GSEB Class 12 Chemistry Solutions Chapter 9 Coordination Compounds

GSEB Class 12 Chemistry Coordination Compounds InText Questions and Answers

 

Question 1. Write the formulas for the following coordination compounds?
1. Tetraamminediaquacobalt(III) chloride
2. Potassium tetracyanidonickelate(II)
3. Tris(ethane-1,2-diamine) chromium(III) chloride
4. Amminebromidochloridonitrito-N-platinate(II)
5. Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate
6. Iron(III) hexacyanoferrate(II)
Answer:
1. \( [\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})_2]\text{Cl}_3 \)
2. \( \text{K}_2[\text{Ni}(\text{CN})_4] \)
3. \( [\text{Cr}(\text{en})_3]\text{Cl}_3 \)
4. \( [\text{Pt}(\text{NH}_3)\text{BrCl}(\text{NO}_2)]^- \)
5. \( [\text{PtCl}_2(\text{en})_2](\text{NO}_3)_2 \)
6. \( \text{Fe}_4[\text{Fe}(\text{CN})_6]_3 \)
In simple words: This question requires recalling the IUPAC nomenclature rules to correctly translate coordination compound names into their chemical formulas, considering oxidation states and ligand types.

🎯 Exam Tip: Mastering IUPAC nomenclature for coordination compounds is crucial. Pay attention to oxidation states, ligand prefixes, and the charge of the complex ion.

 

Question 2. Write the IUPAC names of the following coordination compounds?
1. \( [\text{CO}(\text{NH}_3)_6]\text{Cl}_3 \)
2. \( [\text{CO}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2 \)
3. \( \text{K}_3[\text{Fe}(\text{CN})_6] \)
4. \( \text{K}_3[\text{Fe}(\text{C}_2\text{O}_4)_3] \)
5. \( \text{K}_2[\text{PdCl}_4] \)
6. \( [\text{Pt}(\text{NH}_3)_2\text{Cl}(\text{NH}_2\text{CH}_3)]\text{Cl} \)
Answer:
1. Hexaamminecobalt(III) chloride
2. Pentaamminechloridocobalt(III) chloride
3. Potassium hexacyanidoferrate(III)
4. Potassium trioxalatoferrate(III)
5. Potassium tetrachloridopalladate(II)
6. Diamminechlorido(methanamine)platinum(II) chloride
In simple words: This task involves applying IUPAC rules to name coordination compounds given their chemical formulas, correctly identifying ligands, central metal, and oxidation state.

🎯 Exam Tip: Accurately determining the oxidation state of the central metal ion and correctly identifying all ligands (and their prefixes) is key to proper IUPAC naming. Practice with various complex types.

 

Question 3. Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers:
1. \( \text{K}[\text{Cr}(\text{H}_2\text{O})_2(\text{C}_2\text{O}_4)_2] \)
2. \( [\text{Co}(\text{en})_3]\text{Cl}_3 \)
3. \( [\text{CO}(\text{NH}_3)_5(\text{NO}_2)](\text{NO}_3)_2 \)
4. \( [\text{Pt}(\text{NH}_3)(\text{H}_2\text{O})\text{Cl}_2] \)
Answer:
1. For \( \text{K}[\text{Cr}(\text{H}_2\text{O})_2(\text{C}_2\text{O}_4)_2] \), both geometrical (cis-, trans-) and optical isomers are possible.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र \(\text{K}[\text{Cr}(\text{H}_2\text{O})_2(\text{C}_2\text{O}_4)_2]\) के ज्यामितीय समावयवों को दर्शाता है। 'cis' रूप में, \(\text{H}_2\text{O}\) लिगैंड एक दूसरे के निकट स्थित होते हैं, जबकि 'trans' रूप में, वे केंद्रीय क्रोमियम परमाणु के विपरीत दिशा में होते हैं। ऑक्सलेट लिगैंड (\(\text{C}_2\text{O}_4\)) द्विदंतुर होते हैं और केंद्रीय धातु से दो बिंदुओं पर जुड़ते हैं।
2. \( [\text{Co}(\text{en})_3]\text{Cl}_3 \) के लिए, दो ऑप्टिकल समावयव मौजूद हो सकते हैं।

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख \([\text{Co}(\text{en})_3]^{3+}\) के दो ऑप्टिकल समावयवों को दर्शाता है। एथिलीनडाइमाइन ('en') एक द्विदंतुर लिगैंड है। ये संरचनाएँ गैर-अध्यारोपण योग्य दर्पण छवियों का प्रतिनिधित्व करती हैं, जो ऑप्टिकल समावयवता की विशेषता है, जहाँ एक संरचना दूसरे की दर्पण छवि होती है लेकिन उन्हें एक दूसरे पर पूरी तरह से संरेखित नहीं किया जा सकता।
3. \( [\text{CO}(\text{NH}_3)_5(\text{NO}_2)](\text{NO}_3)_2 \) के लिए, कुल 10 संभावित समावयव होते हैं, जिनमें ज्यामितीय, आयनीकरण और लिंकेज समावयव शामिल हैं। लिंकेज समावयव नीचे दिए गए हैं:
\( [\text{CO}(\text{NH}_3)_5\text{NO}_2] (\text{NO}_3)_2 \) और \( [\text{CO}(\text{NH}_3)_5\text{ONO}] (\text{NO}_3)_2 \)
4. \( [\text{Pt}(\text{NH}_3)(\text{H}_2\text{O})\text{Cl}_2] \) के लिए, ज्यामितीय (cis-, trans-) समावयव मौजूद हो सकते हैं।

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख \([\text{Pt}(\text{NH}_3)(\text{H}_2\text{O})\text{Cl}_2]\) के ज्यामितीय समावयवों को दर्शाता है। 'cis' रूप में, समान लिगैंड एक दूसरे के निकट होते हैं, जबकि 'trans' रूप में, वे केंद्रीय प्लैटिनम परमाणु के विपरीत दिशा में स्थित होते हैं। यह वर्ग समतलीय ज्यामिति को दर्शाता है।
In simple words: Isomerism in coordination compounds involves complexes with the same formula but different arrangements of atoms. This question explores various types like geometrical, optical, ionization, and linkage isomerism, each arising from distinct structural differences.

🎯 Exam Tip: When identifying isomerism, first determine the coordination number and geometry. Then, look for possibilities of cis-trans (geometrical), enantiomers (optical), different counter ions (ionization), or different donor atoms in ambidentate ligands (linkage).

 

Question 4. Give evidence to show that \( [\text{CO}(\text{NH}_3)_5\text{Cl}]\text{SO}_4 \) and \( [\text{CO}(\text{NH}_3)_5\text{SO}_4]\text{Cl} \) are ionisation isomers?
Answer:
Ionisation isomers, when dissolved in water, produce distinct ions that can be identified through chemical tests. For instance, when \( [\text{CO}(\text{NH}_3)_5\text{SO}_4]\text{Cl} \) is dissolved and treated with silver nitrate (AgNO\(_3\)), it forms a white precipitate of AgCl, indicating the presence of a chloride ion outside the coordination sphere:
\( [\text{CO}(\text{NH}_3)_5\text{SO}_4]\text{Cl} + \text{AgNO}_3 \longrightarrow [\text{CO}(\text{NH}_3)_5\text{SO}_4]\text{NO}_3 + \text{AgCl} \downarrow \)
This white precipitate confirms that the isomer has a chloride ion as the counter ion, demonstrating its ionization isomer characteristic.
In simple words: Ionization isomers release different ions when dissolved in water. We can test this by adding a reagent; if one complex precipitates a specific salt (like AgCl from Ag\(\text{NO}_3\)), it confirms the presence of the corresponding ion outside the coordination sphere.

🎯 Exam Tip: To differentiate ionization isomers, perform precipitation reactions. The ion outside the coordination sphere will react and precipitate, while those inside will not, serving as definitive evidence.

