Coordination Compounds

Paramagnetism is a form of magnetism whereby some materials are attracted by an externally applied magnetic field, and form internal, induced magnetic fields in the direction of the applied magnetic field. So, magnetic balance shows downward deflection. While diamagnetic substance shows repulsion in magnetic field and magnetic balance shows upward deflection.

(a) X = H₂O(l)
(Water has no unpaired electrons and is thus diamagnetic). Hence, statement (a) is correct.

(b) X = K₄[Fe(CN)₆](s)
(CN is a strong field ligand which forces the d-orbital electrons to pair up ($t_{2g}^6e_g^0$) and making it diamagnetic. Hence, statement (b) is correct.

(c) X = O₂ (g)
Here, O₂(g) is paramagnetic due to two-unpaired electrons present in $\pi $* (antibonding orbitals). Hence, statement (c) is correct.

(d) X = C₆H₆(l)
(Here, C₆H₆ is diamagnetic due to presence of 0 unpaired electrons. Hence, statement (d) is incorrect.

(a) In [FeCl₄]^-, oxidation number of Fe atom = +3

Electronic configuration of Fe in ground state = 3d⁶4s²

Electronic configuration of

Fe³⁺ = 3d⁵4s⁰4p⁰



Thus, [FeCl₄]^- has tetrahedral geometry.

(b) [Co(en)(NH₃)₂Cl₂]⁺ have three geometrical isomers. Thus, statement (b) is incorrect.



(c) Fe³⁺ in [FeCl₄]^- is sp³-hybridised with 5 unpaired electrons. (higher spin-only magnetic moment

= $\sqrt {n(n + 2)} $ = 5.92 BM). While Co³⁺ in [Co(en)(NH₃)₂Cl₂]⁺ is d²sp³-hybridised with zero unpaired electrons (low spin-only magnetic moment = $\sqrt {n(n + 2)} $ = 0 BM).

Thus, the statement (c) is correct.

(d) Co³⁺[Co(en)(NH₃)₂Cl₂]⁺



Co³⁺ in [Co(en)(NH₃)₂Cl₂]⁺ is d²sp³-hybridised and has octahedral geometry with 0 unpaired electron. Thus, statement (d) is incorrect.

When a ligand are arranged in order of the magnitude of crystal field splitting, the arrangement, thus obtained is called spectrochemical series, the series is as follows:

$ \mathrm{I}^{-}<\mathrm{Br}^{-}<\mathrm{Cl}^{-}<\mathrm{NO}_3^{-}<\mathrm{F}^{-}<\mathrm{H}_2 \mathrm{O}<\mathrm{en} $

Hence, among the given options, $\mathrm{I}^{-}<\mathrm{Cl}^{-}<\mathrm{H}_2 \mathrm{O}<$ en

Thus, option (d) is the correct answer.

$ \begin{aligned} & \text {} \left.\stackrel{\mathrm{III}}{[\mathrm{Ti}}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right.] \mathrm{Cl}_3 \xrightarrow[\mathrm{O}_2]{250^{\circ} \mathrm{C}} \stackrel{\mathrm{IV}}{\mathrm{Ti}} \\ & \left(\mathrm{Ti}^{3+}: 3 d^1\right) \quad\left(\mathrm{Ti}^{4+}: 3 d^0\right) \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \text { Colourless } \end{aligned} $

Violet or purple colours is due to $d-d$ transition of one unpaired $d$-electron.

$\left[\stackrel{0}{\mathrm{Ni}}(\mathrm{CO})_4\right] \Rightarrow \mathrm{CO}$ is a neutral ligand. So, in metal carbonyls metal is present in their zero valent state.

$ \text { In }\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+} $

In $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}=$ show rearrangement and has $d^2 s p^3$ (inner) hybridisation and will be diamagnetic, i.e.

$ \text { Hence, option (b) is the correct answer. } $

$ \begin{aligned} &\text { }\\ &\begin{aligned} A_{x-2}\left[B(C)_x\right]_2 & \longrightarrow(x-2) A^{+2}+\left[B(C)_x\right]^{-(x-2)} \\ A_{x-2}\left[B(C)_x\right] \cdot & A^{+2} \text { and }\left[B(C)_2\right]^{-(x-2)} \\ i & =1-\alpha+x \alpha-2 \alpha+2 \alpha \\ (x-1) \alpha & =3 \text { or } x-1=3 / \alpha \\ \alpha & =0.75, i=4, x=5 \end{aligned} \end{aligned} $

Hydrated copper sulphate $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ is blue in colour. The ligand (water) molecules causes splitting of $d$-orbitals. This facilitated $d-d$ transition and colour.

