Chapter 9 Coordination Compounds

_{K is called the equilibrium constant( or binding constant) and tells us how strongly the ligands bind to the metal centre. So, the higher the value of K, the greater the stability of the complex formed.}

K=

Greater the value of K, the equilibrium is favoured more towards the side of the formation of complex, which is possible only if it is very stable.

Thus, Option B is the correct answer.

According to the spectrochemical series the increasing order of field strength is H₂O<NH₃<CN^-. So, greater the field strength, greater is the crystal field splitting and the energy difference between t_2g and e_g orbitals will be higher. If the energy is higher, then the wavelength will be smaller.

E=

Energy(E) and wavelength(λ) are inversely related. So, the order in which the ligands absorb wavelength will be the reverse of the spectrochemical series.

H₂O>NH₃>CN^-

So, from the above explanation we can conclude that Option C is the correct answer.

When 0.1 mole of CoCl₃(NH₃)₅ was reacted with excess of AgNO₃, we get 0.2 moles of AgCl. So, there are two chloride ions that are free and not part of the complex. The formula for complex has to be [Co(NH₃)₅Cl]Cl₂.

[Co(NH₃)₅Cl]Cl₂→[Co(NH₃)₅Cl]⁺ + 2Cl^-

Therefore, the conductivity of the solution will be 1:2 electrolyte

If we get 3 moles of AgCl, then it means that all three chloride ions are free and not bound to the complex because ligands bound to the complex do not separate out and form precipitates. So, the formula of the complex has to be [Cr(H₂O)₆]Cl₃.

1) Cations are named first and anions are named later. In this case, there are no anions.

2) The ligands are named in alphabetical order and their prefixes carry no significance. Anionic ligands have a suffix –o attached in the end. So, ammine comes first followed by chlorido

3) This is followed by the name of the central metal and its oxidation state which is platinum(II).

So, the IUPAC name for [Pt(NH₃)₂Cl₂] is Diamminedichloridoplatinum (II).

Chelate effect is predominant in polydentate ligands as they form stronger complexes. A polydentate ligand has more than one place of attachment which makes it very stable. The only polydentate ligand in the given options is oxalate(bidentate) so it will form the most stable complex.

Octahedral complexes of the type [MX₂L₄] where X and L are unidentate ligands or

[MX₂(L-L)₂] where X is unidentate and L-L represents a bidentate ligand show geometrical isomerism. They exist as cis or trans. The other options given do not fall under this category and hence do not show geometrical isomerism

Crystal field splitting energy for tetrahedral is times that for octahedral

∆_t= ∆_o

∆_t= ×18,000 cm^–1=8,000 cm^–1

Thus, Option C is the correct answer.

Ligands that have two different bonding sites are called ambident ligands. Few examples are SCN^-,NO₂ etc…

SCN^-→ Metal or NCS→Metal

So, [Pd(C₆H₅)₂(SCN)₂] and [Pd(C₆H₅)₂(NCS)₂] are linkage isomers because they differ in the binding site of the ligands which leads to different linkage.

Whereas, a coordination isomer is a form of structural isomerism in which the composition of the complexion varies. In a coordination isomer the total ratio of ligand to metal remains the same, but the ligands attached to a specific metal ion change.

For Example: A solution containing ([Co(NH₃)₆]³⁺) and [Cr(CN)₆]^3- is a coordination isomer with a solution containing ([Cr(NH₃)₆] and [Co(CN)₆]).

Ionisation Isomers: These are identical except for a ligand has exchanged places with an anion or neutral molecule that was originally outside the coordination complex. The central ion and the other ligands are identical.

The two compounds are completely different and therefore will not show isomerism because for showing isomerism the basic criteria is both of the compound should have same molecular formula and in the given question both of the compounds have different formula.

