Coordination Compounds

[Fe(CO)₄(C₂O₄)]⁺



One unpaired electron Spin only magnetic moment = $\sqrt 3 $ B.M. = 1.73 BM

Biuret ligand is

It forms complexes like



The denticity of organic ligand is 2.

(a) [CoCl₂(en)₂] show Cis-trans isomerism

(b) [Co(CN)₅(NC)]^3$-$ can't show G.I.

(3) [Co(NH₃)₃(NO₂)₃] show fac & mer isomerism

(d) [Co(NH₃)₄Cl₂]⁺ show cis & trans isomerism

(i) CFSE $\propto$ charge or oxidation no. of central metal ion.

(ii) CFSE $\propto$ strength of ligand

en > NH₃ > H₂O > F^$-$

$\therefore$ order of CFSE

$\mathop {{{[Co{{(en)}_3}]}^{ + 3}}}\limits^{III} > \mathop {{{[Co{{(N{H_3})}_6}]}^{ + 3}}}\limits^{III} > \mathop {{{[Co{F_6}]}^{ - 3}}}\limits^{III} > \mathop {{{[Co{{({H_2}O)}_6}]}^{ + 2}}}\limits^{III} $

Hybridization of all the given complexes are given in following table.

${[Mn{(CN)_6}]^{3 - }} \Rightarrow M{n^{3 + }}$

CN^$-$ is a strong field ligand.

$\therefore$ Pairing of electron takes place.

${[Fe{(CN)_6}]^{3 - }} \Rightarrow F{e^{3 + }}$

CN^$-$ is a strong ligand.

$\therefore$ Pairing of electron takes place.

${[Co{({C_2}{O_4})_3}]^{3 - }} \Rightarrow C{o^{3 + }}$

${C_2}O_4^{2 - }$ acts as strong field ligand with Co³⁺.

$\therefore$ Pairing of electrons takes place.

${[Fe{F_6}]^{3 - }} \Rightarrow F{e^{3 + }}$

F is a weak field ligand.

$\therefore$ Pairing of electron does not occurs.

${[MnC{l_6}]^{3 - }} \Rightarrow M{n^{3 + }}$

Cl^$-$ is a weak field ligand.

$\therefore$ Pairing of electron does not occurs.

${[Fe{F_6}]^{3 - }}$ have 5 unpaired electrons and ${[MnC{l_6}]^{3 - }}$ have 4 unpaired electrons.

So, both statement I and statement II are true.

[MnCl₆]^3$-$



Paramagnetic and having 4 unpaired electrons.

Complex [Co(en)₃]Cl₂ is most stable complex among the given complex compounds because more number of chelate rings are present in this complex as compare to others.

(1) [Co(en)(NH₃)₄]Cl₂ 1 chelate ring

(2) [Co(en)₃]Cl₂ 3 chelate ring

(3) [Co(en)₂(NH₃)₂]Cl₂ 2 chelate ring

(4) [Co(NH₃)₆]Cl₂ 0 chelate ring

1. ${[\mathop {Fe}\limits^{ + 3} {({H_2}O)_6}]^{3 + }}$

$F{e^{3 + }}:[Ar]3{d^5}$

Hybridisation : $s{p^3}{d^2}$

Magnetic nature : Paramagnetic (so this complex response to external magnetic field)

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

$N{i^{2 + }}:[Ar]3{d^8}$

Hybridisation : dsp²

Magnetic nature : diamagnetic

3. ${[Co{(CN)_6}]^{3 - }}$

$C{o^{3 + }}:[Ar]3{d^6}$

Hybridisation : ${d^2}s{p^3}$

Magnetic nature : diamagnetic

4. $[Ni{(CO)_4}]$

$Ni:[Ar]3{d^8}4{s^2}$

Hybridisation : $s{p^3}$

Magnetic nature : diamagnetic

In presence of SFL $\Delta$₀ > P means pairing occurs therefore

For Fe⁺² $\to$ 3d⁶



$\therefore$ No of unpaired e^-(s) = 0

$\therefore$ $\mu$ = $\sqrt {n(n + 2)} $BM = 0

[n = No of unpaired e^$-$(s)]

