Class XII Chemistry Co-Ordination compounds.pptx

Class XII Chemistry

1. Coordination Chemistry

Coordination chemistry is the study of compounds that have a central atom (often metallic)
surrounded by molecules or anions, known as ligands. The ligands are attached to the
central atom by dative bonds, also known as coordinate bonds, in which both electrons in
the bond are supplied by the same atom on the ligand.

2. #Coordinatiow Chemistry
a The Peccony (Yogesh Sharma) Metal compound having two or more ions that cannot

dissociate completely in aq solution is known as
* Coordination compounds are the compounds in which a central metal _¢omplex compound
atom or ion is linked to a fixed no of ions or molecules through coordinate bonds.

Donor of e- to CMI >

Metal that Gain e- from ligand Ligand —~ (coordinate bond) set
in a geometrical structure / 2

Co[{NH3'
ion ae
*If any element is present Coordination

sphere is Number —

No of ligand x ch )
counter ion o Coordination sphere — ké

* Ifall the ligands are removed the the remaining charge is the Oxidation number *the no of ligating atoms

present in a ligand is called
* Tf Metal is bound to only one type of ligend it is called Homoleptic Complex denticity Uni/Bi/Tri dentate

& Ambidentate ligand

+ IF Metal is bound to two types of ligend it is called Heteroleptic Complex

3. NAMING OF MONONUCLEAR COORDIBATION COMPOUNDS

1. Cation is first written (In both the cases +ve and —ve ) =
EXAMPLE:

Alphabetically
2, In case( +ve, & neutral, Common name)
LEGAND ‘+ In case (- ve), ligand ends in “ido” [ Cr(NHs)s (H20)s 1Cls
<> If Ligands include neumerical prefix the Bi, Tri used Triamminetriaquachromium (LI)
N If Complex charge +ve & neutral CMI Comman name chloride
If Complex charge is negative CMI ends to “ate” _ __ — —

VRITING FORMULA OF MONONUCLEAR COORDIBATION COMPOUNDS

1. Cation is first written (In both the cases Counter ion/Sphere)

EXAMPLE:
CMI written first
> Ligands alphabetically Potassium trioxalatoaluminate (III)
à en,ox, are written first in abb. ligand [AL (C200)s] -?
IN SPHERE —> If Polyatomic ligand(formula in parenthesis)
No space b/w the letters Triamminetriaquachromium (III)
If no counter ion (Charge on the right top ) chloride
[ Cr(NHs)s (H20)s ] Cl:

A +ve and charges dhould be balanced FT

4. On the basis of the following observations made with aqueous solutions,
assign secondary valences to metals in the following compounds:

5. Write the formulas for the following coordination compounds:
(a) Tetraammineaquachloridocobalt(II) chloride

(b) Potassium tetrahydroxidozincate(I)

(© Potassium trioxalatoaluminate(Ill)

(d) Dichloridobis(ethane- 1,2-diamine)cobalt(Ill)

(e) Tetracarbonvinickel(O)

6. Write the IUPAC names of the following coordination compounds:
(a) [Pt(NH,),CI(NO,)] (b) Ks[Cr(C,0,)s] (c) [CoCl,{en),]Cl
(d) [Co(NH¿)s(CO¿)]CI (e) Hg[Co(SCN),]

7. Structure Different

Types:
STRUCTURAL ISOMERISM
IONISATION SOLVATE / HYDRATE LINKAGE COORDINATION
ISOMERISM ISOMERISM ISOMERISM ISOMERISM

Complex with the same
composition give
different ions in solution

[Co(NHs)sBr] &
[Co(NHs)s SO:]Br

If the water is present in
the complex

[Cr(H20)6]Cls &
[Cr(H20):Cl2]C1.2H20

If the ambidentate
ligand is attached to the
CMI in the complex

[Co(NHs)s(ONO)]2* &
[Co(NHs)s(NO2)}?*

If two Coordination
spheres are present in
one complex

[Co(en)a][Cr(CN)o] &
[Cr(en)s][Co(CN)e]

8. u STEREOISOMERISM TAN

Geometrical Isomerism

Optical Isomerism

They are molecules that are locked into their spatial
positions wrt one another due to a double bond or a ring
structure. It arises in heteroleptic complexes having
coordination no. 4 and 6.

