NCERT Solutions for Class 12 Science Chemistry Chapter 9 - Coordination Compounds

Answer:

The stability of coordination compound increases with the increase in log K value. The value of log K is highest for the reaction.
Cu²⁺ + 4CN^– → [Cu(CN)₄]^2– 
           log K = 27.3
Hence, the correct answer is option (ii).

Answer:

The crystal field splitting energy depends upon the strength of the ligands. Strong field ligands split the five degenerate energy level of d orbitals with more energy separation as a compound to weak-field ligands.

$$\begin{gathered} ∆\mathrm{E}=\frac{\mathrm{hc}}{\lambda } \\ ∆\mathrm{E} \propto \frac{1}{\lambda } \end{gathered}$$

As ∆ E increases, the absorption of the wavelength of light decreases. Among the given ligands in the complexes, the order of splitting power is:
H₂O < NH₃ < CN^–
As the strength of ligand increases, ∆E increases, the value of CFSE increases and therefore, the value of wavelength absorbed decreases.
Therefore, the correct decreasing order for wavelength will be
[Co(H₂O)₆]³⁺ > [Co(NH₃)₆]³⁺ > [Co(CN)₆]^3–
Hence, the correct answer is option (iii).

Answer:

When 0.1 mol of CoCl₃(NH₃)₅ is treated with excess of AgNO₃, 0.22 mol of AgCl are obtained. This means when 1 mole of CoCl₃(NH₃)₅ is treated with excess of AgNO₃, 2 moles of AgCl will be obtained.
Thus, there must be two free chloride ions in the solution of electrolyte. So, the molecular formula of the complex is [Co(NH₃)₅Cl]Cl₂.
[Co(NH₃)₅Cl]Cl₂ → [Co(NH₃)₅Cl]²⁺ + 2Cl–
The electrolytic solution contains [Co(NH₃)₅Cl]²⁺ and 2Cl^– ions in the ratio 1 : 2 electrolyte.
Hence, the correct answer is option (ii).

Answer:

1 mol of AgCl is precipitated to produce 1 mol of Cl^– ion. Therefore, 3 mol of AgCl will produce 3 mole of Cl^– ions. So, the formula of the complex can be [Cr(H₂O)₆]Cl₃.
Hence, the correct answer is option (iv).

Answer:

The given complex compound is [Pt(NH₃)₂Cl₂]. It contains ligands – NH₃ and Cl. The correct IUPAC name is Diamminedichloridoplatinum (II).
Hence, the correct answer is option (i).

Answer:

Chelation is a process in which didentate or polydentate ligand uses two or more donor atom to bind to the central metal atom or ion in order to form a ring structure. It stabalises the coordination compound. The ligands - which forms the chelation are called chelating ligands. Among all the given complexes, [Fe(C₂O₄)₃]^3– contains chelating ligand C₂O₄ (oxalate ion) which stabalises the coordination compound by chelating the central metal Fe³⁺ ion.
Hence, the correct answer is option (iii).

Answer:

MA₄B₂ type of coordination compound shows geometrical isomerism. So, out of the given complex ion, [Cr(H₂O)₄Cl₂]⁺ shows geometrical isomerism. It contains two set of equivalent ligands, four H₂O and 2Cl.
The possible isomers are as follows:



Hence, the correct answer is option (i).

Answer:

The CFSE of octahedral and tetrahedral complexes are related to each other by the formula 

${∆}_t=\frac{4}{9} {∆}_o$

Where ∆_t = CFSE of tetrahedral complex
           ∆_o = CFSE of octahedral complex
Given, ∆_o = 18,000 cm^–1
            ∆_t = $\frac{4}{9}{∆}_o=\frac{4}{9}\times 18,000 {\mathrm{cm}}^{-1}$
                = 4 × 2,000 cm^–1
                = 8,000 cm^–1
Hence, the correct answer is option (iii).

Answer:

Ambidentate ligands are the ligands that have two different bonding sites.
Example: NO₂, NCS etc.
Thiocyanate (SCN) has two binding sites at nitrogen and sulphur and it can be represented as
M  ← NCS or M → SNC
Therefore, the coordination compound containing SCN as a ligand show linkage isomerism as [Pd(C₆H₅)₂ (SCN)₂] and [Pd(C₆H₅)₂(NCS)₂].
Hence, the correct answer is option (i).

