Important Questions for CBSE Class 12 Chemistry Chapter 5 - Coordination Compounds 2024-25

 Very Short Answer Questions                                                            1 Mark 

1. Define the term coordination compound? 

Ans. Coordination compounds are those that contain complicated ions and have  a core metal atom or ion that is linked to a number of ions or neutral molecules  by coordinate bonds. 

Examples:$\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right],\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4}, \mathrm{Ni}(\mathrm{CO})_{4}$

2. Write the names of counter ions in  

(i)Hg[Co(SCN) ]4 

Ans. The coordination complex in the given species is   [Co(SCN) 4  ]−so, its  

counter ion will be Hg+  .  

(ii) [Pt(NH3 ) 4  ]Cl 2 

Ans. The coordination complex in the given species is [Pt(NH3 ) 4  ]+ so, its counter  

ion will be Cl−. 

3. Write the oxidation state of nickel in [Ni(CO) 4 ]. 

Ans. The oxidation state of nickel in the given complex 4 [Ni(CO) ]is zero. 

4. What is the coordination number of central atom in $\mathbf{\left[\mathrm{Co}\left(\mathrm{C}_{2}\mathrm{O}_{4}\right)_{3}\right]^{3-}}$?

Ans. The coordination number of the central atom cobalt in the given  coordination species $\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$ is six (6). 

5. What is the coordination number of iron in[Fe(EDTA)] - ? 

Ans. The coordination number of iron in the given coordination species  [Fe(EDTA)]−is six (6). 

6. Write the name of a complex compound used in chemotherapy. 

Ans. The complex compound used in chemotherapy is cis-platin. 

7. Name the compound used to estimate the hardness of water  volumetrically. 

Ans. The compound which is used in the estimation of the hardness of water  using volumetric techniques is ethylenediaminetetraacetate (EDTA). 

8. Give the IUPAC name of $\mathbf{\left[\mathrm{PtCl}_{2}\left(\mathrm{NH}_{2} \mathrm{CH}_{3}\right)\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}}$? 

Ans. The IUPAC of the given coordination compound is  diaminechloridomethylamine platinum (II) chloride. 

9. How many geometrical isomers are possible for the tetrahedral complex  [Ni(CO)4 ]? 

Ans. Because the relative locations of the unidentate ligands linked to the  central metal atom are the same with respect to each other, no geometrical  isomers will develop for the specified coordination complex.

10. Arrange the following in the increasing order of conductivity in  solution $\mathbf{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]\mathrm{Cl}_{2};\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]\mathrm{Cl}_{3} \text { and }\left[\mathrm{CoCl}_{2}(\mathrm{en})_{2}\right] \mathrm{Cl}}$

Ans. The increasing order for the conductivity of the given coordination  complexes in the solution is: 

$\left[\mathrm{CoCl}_{2}(\mathrm{en})_{2}\right]\mathrm{Cl}<\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]\mathrm{Cl}_{2}<\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$

11. Arrange the following ligands in increasing order of Δo (Crystal field  splitting energy) for octahedral complexes with a particular metal ion.  Cl -  , NH3, I- ,CO , en . 

Ans. The increasing order of the crystal field splitting energy for the given  ligands is:

I- < Cl- < NH3 < en < CO. 

12. Write IUPAC name of Tollens’ reagent. 

Ans. The chemical formula of tollens reagent is Ag(NH ) OH 3 2and its IUPAC  name is ammoniacal silver nitrate. 

13. Which is more stable? K3 [Fe(CN)6 ] or K4  [Fe(CN)6 ]?

Ans. The oxidation state of iron in  K3 [Fe(CN)6 ] is +3 and in K4  [Fe(CN)6 ] is +2.  So,  K3 [Fe(CN)6 ] is more stable due to the  Fe3+   ion being smaller in size and  higher in charge.

14. Calculate the overall dissociation equilibrium constant for the  [Cu(NH3 )4 ]2+  . ion. Given that overall stability constant (β )for this    complex is 2.1×1013 ? 