Anhydrous copper sulphate $\mathrm{CuSO}_4$ is colourless. In the absence of ligand (water) molecules, splitting of $d$-orbitals is not possible. Hence, $d$ - $d$ transition is not possible. Hence, option (b) is correct.

IUPAC name of sodiumnitroprusside Na₂[Fe(CN)₅NO] is sodiumpentacyanonitrosyl ferrate (III) because in it NO is neutral ligand and the oxidation number of Fe is III.

On passing H₂S gas in the aqueous solution of salt in the presence of dilute HCl, black ppt of CuS is formed.

$3CuS + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 3S + 4{H_2}O$

$C{u^{2 + }} + 4N{H_4}OH \to \mathop {{{[Cu{{(N{H_3})}_4}]}^{2 + }}}\limits_{Deep\,blue\,complex} + 4{H_2}O$

$\underbrace {[Cr{{(N{H_3})}_6}]}_{Coordination\,sphere}\,\,\underbrace {C{l_3}}_{Ionisable}$ $\rightleftharpoons$ ${[Cr{(N{H_3})_6}]^{3 + }} + 3C{l^ - }$

$[Cr{(N{H_3})_6}]C{l_3}\buildrel {AgN{O_3}} \over \longrightarrow \mathop {AgCl \downarrow }\limits_{(White\,ppt.)} $

Config. of Cr (24) = 3d⁵ 4s¹

Config. Cr³⁺ = 3d³ 4s⁰

Thus option (d) is correct.

(a) d²sp³ hybridisation, octahedral. Thus, option (a) is correct.

(b) here are three unpaired electrons, hence it is paramagnetic. Thus, option (b) is correct.

(c) d²sp³ inner orbital complex. Thus option (c) is as incorrect.

(d) Due to ionisable Cl^$-$ ions, white ppt with AgNO₃. Thus, option (d) is correct.

So, the incorrect statement is (c).

EDTA is used in treatment of lead poisoning.

Here in [Co(Cl)(en)₂]Cl
'Cl' is monodentate so one coordinate linkage will be made with Co.
'en' is bidentate so two coordinate linkage will be made with Co. There are two 'en ' present so 4 coordinate linkage will be made with Co.
$ \therefore $ Total 5 coordinate linkage will be made with Co by Cl and en. So C.N. of Co is 5.

Here in K₃[Al(C₂O₄)₃]
C₂O₄^-2 is bidentate so two coordinate linkage will be made with Al. There are three C₂O₄^-2' present so 6 coordinate linkage will be made with Al.
$ \therefore $ So C.N. of Al is 6.

The field becomes square planar and the order of energy is
${d_{{x^2} - {y^2}}} > {d_{xy}} > {d_{{z^2}}} > {d_{zx}} = {d_{yz}}$

[Fe(phen)₃]²⁺ $ \to $ [Fe(phen)₃]³⁺

Here for Fe⁺²(3d⁶) :

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

           = -0.4$\Delta $₀

for Fe⁺³(3d⁵) :

CFSE = (-0.4 $ \times $ 3 + 0.6 $ \times $ 2)$\Delta $₀

           = 0

Here crystal field stabilization energy(CFSE) decreses.

CN of [Fe(H₂O)₆]Cl₂ = 6

For CN = 6, CFSE = $\left[ {{3 \over 5} \times n - {2 \over 5} \times {n_1}} \right]{\Delta _0}$

Here, n = number of electron in e_g and
n₁ = number of electron in t_2g

Electronic configuration of Fe⁺² according to crystal field theory = $t_{2g}^4e_g^2$

$ \therefore $ CFSE = $\left[ {{3 \over 5} \times 2 - {2 \over 5} \times 4} \right]{\Delta _0}$ = - 0.4 ${\Delta _0}$

CN of K₂[NiCl₄] = 4

For CN = 4, CFSE = $\left[ {{2 \over 5} \times n - {3 \over 5} \times {n_1}} \right]{\Delta _t}$

Here, n = number of electron in t_2g and
n₁ = number of electron in e_g

Electronic configuration of Ni⁺² according to crystal field theory = $e_g^4t_{2g}^4$