 

Question 5. Explain on the basis of valence bond theory that \( [\text{Ni}(\text{CN})_4]^{2-} \) ion with square planar structure is diamagnetic and the \( [\text{NiCl}_4]^{2-} \) ion with tetrahedral geometry is paramagnetic.
Answer:
According to valence bond theory, the \( \text{CN}^- \) ion acts as a strong field ligand. When it interacts with the \( \text{Ni}^{2+} \) ion, it forces the two unpaired 3d electrons to pair up. This results in \( \text{dsp}^2 \) hybridization, leading to a square planar structure. Since there are no unpaired electrons, the \( [\text{Ni}(\text{CN})_4]^{2-} \) complex is diamagnetic.
Conversely, the \( \text{Cl}^- \) ion is a weak field ligand and cannot cause the pairing of the two unpaired 3d electrons in \( \text{Ni}^{2+} \). This leads to \( \text{sp}^3 \) hybridization, forming a tetrahedral geometry. Due to the presence of unpaired electrons, the \( [\text{NiCl}_4]^{2-} \) complex is paramagnetic.
In simple words: Strong ligands like \(\text{CN}^-\) force electron pairing in nickel, resulting in no unpaired electrons, dsp\(\text{^2}\) hybridization, square planar geometry, and diamagnetism. Weak ligands like \(\text{Cl}^-\) don't force pairing, leaving unpaired electrons, sp\(\text{^3}\) hybridization, tetrahedral geometry, and paramagnetism.

🎯 Exam Tip: Remember that strong field ligands cause electron pairing in central metal ions, leading to lower spin complexes and often diamagnetism, while weak field ligands do not, resulting in high spin complexes and paramagnetism.

 

Question 6. \( [\text{NiCl}_4]^{2-} \) is paramagnetic while \( [\text{Ni}(\text{CO})_4] \) is diamagnetic though both are tetrahedral. Why?
Answer:
In \( [\text{NiCl}_4]^{2-} \), the \( \text{Ni}^{2+} \) ion has a \( \text{3d}^8 \) configuration. Since \( \text{Cl}^- \) is a weak field ligand, it does not induce pairing of the two unpaired electrons in the 3d orbitals, making the complex paramagnetic.
In contrast, for \( [\text{Ni}(\text{CO})_4] \), nickel is in a zero oxidation state with a \( \text{3d}^8\text{4s}^2 \) electronic configuration. Carbon monoxide (CO) is a strong field ligand that forces the 4s electrons to pair up with the 3d electrons, resulting in a \( \text{3d}^{10} \) configuration with no unpaired electrons. Hence, despite having a tetrahedral geometry similar to \( [\text{NiCl}_4]^{2-} \), \( [\text{Ni}(\text{CO})_4] \) is diamagnetic.
In simple words: Both complexes are tetrahedral, but \( [\text{NiCl}_4]^{2-} \) is paramagnetic because the weak \(\text{Cl}^-\) ligand doesn't pair up nickel's two unpaired d-electrons. \( [\text{Ni}(\text{CO})_4] \) is diamagnetic because the strong CO ligand forces all electrons to pair up, leaving no unpaired electrons.

🎯 Exam Tip: The magnetic property (paramagnetic vs. diamagnetic) of a complex is primarily determined by the presence or absence of unpaired electrons, which is influenced by the nature of the ligand (strong vs. weak field) and the oxidation state of the central metal ion.

 

Question 7. \( [\text{Fe}(\text{H}_2\text{O})_6]^{3+} \) is strongly paramagnetic whereas \( [\text{Fe}(\text{CN})_6]^{3-} \) is weakly paramagnetic. Explain?
Answer:
For \( [\text{Fe}(\text{H}_2\text{O})_6]^{3+} \), the central iron ion is in a \( +3 \) oxidation state, meaning its electron configuration is \( \text{3d}^5 \). Since \( \text{H}_2\text{O} \) is a weak field ligand, it does not cause the 3d electrons to pair up. This results in five unpaired electrons, leading to a strongly paramagnetic complex with \( \text{sp}^3\text{d}^2 \) hybridization, forming an outer orbital complex.
In the case of \( [\text{Fe}(\text{CN})_6]^{3-} \), the iron ion is also in a \( +3 \) oxidation state (\( \text{3d}^5 \)). However, \( \text{CN}^- \) is a strong field ligand. It causes the 3d electrons to pair up, leaving only one unpaired electron. This leads to \( \text{d}^2\text{sp}^3 \) hybridization, forming an inner orbital complex with weak paramagnetism.
In simple words: \( [\text{Fe}(\text{H}_2\text{O})_6]^{3+} \) is strongly paramagnetic because water, a weak ligand, doesn't pair up iron's five d-electrons. \( [\text{Fe}(\text{CN})_6]^{3-} \) is weakly paramagnetic because cyanide, a strong ligand, forces most d-electrons to pair, leaving only one unpaired electron.

🎯 Exam Tip: The strength of the ligand directly impacts the pairing of electrons and thus the magnetic moment. Strong field ligands cause pairing (low spin, fewer unpaired electrons), while weak field ligands do not (high spin, more unpaired electrons).

 

Question 8. Explain \( [\text{Co}(\text{NH}_3)_6]^{3+} \) is an inner orbital complex whereas \( [\text{Ni}(\text{NH}_3)_6]^{2+} \) is an outer orbital complex.
Answer:
In \( [\text{Co}(\text{NH}_3)_6]^{3+} \), the cobalt ion is in a \( +3 \) oxidation state with a \( \text{3d}^6 \) electron configuration. Ammonia (\(\text{NH}_3\)) is a strong field ligand, which causes the 3d electrons to pair up, leaving two d-orbitals vacant. This allows for \( \text{d}^2\text{sp}^3 \) hybridization, utilizing inner 3d orbitals, and thus forming an inner orbital complex.
For \( [\text{Ni}(\text{NH}_3)_6]^{2+} \), the nickel ion is in a \( +2 \) oxidation state, also with a \( \text{3d}^8 \) configuration. Although ammonia is a strong field ligand, in the case of \( \text{Ni}^{2+} \), the 3d electrons do not pair up sufficiently to provide two empty d-orbitals for inner orbital hybridization. Instead, the hybridization involves outer orbitals, specifically \( \text{sp}^3\text{d}^2 \), leading to the formation of an outer orbital complex.
In simple words: \( [\text{Co}(\text{NH}_3)_6]^{3+} \) is an inner orbital complex because strong ligand ammonia forces \( \text{Co}^{3+} \) (3d\(\text{^6}\)) electrons to pair, freeing inner d-orbitals for d\(\text{^2}\)sp\(\text{^3}\) hybridization. \( [\text{Ni}(\text{NH}_3)_6]^{2+} \) is an outer orbital complex because \( \text{Ni}^{2+} \) (3d\(\text{^8}\)) electrons do not pair enough to use inner d-orbitals, leading to sp\(\text{^3}\)d\(\text{^2}\) hybridization.

🎯 Exam Tip: The formation of inner or outer orbital complexes depends on the availability of inner d-orbitals for hybridization, which is influenced by the central metal ion's electron configuration and the ligand field strength.

 

Question 9. Predict the number of unpaired electrons in the square planar \( [\text{Pt}(\text{CN})_4]^{2-} \) ion?
Answer:
For a square planar geometry, the hybridization involved is \( \text{dsp}^2 \). In the \( [\text{Pt}(\text{CN})_4]^{2-} \) ion, platinum is in the \( +2 \) oxidation state, corresponding to a \( \text{5d}^8 \) electron configuration. Since \( \text{CN}^- \) is a very strong field ligand, it causes the 5d electrons to pair up, making one 'd' orbital available for \( \text{dsp}^2 \) hybridization. Consequently, there are no unpaired electrons in the \( [\text{Pt}(\text{CN})_4]^{2-} \) ion.
In simple words: In square planar \( [\text{Pt}(\text{CN})_4]^{2-} \), platinum(II) has a 5d\(\text{^8}\) configuration. Cyanide is a strong ligand that forces all d-electrons to pair, making an inner d-orbital available for dsp\(\text{^2}\) hybridization. Thus, there are zero unpaired electrons.

🎯 Exam Tip: For square planar complexes, especially those of \( \text{d}^8 \) ions with strong field ligands, always anticipate \( \text{dsp}^2 \) hybridization and no unpaired electrons, leading to diamagnetism.