Thiosulphato or S₂O₃^- is a monodentate ligand and is therefore not a chelating agent. A chelating agent has to have more than one point of attachment(it has to be polydentate)

Ligands should have a lone pair of electrons that they can donate to the central metal atom to form a complex. NH₄⁺ has no lone pair that it can donate to the central metal atom.

The structure of NH₄⁺ is given below and we can clearly see that there is no lone pair on the central atom Nitrogen;

The number of water molecules inside and outside the coordination complex differs. So, it has to be solvate isomerism. Solvate isomers tell us if the solvent is attached to the central atom or is outside the coordination sphere and free.

Ligands that have two different bonding sites are called ambident ligands. Few examples are SCN^-,NO₂ etc…

SCN^-→ Metal or NCS→Metal

So, [Pd(C₆H₅)₂(SCN)₂] and [Pd(C₆H₅)₂(NCS)₂] are linkage isomers because they differ in the binding site of the ligands which leads to different linkage.

Whereas, a coordination isomer is a form of structural isomerism in which the composition of the complexion varies. In a coordination isomer the total ratio of ligand to metal remains the same, but the ligands attached to a specific metal ion change.

For Example: A solution containing ([Co(NH₃)₆]³⁺) and [Cr(CN)₆]^3- is a coordination isomer with a solution containing ([Cr(NH₃)₆] and [Co(CN)₆]).

Ionisation Isomers: These are identical except for a ligand has exchanged places with an anion or neutral molecule that was originally outside the coordination complex. The central ion and the other ligands are identical.

1) Cations are named first and anions(counter ions) are named later. In this case, there are no anions.

2) The ligands are named in alphabetical order and their prefixes carry no significance. Anionic ligands have a suffix –o attached in the end. So, ammine comes first followed by chloride and then nitrito-N. Since, NO₂ is an ambidentate ligand. The letter N shows that the Nitrogen end of NO₂ is bonded to the central metal atom and not the oxygen end.

3) This is followed by the name of the central metal and its oxidation state which is platinum(II)

So, the IUPAC name for Diamminechloridonitrito-N-platinum (II)

Strong field ligands cause pairing of electrons

Electronic configuration of

1) Co³⁺ - [Ar]

Number of unpaired electrons =0

So, [Co(NH₃)₆]³⁺ is diamagnetic.

2) Mn³⁺

As, Mn forms low spin compound with CN lignad, so

[Ar]

Number of unpaired electrons=2

So, [Mn(CN)₆]^3- is paramagnetic

3) Fe²⁺

[Ar] 3d⁴ 4s⁰ and in d orbital there will be pairing because CN is a strong ligand.

Number of unpaired electrons=0

So, [Fe(CN)₆]^4- is diamagnetic

4) Fe³⁺

[Ar]

Number of unpaired electrons =1 because as 3d orbital is half filled and we all know that half filled orbital is more stable than pairing.

So, [Fe(CN)₆]^3- is paramagnetic

[MnCl₆]^3–

Mn³⁺(3d⁴)

[CoF₆]^3–

Co³⁺ (3d⁶)

Both Co³⁺ and Mn³⁺ have the same number of unpaired electrons=4

Thus, Options A & C are the correct answers.

Fe³⁺

[Ar]

Number of unpaired electrons =1

So, [Fe(CN)₆]^3– is paramagnetic.

Thus,

Pink colour is due to transition of electrons from t_2g to e_g orbitals. When an excess of HCl is added, [Co(H₂O)₆]²⁺ is transformed into [CoCl₄]^2–. Also, tetrahedral complexes have smaller crystal field splitting than octahedral complexes, so the crystal field splitting energy is low, the wavelength absorbed will be higher(towards red) but the complementary colour observed will be blue.

Thus, Options B & C are the correct answers.

Homoleptic complexes are attached to only one kind of ligand. Here, in option (i)[Co(NH₃)₆]³⁺ and (iii)[Ni(CN)₄]^2– are bonded to only one ligand which is NH₃ and CN^- respectively so these two complexes are homoleptic.