Answer:

The compound having same molecular formula but different structural formula are known as isomers. [Co(SO₄)(NH₃)₅]Br and [Co(SO₄)(NH₃)₅]Cl contains different molecular formula, thus, they are not isomers.
Hence, the correct answer is option (iv).

Answer:

A chelating agent has two or more than two donor atoms to bind to a single metal ion. The structure of the given compounds are as follows:



Thiosulphato is not a chelating ligand as it is not favourable for S₂O₃^2– to chelate a metal ion geometrically.
Hence, the correct answer is option (i).

Answer:

The atoms or groups that must donate a pair of electron or loosely held electron pair to metal to form a M - L bond.
e.g. -  $:\stackrel{..}{\mathrm{N}}=\mathrm{O}, :\mathrm{CO} , \stackrel{..}{\mathrm{N}}{\mathrm{H}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2 \stackrel{}{\stackrel{..}{\mathrm{N}}{\mathrm{H}}_2}$
Among the given species, NH₄⁺ does not contain any pair of electron, thus, it does not act as a ligand.
Hence, the correct answer is option (ii).

Answer:

Solvate isomerism is shown when two compounds have same molecular formula but differ on how the solvate molecules are bonded to the metal ion, whether they are directly bonded to metal ion or present as free solvent molecules in the crystal lattice.
Therefore, coordination compounds [Cr(H₂O)₆]Cl₃ (violet) and [Cr(H₂O)₅Cl]Cl₂⋅H₂O (greyish-green) shows solvate isomerism.
Hence, the correct answer is option (ii).

Answer:

The IUPAC name of [Pt (NH₃)₂Cl(NO₂)] is diaminechloridonitrito–N–platinum (II).
Hence, the correct answer is option (iii).

Answer:

(i) Molecular orbital electronic configuration of Co³⁺ in [Co(NH₃)₆]³⁺ is



Number of unpaired electron = 0
∴ Diamagnetic

(ii) Molecular orbital electronic configuration of Mn³⁺ in [Mn(CN)₆]^3– is



Number of unpaired electron = 2
∴ Paramagnetic

(iii) Molecular orbital electronic configuration of Fe²⁺ in [Fe(CN)₆]^4– is



Number of unpaired electron = 0
∴ Diamagnetic

(iv) Molecular orbital electronic configuration of Fe³⁺ in [Fe(CN)₆]^3–



Number of unpaired electron = 1
∴ Paramagnetic
Hence, the correct answer are options (i) and (ii).

 

Answer:

(i) Molecular orbital electronic configuration of Mn³⁺ in [MnCl₆]^3– is



Number of unpaired electrons = 4
∴ Paramagnetic

(ii) Molecular orbital electronic configuration of Fe³⁺ in [FeF₆]^3– is



Number of unpaired electrons = 5
∴ Paramagnetic

(iii) Molecular orbital electronic configuration of Co³⁺ in [CoF₆]^3– is


Number of unpaired electrons = 4
∴ Paramagnetic

(vi) Molecular orbital electronic configuration of Ni²⁺ in [Ni(NH₃)₆]²⁺ is



Number of unpaired electrons = 2
∴ Paramagnetic
Among the given complexes, [MnCl₆]^3– and [CoF₆]^3– contain 4 unpaired of electrons.
Hence, the correct answer are options (i) and (iii).





 

Answer:

According to VBT, molecular orbital electronic configuration of Fe3– in [Fe(CN)₆]^3– is


Number of unpaired electron = 1
∴ Paramagnetic
Hence, the correct answers are options (i) and (iii).

Answer:

An aqueous pink solution of cobalt (II) chloride is due to the electronic transition of electrons from t_2g to e_g energy level of the complex [Co(H₂O)₆]²⁺. On addition of excess of HCl to the above solution, [Co(H₂O)₆]²⁺ is transformed into [CoCl₄]^2–. The tetrahedral complexes have smaller crystal field splitting than octahedral complexes as ${∆}_t=\frac{4}{9} {∆}_o$
Hence, the correct answer are options (ii) and (iii).

Answer:

The complexes that contains only one kind of ligands are known as homoleptic complex. Among the given complexes, [Co(NH₃)₆]³⁺ and  [Ni(CN)₄]^2– are homoleptic complexes containing NH₃ and CN as ligands.
Hence, the correct answers are options (i) and (ii).

Answer:

Heteroleptic complexes are those complexes which contain more than one type of ligands. Among the given complexes, [Fe(NH₃)₄Cl₂]⁺ and [Co(NH₃)₄Cl₂] are heteroleptic complexes.
Hence, the correct answer is options (ii) and (iv).