 

Question 10. The hexaquamanganese(II) ion contains five unpaired electrons, while the hexacyanidomanganese (II) ion contains only one unpaired electron. Explain using Crystal Field Theory?
Answer:
In the hexaquamanganese(II) ion, \( [\text{Mn}(\text{H}_2\text{O})_6]^{2+} \), manganese is in the \( +2 \) oxidation state, resulting in a \( \text{3d}^5 \) electron configuration. Water (\(\text{H}_2\text{O}\)) is a weak field ligand, which causes only a small splitting of the d-orbitals. The crystal field splitting energy (\(\Delta_o\)) is not large enough to force electron pairing. Therefore, the electrons occupy the orbitals according to Hund's rule, giving a \( \text{t}_{2\text{g}}^3\text{e}_{\text{g}}^2 \) configuration with five unpaired electrons.
Conversely, in the hexacyanidomanganese(II) ion, \( [\text{Mn}(\text{CN})_6]^{2-} \), manganese is also in the \( +2 \) oxidation state with a \( \text{3d}^5 \) electron configuration. Cyanide (\(\text{CN}^-\)) is a strong field ligand, leading to a large crystal field splitting energy (\(\Delta_o\)). This large splitting forces the electrons to pair up in the lower energy \( \text{t}_{2\text{g}} \) orbitals, resulting in a \( \text{t}_{2\text{g}}^5\text{e}_{\text{g}}^0 \) configuration with only one unpaired electron.
In simple words: Water, being a weak ligand, causes small d-orbital splitting in \( [\text{Mn}(\text{H}_2\text{O})_6]^{2+} \), leading to five unpaired electrons (high spin). Cyanide, a strong ligand, causes large d-orbital splitting in \( [\text{Mn}(\text{CN})_6]^{2-} \), forcing electrons to pair, resulting in only one unpaired electron (low spin).

🎯 Exam Tip: Crystal Field Theory effectively explains magnetic properties based on ligand strength. Strong field ligands cause large splitting (\(\Delta_o\)) and electron pairing (low spin), while weak field ligands cause small splitting and no pairing (high spin).

 

Question 11. Calculate the overall complex dissociation equilibrium constant for the \( [\text{Cu}(\text{NH}_3)_4]^{2+} \) ion, given that \( \beta_4 \) for this complex is \( 2.1 \times 10^{13} \).
Answer:
The overall dissociation constant for a complex is simply the reciprocal of its overall stability constant.
Given \( \beta_4 = 2.1 \times 10^{13} \).
Overall dissociation constant \( \text{K}_{\text{diss}} = \frac{1}{\beta_4} \)
\( \text{K}_{\text{diss}} = \frac{1}{2.1 \times 10^{13}} \)
\( \text{K}_{\text{diss}} \approx 4.7 \times 10^{-14} \)
In simple words: The dissociation constant is the inverse of the stability constant. Given the stability constant \(\beta_4\) as \(2.1 \times 10^{13}\), the dissociation constant is calculated as \(1 / (2.1 \times 10^{13})\), which approximates to \(4.7 \times 10^{-14}\).

🎯 Exam Tip: Remember the fundamental relationship: the overall dissociation constant is the inverse of the overall stability constant. Ensure correct handling of exponents in calculations.

 

GSEB Class 12 Chemistry Coordination Compounds Text Book Questions and Answers

 

Question 1. Explain the bonding in coordination compounds in terms of Werner's postulates?
Answer:
Coordination compounds were known since the 18th century, but a comprehensive theory to explain their observed properties was lacking. Alfred Werner, in 1893, proposed his coordination theory, which provided a correct explanation for these characteristics. The fundamental postulates of Werner's theory are summarized below:

1. Metals exhibit two types of valencies: primary and secondary. The primary valency is ionizable, while the secondary valency is non-ionizable.
2. Each metal atom or ion possesses a fixed number of secondary valencies, which is equal to its coordination number.
3. Primary valencies are satisfied by negative ions, whereas secondary valencies are satisfied by negative ions or neutral groups (ligands).
4. Ligands satisfying secondary valencies are oriented towards specific positions in space, giving the complex a definite geometry. Primary valencies, however, are non-directional.

For instance, in a series of cobalt(III) chloride compounds with ammonia, it was observed that some chloride ions precipitated as AgCl upon adding excess silver nitrate solution in the cold, while others remained in solution. For example, when \( \text{CoCl}_3 \cdot \text{6NH}_3 \) is treated with excess \( \text{AgNO}_3 \), all three chlorine atoms precipitate as AgCl. However, with \( \text{CoCl}_3 \cdot \text{5NH}_3 \), only two chlorine atoms precipitate. Similarly, for \( \text{CoCl}_3 \cdot \text{4NH}_3 \), one chloride ion precipitates, and for \( \text{CoCl}_3 \cdot \text{3NH}_3 \), none precipitate. These observations support the structural formulas as described in the following table:

Molecular Formula of Complex	Number of Chloride Atoms Precipitated	Formula of the Complex	Colour
CoCl3.6NH3	3	[Co(NH3)6]Cl3	Yellow
CoCl3.5NH3	2	[Co(NH3)5Cl]Cl2	Purple
CoCl3.4NH3	1	[Co(NH3)4Cl2]Cl	Green
CoCl3.3NH3	0	[Co(NH3)3Cl3]	Blue-green

In modern chemistry, the spatial arrangements are termed coordination polyhedra. Species within the square brackets are coordination complexes, and ions outside are counter ions. Based on Werner's theory, the complex \( \text{CoCl}_3 \cdot \text{6NH}_3 \) or \( [\text{Co}(\text{NH}_3)_6]\text{Cl}_3 \) (hexaamminecobalt(III) chloride) can be visually represented.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र \([\text{Co}(\text{NH}_3)_6]\text{Cl}_3\) कॉम्प्लेक्स की संरचना को दर्शाता है, जिसमें कोबाल्ट केंद्रीय परमाणु है। तीन बिंदीदार रेखाएँ कोबाल्ट की प्राथमिक संयोजकता (ऑक्सीकरण अवस्था) को इंगित करती हैं, जबकि छह मोटी रेखाएँ इसकी द्वितीयक संयोजकता (समन्वय संख्या) को दर्शाती हैं, जहाँ अमोनिया लिगैंड के रूप में कार्य करता है।

Differences between a double salt and a complex:
Both double salts and complexes are formed by combining two or more stable compounds in a stoichiometric ratio. However, they differ in their behavior when dissolved in water. Double salts, such as carnallite (\(\text{KCl}\cdot\text{MgCl}_2\cdot\text{6H}_2\text{O}\)), Mohr's salt (\(\text{FeSO}_4\cdot(\text{NH}_4)_2\text{SO}_4\cdot\text{6H}_2\text{O}\)), and alums, dissociate completely into simple ions in aqueous solution. In contrast, complex ions, like \( [\text{Fe}(\text{CN})_6]^{3-} \) from \( \text{K}_3[\text{Fe}(\text{CN})_6] \), do not dissociate into \( \text{Fe}^{3+} \) and \( \text{CN}^- \) ions.
In simple words: Werner's theory explains coordination compounds have two valencies: primary (ionizable, oxidation state) and secondary (non-ionizable, coordination number). Ligands satisfy secondary valencies and define geometry. This explains why some chloride ions precipitate and others don't, distinguishing complexes from simple salts.

🎯 Exam Tip: Werner's theory is foundational. Focus on the distinction between primary and secondary valencies, their nature (ionizable vs. non-ionizable), and how they dictate the complex's formula and geometry. The precipitation experiments are classic evidence.

 

Question 2. \( \text{FeSO}_4 \) solution mixed with \( (\text{NH}_4)_2\text{SO}_4 \) solution in 1:1 molar ratio gives the test of \( \text{Fe}^{2+} \) ion. Explain why?
Answer:
When \( \text{FeSO}_4 \) is combined with \( (\text{NH}_4)_2\text{SO}_4 \) in a 1:1 molar ratio, it forms a double salt, specifically Mohr's salt (\(\text{FeSO}_4\cdot(\text{NH}_4)_2\text{SO}_4\cdot\text{6H}_2\text{O}\)). Double salts completely ionize in solution to yield their constituent simple ions. Thus, Mohr's salt in solution dissociates to produce \( \text{Fe}^{2+} \) ions, which will readily give positive tests for \( \text{Fe}^{2+} \).
Conversely, if \( \text{CuSO}_4 \) solution is mixed with aqueous ammonia in a 1:4 molar ratio, it forms a complex, \( [\text{Cu}(\text{NH}_3)_4]\text{SO}_4 \). The complex ion \( [\text{Cu}(\text{NH}_3)_4]^{2+} \) does not further dissociate into \( \text{Cu}^{2+} \) ions, and therefore, it will not give the characteristic tests for free \( \text{Cu}^{2+} \) ions.
In simple words: Mixing \(\text{FeSO}_4\) and \((\text{NH}_4)_2\text{SO}_4\) creates a double salt (Mohr's salt) that fully dissociates into simple \(\text{Fe}^{2+}\) ions, so it tests positive for \(\text{Fe}^{2+}\). In contrast, mixing \(\text{CuSO}_4\) with ammonia forms a stable complex, \( [\text{Cu}(\text{NH}_3)_4]^{2+} \), which does not release free \(\text{Cu}^{2+}\) ions, so it won't test positive for them.