Whereas, in options (ii) and (iv), there are two types of ligands are connected with Central Atom and thus these are not correct.

Thus, Options A and C are the correct answers.

Heteroleptic complexes are attached to more than one kind of ligand. Here, in Option B[Fe(NH₃)₄ Cl₂]⁺ and Option D[Co(NH₃)₄Cl₂] are attached to two different ligands each and hence they are heteroleptic complexes.

Whereas, in option A and C, there is only one type of ligand is connected with Central Atom.

Thus, Options B & D are the correct answers.

Optically active compounds have non-superimposable mirror images.

So, only [Co(en)₃]³⁺ and cis– [Co(en)₂ Cl₂]⁺ have mirror images that are non-superimposable and hence optically active.

The structure of ethane-1,2-diamine is given below;

Ethane- 1,2- diamine( -NH₂-CH₂-CH₂-NH₂-) is a neutral ligand and it is bidentate.

The point of attachment is the –NH₂ group with a lone pair on it. It is a chelating ligand because it has more than one point of attachment and therefore stabilizes the complex it forms with the central metal atom.

Ambidentate ligands have more than one binding site. Both NO₂ and SCN are ambidentate ligands and will therefore show linkage isomerism.

1) SCN^-→ M NCS→M

Isothiocyanato Thiocyanato

2) MßNO₂ MßO-N=O

Nitrito-N Nitrito-O

The increasing order of conductivity is as follows:

[Co(NH₃)₃Cl₃]<[Co(NH₃)₄Cl₂]Cl< [Cr(NH₃)₅Cl]Cl₂<[Co(NH₃)₆]Cl₃

These complexes have 0,1,2 and 3 free chlorine ions outside the coordination sphere respectively. More free ions not bound to a complex, greater will be the conductivity.

If it forms silver chloride then there is one free chlorine atom outside the coordination sphere. The structural formula has to be [Cr(H₂O)₄Cl₂]Cl.

There are one cation and one anion which corresponds to a total of two ions which show conductivity. The name of this complex is tetraaquadichlorido chromium(III) chloride.

The structure has to be a cis-octahedral structure.

Example: [Co(en)₂Cl₂]⁺ is optically active

A magnetic moment of 5.92 BM means there are 5 unpaired electrons because

Magnetic Moment = √ n(n+2)

There are four ligands attached to Mn²⁺ so the geometry has to be either square planar(dsp²) or tetrahedral(sp³). It can’t be square planar because all the five d orbitals in Mn²⁺ are singly filled and not vacant. So, this means the geometry is tetrahedral with 5 unpaired electrons giving a magnetic moment of 5.92 BM.

Weak field ligands do not cause much crystal field splitting, so the crystal field splitting energy is lesser than pairing energy ( ∆_o<P ). Since pairing energy is higher than crystal field splitting energy, the electron will prefer not to pair up, therefore they will form paramagnetic(have unpaired electrons) octahedral complexes but for strong-field ligands, pairing of electrons is preferred because pairing energy is lower than crystal field splitting energy(∆_o>P). A route where lesser energy is spent will always be preferred because the complex will be more stable in this case.

For tetrahedral complexes, the crystal field splitting energy is too low. It is lower than pairing energy so, the pairing of electrons is not favoured and therefore the complexes cannot form low spin complexes.

∆_t= ∆_o

So, the tetrahedral complexes don’t form in low spin complexes.

The crystal field splitting energy is lesser than pairing energy( ∆_o<P) for complexes containing weak field ligands. So, pairing here is not preferred(higher energy, unstable)

The crystal field splitting energy is greater than pairing energy( ∆_o>P) for complexes containing strong field ligands. So, pairing is preferred here(lower energy, stable)

[Three of the t_2gorbitals can hold two electrons each so they can hold a total of 6 electrons.

Two of the e_g orbitals can hold two electrons each so they can hold a total of 4 electrons.]

