Coordination Compounds

${[Cr{(N{H_3})_6}]^{3 + }}$

Atomic number of Cr = 24

Electron configuration of Cr = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹

Electronic configuration of Cr³⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s⁰

NH₃ is a weak field ligand. The splitting of d-orbital occur.


Number of unpaired electrons, n = 3

Magnetic moment (spin only), ${\mu _s} = \sqrt {n(n + 2)} BM$

$ = \sqrt {3(3 + 2)} = \sqrt {3 \times 5} = \sqrt {15} = 3.87BM$

${[Cu{F_6}]^{3 - }}$ Atomic number of Cu = 29

Electronic configuration of Cu = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹

Electronic configuration of Cu²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s⁰

F^$-$ is a weak field ligand, so splitting of d-orbitals occur as follows:


Number of unpaired electrons, n = 2

Spin only magnetic moment, ${\mu _s} = \sqrt {n(n + 2)} BM$

$ = \sqrt {2(2 + 2)} = \sqrt {2 \times 4} = 2\sqrt 2 = 2.84BM$

(a) ${[FeC{l_4}]^ - }$

Atomic number of Fe = 26

Oxidation number of Fe = x + 4 ($-$1) = $-$1

$\Rightarrow$ x = 3

Electronic configuration of Fe = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²

Electronic configuration of Fe³⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s⁰

Cl^- is a weak field ligand, so no pairing of electrons occurs.


Hybridisation is sp³

Geometry is tetrahedral.

${[Fe{(CO)_4}]^{2 - }}$ Oxidation number of Fe = x + 4(0) = - 2 $\Rightarrow$ x = - 2

Electronic configuration of Fe^2- = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s²



CO is a strong field ligand and cause electrons of 3d orbitals to pair up and shift electrons of 4s to 3d orbital.



One 4s and three 4p orbitals undergoes hybridisation to form four sp³ hybrid orbitals which arrange into a tetrahedral geometry.



Each of the four sp³ hybrid orbitals accept a pair of electrons from carbonyl ligand.



Geometry is tetrahedral.

(b) ${[Co{(CO)_4}]^ - }$ Atomic number of Co = 27

Oxidation number of Co = x + 4(0) = $-$1 $\Rightarrow$ x = $-$1

Electronic configuration of Co = 1s²2s²2p⁶3s²3p⁶3d⁷4s²

Electronic configuration of Co^- = 1s²2s²2p⁶3s²3p⁶3d⁸4s²

CO is a strong field ligand, so pairing of electrons occurs.


Hybridisation is sp³

Geometry is tetrahedral.

${[CoC{l_4}]^{2 - }}$

Oxidation number of Co = x + 4($-$1) = $-$2

$\Rightarrow$ x = + 2

Electronic configuration of Co²⁺ = 1s²2s²2p⁶3s²3p⁶3d⁷4s⁰

Cl^- is a weak field ligand, so pairing of electrons do not occur.


Hybridisation is sp³

Geometry is tetrahedral

(c) $[Ni{(CO)_4}]$ Oxidation number of Ni = x + 4(0) = 0 $\Rightarrow$ x = 0

Atomic number of Ni = 28

Electronic configuration of Ni = 1s²2s²2p⁶3s²3p⁶3d⁸4s²

CO is a strong field ligand, so pairing of electrons occur.


Hybridisation is sp³

Geometry is tetrahedral.

${[Ni{(CN)_4}]^{2 - }}$

Oxidation number of Ni = x + 4($-$1) = $-$2 $\Rightarrow$ x = + 2

Electronic configuration of Ni²⁺ = 1s²2s²2p⁶3s²3p⁶3d⁸4s⁰

CN is a strong field ligand, so pairing of electrons occur.


Hybridisation is dsp².

Geometry is square planar.

(d) ${[Cu{(py)_4}]^ + }$

Oxidation number of Cu = x + 4(0) = + 1 $\Rightarrow$ x = + 1

Atomic number of Cu = 29

Electronic configuration of Cu = 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹

Electronic configuration of Cu⁺ = 1s²2s²2p⁶3s²3p⁶3d¹⁰4s⁰


Hybridisation is sp³

Geometry is tetrahedral.