Ans. Given in the question is

$\beta_{4}=2.1 \times 10^{13}$

The overall dissociation constant is the reciprocal of the given stability constant:

$\begin{aligned}&\frac{1}{\beta_{4}}=\frac{1}{2.1 \times 10^{13}} \\&\frac{1}{\beta_{4}}=4.7 \times 10^{-14}\end{aligned}$

15. What is a chelate ligand? Give one example?

Ans. Chelating ligands are ligands with numerous donor sites that may form  stable five-to six-membered rings with metals. Ethylenediamine, acetylacetone,  and ethylenediaminetetraacetic acid (EDTA) are some examples of chelating  ligands. 

16. Write the IUPAC name of Li[AlH ]4. 

Ans. The IUPAC of the complex is lithium tetrahydridoaluminate. 

17. Name one homogeneous catalyst used in hydrogenation of alkenes. 

Ans. The four-coordinated rhodium complex $\operatorname{Rh}\left[\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \mathrm{P}\right]_{3} \mathrm{Cl}$ is the most fully studied homogeneous hydrogenation catalyst used in the hydrogenation of alkenes. After its discoverer, G. Wilkinson, this catalyst is known as Wilkinson's catalyst.

18. Name the types of isomerism shown by coordination entity: $\left[\mathrm{CrCl}_{2}(\mathrm{Ox})_{2}\right]^{3-}$

Ans. The type of isomerism shown by the coordination entity $\left[\mathrm{CrCl}_{2}(\mathrm{Ox})_{2}\right]^{3-}$ is  stereoisomerism. 

19.$\left[\mathrm{Ti}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}$ is coloured but on heating becomes colourless. Why? 

Ans. When the coordination compound $\left[\mathrm{Ti}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}$ is heated, the water  molecule is removed from the coordination sphere. As a result, no crystal field  splitting occurs. As a result, no colour is seen. 

20.Write the IUPAC name of ionization isomer of [Co(NH3  )5 (SO34 )]Br . 

Ans. The IUPAC name of the ionization isomer given is  pentaminesulphatocobalt[III] bromide. 

21.Write the formula and the name of the coordinate isomer of   [Co(en)3  ][Cr(CN)6 ]. 

Ans. The formula of the coordinate isomer of  [Co(en)3  ][Cr(CN)6 ] is  [Cr(en)3  ][Co(CN)6 ]and its name is Tris- (ethane –1, 2, diammine) chromium (III)  hexacyanocobaltate (III). 

Short Answer Type Questions 2 Marks 

1. Write two differences between a double salt and a coordination  compound with the help of an example of each. 

Ans. The difference between double salt and a coordination compound is:

Double Salt: A compound made up of two distinct salt components is known  as a double salt. They completely dissociate into its ions in the water. Example:  Potash Alum K2 SO4 .Al2 (SO4 )3  .24H2 O .

Coordination Compound: A complex salt is a chemical made up of a core  metal atom and ligands that form coordination bonds with it. In water  coordination compounds do not completely dissociate into its ions. Example:  Potassium ferro cyanide K [Fe(CN) ] 4 6. 

2. Mention the main postulates of Werner’s Theory. 

Ans. The main postulates of Werner’s Theory are: 

(i) Two types of valencies are shown by the central metal ion, the primary  and the secondary valency. 

(ii) The oxidation state of the compound is its primary valency and the  coordination number of a complex is its secondary valency. 

(iii) The secondary valency for every species is fixed, it means that the  coordination number is fixed. 

(iv) The primary and secondary valencies of a metal atom are both satisfied  by the metal atom. The main valency is satisfied by a negative ion.  Negative ions and neutral molecules, on the other hand, fulfil secondary  valencies. 

3. Define Homoleptic and Heteroleptic complexes with the help of one  example of each. 

Ans. Homoleptic complexes are those in which only one kind of ligand is  bound to the central metal ion. Example:  [Co(NH3  )6 ]3+  .