[CoF₆]^3–

F is a weak field ligand. Pairing is not preferred. so the two electrons instead of pairing in t_2g

Co³⁺(d⁶)has electronic configuration t_2g⁴ e_g²

So, the two electrons instead of pairing in t_2g go to e_g orbitals.

[Fe(CN)₆]^4–

CN is a strong field ligand. Pairing is preferred.

Fe²⁺(d⁶) has electronic configuration t_2g⁶ e_g⁰

So, all the six electrons get paired up in t_2g and no electron enters the e_g orbitals

[Cu(NH₃)₆]²⁺

NH₃ is a strong field ligand. Pairing is preferred.

Cu²⁺ (d⁹) has electronic configuration t_2g⁶ e_g³

The six electrons first get paired up in t_2g and the remaining 3 electrons enter the e_g orbitals and they too get paired up leaving behind only one unpaired e_g electron.

[Fe(CN)₆]^3–

Fe³⁺ has six unpaired electrons. CN^- is a strong field ligand which will cause pairing of all the electrons. So, the electrons will start pairing leaving behind one unpaired electron. So, the magnetic moment will be = = 1.74 BM

For [Fe(H₂O)₆]³⁺, H₂O being a weak field ligand won’t cause pairing of electrons. So, the number of unpaired electrons will be 5. The magnetic moment will be = 5.92 BM

The increasing order of crystal field energy is

[Cr(Cl)₆]^3–<[Cr(NH₃)₆]³⁺ <[Cr(CN)₆]^3–

This is also the order of field strength of the ligands according to the spectrochemical series. A ligand with higher field strength will split the degenerate d orbitals more than a ligand with lower field strength.

Even if two compounds have similar geometry, they may have a different number of unpaired electrons depending on the kind of ligand present. A strong field ligand will cause pairing of electrons while a weak field ligand will not cause pairing. Pairing or not pairing will change the number of unpaired electrons, thereby affecting the magnetic moment.

In CuSO₄.5H₂O, there are water molecules that act as ligands. These ligands cause crystal field splitting and the d orbitals are no longer degenerate. The electrons can now be excited to a higher d orbital and therefore show colour. Whereas, in CuSO₄, there are no water molecules to act as ligands, so no crystal field splitting happens.

Ambidentate ligands are monodentate ligands that have two different binding sites. Few examples of ambidentate ligands are given below. M denotes the central metal ion.

1) SCN^-→ M NCS→M

Isothiocyanato Thiocyanato

In Isothiocynato, the binding atom is Nitrogen, while Thiocyanato, the binding atom to the central metal ion is Sulphur.

2) MßNO₂ MßO-N=O

Nitrito-N Nitrito-O

Similarly in Nitrito-N, the binding atom is Nitrogen, whereas in Nitrito-O, the binding atom is Oxygen.

The type of isomerism when ambidentate ligands are attached to central metal ion is called linkage isomerism, because they only differ in the atom that is linked to the central metal ion.

The generation of colours in coordination complexes indicates that some of the portions of the visible spectrum are being absorbed from the white light when it passes through a sample. Hence, the light emerges is no longer white. The observed colour of the complex is complementary to which is absorbed.(and this colour is the colour that is generated from the wavelength that is left over after absorption.)

• The correct answer will be the option (ii) A (4) B (3) C (2) D (1).

Hence, matching the two columns the right complex ion-colour arrangement will be:

A. [Co(NH₃)₆]³⁺ -4.Yellowish orange

B.[Ti(H₂O)₆]³⁺-3. Pale blue

C.[Ni(H₂O)₆]²⁺ -2. Green

D.[Ni (H₂O)₄(en)]²⁺(aq.) -1.Violet.

The correct answer will be the option - (i)A(5) B (4) C (1) D (2).