${[Cu{(CN)_4}]^{3 - }}$

Oxidation number of Cu = x + 4($-$1) = $-$ 3 $\Rightarrow$ x = + 1


Hybridisation is sp³

Geometry is tetrahedral.

$\mathrm{\mathop {Al{F_3}(s)}\limits_{Alu\min ium\,trifluoride} \buildrel {HF} \over \longrightarrow Al{F_3}(HF)\buildrel {KF} \over \longrightarrow \mathop {{K_3}(Al{F_6})}\limits_{Potassium\,hexafluorido\,alu{\mathop{\rm minate}\nolimits} \,(III)}}$

$\mathrm{AlF}_3$ is soluble in HF only in presence of KF. It is due to formation of $\mathrm{K}_3\left[\mathrm{AlF}_6\right]$.

$\mathrm{AlF}_3$ is insoluble in anhydrous $\mathrm{HF}$ because $\mathrm{F}^{-}$ ion is not available for $\mathrm{H}$-bonding.

KCN $\mathrm{\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{{S_2}O_3^{2 - }}^{NaOH}}}$ $\mathrm{\mathop {KSCN}\limits_{Potassium\,thiocyanate}}$

$\mathrm{FeCl_3+3KSCN\rightarrow}$ $\mathrm{\mathop {Fe{{(SCN)}_3}}\limits_{Blood\,red\,colour}}$ + 3KCl

Reaction of FeCl$_3$ with KSCN gives a blood red solution.

(a)

$\begin{aligned} & {\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}} \\ & \mathrm{Ni}^{+2}=3 d^8 \\ & 2 \text { unpaired electrons are present. So, } \\ & \therefore \text { Nature = paramagnetic } \\ & \text { Shape }=\text { square planar } \end{aligned}$

(b)

$\begin{aligned} & {\left[\mathrm{Ni}(\mathrm{CO})_4\right]} \\ & \mathrm{Ni}=3 d^8 4 s^2 \\ & \text { Shape }=\text { tetrahedral } \end{aligned}$

As CO is strong ligand, it causes pairing of electrons, therefore nature is diamagnetic.

(c)

$\begin{aligned} & {\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}} \\ & \mathrm{Ni}^{2+}=3 d^8 \\ & \text { Shape }=\text { tetrahedral } \end{aligned}$

As $\mathrm{Cl}^{-}$ is weak ligand, no pairing occurs therefore, nature is paramagnetic.

(d)

$\begin{aligned} & {\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}} \\ & \mathrm{Ni}^{2+}=3 d^8 4 s^0 \end{aligned}$

Shape $=$ octahedral and nature is paramagnetic.

Coordination number is defined as number of coordinate bonds of central metal atom with the ligand. Here, en is bidentate ligand and Cl$^-$ is monodentate ligand.

In the given complex, Co binds with two a bidentate ligand en (2 $\times$ 2 = 4) and two monodentate ligand (2 $\times$ 1 = 2).

$\therefore$ Its coordination number is, 4 + 2 = 6.

In MnO$_4^ - $

x + ($-$2)4 = $-$1 $\Rightarrow$ x = + 7

${}_{25}$Mn⁺⁷ = [Ar], no unpaired elecrons.

Thus, it will now show d-d transition. It is dark purple coloured due to charge transfer from ligand to metal.

The oxidation state of Ni in [Ni(CO)₄] is 0.

The electronic configuration of Ni is 3d⁸ 4s².

As CO is a strong ligand, pairing of electrons occur.

As, it has sp³-hybridisation, so the geometry is tetrahedral. Since, it has no unpaired electrons. So, the complex is diamgnetic.

Oxidation number of Co,

x + ($-$1 $\times$ 6) = $-$2

$\therefore$ x = 4

$\therefore$ Electronic configuration of Co⁴⁺,

Co : [Ar] 3d⁵

$\therefore$ Number of unpaired electrons = 5

$\therefore$ Magnetic moment = $\sqrt {n(n + 2)} = \sqrt {5 + 7} = \sqrt {35} $ BM

back pairing is not possible
because pairing energy > $\Delta $₀.

$ \therefore $ Unpaired electrons = 5

$ \therefore $ $\mu $ = $\sqrt {5\left( {5 + 2} \right)} $ = 5.9 BM

trans-[Co(en)₂Cl₂]⁺ contain a plane of symmetry. It cannot be optically active.

cis-[Co(en)₂Cl₂]⁺ does not contain any plane of symmetry. Hence the compound (B) is optically active.