🎯 Exam Tip: The key difference lies in whether the substance is a double salt (which dissociates completely) or a coordination complex (where the central metal ion is part of a stable, non-dissociating complex ion).

 

Question 3. Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic?
Answer:
1. **Central ion or center of coordination:** This refers to the cation to which one or more neutral molecules or anions are directly bonded. These central atoms/ions also act as Lewis acids.
*Examples:* In \( [\text{Ni}(\text{NH}_3)_6]^{2+} \), \( \text{Ni}^{2+} \) is the central ion. In \( [\text{Fe}(\text{CN})_6]^{3-} \), \( \text{Fe}^{3+} \) is the central ion.

2. **Ligand:** An atom or group of atoms that binds to the central metal atom through its lone pair of electrons (donor atom). Ligands are classified by the number of donor atoms: unidentate (one), bidentate (two), and polydentate (more than two).
*Monodentate ligands:* Donate one electron pair per molecule or ion.
*Examples:* \( \text{NH}_3 \), \( \text{H}_2\text{O} \), \( \text{CO} \), \( \text{Cl}^- \), \( \text{F}^- \), \( \text{OH}^- \).
*Bidentate ligands:* Furnish two electron pairs, bonding to a metal ion through two sites.
*Examples:* Ethylenediamine (\(\text{en}\)) \( (\text{NH}_2\text{-CH}_2\text{-CH}_2\text{-NH}_2) \). Oxalate ion (\(\text{C}_2\text{O}_4^{2-}\)) \( ([\text{OOC-COO}]^{(-)}) \).

3. **Ambidentate ligands:** These are monodentate ligands that can ligate through two different atoms.
*Examples:* The \( \text{NO}_2^- \) ion can coordinate via nitrogen (nitrito-N, \( \text{M-NO}_2 \)) or oxygen (nitrito-O, \( \text{M-ONO} \)). The \( \text{SCN}^- \) ion can coordinate via sulfur or nitrogen. Ambidentate ligands are not bidentate as they cannot coordinate simultaneously from two atoms.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख नाइट्राइटो लिगैंड के दो अलग-अलग बंधन विधियों को दर्शाता है। एक में, केंद्रीय धातु परमाणु (M) नाइट्रोजन परमाणु (नाइट्राइटो-N) के माध्यम से लिगैंड से बंधता है। दूसरे में, धातु परमाणु ऑक्सीजन परमाणु (नाइट्राइटो-O) के माध्यम से जुड़ता है।

4. **Chelating ligand:** A bidentate or polydentate ligand that binds to a central atom, forming a ring-like structure called a 'chelate'. These ligands form more stable coordination compounds than their monodentate counterparts.
*Example:* Ethylene diamine forming a chelate ring with \( \text{Cu}^{2+} \). EDTA (ethylenediaminetetraacetic acid) is a hexadentate ligand with six donor groups (two nitrogen and four oxygen atoms from carboxylate groups).

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एथिलीनडाइमाइन (en) लिगैंड द्वारा केंद्रीय कॉपर (II) आयन के साथ बनाए गए दो किलेट वलयों को दर्शाता है। इसमें दो नाइट्रोजन परमाणु कॉपर से जुड़ते हैं, जिससे पाँच-सदस्यीय वलय बनता है, जो जटिल यौगिकों को अतिरिक्त स्थिरता प्रदान करता है।

5. **Coordination sphere:** This consists of the central atom and its surrounding ligands, enclosed within square brackets \( [] \). Atoms, ions, or molecules within this sphere are non-ionizable. Ionizable groups, known as counter ions, are written outside the brackets.
*Example:* In \( \text{K}_4[\text{Fe}(\text{CN})_6] \), \( [\text{Fe}(\text{CN})_6]^{4-} \) is the coordination sphere, and \( \text{K}^+ \) is the counter ion.

6. **Coordination number:** The total number of ligands directly bonded to the central metal atom in the coordination sphere.
*Examples:* In \( [\text{Ni}(\text{NH}_3)_6]\text{Cl}_2 \), \( \text{Ni}^{2+} \) forms six bonds with six \( \text{NH}_3 \) molecules, so its coordination number is six. In \( [\text{Cu}(\text{NH}_3)_4]^{2+} \), \( \text{Cu} \) has a coordination number of four.

7. **Charge on complex:** This is the sum of the oxidation numbers of all atoms/molecules or ions within the coordination sphere.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक समन्वय परिसर के घटकों को दर्शाता है। केंद्रीय धातु आयन लिगैंड से बंधा होता है, जो समन्वय संख्या को परिभाषित करता है। समन्वय गोले के भीतर सभी घटकों के ऑक्सीकरण संख्याओं का योग परिसर पर कुल आवेश को दर्शाता है।

8. **Coordination polyhedron:** This describes the spatial arrangement of the ligand atoms directly attached to the central atom/ion, defining the complex's geometry.
*Common shapes:* Tetrahedral, square planar, octahedral, square pyramidal, and trigonal bipyramidal.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख विभिन्न समन्वय बहुफलकों को दर्शाता है, जो केंद्रीय धातु परमाणु के चारों ओर लिगैंड परमाणुओं की स्थानिक व्यवस्था का प्रतिनिधित्व करते हैं। इनमें अष्टफलकीय, चतुष्फलकीय, वर्ग समतलीय, त्रिकोणीय द्विपिरामिडी और वर्ग पिरामिडी जैसी सामान्य ज्यामितियाँ शामिल हैं।

9. **Homoleptic and heteroleptic complexes:**
*Homoleptic complexes:* Complexes where the metal is bound to only one type of ligand.
*Example:* \( [\text{Co}(\text{NH}_3)_6]^{3+} \).
*Heteroleptic complexes:* Complexes where the metal is bound to more than one kind of ligand.
*Example:* \( [\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+ \).
In simple words: This question defines core coordination chemistry terms: central ion (metal), ligand (donor molecule/ion), coordination number (ligands bonded), coordination polyhedron (3D shape), and classifies complexes as homoleptic (one ligand type) or heteroleptic (multiple ligand types).

🎯 Exam Tip: Be ready to define each term and provide clear, distinct examples. Understanding these fundamental concepts is essential for grasping the complexities of coordination chemistry.

 

Question 4. What is meant by unidentate, didentate, and ambidentate ligands? Give two examples for each?
Answer:
1. **Ligand:** A ligand is an atom or group of atoms that binds to a central metal atom through the lone pair of electrons present in its donor atom. Ligands are classified based on the number of donor atoms they possess:

* **Unidentate ligands:** These ligands furnish one electron pair and coordinate through a single donor atom to the central metal ion.
*Examples:* \( \text{NH}_3 \) (ammonia), \( \text{H}_2\text{O} \) (water), \( \text{CO} \) (carbon monoxide), \( \text{Cl}^- \) (chloride ion), \( \text{F}^- \) (fluoride ion), \( \text{OH}^- \) (hydroxide ion).

* **Bidentate ligands:** These ligands furnish two electron pairs and coordinate through two donor atoms to the central metal ion, forming a chelate ring.
*Examples:* Ethylenediamine (\(\text{en}\)): \( \text{NH}_2\text{-CH}_2\text{-CH}_2\text{-NH}_2 \) and Oxalate ion (\(\text{C}_2\text{O}_4^{2-}\)): \( [\text{OOC-COO}]^{(-)} \).