$ \therefore $ CFSE = $\left[ {{2 \over 5} \times 4 - {3 \over 5} \times 4} \right]{\Delta _t}$ = - 0.8 ${\Delta _t}$

As in a co-ordination compound, the strong field ligand causes higher splitting of the d-orbitals

Also we know,

strength of ligand $ \propto $ ${1 \over {{\lambda _{absorbed}}}}$

Order of strength of ligand

NH₃ > H₂O > Cl^-

Therefore decreasing order of wavelength absorbed is (I) > (II) > (III)

The maximum possible denticities of the given ligand towards transition metal ion is 6.

The maximum possible denticities of the given ligand towards inner transition metal ion is 8.

To determine which statements are correct, let's analyze each one individually in the context of Valence Bond Theory (VBT) as it applies to transition metal complexes.

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Statement (I):

"Valence bond theory cannot explain the color exhibited by transition metal complexes."

Analysis:

Color in Transition Metal Complexes:

The colors of transition metal complexes arise from electronic transitions between different energy levels of the d-orbitals, specifically d-d transitions.

These transitions occur when an electron absorbs light energy and moves from a lower-energy d-orbital to a higher-energy d-orbital.

Valence Bond Theory Limitations:

VBT focuses on the hybridization of atomic orbitals to form covalent bonds.

It does not account for the splitting of d-orbitals into different energy levels in the presence of ligands (known as crystal field splitting).

Therefore, VBT cannot explain the origin of color in these complexes because it doesn't address the electronic transitions responsible for color.

Conclusion:

Statement (I) is correct.

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Statement (II):

"Valence bond theory can predict quantitatively the magnetic properties of transition metal complexes."

Analysis:

Magnetic Properties:

The magnetic behavior of a complex depends on the number of unpaired electrons in the metal ion.

Quantitative prediction requires calculating the magnetic moment, often using the formula:

$ \mu = \sqrt{n(n+2)} \ \text{Bohr Magnetons (BM)} $

where $ n $ is the number of unpaired electrons.

Valence Bond Theory Capabilities:

VBT can provide a qualitative idea about the magnetic properties by indicating whether a complex is paramagnetic (unpaired electrons present) or diamagnetic (no unpaired electrons).

However, VBT does not offer the tools to quantitatively predict the exact magnetic moment.

Accurate quantitative predictions require more advanced theories like Crystal Field Theory (CFT) or Ligand Field Theory (LFT).

Conclusion:

Statement (II) is incorrect.

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Statement (III):

"Valence bond theory cannot distinguish ligands as weak and strong field ones."

Analysis:

Weak and Strong Field Ligands:

Ligands are classified based on their ability to split the d-orbitals of the metal ion, influencing the pairing of electrons.

Strong field ligands cause a large splitting, often leading to low-spin complexes.

Weak field ligands cause small splitting, leading to high-spin complexes.

Valence Bond Theory Limitations:

VBT does not address the energy splitting of d-orbitals.

It assumes that all bonds are formed via overlap of orbitals without considering the effect of ligands on d-orbital energies.

Therefore, VBT cannot distinguish between weak and strong field ligands because it doesn't involve the spectrochemical series or orbital splitting concepts.

Conclusion:

Statement (III) is correct.

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Final Answer:

Statements (I) and (III) are correct.

Statement (II) is incorrect.

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Answer: Option D

(I) and (III) only

Degenerate orbitals means orbitals of equal energy.

Cr³⁺ forms an octahedral inner orbitals complex and its d orbital get splitted into two differnt energy level.

The two set of degenerate orbitals are

(1) d_xy, d_yz and d_xz

(2) d_{x² - y²} and d_z²

Cis-platin or cis-[Pt(Cl)₂(NH₃)₂] is used as an anti-cancer drug.

Compount is Fe(H₂O)₆]₂ [Fe(CN)₆]

Cation is Fe(H₂O)₆]²⁺

Anion is [Fe(CN)₆]^4-

Here number of unpaired electrons = 3

$ \therefore $ Spin only magnetic moment ($\mu $) = $\sqrt {3\left( {3 + 2} \right)} = \sqrt {15} $ B.M

Note : Energy of t_2g is less than e_g. As electrons always go to the lower energy level orbitals first, that is why electrons goes to the t_2g orbital first.
Here number of unpaired electrons = 0

$ \therefore $ Spin only magnetic moment ($\mu $) = 0 B.M

Note : (1) As CN^- is a strong field ligand so Energy gap between e_g and t_2g orbital is very high.