** When similar groups are present in adjacent position
, it is cis. When they are present in opposite position , it is
trans.
Structure ( MX:L:)( X and L are unidentate)

a ns Bing Zit
a “NH nag) Sa
cis trans

Optical isomers are mirror images that cannot be
superimposed on one another. These are called as
enantiomers. The molecules or ions that cannot be
superimposed are called chiral (all the valances should be
completed by different atoms).

The two forms are called dextro
(d) and laevo (1) depending upon the direction they rotate
the plane of polarised light in a polarimeter (d rotates to the
right, / to the left). Optical isomerism is common in
octahedral complexes involving bidentate ligands .In a
coordination entity of the type PtC12(en)2]2+, only thecis-
isomer shows optical activity.

9. Structure ( MX:L:)( X and L are unidentate) in which | Structure [ M(AA)s]

the two ligends x may be oriented cis or trans to each

other. si 3+
> +
a cl oR en
al Jan (3 )
co ER Pa:
NH, Zu Sm, NH, | NH, rs x en
NH, a C
cis trans en : ER
dextro mirror laevo

Structure[MX2(L - L )2] (NH2 CH2 CH2 NH2)

Qu A el A Structure [ M(AA): Be]
en Co,
YO RATS
en a

10. Structure[Ma3b3] [(Co(NH3)3(NO2)3]3
NH NH

©; de H ee
O. NH ON | NO,

ÑO, NH

Sac- mer-
*If the three donor atoms * If the three donor atom’s
Of the same ligand occupy positions are around the
adjacent positions of an meridian of the octahedron,
octahedral face, then it then it is meridional (mer)
is facial (fac) isomer. isomer

2%

** Tetrahedral complexes [MAs or MAsBs] do not show
geometrical isomerism because the relative positions of
unidentate ligands attached with central atom are the
same with respect to each other

11. Why is geometrical isomerism not possible in tetrahedral complexes
having two different types of unidentate ligands coordinated with
the central metal ion ? o

12. Draw structures of geometrical isomers of [Fe(NH.).(CN)¿J”

13. Out of the following two coordination entities which is chiral
(optically active)?

(a) cis-[CrCL(ox),]” (b) trans-[CrCl,(0x),J"

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

(1) K[Cr(H,0)2(C,0,)2 ii) [Cofen),]Cl,
(iii) [Co(NH,);(NO,)](NO3). (iv) [Pt(NH,)(H,0)Cl]

15. Give evidence that [Co(NH;),C1]SO, and [Co(NH,),(SO,)]Cl are ionisation
isomers.

16. * BONDING IN COORDINATION COMPOUND
*Werner's Theory (According to Werner's )
CMI consist of two types of valencies

* Primary Valancy (Metal's Oxidation Sate)

* Secondary Valaney (satisfied towards ligand, eq to coordination no)

*Valance Bond Theory (According to Linus Pauling & C. slater , depends on Magnetic behavior )

Metal ion must have vacant orbitals of equal energies, equal to the number of ligands to be attached.

In strong field ligand forces, the d-electrons of
central metal are pair up against Hund’s rule.
The strong field ligand which causes this are
CO, NO, CN-, NOz, NH, en. Some weak field
ligands are H20, X-, NO-3, ROH, ete

*Unpaired electron present (Paramagneetic)

1. Metal — ligand bond is formed by the donation of electrons by ligand to metal.
2.
3. Sometimes unpaired (n — 1)d electrons pair upto create empty orbitals.
4. Depending upon the types of Hybridisation the shape of the complexes are :
4 Sp? Tetrahedral
4 dsp? Square planner
5 Sp’d Trigonal bipyramidal
6 Spd? Octahedral
6 disp? Octahedral

*All paired electrons present (Diamagnetic)

17. Cucl53- J
Tetrahedral Complexes (sp?)

orbitals ofneion [etait]
3d As =
sp'hybridised
orbitals of Ni” Man
ag sp" hybrid
we ncomin PUN tt] Cape
Four pairs of electrons
from 4 Cr

Square planner Complexes (dsp? )

Orbitals of Nion [ti[tiftit]?