Answer:

[Co(en)₃]³⁺ and cis– [Co(en)₂ Cl₂]⁺ are optically active compounds because their mirror images are non-superimposable in nature.



Hence, the correct answers are options (i) and (iii).

Answer:

Ethane – 1, 2 – diamine
Structure: 

(i) It is a neutral ligand due to absence of any change.
(ii) It is a didentate ligand due to presence of two donor sites one at each nitrogen atom of amino group.
(iii) It has the ability to chelate with the central metal atom, and thus, acts as a chelating ligand.
Hence, the correct answers are options (i), (ii) and (iii).

Answer:

The coordination compounds that contains ligand with more than one non-equivalent binding position show linkage isomerism.
[Co(NH₃)₅(NO₂)]²⁺ contains NO₂ which have two donor sites N and O.


[Cr(NH₃)₅SCN]²⁺ contains SCN which have two donor sites S and N.
→ S – C ≡ N ←
Therefore, [Co(NH₃)₅(NO₂)]²⁺ and [Cr(NH₃)₅SCN]²⁺ show linkage isomerism.
Hence, the correct answers are options (i) and (iii).

Answer:

Molecules or ions that are present outside the coordination sphere are ionisable in nature. The complex which give more ions on dissolution is more conducting.
[Co(NH₃)Cl₃] < [Co(NH₃)₄Cl₂] Cl < [Cr (NH₃)₅ Cl]Cl₂ < [Co(NH₃)₆]Cl₃
As the number of ions increases the conductivity also increases.
 

Answer:

Since only two ions are available for the conductance of the solution, out of which there should be a cation and an anion. Cl^– can precipitate silver nitrate. Therefore, it will act as anion. This is only possible if it is present outside the coordination sphere (ionisation sphere). Thus, the formula of the complex is [Cr(H₂O)₄ Cl₂] Cl and its IUPAC name is Tetraaquadichloridochromium(III)chloride.

Answer:

An optically active complex of type [M (AA)₂ X₂]ⁿ⁺ shows that cis-octahedral structure, example - cis - [Pt (en)₂ Cl₂]²⁺ or cis-[Cr(en)₂Cl₂]⁺. The mirror image of such isomers are non-super impossible.

Answer:

The magnetic moment of 5.92 BM shows that there are five unpaired electrons present in the d orbitals of Mn²⁺ ion. The hybridisation involved is sp³ rather than dsp² because Cl is a weak field ligand. Therefore, the tetrahedral structure of [MnCl₄]^2– complex will have a magnetic moment of 5.92 BM.

Answer:

In weak field ligands, ∆₀ < P (pairing energy), such that, the electronic configuration of Co (III) becomes ${\mathrm{t}}_{2g}^4\mathrm{e}{g}^2$

It has four unpaired electrons and is paramagnetic in nature. In strong field ligands, ∆₀ > P (pairing energy), so pairing occurs. The electronic configuration becomes ${\mathrm{t}}_{2g}^6{\mathrm{e}}_g^0$

 

Answer:

The d-orbital splitting in tetrahedral complexes is smaller in comparison with the octahedral complexes. For same metal and same ligand, ${∆}_t=\frac{4}{9}{∆}_{\mathit{0}}$. Thus, the orbital splitting energies are not enough to force pairing, such that, low spin configurations are rarely formed in tetrahedral complexes.

Answer:

The increasing field strength of the given ligands as per the spectro chemical series is as follows:
F^– < NH₃ < CN^– 
CN^– and NH₃ being strong field ligand pair up the t_2g electrons before filling the e_g set.
(i) [CoF₆]^3–  
The Co metal is in +3 oxidation state.
${\mathrm{Co}}^{+3}=3d^6={\mathrm{t}}_{2g}^4{\mathrm{e}}_g^2$

(ii) [Fe (CN)₆]^4– 
The Fe metal is in +2 oxidation state.
${\mathrm{Fe}}^{2+}=3d^6= {\mathrm{t}}_{2g}^6{\mathrm{e}}_g^0$

(iii) [Cu (NH₃)₆]²⁺
The Cu metal is in +2 oxidation state.
Cu²⁺ = 3d⁹= ${\mathrm{t}}_{2g}^6$eg³ 

 

Answer:

We know, the magnetic moment can be calculated by the formula
${\mathrm{\mu }}_m=\sqrt{n(n+2)} \mathrm{BM}$
where n = number of unpaired electrons [Fe(H₂O)₆]³⁺
The fe metal is in +3 oxidation state.
Fe³⁺ = d⁵

