetic and optical properties shown by coordination compounds.

29
Q

How many geometrical isomers are possible in . the following coordination entities?
(i) [Cr(C2O4)3]3- (ii) [CoCl3(NH3)3]

A

(i) [Cr(C2O4)3]3- => No geometrical isomers

(ii) [Co(NH3)3 Cl3] => Two geometrical isomers are possible (fac and mer) in this coordination entity.

30
Q

Indicate the types of isomerism exhibited by the following complexes

(i)K[Cr(H2O)2(C2O4)2]
(ii)[CO(en)3]Cl3
(iii)[CO(NH3)5(NO2)(NO3)2], .
(iv)[Pt(NH3)(H2O)Cl2]

A
  1. a. geometrical isomerism of cis and trans
    b. optical isomerism of cis
  2. optical isomerism
  3. ionisation and linkage
  4. geometric
31
Q

why is geometrical isomerism for tetrahedral complexes not possible

A

bcz the relative positions of unidentate ligands with CMA are the same with respect to each other

32
Q

enantiomer

[Co(en)3]3+
[PtCl2(en)2]2+

A

non superimposable image
optically active

33
Q

limitations of VALENCE BOND THEORY

A
  1. involve no. of assumptions
  2. no quantitative interpretation of magnetic data
  3. no explanation about colour exhibited by coordination compounds
  4. it doesnt explain about the thermodynamic or kinetic stabilities of CC
  5. no exact prediction btwn tetrahedral and square planar structures of 4 cordinate complexes
  6. doesnt distinguish btwn weak and strong ligand
34
Q

sp3d2

A
  1. weak ligand
  2. outer orbital
  3. high spin
  4. presence of unpaired electron
  5. paramagnetic
  6. OCTAHEDRAL
35
Q

d2sp3

A
  1. strong
  2. low spin
  3. diamagnetic
  4. inner orbital
  5. OCTAHEDRAL
  6. No unpaired e-
36
Q

sp3

A
  1. tetrahedral
  2. outer orbital complex
  3. high spin
  4. paramagnetic
37
Q

dsp2

A
  1. square planar
  2. low spin
  3. diamagnetic
  4. inner orbital complex
38
Q

SPECTROCHEMICAL SERIES

I BRought Some CoLourful Sweets From Office COntaining H2O. Nisha EDiTs Nine ENglish Cyan COupons

A

weak ligand:
1. I
2. Br
3. SCN
4. Cl
5. S
6. F
7. OH
8. C2O42-
9. H2O
strong ligands
1. NCS
2. EDTA4-
3. NH3
4. en
5. CN
6. CO

39
Q

CRYSTAL FIELD SPLITTING

A

Thus, the repulsions in octahedral coordination compound yield two energy levels:

t2g– set of three orbitals (dxy, dyz and dxz) with lower energy
eg – set of two orbitals (dx2-y2 and dz2) with higher energy

This splitting of degenerate level in the presence of ligand is known as crystal field splitting.

40
Q

IN OCTAHEDRAL COMPLEX WHAT DOES DEGENERATE ORBITALS SPLIT INTO?

A

t2g– set of three orbitals (dxy, dyz and dxz) with lower energy
eg – set of two orbitals (dx2-y2 and dz2) with higher energy

41
Q

WHY does CRYSTAL SPLITTITNG OCCUR IN OCTAHEDRAL CC

A

😎repulsion between the electrons in d orbitals and ligand electrons.

This repulsion is experienced more in the case of** dx2-y2 and dz2 orbitals as they point towards the axes along the direction of the ligand.
Hence, they have higher energy than average energy in the spherical crystal field.
On the other hand, dxy, dyz, and dxz **orbitals experience lower repulsions as they are directed between the axes.
Hence, these three orbitals have* less energy *than the average energy in the spherical crystal field.

42
Q

WHAT does CFS depend on

A

field produced by ligand
and
charge on metal ion

43
Q

how would u determine where the fourth e- entering the eg or pair with t2g e-

A

relative magnitude of cfs delta0
and
pairing energy

pairing energy: energy required for e- pairing in single orbital

44
Q

if relative magnitude of CFS< PAIRING ENERGY

A

4TH e- enters eg
ligands are weak
form high spin complex
config: t23g eg1

45
Q

if relative magnitude of CFS > PAIRING ENERGY

A

4th e- occupy t2g
config: t2g4 eg0
strong field ligands
low spin complexes

d4 to d7

46
Q

treatment of LEAD POISONING

A

EDTA

47
Q

CIS PLATIN

A

PT
INHIBIT growth of tumours

48
Q

wilkinson catalyst

A

rhodium complex

49
Q

bonding in carbonyl complex

A
  1. synergic bonding provides stability
  2. metal carbon bond contain both pi and sigma character
  3. M-C sigma bond is formed by donation of lone pair of e- on CO into a vacant orbital of metal
  4. M-C pi bond is formed by the donation of a pair of e- from a filled d orbital of metal into vacant antibonding pi orbital of CO
50
Q

THE LIGAND TO METAL IS WHICH BOND IN METAL CARBONYLS

A

σ

51
Q

THE METAL TO LIGAND IS WHICH BOND IN METAL CARBONYLS

A

π

Chemistry (9 decks)

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