3d ES ap
dsp" hybridised
orbitals of Ni" tit IO LEZ |
3d dsp' hydrid ap
INiCN).
(low spin comptes) (Lt Aro] te raft slr ity)
3d Four pairs of electrons 4p

from 4 CN groups

18. edral C lexes (sp*d?

Orbitals of Co*ion [tt tır|t
E ES 1 td
sp'd hybridised Ya
orbitals of Co” tt tlt
3d sp'd hybrid 4d
Ca ©
(outer orbital or ttt) t]t AUNT LT O LEA
high spin complex) 3d Six pairs of electrons 4d

from six F ions

Octahedral Complexes (d?sp* )

Orbitals of co*ion [ti] t tt Ir :
3d ns nd
dsp hybridised Ya
orbitals of Co" [ty nn |
d’sp hybrid
{Co{NH,),I” G
finner orbital or [nm] one

low spin complex}

Six pairs of electrons
from six NH, molecules

19. LIMITATION OF VALENCE BOND THEORY

1. It fails to distinguish between weak and strong ligands

2. It could not give any satisfactory explanation for the colour of the complexes

3. It does not give an exact explanation of thermodynamics or kinetic tability of coordination compounds.
4. It does not give quantitative interpretations of magnetic moment data.

5. It fails to explain relative energies of different shapes.

6. It fails to predict tetrahedral and square planar structure of some 4 coordinate complexes .

7. It is based on the number of assumption.

**To know the solutions of these limitations physicist Hans Bethe & John Hasbrouck developed new theory called.

20. The spin only magnetic moment of [MnBr,]” is 5.9 BM. Predict the
geometry of the complex ion ?

21. Explain on the basis of valen: bond theory that [Ni(CN),]” ion with square
planar structure is di c and the [NiCL]” ion with tetrahedral
geometry is en : AS

22. [NiCl,]* is paramagn while [Ni(CO) ] is diamagnetic though both are
tetrahedral. Why? © Ñ -

23. [Fe(H,0),]** is sen isma gueto whereas [Fe(CN)¿”” is weakly
paramagnetic. Explain.

24. Explain [Co(NH.); oe an inner orbital complex whereas [Ni(NH,),]” is an

outer orbital co:

25. Predict the number of unpaired electrons in the square planar [Pt(CN),]* ion.

26. CRYSTAL FIELD THEORY

1. CFT describes the breaking of degeneracy of electron orbital state, usually d or f orbital, due to a
static electric field produced by surroundings charge distribution .

2. lit accounts for some magnetic properties, colors, hydration enthalpies & spinal structure of
transition metal complexes

3. Inan isolated gaseous atom/ion the five d orbitals contains the same energy. They are degenerate.

4, When the negative field is due to ligands in a complex, it become asymmetrical and the degeneracy
of the d orbitals is lifted. It results in splitting of the d orbitals.

27. Crystal field splitting in Octahedra & Tetrahedral coordination entities :
Tetrahedral

Octahedral
5 3/5 Ao z El
E Ei 5 A
a dl By 5 \
The higher energy set of orbitals (dxz, dyz, dxy) is labeled

eg - The higher energy set of orbitals (dz? and dx?-y2)
tug - The lower energy set of orbitals (dxy, dyz and dx2)

40 or 10 Dq - The energy separation between the two | ase.
levels The crystal field splitting in the tetrahedral field is

The ez orbitals are repelled by an amount of 0.6 Ao intrinsically smaller than in the octahedral field. For

The t.g orbitals to be stabilized to the extent of 0.4 Ao. most purposes the relationship may be represented as
At = 4/9 Ao

as tz and the lower 2 energy set (dz? and dx2-y’) is labeled

28. Electron Configuration in ‘Tetrahedral vs Octahedral’

= te Octahedral
Tetrahedral 45 PAR.