In octahedral

CFSE = [-0.4n_t2g + 0.6n_eg]$\Delta $₀

= [-0.4$ \times $4 + 0.6$ \times $2]$\Delta $₀

= -0.4 $\Delta $₀

In tetrahedral fields

CFSE = [-0.6n_eg + 0.4n_t2g]$\Delta $_t

= (– 0.6 × 3 + 0.4 × 3)$\Delta $_t

= (–1.8 + 1.2)$\Delta $_t

= –0.6$\Delta $_t

[Co(OX)₂(OH)₂]^–

(As here $\Delta $₀ > P so those ligands act as strong field ligand and pairing happens.)

₂₇Co⁺⁵ = [₁₈Ar] 3d⁴ 4s⁰( $t_{2g}^{2,1,1},e_g^{0,0}$ )

It has highest number of unpaired e^– s. so it is most paramagnetic.

As $\Delta $₀ < P, so all ligands behaves as weak field ligands.

For octahedral

Crystal field stabilization energy (CFSE)

= (-0.4$\Delta $₀) $ \times $ n_t2g + (+0.6$\Delta $₀) $ \times $ n_eg + np

n_t2g = number of electrons in t_2g orbital

n_eg = number of electrons in e_g orbital

n = number of extra pairs

p = Pairing energy

Here n_t2g = 1

n_eg = 0

n = 0 ( For weak field ligands n always zero. Here H₂O is weak field ligand)

$ \therefore $ Crystal field stabilization energy (CFSE)

= (-0.4$\Delta $₀) $ \times $ 4 + (+0.6$\Delta $₀) $ \times $ 2 + 0$ \times $P

= -1.6$\Delta $₀ + 1.2$\Delta $₀

= -0.4$\Delta $₀

XeF₄ = sp³d²

[Ni(CN)₄]^2– = dsp²

[CrF₆]^3– = d²sp²

BrF₅ = sp³d²

Species with same number of unpaired electrons have equal magnetic moment.

Complex	Ligand	Ligand Type	Number of unpaired electrons
[Mn(H2O)6]2+	H2O	Weak Field Ligand	5
[Cr(H2O)]2+	H2O	Weak Field Ligand	4
[CoCl4]2–	Cl	Weak Field Ligand	3
[Fe(H2O)6]2+	H2O	Weak Field Ligand	4
[Co(OH)4]2–	OH	Weak Field Ligand	3
[Fe(NH3)6]2+	NH3	Weak Field Ligand	4

Total 3 geometrical isomers are possible.

[Ru(en)₃ ]Cl₂ :

Here CN = 6 so octahedral splitting happens.

'en' is strong field ligand so $\Delta $₀ > P(pairing energy). That is why pairing of electrons happens.

$\Delta $₀ = Energy gap between e_g and t_2g orbital.


[Fe(H₂O)₆]Cl₂ :

Here CN = 6 so octahedral splitting happens.

H₂O is weak field ligand so $\Delta $₀ < P(pairing energy). That is why no pairing of electrons happens.

$\Delta $₀ = Energy gap between e_g and t_2g orbital.

Let x molecule of water are lost then

13.5 = $\left[ {{{x \times 18} \over {6 \times 18 + 3 \times 35 + 52}}} \right] \times 100$

$ \Rightarrow $ x = 1.99 $ \simeq $ 2

$ \therefore $ Around two moles of water are lost during heating.

$ \therefore $ Formula of complex could be

[Cr(H₂O)₄Cl₂]Cl.2H₂O

It does not have symmetry, so, optically active.

[Ti(H₂O)₆]³⁺

Ti = [Ar]3d²4s²

Ti⁺³ = [Ar]3d¹


For octahedral

Crystal field stabilization energy (CFSE)

= (-0.4$\Delta $₀) $ \times $ n_t2g + (+0.6$\Delta $₀) $ \times $ n_eg + np

n_t2g = number of electrons in t_2g orbital

n_eg = number of electrons in e_g orbital

n = number of extra pairs

p = Pairing energy

Here n_t2g = 1

n_eg = 0

n = 0 ( For weak field ligands n always zero. Here H₂O is weak field ligand)

$ \therefore $ Crystal field stabilization energy (CFSE)

= (-0.4$\Delta $₀) $ \times $ 1 + (+0.6$\Delta $₀) $ \times $ 0 + 0$ \times $P

= -0.4$\Delta $₀ = –0.4 × 20300 = –8120 cm^–1

CFSE (in kJ) = ${{8120} \over {83.7}}$ = 97 kJ/mol

[Pt(NH₃)₂Cl₂] is dsp² hybridisation and shows geometrical isomerism.