Heteroleptic complexes are those in which different types of ligands are bound  to the central metal ion. Example: [Co(NH3 )4 Cl2 ]+  . 

4. In the following coordination entity: [Cu(en)2 ]2+ 

1. Identify the ligand involved. 

Ans. The ligand involved in the given coordination entity is ethylenediamine.

2. Oxidation state of copper metal. 

Ans. As ethylenediamine is a neutral ligand, the charge on the copper metal is  +2. 

5. Calculate the magnetic moments of the following complexes:

 (i)[Fe(CN)6 ]4- 

Ans. The oxidation state of central iron is +2. So, the Fe2+ is a  3d6  complex. As,  

CN− is a strong field ligand, the electron gets paired up and hence no unpaired  electron is present. The magnetic moment will be: 

$\begin{aligned}&\mu_{B H}=\sqrt{\mathrm{n}(\mathrm{n}+2)} \\&\mu_{B H}=0\end{aligned}$

(ii) [CoF6]3-

Ans. The oxidation state of central cobalt is +3. So, the  Co+3  is a 3d5  complex.  

As fluoride is a weak field ligand, there will be no pairing in the electrons. So, n = 5

$\begin{aligned}&\mu_{B H}=\sqrt{\mathrm{n}(\mathrm{n}+2)} \\&\mu_{B H}=\sqrt{5(5+2)} \\ &\mu_{B H}=5.9 \mathrm{~B} \cdot\mathrm{M}\end{aligned}$

6. Explain the following: 

1. $\mathbf{\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}}$ is an inner orbital complex whereas $\mathbf{\left[\mathrm{FeF}_{6}\right]^{3-}}$ is an outer orbital complex.

Ans. According to the valence bond theory:

For the complex $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ :

The electronic configuration of $\mathrm{Fe}$ is $\operatorname{Ar}[18] 4 \mathrm{~s}^{2} 3 \mathrm{~d}^{6}$.

The electronic configuration of $\mathrm{Fe}^{3+}$ is $\operatorname{Ar}[18] 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{0}$

Due the CN− being a strong field ligand, d-orbital that is the 3d-orbital is used  for the hybridization of the complex, 3 

[Fe(CN)6  ] −is an inner orbital octahedral  complexes.

For the complex $\left[\mathrm{FeF}_{6}\right]^{3-}:$

The electronic configuration of $\mathrm{Fe}$ is $\operatorname{Ar}[18] 4 \mathrm{~s}^{2} 3 \mathrm{~d}^{6}$.

The electronic configuration of $\mathrm{Fe}^{3+}$ is $\operatorname{Ar}[18] 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{0}$

As fluoride is a weak field the hybridization in $\left[\mathrm{FeF}_{6}\right]^{3-}$ will be due to the outer dorbital that is the $4 \mathrm{~d}$-orbitals. Hence, $\left[\mathrm{FeF}_{6}\right]^{3}$ is an outer orbital octahedral complex.

2. NH3 acts as complexing agent but NH4+ does not. 

Ans. Because ammonia only has one lone pair of electrons, it can accept  electrons from coordination complexes. Ammonia is transformed into an ion  called ammonium, which lacks a lone pair to donate and form complexes. As a  result, whereas ammonia can be an excellent ligand, the ammonium ion does  not form complexes. 

7. What type of structural isomerism is represented by the following  complexes: 

(i) [Mn(CO)5 (SCN)]and [Mn(CO)5 (NCS)]

Ans. Linkage type of structural isomerism is represented in the given  complexes. The presence of coordination compounds with the same  composition but different connectivity of the metal to a ligand is known as  linkage isomerism. The ligands showing linkage isomerism are SCN−and NCS− .

(ii)[Co(NH3 )5 (NO3 )] SO4     and [Co(NH3 )5 (SO4 )]NO3 

Ans. Ionization isomers are isomers that produce distinct ions in solution  despite having the same chemical composition. The nitrate ion is coordinated to  the cobalt ion in the former, while the sulphate ion is outside the coordination  sphere in the later; the sulphate ion is inside the coordination sphere in the  latter. As a result, it's ionisation isomers. 