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एथिलीनडाइमाइन (\(\text{NH}_2\text{-CH}_2\text{-CH}_2\text{-NH}_2\)) को एक द्विदंतुर लिगैंड के रूप में दिखाता है, जो अपने दो नाइट्रोजन परमाणुओं के माध्यम से केंद्रीय धातु से जुड़ सकता है। ऑक्सलेट आयन (\([\text{OOC-COO}]^{(-)}\)) एक और द्विदंतुर लिगैंड है, जो अपने दो ऑक्सीजन परमाणुओं के माध्यम से धातु से जुड़ता है।

* **Polydentate ligands:** These ligands supply several pairs of electrons and have multiple donor sites, binding to the central metal ion through more than two points.
*Example:* EDTA (Ethylenediaminetetraacetic acid) is a hexadentate ligand with six donor groups: two nitrogen atoms and four oxygen atoms from four carboxylic acid groups.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख EDTA (एथिलीनडाइमाइनटेट्राएसिटिक एसिड) आयन की संरचना को दर्शाता है। यह एक हेक्साडेंटेट लिगैंड है, जिसमें दो नाइट्रोजन परमाणु और चार ऑक्सीजन परमाणु (कार्बोक्सिलिक समूहों से) केंद्रीय धातु आयन के साथ समन्वय बंधन बनाने में सक्षम हैं, जो इसे एक शक्तिशाली किलेटिंग एजेंट बनाता है।

2. **Ambidentate ligands:** These are monodentate ligands that possess two different potential donor atoms but typically coordinate through only one at a time.
*Examples:* The \( \text{NO}_2^- \) ion can coordinate either through nitrogen (nitrito-N, \( \text{M-N=O} \)) or through oxygen (nitrito-O, \( \text{M-O-N=O} \)). Similarly, the \( \text{SCN}^- \) ion can coordinate through sulfur or nitrogen. It's important to note that ambidentate ligands are not bidentate; they cannot coordinate to the central metal ion simultaneously from two atoms.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख नाइट्राइटो लिगैंड के दो अलग-अलग बंधन विधियों को दर्शाता है। एक में, केंद्रीय धातु परमाणु (M) नाइट्रोजन परमाणु (नाइट्राइटो-N) के माध्यम से लिगैंड से बंधता है। दूसरे में, धातु परमाणु ऑक्सीजन परमाणु (नाइट्राइटो-O) के माध्यम से जुड़ता है।
In simple words: Ligands bond to a central metal atom. Unidentate ligands have one donor atom (like \(\text{NH}_3\), \(\text{Cl}^-\)). Bidentate ligands have two donor atoms (like ethylenediamine, oxalate). Ambidentate ligands have two potential donor atoms but use only one at a time (like \(\text{NO}_2^-\), \(\text{SCN}^-\)).

🎯 Exam Tip: Differentiate between bidentate (multiple *simultaneous* donor sites) and ambidentate (multiple *potential* donor sites, but only one used at a time). Providing clear, correct chemical examples for each type is crucial for full marks.

 

Question 5. Specify the oxidation numbers of the metals in the following coordination entities:
1. \( [\text{Co}(\text{H}_2\text{O})(\text{CN})(\text{en})_2]^{2+} \)
2. \( [\text{PtCl}_4]^{2-} \)
3. \( [\text{PtCl}_4]^{2-} \)
4. \( \text{K}_3[\text{Fe}(\text{CN})_6] \)
5. \( [\text{Cr}(\text{NH}_3)_3\text{Cl}_3] \)
Answer:
1. \( +3 \)
2. \( +2 \)
3. \( +2 \)
4. \( +3 \)
5. \( +3 \)
In simple words: To find the oxidation number, sum the charges of all ligands (water is neutral, cyanide is -1, ethylenediamine is neutral, chloride is -1) and equate to the complex's overall charge. For example, in \( [\text{Co}(\text{H}_2\text{O})(\text{CN})(\text{en})_2]^{2+} \), Co + 0 + (-1) + 0 = +2, so Co = +3.

🎯 Exam Tip: Always remember the charges of common ligands (e.g., neutral for \(\text{H}_2\text{O}\), \(\text{NH}_3\), \(\text{en}\); -1 for \(\text{Cl}^-\), \(\text{CN}^-\), \(\text{NO}_2^-\); -2 for \(\text{C}_2\text{O}_4^{2-}\)) and the overall charge of the complex to correctly determine the central metal's oxidation state.

 

Question 6. Using IUPAC norms write the formulas for the following:
i. Tetrahydroxidozincate(II)
ii. Potassium tetrachloridopalladate(II)
iii. Diamminedichloridoplatinum(II)
iv. Potassium tetracyanidonickelate(II)
v. Pentaamminenitrito-O-cobalt(III)
vi. Hexaamminecobalt(III) sulphate
vii. Potassium tri(oxalato)chromate(III)
viii. Hexaammineplatinum(IV)
ix. Tetrabromidocuprate(II)
x. Pentaamminenitrito-N-cobalt(III)
Answer:
i. \( [\text{Zn}(\text{OH})_4]^{2-} \)
ii. \( \text{K}_2[\text{PdCl}_4] \)
iii. \( [\text{Pt}(\text{NH}_3)_2\text{Cl}_2] \)
iv. \( \text{K}_2[\text{Ni}(\text{CN})_4] \)
v. \( [\text{Co}(\text{NH}_3)_5(\text{ONO})]^{2+} \)
vi. \( [\text{Co}(\text{NH}_3)_6]_2(\text{SO}_4)_3 \)
vii. \( \text{K}_3[\text{Cr}(\text{C}_2\text{O}_4)_3] \)
viii. \( [\text{Pt}(\text{NH}_3)_6]^{4+} \)
ix. \( [\text{CuBr}_4]^{2-} \)
x. \( [\text{Co}(\text{NH}_3)_5(\text{NO}_2)]^{2+} \)
In simple words: This task requires translating coordination compound names into their chemical formulas by correctly identifying the central metal, its oxidation state, and the ligands, then arranging them according to IUPAC rules, including counter ions.

🎯 Exam Tip: When writing formulas from IUPAC names, carefully identify the central metal's oxidation state, the number and type of ligands, and whether the complex itself is charged or neutral. Also, remember the correct placement of counter ions.

 

Question 7. Using IUPAC norms write the systematic names of the following:
1. \( [\text{CO}(\text{NH}_3)_6]\text{Cl}_3 \)
2. \( [\text{Pt}(\text{NH}_3)_2\text{Cl}(\text{NH}_2\text{CH}_3)]\text{Cl} \)
3. \( [\text{Ti}(\text{H}_2\text{O})_6]^{3+} \)
4. \( [\text{CO}(\text{NH}_3)_4\text{Cl}(\text{NO}_2)]\text{Cl} \)
5. \( [\text{Mn}(\text{H}_2\text{O})_6]^{2+} \)
6. \( [\text{NiCl}_4]^{2-} \)
7. \( [\text{Ni}(\text{NH}_3)_6]\text{Cl}_2 \)
8. \( [\text{CO}(\text{en})_3]^{3+} \)
9. \( [\text{Ni}(\text{CO})_4] \)
Answer:
1. Hexaamminecobalt(III) chloride
2. Diamminechlorido(methanamine)platinum(II) chloride
3. Hexaaquatitanium(III) ion
4. Tetraamminechloridonitrito-N-cobalt(III) chloride
5. Hexaaquamanganese(II) ion
6. Tetrachloridonickelate(II) ion
7. Hexaamminenickel(II) chloride
8. Tris(ethylenediamine)cobalt(III) ion
9. Tetracarbonylnickel(0)
In simple words: This task involves applying IUPAC naming conventions to assign systematic names to various coordination compounds based on their chemical formulas, ensuring correct identification of central metal, ligands, and oxidation states.

🎯 Exam Tip: For naming, identify the central metal, its oxidation state, and all ligands. Arrange ligands alphabetically, use appropriate prefixes for multiple ligands, and apply specific endings for anionic or neutral complexes.