d elt? 064 # a tae? 0.440 ttes 0.440
& et? 064 Fi ask tee? 0840 tafes 0.840
e et? 064 days yes ha di tre? 1240 tae? 1.240
deft” 0.64 044 d' te 0.640 tage? 1.6 Ao
eet? 064 | |. #1. h $ tyeÿ 0040 tafes 2.0.40
d et? 064 d tates? 0.4 Ao tag°@q? 2.4 40
d@ et? 064 06m wake d tae? 0.8 Ao te! 1.8.40
d et 064 de tae? 1.240 tae? 1.2 Ao
d et? 0.64: dep de EA Pte” 0.6 Ao tes? 0.6 do
d et? 0.64 5 SS dy de hz 0 tae 0.040 tes‘ 0.040

& Tetrahedral Spherical Octahedral ee

Field Field Field

From d! to d‘ e: config. Is same
but from d to d% is diff. due to

Spectrochmical Series :

Ju — =

CO > CN> en > NH3 > edtat > NCS > H20 > C0 > It create two conditions :

OH: > F > S? > Cl > SCN: > I If Ao( splitting eneray ) > P ( Pairing energy )[weak field]

4% e- occupy tx orbital (t'es!)
If Ao (splitting energy) < PX Pairing energy [strong field]

Mm 4th e- occupy es orbital (tssÿest ) u

29. Colouring properrties of Transition Metals

* When light of vertain frequency falls on the ESA A

u 3 me 400 — 435 Yello - Green Violet
complex, it bserbs light from visible range for 255460 Yellow Blue
transition of electrons from lower d — energy 180 — 490 Oar Ca Bias
level to higher d — energy level. Colour of the 490 — 500 Red Blue Green
compoundis the complementary colour of the [500-560 | Purple | Green |
absorbed light.This is called d- d* transition. 560 - 580 Violet Yellow - Green
Ruby is aluminium oxide (Al2O3) containing Cr’* ‚580 — 595 Blue Yellow
Ion (d*) which is randomly distributed in between = Creal bis Drang
the AP* transmit red colour. Similarly in emerald S05 = 700) DCE pes

Cr** ions occupy some place in mineral beryl ( BesAleSicO1s) causing emerald to transmit light in the
green region.

30. Limitations of Crystal Field Theory

1.

The crystal field model is successful in explaining the formation, structures, colour and magnetic
properties of coordination compounds to a large extent. However, from the assumptions that the ligands
are point charges, it follows that anionic ligands should exert the greatest splitting effect. The anionic
ligands actually are found at the low end of the spectrochemical series. Further, it does not take into
account the covalent character of bonding between the ligand and the central atom. These are some of
the weaknesses of CFT, which are explained by ligand field theory (LFT) and molecular orbital theory
which are beyond the scope of the present study.

This theory take only — orbitals of a central atom into account. The S and P orbits are not considered for
the study.

The theory fails to explain the behavior of certain metals which cause large splitting while others show
small splitting. For example the theory has no explanation as to why H20 is a stronger ligand as
compared to OH-.

The theory rules out the possibility of having p bonding. This is a serious drawback because p bonding
is found in many complexes.

31. Compounds which have at least one metal — carbon bond are called organometallic compounds.
The metal-carbon bond in metal carbonyls possess both ( and (| character. The M-C [ bond
is formed by the donation of lone pair of electrons

on the carbonyl carbon into a vacant orbital of the = =

metal. The M-C bond is formed by the donation Q = DO
of a pair of electrons from a filled d orbital of metal M a c=00D)
into the vacant anti bonding orbital of carbon J Y
monoxide. The metal to ligand bonding creates a Fr

synergic effect which strengthens the bond between CO Synergic bonding

and the metal. In these the oxidation state of metal is zero. nens

32. Consider the following equilibrium between undissociated complex ion and dissociated ion.
[MLn]>+ Me + nl

The equilibrium constant Ke= [Me*][L=]"
[MLn)®]
The smaller the value of Ke, the grater is the stabilityof complexes ion and vice versa,
The reciprocal of equilibrium constant is called stability constant.
Ks = 1 = [ML]
Ke MIEL 7"
The higher the value of Ks, the more is the stability of complex ion.

33. The hexaquo manganese(Il) ion contains five unpaired electrons, while the
hexacyanoion contains only one unpaired electron. Explain using Crystal
Field Theory.