Coordination Compounds

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.

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.

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.

---

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.

---

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^$-$.