(2) [Fe(CN)₆]^4– is an octahedral complex. And for octahedral complex Energy gap between e_g and t_2g orbital is called $\Delta $₀ or Crystal Field splitting Energy.

(3) Energy required to pair up the electron in same orbital is called Pairing Energy(P).

(4) For strong field ligand, $\Delta $₀ is very high and for weak field ligand, $\Delta $₀ is very low.

(5) For strong field ligand, $\Delta $₀ > P, so when electron gets energy, pairing of electrons happens as for pairing of electrons very low energy requied.

(6) For weak field ligand, $\Delta $₀ < P, so when electron gets energy, pairing of electrons does not happens as the energy required to enter into the e_g orbital is less than pairing energy. That is why electron go to e_g orbital first.

(7) As CN^- is a strong field ligand so pairing of electron occurs.
Here number of unpaired electrons = 1

$ \therefore $ Spin only magnetic moment ($\mu $) = $\sqrt {1\left( {1 + 2} \right)} = \sqrt {3} $ B.M

Note : (1) In [Ru(NH₃)₆]³⁺ complex, NH₃ is a strong field ligand so Energy gap between e_g and t_2g orbital is very high. That is why pairing of electrons occurs.
Here number of unpaired electrons = 2

$ \therefore $ Spin only magnetic moment ($\mu $) = $\sqrt {2\left( {2 + 2} \right)} = \sqrt {8} $ B.M

$ \therefore $ Correct order of the spin-only magnetic moment of metal ions

V²⁺ > Cr²⁺ > Ru³⁺ > Fe²⁺

Here two O^- + two N = 4 donar atoms present which can donate total 4 lone pair of electrons.

$ \therefore $ It is tetradentate ligand.

Given, $\mu $ = 5.9 BM

$ \therefore $ n = number of unpaired electron = 5


For Mn⁺² with 5 unpaired electronic configuration only possible when ligand is weak field ligand.

Here weak field ligand is NCS^$-$.

Due to presence of strong field ligand (CN^–) pairing occurs in which two d-orbitals i.e., d_x²–y² and d_z² directly face the CN^- ligand.

Compounds that contain at least one carbon-metal bond are called organometallic compounds.

Magnetic moment = $\sqrt {n\left( {n + 2} \right)} $ BM

$ \therefore $ $\sqrt {n\left( {n + 2} \right)} $ = 3.9

$ \Rightarrow $ n = 3

$ \therefore $ no. of unpaired electron = 3

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

Fe²⁺ = t_2g⁴e_g²

Co²⁺ = t_2g⁵e_g²

V²⁺ = t_2g³e_g⁰

$ \therefore $ M is V, Co.

Oxalato (C₂O₄^2–) is a bidentate and H₂O is unidentate ligand.

4C₂O₄^2– creates 8 covalent bonds.

2H₂O creates 2 covalent bonds.

$ \therefore $ Around Th 10 coordinate covalent bonds will be present.

Co $ \to $ Vitamin B₁₂

Zn $ \to $ Carbonic anhydrase

Rh $ \to $ Wilkinson catalyst

Mg $ \to $ Chlorophyll

CoCl₃ + en $ \to $ [Co(en)₂Cl₂]Cl
   1      :   2

A and B are Geometrical isomers.

$ \therefore $ The difference in the number of unpaired electrons = 3 - 1 = 2

So, The total number of isomers for a square planar complex [M(F)(Cl)(SCN)(NO₂ )] is 12

Stronger the ligand, absorption of light having lower wavelength is more.

Order of $\lambda $ : Red $>$ Green $>$ Blue

Hence, ligand strength is L₃ $<$ L₁ $<$ L₂

As complex K₃[Co(CN)₆] have CN^$-$ ligand which is strongfield ligand amongst the given ligands in other complexes.

NH₃ is a strong field ligand.

H₂O is weak field ligand.

Let $\Delta $_violet is the crystal field splitting energy of complex A and $\Delta $_yellow is for complex B.

$ \therefore $   $\Delta $_yellow > $\Delta $_violet.

We know, $\Delta $ = ${{hc} \over {{\lambda _{absorb}}}}$

Violet color will absorb yellow color and yellow color will absorb violet color.