Hence, matching the two columns the right coordination compound-central metal atom combination will be:

A. Chlorophyll -5. magnesium

B. Blood pigment -4. iron

C. Wilkinson catalyst -1. rhodium

D. Vitamin B₁₂ -2.Cobalt.

• In the case of A. [Cr(H₂O)₆]^{3 +}: The central metal atom is Cr and oxidation state x=+3 as water molecules are neutral.

Therefore , having a d³ configuration (1s² 2s² 2p⁶ 3s² 3p⁶ 3d³).

Cr has +3 oxidation state with d³ configuration and left with 2 empty d orbitals and 1 empty 4s orbital which in interaction with 3,4p orbitals and attain the hybridisation d²sp³ and the 6 H₂O electrons will occupy these 6 hybridised orbitals. And the hybridisation scheme will be :

The orbital energy diagram of [Cr(H₂O)₆]³⁺is :

Hence, there will be 3 unpaired electrons.

• In case of B. [Co(CN)₄]^2– : The central metal atom is Co and oxidation state x= -4 (-1) – 2=+2,

Therefore, having a d⁷configuration (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷).Hence, the Co²⁺ is left with 1 3d orbital,1 4s orbital which in interaction with 2, 4p orbitals and attains hybridisation of dsp²and the 4CN^- electrons will occupy these 4 hybridised orbitals and the scheme will be:

The orbital energy diagram will be:

hence, there will be 1 unpaired electron.

• In the case ofC. [Ni(NH₃)₆]²⁺: The central metal atom is Ni and oxidation state x=+2,

Therefore, having a d⁸ configuration (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸).Hence, the Ni²⁺ is left with 2, 3d orbital,1 4s orbital which in interaction with 2, 4p orbitals and attain hybridisation of sp³d²and the 6, NH₃ electrons will occupy these 6 hybridised orbitals and the scheme will be:

The orbital energy diagram will be:

hence, there will be 2 unpaired electrons.

• In the case of D. [MnF₆]^4–: The central metal atom is Mnand oxidation state x=-6(-1) – 4 =+2

Therefore, having a d⁵configuration (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵).Hence, the Mn²⁺ is left with 1 4s, 2 of 4d (outer orbital complex) which in interaction with 3, 4p orbitals and attain hybridisation of sp³d²and the 6, F^-electrons will occupy these 6 hybridised orbitalsand the scheme will be:

hence, there will be 5 unpaired electrons as it a high spin complex.

The correct answer will be the option (i)A (3) B (1) C (5) D (2)

Hence, matching the two columns the right combination will be:

A. [Cr(H₂O)₆]³⁺ - 3. d²sp³, 3

B. [Co(CN)₄]^2–-1. dsp², 1.

C. [Ni(NH₃)₆]²⁺-5. sp³d², 2

D. [MnF₆]^4– -2. sp³d², 5

• A.[Co(NH₃)₄Cl₂]⁺ shows geometrical isomerism due to different geometric arrangements of the ligands.

• B Cis-[Co(en)₂Cl₂]⁺ shows optical isomerism because the two (Dextro d and laevo l) mirror images formed cannot be superimposed on one another.

• C.[Co(NH₃)₅(NO₂)]Cl₂ shows linkage isomerism because it contains ambidentate ligand NO₂.

• D.[Co(NH₃)₆][Cr(CN)₆] shows coordination isomerism due to interchanging of ligands between ionic entities.

The correct answer will be the option (iii) A (4) B (1) C (5) D (3).