 

Question 8. List various types of isomerism possible for coordination compounds, giving an example of each?
Answer:
Isomerism describes the phenomenon where compounds have the same chemical composition but differ in the structural arrangement of their atoms. Coordination compounds display two main types: structural isomerism and stereoisomerism, each with further subdivisions.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख समन्वय यौगिकों में समावयवता के वर्गीकरण को दर्शाता है। इसे दो मुख्य प्रकारों में विभाजित किया गया है: संरचनात्मक समावयवता (जिसमें आयनीकरण, हाइड्रेट, लिंकेज और समन्वय समावयवता शामिल है) और त्रिविम समावयवता (जिसमें ज्यामितीय और ऑप्टिकल समावयवता शामिल है)।

**Structural isomerism:** This type arises from differences in the structures of coordination compounds.
1. **Ionization isomerism:** Occurs when coordination compounds with the same molecular formula yield different ions in solution.
*Example:* Violet-colored \( [\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4 \) (pentaamminebromidocobalt(III) sulphate) and red-colored \( [\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br} \) (pentaamminesulphatocobalt(III) bromide) are ionization isomers.

2. **Linkage isomerism:** Found in complexes containing ambidentate monodentate ligands with more than one donor atom, where the ligand can attach through different atoms.
*Example:* The \( \text{NO}_2^- \) ligand can bond to a metal through nitrogen (nitrito-N, \( \text{M-NO}_2 \)) or oxygen (nitrito-O, \( \text{M-ONO} \)). For example, \( [\text{Co}(\text{NH}_3)_5\text{NO}_2]\text{Cl}_2 \) (pentaamminenitrito-N-cobalt(III) chloride, yellow) and \( [\text{Co}(\text{NH}_3)_5\text{ONO}]\text{Cl}_2 \) (pentaamminenitrito-O-cobalt(III) chloride, red) are linkage isomers. Another common ambidentate ligand is \( \text{CNS}^- \), as seen in \( [\text{Cr}(\text{H}_2\text{O})_5\text{SCN}]^{2+} \) and \( [\text{Cr}(\text{H}_2\text{O})_5\text{NCS}]^{2+} \).

3. **Coordination isomerism:** This isomerism is possible when both positive and negative ions of a salt are complex ions. It involves the exchange of ligands between the complex cation and complex anion (i.e., between the coordination spheres in a compound).
*Example:* \( [\text{Cr}(\text{NH}_3)_6][\text{Co}(\text{CN})_6] \) (hexaamminechromium(III) hexacyanidocobaltate(III)) and \( [\text{Co}(\text{NH}_3)_6][\text{Cr}(\text{CN})_6] \) (hexaamminecobalt(III) hexacyanidochromate(III)) are coordination isomers.

4. **Hydrate isomerism:** Arises in complexes due to a different number of water molecules present both inside and outside the coordination spheres.
*Example:* \( [\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3 \) (violet, hexaaquachromium(III) chloride) and \( [\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2\cdot\text{H}_2\text{O} \) (grey-green, pentaaquachloridochromium(III) chloride monohydrate) are hydrate isomers.

**Stereoisomerism:** Occurs when compounds have the same ligands coordinated to the same central ion, but these ligands have different spatial arrangements around the central atom.
1. **Geometrical isomerism (cis-trans isomerism):** Arises when ligands occupy different positions (adjacent or opposite) around the central ion. This is significant for coordination numbers 4 (square planar) and 6 (octahedral).
*Square planar complexes:* \( \text{Ma}_2\text{b}_2 \) and \( \text{Ma}_2\text{bc} \) types show cis-trans isomerism.
*Example:* \( [\text{Pt}(\text{NH}_3)_2\text{Cl}_2] \).

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख \([\text{Pt}(\text{NH}_3)_2\text{Cl}_2]\) के दो ज्यामितीय समावयवों को दर्शाता है। 'cis' रूप में, दो समान लिगैंड (\(\text{NH}_3\) और \(\text{Cl}\) दोनों) एक दूसरे के निकट स्थित होते हैं, जबकि 'trans' रूप में, वे केंद्रीय प्लैटिनम परमाणु के विपरीत दिशा में स्थित होते हैं। यह वर्ग समतलीय ज्यामिति को दर्शाता है।

*Octahedral complexes:* \( \text{Ma}_4\text{b}_2 \) and \( \text{Ma}_3\text{b}_3 \) types show cis-trans isomerism.
*Example:* \( [\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+ \) and \( [\text{Co}(\text{en})_2\text{Cl}_2]^+ \). Another type is facial (fac) and meridional (mer) isomers for \( \text{Ma}_3\text{b}_3 \) complexes like \( [\text{Co}(\text{NO}_2)_3(\text{NH}_3)_3] \).

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख अष्टफलकीय समन्वय संख्या 6 वाले परिसरों के ज्यामितीय समावयवों को दर्शाता है। शीर्ष पंक्ति \([\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+\) के cis और trans रूपों को दिखाती है। निचली पंक्ति में, \([\text{Co}(\text{NO}_2)_3(\text{NH}_3)_3]\) के लिए फेशियल (fac) और मेरिडियनल (mer) समावयव दिखाए गए हैं। 'fac' रूप में, समान लिगैंड एक अष्टफलक के एक फलक पर स्थित होते हैं, जबकि 'mer' रूप में, वे अष्टफलक के एक भूमध्यरेखीय तल पर स्थित होते हैं।

2. **Optical isomerism:** The ability of a compound to rotate the plane of plane-polarized light. Optically active compounds exist as d- and l-isomeric forms (enantiomers), which are non-superimposable mirror images.
*Example:* Chelates like \( [\text{Co}(\text{en})_3]\text{Br}_3 \) (tris(ethylenediamine)cobalt(III) bromide) lack a plane of symmetry and exhibit optical isomerism, producing non-superimposable mirror images.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख \([\text{Co}(\text{en})_3]^{3+}\) के ऑप्टिकल समावयवों को दर्शाता है, जहाँ 'en' एथिलीनडाइमाइन लिगैंड है। ये संरचनाएँ एक दूसरे की गैर-अध्यारोपण योग्य दर्पण छवियाँ हैं, जो ऑप्टिकल समावयवता की विशेषता है, जहाँ एक संरचना ध्रुवीकृत प्रकाश को दक्षिणावर्त और दूसरी वामावर्त दिशा में घुमाती है।
Additionally, for the coordinate entity \( [\text{PtCl}_2(\text{en})_2]^{2+} \), only the *cis* isomer exhibits optical isomerism.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख \([\text{PtCl}_2(\text{en})_2]^{2+}\) के cis और trans ज्यामितीय समावयवों को दर्शाता है। विशेष रूप से, केवल cis समावयव ऑप्टिकल समावयवता प्रदर्शित करता है क्योंकि इसमें एक गैर-अध्यारोपण योग्य दर्पण छवि होती है, जबकि trans समावयव में समरूपता का एक तल होता है, जिससे यह ऑप्टिकली निष्क्रिय हो जाता है।
In simple words: Coordination compounds show isomerism, meaning they have the same formula but different atom arrangements. Structural isomers include ionization (different ions released), linkage (different donor atoms), coordination (ligand exchange between complex ions), and hydrate (different water positions). Stereoisomers include geometrical (cis-trans/fac-mer, different spatial positions) and optical (non-superimposable mirror images).

🎯 Exam Tip: When listing types of isomerism, define each clearly and provide a distinct, correct chemical example. For stereoisomerism, be prepared to explain the conditions for cis-trans and optical activity, especially for common geometries (square planar, octahedral).