[Ni(NH₃)₄(H₂O)₂]²⁺ is Octahedral, show geometrical isomerism.

[Ni(en)₃]²⁺ is Octahedral and shows optical isomerism.

[Ni(NH₃)₂Cl₂] is sp³ hybridisation and tetrahedral complex, therefore does not show geometrical and optical isomerism and structural isomerism.

Stronger the ligand greater is splitting of d orbitals and smaller will be wavelength of light absorbed.

The splitting of ligand is F^– < NCS^– < NH₃

So order of wavelength of light absorbed is

($\lambda $_max)_NH3 < ($\lambda $_max)_NCS^– < ($\lambda $_max)_F^–

(I) Under weak field ligand, octahedral Mn(II) and tetrahedral Ni(II) both the complexes are high spin complex.

(II) Tetrahedral Ni(II) complex can very rarely be low spin because square planar (under strong ligand) complexes of Ni(II) are low spin complexes.

(III)With strong field ligands Mn (II) complexes can be low spin because they have less number of unpaired electron (unpaired electron = 1) While with weak field ligands Mn(II) complexes can be high spin because they have more number of unpaired electron (unpaired electron = 5)

(IV) Aqueous solution of Mn(II) ions is pink in colour.

Spin only magnetic moment = 4.9 = $\sqrt {n\left( {n + 2} \right)} $

From this, n = 4 (unpaired electrons)

(A) In octahedral complex: [M(H₂O)₆] ²⁺

C.F.S.E. = (–0.4 $\Delta $₀) × 4 + (+0.6 $\Delta $₀) × 2 + 0 × P
= –0.4 $\Delta $₀

(B) In tetrahedral complex: [M(H₂O)₄] ²⁺


C.F.S.E. = (–0.6 $\Delta $_t) × 3 + (+0.4 $\Delta $_t) × 3 + 0 × P
= –0.6 $\Delta $_t

(I) [Cr(H₂O)₆]Br₂

H₂O is weak field ligand so it can not pair up all the electrons.

Cr²⁺ : [Ar] 4s⁰3d⁴ ($t_{2g}^3e{g^1}$)

Unpaired e^– = 4

Magnetic moment = $\sqrt {24} $ = 4.89 BM

(II) Na₄[Fe(CN)₆]

CN^- is strong field ligand so it pair up all the electrons.

Fe²⁺ = [Ar] 4s⁰3d⁶ ($t_{2g}^6e{g^0}$)

Unpaired e^– = 0

Magnetic moment = 0 BM

(III) Na₃[Fe(C₂O₄)₃]

As $\Delta $₀ $>$ P, so pairing of electrons happens.

Fe³⁺ = [Ar] 4s⁰3d⁵ ($t_{2g}^5e{g^0}$)

Unpaired e^– = 1

Magnetic moment = $\sqrt 3 $ = 1.73 BM

(IV) (Et₄N)₂[CoCl₄]

Cl^-is weak field ligand so it can not pair up all the electrons.

Co²⁺ = [Ar] 4s⁰3d⁷ ($e{g^4}t_{2g}^3$)

Unpaired e^– = 3

Magnetic moment = $\sqrt {15} $ = 3.87 BM

Hence order of magnetic moment is
I > IV > III > II

[Co(NH₃)₄Cl₂] has 2 geometrical isomers.
Among cis and trans isomers, cis isomer has Cl–Co–Cl angle of 90^o.

Given complex Cr(H₂O)₆Cl_n

As the magnetic mement is 3.83 BM then

Spin only magnetic moment = $\sqrt {n\left( {n + 2} \right)} $ BM = 3.83

$ \Rightarrow $ n = 3

Chromium in +3 oxidation state so molecular formula is Cr(H₂O)₆Cl_n.

$ \therefore $ This formula have following isomers

(a) [Cr(H₂O)₆]Cl₃ : react with AgNO₃ but does not show geometrical isomerism.