Hence, matching the two columns the right combination will be:

A. [Co(NH₃)₄Cl₂]⁺ -4. geometrical

B. Cis-[Co(en)₂Cl₂]⁺-1. optical

C. [Co(NH₃)₅(NO₂)]Cl₂-5. linkage

D. [Co(NH₃)₆][Cr(CN)₆] -3. Coordination.

The correct answer will be the option (iii) A (5) B (1) C (4) D (2)

Hence, matching the two columns the right combination will be:

A. [Co(NCS)(NH₃)₅](SO₃) -5. + 3 (oxidation no.of Co, x =-1(-1)-(-2)=+3)

B. [Co(NH₃)₄Cl₂]SO₄ -1. + 4 (oxidation no. Of Co, x =-2x(-2)-1(-2)=+4)

C. Na₄[Co(S₂O₃)₃] -4.+ 2 (oxidation no. Of Co, x=-3(-2) – 4x(+1)=+2 )

D. [Co₂(CO)₈] –2. 0 . (as CO is a neutral ligand, therefore Co here has a zero oxidation state)

Assertion and the reason both are true, and the reason is the correct explanation of assertion.

When a solution of chelating ligand (like D-penicillamine) is added to solution containing toxic metals ligands (like excess iron and copper) the chelating ligand immediately transforms the metal ions by formation of stable complex, that is how toxic metal ions are removed by chelating ligands.

Eg- EDTA is an important chelating ligand.

Assertion and reason both are true, but the reason is not the correct explanation of assertion.

In [Cr(H₂O)₆]Cl₂ the oxidation state of central metal atom Cr is x=-(-2)=+2 and for [Fe(H₂O)₆]Cl₂ also the oxidation state of Fe is +2. Hence, both of them are in their lower oxidation states and that means they can further be oxidised and therefore they are reducing in nature. Also, they have enough d-electrons to donate to other ligands and have more compounds formed.

And the corresponding structures are:

and

Assertion and reason both true, the reason is the correct explanation of assertion. Ambidentate ligands are those which have two different donor atoms in one single ligand. And these types of ligands give rise to the formation of linkage isomerism where the isomers only differ in the linkage of the ligand atom to the central metal atom.

Eg- this can be observed in the case of [Co(NH₃)₅(NO₂)]Cl₂ complex.

• In one form, the complex appears red in which the nitrite ligand is bound through one of the oxygen atom (-ONO) to the Co atom.

• In the other form, the complex is obtained in the yellow form in which the nitrite ligand is bound through the nitrogen atom (-NO₂).

Assertion and reason both are true, but the reason is not the correct explanation of assertion.

Soln.: In the case of complexes type of MX₆ and MX₅L, plane of symmetry is always present as there is, only one or two types of ligands present. And the necessary condition for showing geometrical isomerism is that the complex must be of MX₄B₂ or [M (AB)₂X₂] type.

the plane of symmetry is aligned to the plane of any M-X.

the plane of symmetry is aligned to the plane of L-M-X.

The assertion is false butthe reason is true.

For [Fe(CN)₆]^4– the central metal atom is Fe and its oxidation state is x=-6 times(-1)-4=+2.

Hence, it has a d⁶ configuration (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶) and as it is left with 2 empty 3d orbitals and 1 empty 4s orbital which in interaction with 3,4p orbitals and attain the hybridisation d²sp³ and the 6 CN^- electrons will occupy these 6 hybridised orbitals. And the hybridisation scheme will be:

And the orbital energy level diagram is:

which clearly shows that it has no unpaired electrons as CN^- is strong field ligand and the energy gap becomes larger, therefore [Fe(CN)₆]^4– ion is diamagnetic and therefore shows no magnetic because of the absence of unpaired electrons.

(i)

• In [CoF₆]^3–, the central metal atom is cobalt Co and oxidation state of Co in hereis: x=-6 times (-1)-3=+3

Hence, Co³⁺has a d⁶ configuration (1s²2s²2p⁶3s²3p⁶3d⁶, while losing 1 3d and 2 4s electrons.) So the orbital energy level diagram is:

And the number of unpaired electrons=4.

Therefore, the magnetic moment of Co= = 4.9 BM. (BM stands for Bohr Magneton)

• For [Co(H₂O)₆]²⁺ the central metal atom is also Co and the oxidation state of Co here is x=2+; as aqua molecules are neutral.