2. Optical Isomerism

Two optical isomers may exist.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [Cr(H2O)2(C2O4)2]- कॉम्प्लेक्स के लिए दो ज्यामितीय आइसोमर, सिस- और ट्रांस- रूप को दर्शाता है। सिस-आइसोमर में दोनों जल अणु एक-दूसरे के पास होते हैं, जबकि ट्रांस-आइसोमर में वे एक-दूसरे के विपरीत होते हैं। ये संरचनाएँ केंद्रीय क्रोमियम परमाणु के चारों ओर अष्टफलकीय समन्वय दिखाती हैं।
The cis-isomer of the complex [Cr(H2O)2(C2O4)2]- is depicted with two water ligands adjacent to each other. The trans-isomer shows the two water ligands positioned opposite each other. These structures illustrate the octahedral coordination geometry around the central chromium atom.
The optical isomers for [Co(en)3]3+ complex are illustrated here. Each diagram shows a central cobalt atom coordinated to three bidentate ethylenediamine (en) ligands, forming an octahedral geometry. The two structures are non-superimposable mirror images of each other, representing the enantiomers.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस आरेख में [Co(en)3]3+ कॉम्प्लेक्स के दो प्रकाशीय आइसोमर दिखाए गए हैं। प्रत्येक आरेख एक केंद्रीय कोबाल्ट परमाणु को तीन द्वि-दंती एथिलीनडाइमाइन (en) लिगेंड के साथ समन्वित दर्शाता है, जो अष्टफलकीय ज्यामिति बनाते हैं। दोनों संरचनाएँ एक-दूसरे की गैर-अध्यारोपण योग्य दर्पण छवियाँ हैं, जो एनांशिओमर (प्रतिबिंब समावयवी) का प्रतिनिधित्व करती हैं।
In the coordinate entity of the type [PtCl2(en)2]2+, only the cis isomer shows optical isomerism.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [PtCl2(en)2]2+ कॉम्प्लेक्स के सिस-आइसोमर के दो प्रकाशीय आइसोमर को दर्शाता है। केंद्रीय प्लैटिनम परमाणु से जुड़े दो क्लोरीन परमाणु और दो एथिलीनडाइमाइन (en) लिगेंड हैं। ये दो संरचनाएँ एक-दूसरे की दर्पण छवियाँ हैं जो गैर-अध्यारोपण योग्य हैं, और केवल सिस-रूप ही प्रकाशीय सक्रियता प्रदर्शित करता है।
The cis-isomer of the [PtCl2(en)2]2+ complex is shown here, exhibiting two optical isomers. The central platinum atom is bonded to two chloride ligands and two bidentate ethylenediamine (en) ligands. These two structures are non-superimposable mirror images of each other, demonstrating that only the cis-form is optically active.
In simple words: Optical isomers are molecules that are mirror images of each other but cannot be perfectly superimposed, like your left and right hands. For complexes, this often happens when the central metal is surrounded by ligands in a way that creates a non-symmetrical structure.

🎯 Exam Tip: Remember to differentiate between geometrical (cis/trans/fac/mer) and optical isomerism. Optical isomers require chirality (non-superimposable mirror images), often seen in octahedral complexes with bidentate ligands or specific arrangements.

 

Question 9. How many geometrical isomers are possible in the following coordination entities?

(1) [Cr(C2O4)3]3-
(2) [Co(NH3)3Cl3]
Answer:(1) For [Cr(C2O4)3]3- : No geometrical isomers are possible.
(2) For [Co(NH3)3Cl3] : Two geometrical isomers (facial and meridional) are possible.
In simple words: Geometrical isomers relate to how atoms are arranged in space around a central atom. The first complex has a symmetrical structure with identical bidentate ligands, so no geometrical isomers exist. The second complex can have two distinct arrangements of its three ammonia and three chloride ligands.

🎯 Exam Tip: Recognizing symmetry is key for geometrical isomerism. Octahedral complexes of the type Ma6, M(AA)3, or M(AA)2XY generally do not show geometrical isomerism, while Ma3b3, Ma4b2, etc., often do.

 

Question 10. Draw the structures of optical isomers of:

(1) [Cr(C2O4)3]3-
(2) [PtCl2(en)2]2+
(3) [Cr(NH3)2Cl2(en)]+
Answer:(i) For [Cr(C2O4)3]3- :
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [Cr(C2O4)3]3- कॉम्प्लेक्स के दो प्रकाशीय आइसोमर (एनान्शियोमर) को दर्शाता है। केंद्रीय क्रोमियम परमाणु तीन द्वि-दंती ऑक्सालेट (C2O4) लिगेंड से जुड़ा हुआ है। दोनों संरचनाएँ एक-दूसरे की गैर-अध्यारोपण योग्य दर्पण छवियाँ हैं, जो इस कॉम्प्लेक्स की चिरलता (chirality) को दर्शाती हैं।
This diagram illustrates the two optical isomers (enantiomers) of the [Cr(C2O4)3]3- complex. The central chromium atom is coordinated to three bidentate oxalate (C2O4) ligands. The two structures are non-superimposable mirror images of each other, demonstrating the chirality of this complex. (ii) For [PtCl2(en)2]2+ :
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [PtCl2(en)2]2+ कॉम्प्लेक्स के सिस-रूप के दो प्रकाशीय आइसोमर को दर्शाता है। केंद्रीय प्लैटिनम परमाणु दो क्लोरीन लिगेंड और दो द्वि-दंती एथिलीनडाइमाइन (en) लिगेंड से जुड़ा है। ये दो संरचनाएँ एक-दूसरे की दर्पण छवियाँ हैं जो गैर-अध्यारोपण योग्य हैं, यह दर्शाती हैं कि केवल सिस-रूप प्रकाशीय सक्रियता प्रदर्शित करता है।
This diagram depicts the two optical isomers of the cis-form of the [PtCl2(en)2]2+ complex. The central platinum atom is bonded to two chloride ligands and two bidentate ethylenediamine (en) ligands. These two structures are non-superimposable mirror images of each other, indicating that only the cis-form exhibits optical activity. (iii) For [Cr(NH3)2Cl2(en)]+ :
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [Cr(NH3)2Cl2(en)]+ कॉम्प्लेक्स के दो संभावित प्रकाशीय आइसोमर को दर्शाता है। केंद्रीय क्रोमियम परमाणु दो अमोनिया लिगेंड, दो क्लोरीन लिगेंड और एक द्वि-दंती एथिलीनडाइमाइन (en) लिगेंड से जुड़ा है। ये संरचनाएँ एक-दूसरे की गैर-अध्यारोपण योग्य दर्पण छवियाँ हैं, जो इस अष्टफलकीय कॉम्प्लेक्स की चिरल प्रकृति को दर्शाती हैं।
This diagram illustrates two potential optical isomers for the [Cr(NH3)2Cl2(en)]+ complex. The central chromium atom is coordinated to two ammonia ligands, two chloride ligands, and one bidentate ethylenediamine (en) ligand. These structures are non-superimposable mirror images of each other, indicating the chiral nature of this octahedral complex.
In simple words: Optical isomers are mirror images that can't be perfectly stacked. We're showing these pairs for three different complexes, highlighting their three-dimensional arrangement around the central metal atom.

🎯 Exam Tip: When drawing optical isomers, always check for a plane of symmetry. If a molecule has a plane of symmetry, it is achiral and will not have optical isomers. Chirality is essential for optical activity.

 

Question 11. Draw all the isomers (geometrical and optical) of:

(1) [CoCl2(en)2]+
(2) [Co(NH3)Cl(en)2]2+
(3) [Co(NH3)2Cl2(en)]+
Answer:(1) For [CoCl2(en)2]+:
ℹ️ चित्र व्याख्या (Diagram Explanation): ये आरेख [CoCl2(en)2]+ कॉम्प्लेक्स के सिस- और ट्रांस- ज्यामितीय आइसोमर को दर्शाते हैं। सिस-आइसोमर में दोनों क्लोरीन लिगेंड एक-दूसरे के निकट होते हैं, जबकि ट्रांस-आइसोमर में वे विपरीत दिशा में होते हैं। सिस-रूप भी प्रकाशीय सक्रियता प्रदर्शित करता है, और यहाँ उसके दो एनान्शियोमर दिखाए गए हैं जो एक-दूसरे की दर्पण छवियाँ हैं।
These diagrams illustrate the cis- and trans-geometrical isomers of the [CoCl2(en)2]+ complex. In the cis-isomer, the two chloride ligands are adjacent, while in the trans-isomer, they are opposite. The cis-form also exhibits optical activity, and its two enantiomeric mirror images are shown. (2) For [Co(NH3)Cl(en)2]2+:
ℹ️ चित्र व्याख्या (Diagram Explanation): ये आरेख [Co(NH3)Cl(en)2]2+ कॉम्प्लेक्स के सिस- और ट्रांस- ज्यामितीय आइसोमर को दर्शाते हैं। सिस-आइसोमर में अमोनिया और क्लोरीन लिगेंड एक-दूसरे के निकट होते हैं, जबकि ट्रांस-आइसोमर में वे विपरीत दिशा में होते हैं। सिस-रूप भी प्रकाशीय सक्रियता प्रदर्शित करता है, और यहाँ उसके दो एनान्शियोमर दिखाए गए हैं जो एक-दूसरे की दर्पण छवियाँ हैं।
These diagrams illustrate the cis- and trans-geometrical isomers of the [Co(NH3)Cl(en)2]2+ complex. In the cis-isomer, the ammonia and chloride ligands are adjacent, while in the trans-isomer, they are opposite. The cis-form also exhibits optical activity, and its two enantiomeric mirror images are shown. (3) For [Co(NH3)3Cl3]: (Note: The question asked for [Co(NH3)2Cl2(en)]+, but the provided answer diagrams are for [Co(NH3)3Cl3].)
ℹ️ चित्र व्याख्या (Diagram Explanation): ये आरेख [Co(NH3)3Cl3] कॉम्प्लेक्स के फेशियल (fac-) और मेरिडियनल (mer-) आइसोमर को दर्शाते हैं। फेशियल आइसोमर में तीन समान लिगेंड (अमोनिया या क्लोरीन) अष्टफलक के एक फलक के शीर्ष पर स्थित होते हैं। मेरिडियनल आइसोमर में, तीन समान लिगेंड केंद्रीय धातु आयन के एक मेरिडियन (भूमध्य रेखा) पर स्थित होते हैं।
These diagrams depict the facial (fac-) and meridional (mer-) isomers of the [Co(NH3)3Cl3] complex. In the facial isomer, the three identical ligands (either ammonia or chloride) occupy vertices on one face of the octahedron. In the meridional isomer, the three identical ligands are positioned along a meridian of the central metal ion.
In simple words: We are visualizing the different spatial arrangements (isomers) of these coordination compounds. This includes showing where specific ligands are placed relative to each other (geometrical isomers) and identifying mirror-image forms that can't be stacked perfectly (optical isomers).