8. How are complex compounds applicable in  

1. Electroplating of silver, gold or other noble metals  

Ans. The formation of complex ions of specific cations serves two purposes: To  keep the cation stable. Some metal cations, such as gold, are not stable in their  pure state. When they are complexed with a ligand, they become significantly  more stable. 

2. In photography 

Ans. Light sensitive silver bromide is a key component of photographic film.  Sodium bromide is added to a silver nitrate solution to make silver salts.  Ordinary film has a coating of gelatin on the sensitive surface that contains  silver bromide crystals.

9. Explain on the basis of Valance Bond Theory that diamagnetic  [Ni(CN)4 ]2- has square planar structure and paramagnetic  [Ni(CN)4 ]2-   ion has tetrahedral geometry ? 

Ans.In the complex $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$, the oxidation state of nickel is $+2$ and it has $\mathrm{d}^{8}$ configuration in its outer shell. Four $\mathrm{CN}^{-}$attached to the centre nickel metal indicates that it can either has a tetrahedral or square planar geometry. As the $\mathrm{CN}^{-}$is a strong field ligand, the unpaired $3 \mathrm{~d}$ electrons gets paired due to the cyanide ion ligand. So, this complex will have a dsp $^{2}$ hybridization, making its shape as square planer.

In the complex $\left[\mathrm{NiCl}_{4}\right]^{2-}$, the chloride ion is a weak field ligand. Hence, the unpaired $3 \mathrm{~d}$ electrons will not be paired and this complex will have a $\mathrm{sp}^{3}$ hybridization making its shape as tetrahedral.

10.Explain as to how the two complexes of nickel [Ni(CN)4 ]2-       and Ni(CO)4   have different structures but do not differ in their magnetic behaviours.  (At. no. of Ni = 28).

Ans. In $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ the ground state valence shell electronic configuration of nickel atom is $\operatorname{Ar}[18] 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$. The oxidation state of nickel is $+2$, so the electronic configuration of $\mathrm{Ni}^{2+}$ is $\operatorname{Ar}[18] 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2}$.

Due to the presence of a strong field ligand $\mathrm{CN}^{-}$the pairing in the electrons starts and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ gets dsp $^{2}$ hybridized having a square planar geometry. Due to the presence of two unpaired electron in $3 \mathrm{~d}$ sub shell, $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ is diamagnetic in nature.

In $\mathrm{Ni}(\mathrm{CO})_{4}$, nickel has zero oxidation state, so the electronic configuration of nickel will be $\operatorname{Ar}[18] 3 \mathrm{~d}^{8} 4 \mathrm{~s}^{2} .$ As, $\mathrm{CO}$ is a strong field ligand, the pairing in the electrons starts, all the ten valence electrons moves to the $3 \mathrm{~d}$ sub shell and are paired. Therefore, $\mathrm{Ni}(\mathrm{CO})_{4}$ is $\mathrm{sp}^{3}$ hybridized having a tetrahedral geometry. As there are no unpaired electrons, $\mathrm{Ni}(\mathrm{CO})_{4}$ is diamagnetic.

11. Draw the structures of geometrical isomers of the coordination  complexes– [Co(NH3  )3  Cl3  ]and  [CoCl2 (en)2 ]+ 

Ans. The structures of the geometrical isomers of the given complexes are: 

The geometrical isomers of  [CoCl2 (en)3  ]+are: 

The geometrical isomers of  [Co(NH3  )3  Cl3  ] are: 

12.Write the IUPAC name of the complexes: 

1. [NiCl2  (PPh3 )2  ]

Ans. The IUPAC name of this complex is  Dichloridobis(triphenylphosphine)nickel (II). 