[Fe(H₂O)₆]³⁺ involves sp³d² hybridisation with five unpaired electrons.
$\begin{array}{rcl}\therefore {\mathrm{\mu }}_m& =& \sqrt{\mathrm{n}(\mathrm{n}+2)} \mathrm{BM}\\ & =& \sqrt{5\left(5+2\right)} \mathrm{BM}\\ & =& \sqrt{35} \mathrm{BM}\\ & =& 5.92 \mathrm{BM}\end{array}$
[Fe(CN)₆]^3–
The Fe metal is in +3 oxidation state.
Fe³⁺ = d⁵

[Fe(CN)₆]^3– involve d²sp³ hybridisation with one unpaired electron.
$\begin{array}{rcl}\therefore {\mathrm{\mu }}_m& =& \sqrt{\mathrm{n}(\mathrm{n} + 2)} \mathrm{BM}\\ & =& \sqrt{1(1+2)}\\ & =& \sqrt{3}\\ & =& 1.74\end{array}$
[Fe(H₂O)₆]³⁺ is an outer orbital complex whereas [Fe(CN)₆]^3– is an inner orbital complex.

Answer:

The crystal field splitting energy (∆₀) increases with the increase in strength of the ligands attached to the central metal atom.
Since, the field strength increases in the order.
Cl^– < NH₃ < CN^– 
Therefore, the increasing order of crystal field splitting energy (∆₀) will be 
[Cr(Cl)₆]^3– < [Cr(NH₃)₆]³⁺ < [Cr(CN)₆]^3–
 

Answer:

The compounds having similar geometry can have a different magnetic moment due to the presence of ligands having different strengths i.e. due to the presence of weak and strong field ligands in complexes. If CFSE is high, the complex will show low value of magnetic moment and if the CFSE is low, the complex will show high value of magnetic moment.
Example: [CoF₆]^3– and [Co(NH₃)₆]³⁺ They both have similar geometry but contains ligands with different strengths. The former is paramagnetic whereas the latter is diamagnetic. This is because F^– is comparatively weak field ligand than NH₃.

Answer:

In CuSO₄.5H₂O, H₂O acts as ligand which causes crystal field splitting. Thus, d-d transition is possible in CuSO₄.5H₂O and therefore, it is blue in colour. In anhydrous CuSO₄, due to absence of H₂O ligand, crystal field splitting is not possible and thus, it is colourless.

Answer:

Ambidentate ligands have the ability to bind with the central metal atom through atoms of two different elements. Example: SCN has two different binding position S and N and NO₂ also has two binding sites, i.e., N and O. So, they are ambidentate ligands.
The coordination compound containing ambidentate ligands shows linkage isomerism due to the presence of two different binding positions.
e.g. [Co(NH₃)₅ SCN]³⁺ and [Co(NH₃)₅ NCS]³⁺ show linkage isomerism.

Answer:

The colour of the coordination compound is closely related to CFSE of the coordination compound. On the basis of CFSE value, the correct matching will be:
 

Answer:

• Magnesium metal ion is present in chlorophyll.
• Iron is a metal that is predominantly present in the red blood cells.
• Wilkinson catalyst is the coordinate compound of the rhodium with triphenylphosphine and chloride.
• Vitamin-B₁₂ is a cobalt containing coordination compound.

Answer:

​

Answer:

​

]⁺ shows geometrical isomerism due to the presence of two types of ligands.



B. 

Answer:

Answer:

Toxic metal ions are removed by the chelating ligands. When these toxic metal ions are treated with a solution containing chelating ligand, stable complexes are formed. These chelate complexes are more stable.
Hence, the correct answer is option (i).

Answer:

[Cr(H₂O)₆]Cl₂and [Fe(H₂O)₆]Cl₂ are reducing in nature due to formation of more stable complex ion after gaining electrons. The unpaired electrons get paired up after reduction.
Hence, the correct answer is option (ii).

Answer:

Linkage isomerism is shown by the ambidentate ligands because they have two different donor atoms.
Example - SCN, NO₂ are ambidentate ligands.
Hence, the correct answer is option (i).

Answer:

The comlexes of MX₆ and MX₅L type (X and L are unidentate) do not show geometrical isomerism. This is because it lacks plane of symmetry. The necessary condition for showing geometrical isomerism is the complex must be of type MA₄B₂ type or [M(AB)₂X₂]. ligand has two different donor atoms. For example, SCN is an ambidentate ligand with two donor atoms, i.e., S and N.