$ \therefore $   $\Delta $_yellow = ${{hc} \over {{\lambda _{violet}}}}$

and $\Delta $_violet = ${{hc} \over {{\lambda _{yellow}}}}$

So, $\Delta $_o of A is calculated from energy of yellow light and $\Delta $_o of B is calculated from energy of violet light.

(a)   [Co(H₂O) Cl(en)₂]Cl  $ \to $  it shows gemetrical isomerism.



(b)   [Cr(H₂O)₅Cl]Cl₂  :   This complex is [MA₅B] type. It does not show geometrical isomerism.

(c)   K[Pt(NH₃)Cl₃]  :  Thiscomplex is [MAB₃] type. It does not show geometrical isomerism.

(d)   K₃[Cr(OX)₃]  :  This complex is [M(AA)₃] type. It does not show geometrical isomerism.

Wilkinson's catalyst is [RhCl(PPh₃)₃]

Here central atom is Rh.

Oxidation state of Rh here is = + 1

$ \therefore $   Electronic configuration of Rh⁺ = [Kr]4d⁸


And the complex is square planar

When reactant is cis isomer then following reaction takes place.


So, if reactant is cis - isomer then two isomers are produced.

When reactant is trans isomer then following reaction takes place.



So, if the reactant is trans isomer then only one isomer is produced.

Assume oxidation state of Cr in all the compounds = x

(i)$\,\,\,$ In [Cr(H₂O)₆] Cl₃ oxidation state of Cr is

x + 0 $ \times $ 6 + ($-$1 $ \times $ ) = O

$ \Rightarrow \,\,\,\,\,$ x + 0 $-$ 3 = O

$ \Rightarrow \,\,\,\,$ x = + 3

(ii)$\,\,\,$ [Cr (C₆ H₆)₂] oxidation state of Cr is

x + 0 $ \times $ 2 = 0

$ \Rightarrow \,\,\,\,$ x = 0

(iii) $\,\,\,$ In K₂ [ Cr(CN)₂ (O)₂ (O₂) (NH₃)] oxidation state of Cr is

1 $ \times $ 2 + x + ($-$ 1 $ \times $ 2) + ($-$2 $ \times $ 2) + ($-$2) + 0 = 0

$ \Rightarrow \,\,\,\,$ x = + 6

$\therefore\,\,\,$ + 3, 0 and + 6 is the correct answer.

Note :

O₂ molecule can have 0, $-$ 1, $-$ 2 oxidation state but in K₂ [ Cr (CN)₂ (O)₂ (O₂) NH₃ ] if we choose zero as the oxidation state of O₂ then for Cr oxidation state will be $+$ 4. But + 4 oxidation state of Cr is unstable and $+$ 6 is most stable that is why we choose $-$ 2 oxidation state of O₂.

The complexes with formula [Mabcd]^n± can have three geometrical isomers. As NO²⁻ and SCN⁻ are ambidentate ligands, therefore 4 × 3 = 12 geometrical isomers will be possible.

We know,

Spin only magnetic moment ($\mu $) = $\sqrt {n\left( {n + 2} \right)} $   B.M

Where, n is the number of unpaired electrons.

So, the complex having higher number of unpaired electrons will have higher value of spin only magnetic moment.

(1)   Zn⁺² : [Ar] 3d¹⁰
Here 0 unpaired electrons present

(2)   Ni⁺² : [Ar] 3d⁸
Here 2 unpaired electrons present.

(3)   Co⁺² : [Ar] 3d⁷
Here 3 unpaired electrons present.

(4)   Mn⁺² : [Ar] 3d⁵
Here 5 unpaired electrons present.

So, correct order is $ \to $
    [MnCl₄]^2$-$ > [CoCl₄]^2$-$ > [NiCl₄]^2$-$ > [ZnCl₄]^2$-$

[NiCl₄]^2– : Oxidation state of Ni in [NiCl₄]^2– = + 2


Cl^– is a weak field ligand and cannot take part in pairing of electrons.
Hence, the complex is tetrahedral and paramagnetic with two unpaired electrons.

[Ni(CO)₄] : Oxidation state of Ni in [Ni(CO)₄] is zero. CO is a strong field ligand thus pairing of electrons takes place in d-orbitals.
Hence, the complex is tetrahedral and diamagnetic.