In NiCl₂, Ni⁺² is having configuration 3d⁸

$\therefore$ Number of unpaired electron = 2

After formation of oxidised product

[Ni(CN)₆]^$-$2$ \Rightarrow $ Ni⁺⁴ is obtained

Ni⁺⁴ $\Rightarrow$ 3d⁶ and CN^$-$ is strong field ligand

$\therefore$ Number of unpaired electrons = 0

$\therefore$ The charge is 2 $-$ 0 = 2

Species does not have a magnetic moment of 1.7 BM means species must not contain single unpaired electron.

(a) Molecular orbital configuration of O$_2^ + $ (15 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $

Unpaired electron = 1

$\mu = \sqrt {n(n + 2)} = \sqrt {1(1 + 2)} = \sqrt 3 = 1.73$

where n = no. of unpaired e^$-$

$\therefore$ $\mu$ = 1.73 BM

(b) Cul $\Rightarrow$ Cu⁺ $-$[Ar]3d¹⁰, Unpaired electron = 0

I^$-$ $\to$ [Xe], unpaired electron = 0

Therefore, $\mu$ = 0

(c) [Cu(NH₃)₄]Cl₂

Cu²⁺ $\to$ [Ar]3d⁹

Unpaired electron = 1,

Therefore, $\mu$ = 1.73 BM

(d) Molecular orbital configuration of $O_2^ - $ (17 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $

Unpaired electron = 1

Therefore, $\mu$ = 1.73 BM

Therefore, Cul does not have magnetic moment of 1.73 BM.

According to VBT i.e. valence bond theory,

Electronic configuration of Ni = [Ar]3d⁸4s².

(a) NiCl₂ . 6H₂O

NiCl₂ . 6H₂O $\rightarrow$ NiCl₂ + 6H₂O

Oxidation number of Ni(x) = x + 2($-$1) = 0; where x, $-$1 and 2 are the oxidation number of Ni, oxidation number of Cl and number of Cl atoms respectively.

NiCl₂ $\Rightarrow$ x + ($-$2) = 0 $\Rightarrow$ x = 2

Electronic configuration of Ni²⁺ = [Ar]3d⁸4s⁰

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

For C.N. = 6


(b) K₂[Ni(CN)₄]

K₂[Ni(CN)₄] $\rightarrow$ 2K⁺ + [Ni(CN)₄]^2$-$

x + 4($-$1) $-$ ($-$2) = 0; where x, 4, $-$1 and $-$2 are the oxidation number of Ni, number of CN ligands, charge on one CN and charge on complex.

[Ni(CN)₄]^2$-$ $\Rightarrow$ x $-$ 4 + 2 = 0 $\Rightarrow$ x = +2

Electronic configuration of Ni²⁺ $\Rightarrow$ [Ar]3d⁸4s⁰

CN^$-$ is a strong field ligand. So, pairing of electrons occur. For C.N. = 4


(c) Ni(CO)₄

CO is neutral and strong field ligand. So, pairing of electrons occur.

Oxidation number of Ni is zero

Electronic configuration of Ni = [Ar]3d¹⁰4s⁰

For C.N. = 4


(d) Na₂[NiCl₄]

Na₂[NiCl₄] $\rightarrow$ 2Na + [NiCl₄]^2$-$

x + 4($-$1) $-$ ($-$2) = 0; where x, 4, $-$1 and $-$2 are the oxidation number of Ni, number of CN ligands, charge on one CN and charge on complex.

[NiCl₄]^2$-$ $\Rightarrow$ x $-$ 4 + 2 = 0 $\Rightarrow$ x = +2

Electronic configuration of Ni²⁺ = [Ar]3d⁸4s⁰

For C.N. = 4


Hence, only K₂[Ni(CN)₄] has dsp² hybridization.