Hence, the correct answer is option (ii).

Answer:

According to VBT, molecular orbital electronic configuration of Fe³⁺ in [Fe(CN)₆]^3– is



Hybridisation = d²sp³
Number of unpaired electrons = 1

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

                              = $$\begin{gathered} =\sqrt{1(1+2)} \\ =\sqrt{3} \\ =1.73 BM \end{gathered}$$

Hence, the correct answer is option (iv).         
 

Answer:

(a) (i) [CoF₆]^3–
F^– is a weak field ligand,




Co³⁺ = 3d⁶ $\left(t_{2g}^4 e{g}_g^2\right)$
Number of unpaired elements = 4
 $\begin{array}{rcl}\mathrm{Magnetic} \mathrm{moment} \left(\mu \right)& =& \sqrt{n(n+2)}\\ & =& \sqrt{4(4+2)}\\ & =& \sqrt{24}\\ & =& 4.9 \mathrm{BM}\end{array}$

(ii) [Co(H₂O)₆]²⁺
H₂O is a weak field ligand.

${\mathrm{Co}}^{2+}=3d^7\left(t_{2g}^5 e_g^2\right)$
Number of unpaired electron = 3
$\begin{array}{rcl}\mathrm{Magnetic} \mathrm{moment} \left(\mu \right)& =& \sqrt{n(n+2)}\\ & =& \sqrt{3(3+2)}\\ & =& \sqrt{15}\\ & =& 3.87 \mathrm{BM}\end{array}$
[Co(CN)₆]^3–
CN^– is a strong field ligand.

${\mathrm{Co}}^{3+}=3d^7\left(t_{2g}^5 e_g^2\right)$
Number of unpaired electrons = 0
$\begin{array}{rcl}\mathrm{Magnetic} \mathrm{moment} \left(\mu \right)& =& \sqrt{n(n+2)}\\ & =& \sqrt{0(0+2)}\\ & =& 0 \mathrm{BM}\end{array}$
It behaves as diamagnetic.

(b) (i) [FeF₆]^3–
 F^– is weak field ligand.

${\mathrm{Fe}}^{3+}=3d^5\left(t_{2g}^3 e_g^2\right)$
Number of unpaired electron = 5
$\begin{array}{rcl}\mathrm{Magnetic} \mathrm{moment} \left(\mu \right)& =& \sqrt{n(n+2)}\\ & =& \sqrt{5(5+2)}\\ & =& \sqrt{5\left(7\right)}\\ & =& \sqrt{35}\\ & =& 5.92 \mathrm{BM}\\ & & \end{array}$

(ii) [Fe(H₂O)₆]²⁺
H₂O is a weak field ligand.


${\mathrm{Fe}}^{2+}=3d^6\left(t_{2g}^4 e_g^2\right)$
Number of unpaired electrons = 4
$\begin{array}{rcl}\mathrm{Magnetic} \mathrm{moment} \left(\mu \right)& =& \sqrt{n(n+2)}\\ & =& \sqrt{4(4+2)}\\ & =& \sqrt{4\left(6\right)}\\ & =& \sqrt{24}\\ & =& 4.98 \mathrm{BM}\\ & & \end{array}$
(ii) [Fe(CN)₆]^4– 
CN^– is a strong field ligand.


${\mathrm{Fe}}^{2+}=3d^6\left(t_{2g}^4 e_g^2\right)$
Number of unpaired electrons = 0
$\begin{array}{rcl}\mathrm{Magnetic} \mathrm{moment} \left(\mu \right)& =& \sqrt{n(n+2)}\\ & =& \sqrt{0(0+2)}\\ & =& 0 \mathrm{BM}\\ & & \end{array}$
It behaves as diamagnetic.