Hence Co²⁺ has a d⁷ configuration (1s²2s²2p⁶3s²3p⁶3d⁷,while losing only the 2 4s electrons.) So the orbital energy level diagram is:

And the number of unpaired electrons=3.

Therefore, the magnetic moment of Co= = 3.87 BM. (BM stands for Bohr Magneton).

• For [Co(CN)₆]^3– the central atom is Co, the oxidation state of Co in here is x=-6 times (-1)-3=+3.

Hence, Co³⁺ has a d⁶ configuration (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶, while losing 1 3d and 2 4s electrons.) So the orbital energy level diagram is:

There will be no unpaired electrons as it is diamagnetic,(as CN^- is a strong field ligand, the energy gap becomes larger and all the electrons get paired) therefore no value of magnetic moment can be found.

(ii).

• In the case of [FeF₆]^3– the central metal atom here is Fe and the oxidation state of Fe x=-6 times (-1)-3=+3.

Hence, Fe³⁺ will have a d⁵ configuration (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵).So the orbital energy level diagram is:

Therefore it is paramagnetic and number of unpaired electrons are =5

Therefore the value of magnetic moment for Fe³⁺ = = 5.92 BM.

• For [Fe(H₂O)₆]²⁺the central metal atom is iron Fe and the oxidation state of Fe in here is x=2+.

Hence, it has a d⁶ configuration (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶).

So the orbital energy level diagram is:

Therefore, it has 4 numbers of unpaired electrons

Hence, the magnetic moment= = 4.9 BM. (BM stands for Bohr Magneton).

• For [Fe(CN)₆]^4– the central metal atom is Fe and its oxidation state is x=-6 times(-1)-4=+2.

Hence, it has a d⁶ configuration (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶).

So the orbital energy level diagram is:

It has no unpaired electrons as CN^-are strong field ligand and the energy gap becomes larger, therefore [Fe(CN)₆]^4– is diamagnetic.

• [Mn(CN)₆]^3–: The central metal atom is Mn and oxidation state is, x=-6(-1)-3=+3.

Therefore , having a d⁴ configuration (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴).

(i)Mn has +3 oxidation state with d⁴ configuration and left with 2 empty d orbitals and 1 empty 4s orbital which in interaction with 3,4p orbitals and attain the hybridisation d²sp³ and the 6 CN^- electrons will occupy these 6 hybridised orbitals. And the hybridisation scheme will be :

(ii) Since, in the formation of the complex [Mn(CN)₆]^3– inner d orbital (3d) is used, it will be an inner orbital complex.

(iii) The orbital energy diagram of [Mn(CN)₆]^3– is :

Hence, there will be 2 unpaired electrons and therefore, it will be paramagnetic.

(iv) The Spin only magnetic moment value for [Mn(CN)₆]^3– will be:

= 2.87 BM.

• [Co(NH₃)₆]^{3 +}: The central metal atom is Co and oxidation state is x=+3;as NH₃ is a neutral molecule.

Therefore, having a d⁶ configuration (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶) .

(i)Co has one +3 oxidation state with d⁶ configuration and left with 2 empty d orbitals and 1 empty 4s orbital which in interaction with 3,4p orbitals and attain the hybridisation d²sp³ and the 6 NH₃ electrons will occupy these 6 hybridised orbitals. And the hybridisation scheme will be :

(ii) Since in the formation of the complex [Co(NH₃)₆], ³⁺ inner d orbital (3d) is used, it will be an inner orbital complex.

(iii) The orbital energy diagram of [Co(NH₃)₆]³⁺ is :

Hence, there will be no unpaired electrons as ammonia is a strong field ligand and therefore, it will be diamagnetic.

(iv)The Spin only magnetic moment value for will be equal to zero because there are no unpaired electrons.

• [Cr(H₂O)₆]^{3 +}: The central metal atom is Cr and oxidation state is x=+3 as water molecules are neutral.