2.  [Co(NH3 )4 Cl(NO2  )]Cl 

Ans. The IUPAC name of this complex is Tetraamminechloridonitrito-N-cobalt  (III) Chloride. 

3. K[Cr(H2 O)2 (C2  O4 )2 ] 

Ans. The IUPAC name of this complex is Potassium  diaquadioxalatochromate(III) trihydrate. 

13.Using IUPAC norms write the formula for the following:  

1. Tetrabromidocuprate (II)  

Ans. The formula for this compound is [CuBr4 ]2− 

2. Pentaamminenitrito–O– Cobalt (III) 

Ans. The formula for this compound is[Co(NH3  )5 (ONO)]2+

14.How does EDTA help as a cure for lead poisoning? 

Ans. We employ a ligand that can create metal complexes with lead to  eliminate lead from our bodies. EDTA is a chemical that is used to treat lead  poisoning. Ethylene diamine tetraacetic acid (EDTA) is an acronym for  ethylene diamine tetraacetic acid. It has two nitrogen atoms and four oxygen  atoms, all of which are capable of donating electrons to metals. Lead replaces calcium in the Ca–EDTA complex in the body. Lead-EDTA, a more soluble  complex, is excreted in urine. 

15.A complex is prepared by mixing CoCl3and NH3in the molar ratio of  1:4. 0.1 m solution of this complex was found to freeze at 0.372°C. What  is the formula of the complex? Kf of water = 1.86°C/m 

Ans. According to the formula, the theoretical value of the change in temperature will be:

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times \mathrm{m}$

$\Delta \mathrm{T}_{\mathrm{f}}=1.86^{\circ} \mathrm{C} / \mathrm{m} \times 0.1\mathrm{~m}$

$\Delta \mathrm{T}_{\mathrm{f}}=0.186^{\circ} \mathrm{C}$

The observed change in temperature is:

$\Delta \mathrm{T}_{\mathrm{f}}=0.372^{\circ} \mathrm{C}$

As, it can be seen that the observed change in temperature is twice the  theoretical change, this shows that the each molecule of the complex dissociates  to form two ions. This is possible if the formula of the complex is  3 4 2 [Co(NH ) Cl ]Cl. 

16.The [Mn(H2 O)6 ]2+ ion contains five unpaired electrons while [Mn(CN)6 ]4-   ion contains only one unpaired electron. Explain using Crystal Field  Theory.

Ans. The electronic configuration of Mn2+is 3d5 4s0. The ligand linked to the  metal in the instance of hexaaquomanganese (II) ion is water, which is a weak  field ligand that is unable to couple up the electrons of the 3d subshell, and the  configuration is t2g3 and eg2. In the case of the hexacyano ion, the CN− is a  2 strong field ligand that causes electron pairing in the d orbitals, leaving just one  unpaired electron in t2g5 eg0 

Short Answer Type Questions             3 Mark 

1. Account for the following: 

(i) [NiCl4 ]2− is paramagnetic while Ni(CO)4 is diamagnetic though both  4 are tetrahedral.  

Ans. Although both are tetrahedral,[NiCl4 ]2− is paramagnetic and Ni(CO)4is diamagnetic. has a structure of 3d8 4s2, indicating that it is in the zero oxidation  state. CO, on the other hand, is a strong field ligand. As a result, it is  diamagnetic in nature and induces the pairing of unpaired 3d electrons, but Cl−  is a weak ligand and is unable to pair up the unpaired electrons, therefore  [NiCl4 ]2− is paramagnetic in nature.  

(ii) $\left[\mathrm{Fe}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{3-}$ is strongly paramagnetic whereas  $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ weakly  paramagnetic.  

Ans. The ligand $\mathrm{NH}_{3}$ is neither a strong field ligand nor a weak field ligand in the $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ complex ion. It is, in fact, a weak strong field ligand. However, because the crystal field stabilisation energy is smaller than the pairing energy, the ligand $\mathrm{NH}_{3}$ acts as a weak field ligand in the $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ ion. As a result, the electronic configuration under the effect of an octahedral crystal field is $\mathrm{t}_{2} \mathrm{~g}^{6} \mathrm{eg}^{2}$. The complex has two unpaired electrons, according to the aforementioned electrical structure. As a result, the $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ complex is weakly paramagnetic.