 

Answer:

(i) [Mn(CN)₆]^3– 
The oxidation state of Mn³⁺ = [Ar] 3d⁴ 

1. Hybridisation = d²sp³
2. It is an inner orbit complex because (n – 1) d-orbitals are used.
3. It is paramagnetic in nature due to two unpaired electrons.
$\begin{array}{rcl}4. \mathrm{Spin} \mathrm{only} \mathrm{magnetic} \mathrm{moment} \left(\mu \right) & =& \sqrt{n(n+2)}\\ & =& \sqrt{n(n+2)}\\ & =& \sqrt{2(2+2)}\\ & =& \sqrt{8}\\ & =& 2.82 \mathrm{BM}\\ & & \end{array}$
(ii)  [Co(NH₃)₆]³⁺
The oxidation state of C0 is +3. Therefore, electronic configuration of Co³⁺ = [Ar] 3d⁶

1. Hybridisation = d²sp³
2. It is an inner orbital complex because (n – 1) d-orbitals are used.
3. It is diamagnetic in nature as there are no unpaired of electrons.
$\begin{array}{rcl}4. \mathrm{Spin} \mathrm{only} \mathrm{magnetic} \mathrm{moment} \left(\mu \right) & =& \sqrt{n(n+2)}\\ & =& \sqrt{0(0+2)}\\ & =& 0 \mathrm{BM}\\ & & \end{array}$
(iii) [Cr(H₂O)₆]³⁺
The oxidation state of Cr is +3.
Therefore, the electronic configuration of Cr³⁺
Cr³⁺ =[Ar] 3d³ 

1. Hybridisation = d²sp³
2. It is an inner orbital complex because (n – 1) d-orbitals are used.
3. It is paramagnetic in nature as three unpaired electrons are present.
$\begin{array}{rcl}4. \mathrm{Spin} \mathrm{only} \mathrm{magnetic} \mathrm{moment} \left(\mu \right) & =& \sqrt{n(n+2)}\\ & =& \sqrt{3(3+2)}\\ & =& \sqrt{3\left(5\right)}\\ & =& \sqrt{15}\\ & =& 3.87 \mathrm{BM}\end{array}$
(iv) [FeCl₆]^{4 –}
The oxidation state of Fe is +2.
Therefore, electronic configuration of Fe²⁺ = [Ar] 3d⁶


1. Hybridisation = sp²d³
2. It is an outer orbital complex because nd - orbits are involved.
3. It is paramagnetic due to four unpaired electrons.
$\begin{array}{rcl}4. \mathrm{Spin} \mathrm{only} \mathrm{magnetic} \mathrm{moment} \left(\mu \right) & =& \sqrt{n(n+2)}\\ & =& \sqrt{4(4+2)}\\ & =& \sqrt{4\left(6\right)}\\ & =& \sqrt{24}\\ & =& 4.9 \mathrm{BM}\end{array}$







 

Answer:

(i) Since, isomer 'A' reacts with AgNO₃ to give white precipitate, so in this Cl is present outside the coordination sphere.
Hence, "A' is [Co(NH₃)₅ SO₄] Cl
Also, isomer 'B' gives white precipitate with BaCl₂ but does not reach with AgNO₃, so in this ${\mathrm{SO}}_4^{2-}$ is present outside the coordination sphere.
Hence, 'B' is [Co(NH₃)₅ Cl] SO₄ 
(ii) It involves ionisation isomerism. 
(iii) IUPAC name of the given isomers:

(A) [Co(NH₃)₅SO₄]Cl: Pentaamminesulphatocobalt(III)chloride

(B) [Co(NH₃)₅Cl]SO_4: Pentaamminechloridocobalt(III)sulphate
 

Answer:

When light falls over the complex, certain wavelengths of visible light is absorbed. The higher the crystal field splitting energy of the complex, lower will be the wavelength of digit observed by it. The observed colour of any complex is the complimentary colour generated from the wavelengths of visible light. Example: If any complex is absorbing green light, it will appear as red.
In accordance with crystal field theory, for an octahedral complex having empty e_g level and unpaired electrons in the t_2g level in ground level. If the electron in the t_2g level absorbs the blue-green region, it will excite to e_g level and the complex will appear violet in colour. Crystal field splitting does not occur in the absence of ligand and thus, the substance appears colourless.
Example: CuSO₄.5H₂O is blue in colour anhydrous CuSO₄ is white in colour.

Answer:

The splitting of d-orbitals are different in octahedral and tetrahedral field. CFSE in octahedral and tetrahedral field are closely related as:
${∆}_t=\left(\frac{4}{9}\right){∆}_o$
where, ${∆}_t$ = Crystal field splitting energy in tetrahedral field 
${∆}_o$ = Crystal field splitting energy in octahedral field
Wavelength of light and CFSE are related to each other by the formula 
$$\begin{gathered} {∆}_o=E=\frac{hc}{\lambda } \\ E\propto \frac{1}{\lambda } \end{gathered}$$
So, higher the wavelength of light is absorbed in octahedral complexes in comparison with tetrahedral complexes for same metal and ligands. This is why different colours are observed.