Therefore , having a d³ configuration (1s² 2s² 2p⁶ 3s² 3p⁶ 3d³).

(i)Cr has one +3 oxidation state with d³ configuration and left with 2 empty d orbitals and 1 empty 4s orbital which in interaction with 3,4p orbitals and attain the hybridisation d²sp³ and the 6 H₂O electrons will occupy these 6 hybridised orbitals. And the hybridisation scheme will be :

(ii) Since in the formation of the complex [Cr(H₂O)₆], ³⁺inner d orbital (3d) is used, it will be an inner orbital complex.

(iii) The orbital energy diagram of [Cr(H₂O)₆]³⁺is :

Hence, there will be 3 unpaired electrons and therefore, it will be paramagnetic.

(iv) The Spin only magnetic moment value for [Mn(CN)₆]^3– will be:

• = 3.87 BM

• [FeCl₆]^4–: The central metal atom is Fe and oxidation state is-

x=-6 (-1)-4=+2.

Therefore, having a d⁶ configuration (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶).

(i)He has a +2 oxidation state with d⁶ configuration and left with 2 empty outer d orbitals (4d not 3d as all the 3d will be occupied) and 1 empty 4s orbital which in interaction with 3,4p orbitals and attain the hybridisation sp³d² and the 6 Cl^- electrons will occupy these 6 hybridised orbitals. And the hybridisation scheme will be :

(ii) Since, in the formation of the complex [FeCl₆]^4–the outer d orbital (4d) is used, it will be an outer orbital complex.

(iii) The orbital energy diagram of [FeCl₆]^4–is:

Hence, there will be 4 unpaired electrons and therefore, it will be paramagnetic.

(iv)The Spin only magnetic moment value for [FeCl₆]^4–will be:

• = 4.9 BM.

(i). Isomer ‘A’ reacts with AgNO₃ and gives white precipitate, as the complex consists of Cl it reacts with the AgNO₃ and gives the precipitate of AgCl and the reaction will be:

White

Hence isomer ‘A’ is

Similarly, in case of isomer ‘B’ it reacts with BaCl₂ and gives white precipitate and it is obviously BaSO₄ and the reaction will be:

White

Hence, the isomer ‘B’ will be

(ii)In this case, the counter ion (Cl^-) in the complex salt is itself a potential ligand and displaces the ligand (SO₄) which can then become the counter ion, therefore, it is a form of ionisation isomerism.

(iii) IUPAC name of isomer ‘A’-Pentaamminesulphatocobalt (III) chloride

And‘B’-Pentaamminechlorocobalt (III) sulphate.

In case of transition metal complexes, wide range of colours are one of the most important properties which means some of the portions of the visible spectrum is being absorbed from the white light when it passes through a transition metal complex. Hence, the light emerges is no longer white. The observed colour of the complex is complementary to which is absorbed. And this complementary colour is the colour that is generated from the wavelength that is left over after absorption.

Eg: if the green light is absorbed by the complex it appears red.

According to the crystal field theory Higher the crystal field splitting, lower will be the wavelength absorbed by thecomplex. And there is no splitting and the ligand is absent the compound will appear colourless.

The CFSE or crystal field spli4tting energy of the octahedral and tetrahedral field is related as:

Where, ∆_t = crystal field splitting energy in the tetrahedral field

Δ_O= crystal field splitting energy in an octahedral field.

• Hence, the CFSE has a higher value in the octahedral field rather than in tetrahedral field which also accounts for greater splitting in the octahedral field for the same metal and the ligand. Higher the CFSE, higher will be the energy radiated in d-d transition which also means lower will be the wavelength of the light absorbed (as ).

• Another reason which can be taken into account for the difference in colour is that, the pairing of the unpaired d electrons in the octahedral complexes in the presence of strong ligand field, which does not occur in case of tetrahedral complexes as they are formed by using outer orbital electrons (low spin complexes) so, there will be no pairing even in the presence of strong ligand field.