In the complex $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ the oxidation state of $\mathrm{Fe}$ is $+3$, having the structure $3 \mathrm{~d}^{5}$. Water $\left(\mathrm{H}_{2} \mathrm{O}\right)$ is a weak ligand, and $3 \mathrm{~d}$ electrons do not couple up in its presence. The $s p^{3} d^{2}$ hybridization results in an outer orbital octahedral complex with 5 unpaired electrons. As a result, it is very paramagnetic.

(iii) $\mathbf{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}}$ is an inner orbital complex whereas  $\mathbf{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}}$ is in  outer orbital complex. 

Ans. In $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ the oxidation state of cobalt is $+3$, so its d-orbital will have six electrons making it a d $^{6}$ system. As $\mathrm{NH}_{3}$ is a strong field ligand, the pairing will start in $3 \mathrm{~d}$ electrons. The hybridization involves the two sub-shells of $3 \mathrm{~d}$, $4 \mathrm{~s}$ and $4 \mathrm{p}$ empty orbitals. Therefore the hybridization is $\mathrm{d}^{2} \mathrm{sp}^{3}$. Hence, $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ is an inner orbital octahedral complex.

In$\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ the oxidation state of nickel is $+2$ and its outer shell electronic configuration is $3 \mathrm{~d}^{8} 4 \mathrm{~s}^{0}$, due to the presence of two unpaired electrons in $3 \mathrm{~d}$ sub-shell. The hybridization involves, empty $4 \mathrm{~s}, 4 \mathrm{p}$ and $4 \mathrm{~d}$ orbitals, thats why it is an outer orbital octahedral complex.

2. Compare the following complexes with respect to their shape, magnetic  behaviors and the hybrid orbitals involved. 

1. [CoF6 ]3- 

Ans. Fluoride ion is a weak field ligand, thus [CoF6 ]3-  is an outer orbital  octahedral complex which is paramagnetic in nature.  

2.  [Cr(NH3 )6 ]3+  

Ans. The NH3is a strong field ligand, thus [Cr(NH3 )6 ]3+ is an inner orbital  octahedral complex which is paramagnetic in nature. It can be shown as:

3. [Fe(CN)6 ]4-

Ans. CN−is a strong field ligand, thus [Fe(CN)6 ]4- is an inner orbital octahedral  6 complex which is diamagnetic in nature. This can be shown as: 

3. Draw the structure of  

1. cis-dichloridotetracyanochromate (II) ion  

Ans. The structure of the complex is:

2. mer-triamminetrichloridocobalt (III)  

Ans. The structure of the complex is: 

3. fac-triaquatrinitrito–N–cobalt (III) 

Ans. The structure of the complex is: 

4. Name the central metal atom/ion present in  

1. Chlorophyll  

Ans. The central metal ion in the structure of chlorophyll is magnesium. 

2. Haemoglobin

Ans. The central metal ion in the structure of haemoglobin is iron. 

3. Vitamin B-12 

Ans. The central metal ion in the structure of vitamin B-12 is cobalt. 

5. A metal complex having composition Cr(NH3 )4 Cl2 Br has been isolated in  two forms ‘A’ and ‘B’. The form ‘A’ reacts with AgNO3 solution to give  white precipitate which is readily soluble in dilute aqueous ammonia,  whereas ‘B’ gives a pale yellow precipitate which is soluble in  concentrated ammonia solution. Write the formula of ‘A’ and ‘B’. Also  mention the isomerism which arises among ‘A’ and ‘B’. 