(b) [Cr(H₂O)₅Cl]Cl₂.H₂O react with AgNO₃ but does not show geometrical isomerism.

(c) [Cr(H₂O)₄Cl₂]Cl.2H₂O react with AgNO₃ & show geometrical isomerism.

(d) [Cr(H₂O)₃Cl₃].3H₂O does not react with AgNO₃ & show geometrical isomerism.

Compound will be [Cr(H₂O)₄Cl₂] Cl.2H₂O

IUPAC NAME : Tetraaquadichlorido chromium(III) chloride dihydrate

Complex [Pd(F)(Cl)(Br)(I)]^2– (square planar geometry)-has 3 geometrical isomers.

$ \therefore $ n = 3

[Fe(CN)₆]^n–6 becomes [Fe(CN)₆]^3-

Here Oxidation state of Fe = +3

$ \therefore $ Fe⁺³ = [Ar]3d⁵4s⁰

CN^- is strong field ligand so it pairing of electrons happens.

E.C. according to CFT = ($t_{2g}^5e{g^0}$)

$ \therefore $ No. of unpaired e^- = 1

Then Magnetic moment = $\sqrt {n\left( {n + 2} \right)} $ = $\sqrt 3 $ = 1.73 B.M

CFSE = [(–0.4 × 5) + (0.6 × 0)] $\Delta $₀ = –2.0 $\Delta $₀

(A) Ni(CO)₄

Ni = 3d⁸4s²

CO is strong field ligand. So pairing of elections happens.

$ \therefore $ Number of unpaired electrons = 0

$ \therefore $ $\mu $_spin = 0

(B) [Ni(H₂O)₆]Cl₂

Ni⁺² = 3d⁸4s⁰

H₂O is weak field ligand. So no pairing of electrons happens.

Number of unpaired electron = 2

$ \therefore $ $\mu $_spin = $\sqrt {n\left( {n + 2} \right)} $ = $\sqrt {2\left( {2 + 2} \right)} $ = $\sqrt 8 $ B.M

(C) Na₂[Ni(CN)₄]

Ni⁺² = 3d⁸4s⁰

CN^- is strong field ligand. So pairing of electrons happens.

Number of unpaired electron = 0

$ \therefore $ $\mu $_spin = 0

(D) PdCl₂(PPh₃)₂

Pd²⁺ = 4d⁸

This is dsp² complex. And shape is square planar.

$ \therefore $ $\mu $_spin = 0

[Pt(NH₃)₂Cl(NO₂)] and [Pt(NH₃)₄ClBr]²⁺ can display geometrical isomerism.

[Ma₃b₃] type complex shows fac and mer isomerism.

So [Co(NH₃)₃(NO₂)₃] is correct answer.

(a) If the complex MA₂B₂ is sp³ hybridised then the shape of this complex is tetrahedral this structure is opticaly inactive due to the presence of plane of symmetry.

(b) If the complex MA₂B₂ is dsp² hybridised then the shape of this complex is square planar.
Both isomers are optically inactive due to the presence of plane of symmetry.

$ \therefore $ Total number of optical isomer is zero in both the cases.

(a) Co³⁺ with strong field complex forms low magnetic moment complex.

(b) If $\Delta $₀ < P configuration of Co³⁺ will be $t_{eg}^4e_g^2$.

(c) Splitting power of ethylenediamine (en) is greater than fluoride (F^–) ligand therefore more energy absorbed by [Co(en)₃]³⁺ as compared to [CoF₆]^3–.

So wave length of light absorbed by [Co(en)₃]³⁺ is lower than that of [CoF₆]^3–

(d) ${\Delta _t} = {4 \over 9}{\Delta _0}$ = ${4 \over 9} \times 18000$ = 8000 cm^-1

$ \therefore $ Statement (a) and (d) are incorrect.

In complex [Ni(CO)₄] decrease in Ni–C bond length and increase in C–O bond length as well as it's magnetic property is explained by MOT.

[Pt(NH₃)₂Cl(NH₂CH₃)]Cl

x + 0 – 1 + 0 = +1

x = +2

So Pt present in Pt⁺² form.

$ \therefore $ Correct IUPAC form

Diamminechlorido (methanamine) platinum(II)chloride.