Correct order of intensity of colours of the compounds is

[NiCl₄]^2$-$ > [Ni(H₂O)₆]²⁺ > [Ni(CN)₄]^2$-$

Ni is in +2 oxidation state in all complexes. The intensity of colour depends on the strength of the ligand attached with the central metal atom because more strong ligand more splitting energy, less is intensity of colour. Strength of ligand is in the order CN^$-$ > H₂O > Cl^$-$.

Splitting energy order [NiCl₄]^2$-$ < [Ni(H₂O)₆]²⁺ < [Ni(CN)₄]^2$-$

$\therefore$ Intensity of colour of compound

[NiCl₄]^2$-$ > [Ni(H₂O)₆]²⁺ > [Ni(CN)₄]^2$-$

CuSO₄.5H₂O $ \Rightarrow $ [Cu(H₂O)₄]SO₄.H₂O

In CuSO₄.5H₂O, Cu is co-ordinated with 4 water molecule and two more oxygen atom from sulphate ion and fifth water molecule is hydrogen bonded.

So secondary valency = 4

No. of hydrogen bond per molecule = 1

(a) [Co(NH₃)₆][Cr(CN)₆] $ \to $ (iii) Coordination isomerism

(b) [Co(NH₃)₃(NO₂)₃] $ \to $ (i) Linkage isomerism

(c) [Cr(H₂O)₆]Cl₃ $ \to $ (ii) Solvate isomerism

(d) cis-[CrCl₂(ox)₂]^3– $ \to $ (iv) Optical isomerism

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

where, n = number of unpaired electrons and $\mu$ $\propto$ n.

(i) ${[Fe{F_6}]^{3 - }} \Rightarrow F{e^{3 + }} = (3{d^5}) , {F^ - }$ (weak field ligand).

Thus, pairing of electron does not take place.

(ii) ${[Co{(N{H_3})_6}]^{3 + }} \Rightarrow C{o^{3 + }} = (3{d^6}), N{H_3}$ (strong field ligand).

Thus, pairing of electron takes place.

(iii) ${[NiC{l_4}]^{2 - }} \Rightarrow N{i^{2 + }} = (3{d^8}), C{l^ - }$ (weak field ligand).

Thus, pairing of electron takes place.

(iv) ${[Cu{(N{H_3})_4}]^{2 + }} \Rightarrow C{u^{2 + }}(3{d^9}), N{H_3}$ (strong field ligand).

Thus, pairing of electron takes place.

So, the decreasing order of $\mu$ is

$\mathop {{\mu _i}}\limits_{(n = 5)} > \mathop {{\mu _{iii}}}\limits_{(n = 2)} > \mathop {{\mu _{iv}}}\limits_{(n = 1)} > \mathop {{\mu _{ii}}}\limits_{(n = 0)} $

${[Mn{(CN)_6}]^{4 - }}$

Mn²⁺ = 3d⁵ 4s⁰

CN^– is a strong field ligand.

$ \therefore $ Pairing will occur.


${[Fe{(CN)_6}]^{3 - }}$

Fe³⁺ = 3d⁵4s⁰

CN^– is a strong field ligand.

$ \therefore $ Pairing will occur.

(i) ${[FeC{l_4}]^{2 - }}\buildrel {} \over \longrightarrow F{e^{2 + }}\buildrel {} \over \longrightarrow [Ar]3{d^6}$

$\therefore$ Cl is weak field ligand so does not pairing occur


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

$ = \sqrt {4(4 + 2)} BM$ (n = Number of total unpaired e^$-$ = 4)

$ = \sqrt {24} BM = 4.90\,BM$

(ii) ${[Co{({C_2}{O_4})_3}]^{3 - }}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} C{o^{3 + }}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} [Ar]3{d^6}$

C₂O₄ is strong field ligand so pairing occur.


All electrons are paired, n = 0

hence, $\mu$ = 0

(iii) $MnO_4^{2 - }\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} M{n^{ + 6}}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} [Ar]3{d^1}$

$n = 1$

$ = \sqrt {n(n + 2)} BM$

$ = \sqrt {1(1 + 2)} BM$

$ = \sqrt 3 BM$

$ = 1.73\,BM$

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

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.

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