Ans. Compound $\mathrm{A}$ is $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{ClBr}\right] \mathrm{Cl}$. Its reaction with $\mathrm{AgNO}_{3}$ :

$\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{ClBr}\right] \mathrm{Cl}+\mathrm{AgNO}_{3} \rightarrow \mathrm{AgCl}+\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{ClBr}\right]^{+} \mathrm{NO}_{3}^{-}$

The white precipitate is of silver chloride $(\mathrm{AgCl})$. This white silver chloride is soluble in ammonia solution as:

$\mathrm{AgCl}+2 \mathrm{NH}_{4} \mathrm{OH} \rightarrow\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}\right]+2 \mathrm{H}_{2} \mathrm{O}$

Compound $\mathrm{B}$ is $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Br}$. Its reaction with $\mathrm{AgNO}_{3}$ :

$\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Br}+\mathrm{AgNO}_{3} \rightarrow \mathrm{AgBr}+\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}\right]^{+} \mathrm{NO}_{3}^{-}$

The pale yellow precipitate of silver bromide $(\mathrm{AgBr})$ is formed which is soluble in concentrated ammonia solution:

$\mathrm{AgBr}+2 \mathrm{NH}_{4} \mathrm{OH} \rightarrow\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}\right]+2 \mathrm{H}_{2} \mathrm{O}$

Therefore compound A is 3 4 [Cr(NH ) ClBr]Cland compound B is 3 4 2 [Cr(NH ) Cl ]Br .  They both are ionization isomers of each other. 

6. Write the limitations of Valence Bond Theory. 

Ans. The limitations of valence bond theory are: 

(i) VBT fails in explaining the tetravalency of carbon. 

(ii) The geometries of ammonia, water, methane, etc cannot be properly  explained by this theory. 

(iii) Bond angles of molecules such as carbon dioxide, water, ammonia, etc  could not properly be given by VBT. 

(iv) VBT does not explain the magnetic properties of complex. 

7. Draw a sketch to show the splitting of d-orbitals in an octahedral crystal  field state for a  d4  ion. How the actual electronic configuration of the  split d-orbitals in an octahedral crystal field is decided by the relative  values of Δo and pairing energy (P)? 

Ans:  The sketch for the splitting of d-orbitals in an octahedral crystal field of  d4  ion is given as:

 The crystal field splitting energy for d4   will be:

CFSE $=-0.4 \mathrm{t}_{2} \mathrm{~g}+0.6 \mathrm{eg} \Delta_{\circ}$

CFSE $=(-3 \times 0.4)+(1 \times 0.6)$

CFSE $=-0.6 \Delta$ 。

8. For the complex [Fe(en)2  Cl2  ]Clidentify  

1. The oxidation number of iron.  

Ans. The oxidation number of iron will be +3. 

2. The hybrid orbitals and the shape of the complex.  

Ans. The complex is  d2 sp3 hybridized and is an inner orbital octahedral  complex. 

3. The magnetic behaviour of the complex.  

Ans. The coordination complex will be paramagnetic due to the presence of  three unpaired electrons. 

4. The number of geometrical isomers. 

Ans. This complex will have two geometrical isomers. The cis- and the trans form.

5. Whether there is an optical isomer also?  

Ans. The cis isomer will show optical activity. 

6. Name of the complex. 

Ans. Dichlorido bis (ethane-1, 2 diamine) iron (III) chloride or Dichloro bis  (ethylenediamine) iron (III) chloride. 

9. A chloride of fourth group cation in qualitative analysis gives a green  coloured complex [A] in aqueous solution which when treated with  ethane –1, 2-diamine (en) gives pale yellow solution [B] which on  subsequent addition of ethane –1, 2-diamine turns to blue/purple [C]  and finally to violet [D]. Identify [A], [B], [C] and [D] complexes. 

Ans. Complex $[\mathrm{A}]$ is $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$

Complex [B] is $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}(\mathrm{en})\right]^{2+}$

Complex [C] is $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}(\mathrm{en})_{2}\right]^{2+}$

Complex [D] is Ni(en) $\left._{3}\right]^{2+}$

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