🎯 Exam Tip: For octahedral complexes, remember that M(AA)2X2 and M(AA)2XY type complexes can exhibit cis-trans isomerism, and often the cis form is chiral and shows optical isomerism. Complexes of type Ma3b3 show fac-mer isomerism.

 

Question 12. Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomerism?

Answer: Three isomers are possible.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [Pt(NH3)(Br)(Cl)(py)] कॉम्प्लेक्स के तीन ज्यामितीय आइसोमर को दर्शाता है, जहाँ केंद्रीय प्लैटिनम परमाणु एक अमोनिया, एक ब्रोमीन, एक क्लोरीन और एक पाइरीडीन लिगेंड से जुड़ा है। ये संरचनाएँ एक वर्ग-समतलीय ज्यामिति में लिगेंड के विभिन्न सापेक्ष व्यवस्थाओं को प्रदर्शित करती हैं। प्रत्येक आइसोमर में लिगेंड की स्थिति अलग-अलग होती है, जो विशिष्ट ज्यामितीय विन्यास बनाती है।
This diagram illustrates the three geometrical isomers of the [Pt(NH3)(Br)(Cl)(py)] complex, with the central platinum atom bonded to an ammonia, a bromine, a chlorine, and a pyridine ligand. These structures show different relative arrangements of the ligands in a square planar geometry. Each isomer displays distinct ligand positions, forming unique geometrical configurations.
These isomers do not show optical isomerism (square planar complexes generally do not exhibit optical activity because they possess a plane of symmetry).
In simple words: For this platinum complex, three different arrangements of the ligands are possible in a flat, square shape. None of these arrangements will have an optical mirror image because square planar complexes are typically symmetrical.

🎯 Exam Tip: Square planar complexes rarely exhibit optical isomerism because they usually possess a plane of symmetry. Focus on identifying cis/trans arrangements for geometrical isomerism in such complexes.

 

Question 13. Aqueous copper sulphate solution (blue in colour) gives:

(1) A green precipitate with aqueous potassium fluoride and
(2) A bright green solution with aqueous potassium chloride. Explain these experimental results?
Answer: Aqueous \( \text{CuSO}_4 \) solution exists as \( \text{[Cu(H}_2\text{O)}_4\text{]SO}_4 \), which has a blue colour due to the presence of \( \text{[Cu(H}_2\text{O)}_4\text{]}^{2+} \) ions.

(1) When potassium fluoride (KF) is added, the weakly coordinating \( \text{H}_2\text{O} \) ligands are replaced by \( \text{F}^- \) ligands, leading to the formation of \( \text{[CuF}_4\text{]}^{2-} \) ions. This results in a green precipitate.
\[\text{[Cu(H}_2\text{O)}_4\text{]}^{2+} + 4\text{F}^- \implies \text{[CuF}_4\text{]}^{2-} + 4\text{H}_2\text{O}\] (2) When potassium chloride (KCl) is added, the \( \text{Cl}^- \) ligands replace the weakly coordinating \( \text{H}_2\text{O} \) ligands, forming \( \text{[CuCl}_4\text{]}^{2-} \) ions. This results in a bright green solution.
\[\text{[Cu(H}_2\text{O)}_4\text{]}^{2+} + 4\text{Cl}^- \implies \text{[CuCl}_4\text{]}^{2-} + 4\text{H}_2\text{O}\]
In simple words: Copper sulfate solution is blue because of water ligands. When fluoride is added, it replaces water, forming a green precipitate. When chloride is added, it also replaces water, forming a different green solution. The colour changes occur because different ligands create different electronic transitions in the copper complex.

🎯 Exam Tip: This question tests your understanding of ligand exchange reactions and how different ligands affect the colour of transition metal complexes. Remember that colour is often due to d-d transitions, which are influenced by the crystal field splitting energy caused by ligands.

 

Question 14. What is the coordination entity formed when an excess aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H2S(g) is passed through this solution?

Answer: When an excess of aqueous KCN is added to a copper sulfate solution, the coordination entity \( \text{[Cu(CN)}_4\text{]}^{2-} \) is formed. The reaction is:
\[\text{[Cu(H}_2\text{O)}_4\text{]}^{2+} + 4\text{CN}^- \implies \text{[Cu(CN)}_4\text{]}^{2-} + 4\text{H}_2\text{O}\] The cyanide \( \text{(CN)}^- \) ion acts as a strong ligand, forming a highly stable complex with the \( \text{Cu}^{2+} \) ion. Because this complex is very stable, virtually no free \( \text{Cu}^{2+} \) ions are available in the solution. Consequently, when hydrogen sulfide \( \text{(H}_2\text{S(g))} \) is passed through this solution, no copper sulfide precipitate \( \text{(CuS)} \) is formed, as the \( \text{Cu}^{2+} \) ions are sequestered within the stable cyanide complex.
In simple words: When a lot of cyanide is added to copper sulfate, it forms a super stable complex with copper, trapping all the copper ions. So, even if you add hydrogen sulfide, which normally forms copper sulfide precipitate, there are no free copper ions to react, hence no precipitate.

🎯 Exam Tip: This illustrates the concept of complex ion stability and its impact on qualitative analysis. Strong field ligands often form very stable complexes, effectively removing the metal ion from solution and preventing typical precipitation reactions.

 

Question 15. Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:

(1) [Fe(CN)6]4-
(2) [FeF6]3-
(3) [Co(C2O4)3]3-
Answer: Valence bond theory (VBT) explains the nature of bonding, geometry, and magnetic properties of coordination complexes based on the hybridization of metal orbitals. Here's an analysis of the given complexes:

Complex	Metal Ion	E.C of metal ion	Type of ligand	Hybridisation	Geometry of complex	No. of electrons	Magnetic property
i. [Fe(CN)6]4-	Fe2+	3d6	Strong	d2sp3	octahedral	0	Diamagnetic
ii. [FeF6]3-	Fe3+	3d5	Weak	sp3d2	octahedral	5	Paramagnetic
iii. [Co(C2O4)3]3-	Co3+	3d6	Strong	d2sp3	octahedral	0	Diamagnetic
iv. [CoF6]3-	Co3+	3d6	Weak	sp3d2	octahedral	4	Paramagnetic

Werner's theory had limitations in explaining specific properties like complex formation, geometry, and magnetic behaviour.

Thus, Linus Pauling (1931) extended valence bond theory (VBT) to understand complexes. The main principles of VBT are:
(1) The central metal ion in a complex provides a sufficient number of vacant orbitals for forming coordinate bonds with the ligands.
(2) The quantity of empty orbitals made available is equivalent to the coordination number of the central metal ion.
(3) These vacant orbitals (s, p, or d) of the central metal ion undergo hybridization to produce an equal number of hybrid orbitals. These hybrid orbitals are oriented towards the ligand positions, determining the complex's geometry.
(4) The empty hybrid orbitals of the metal ion overlap with the fully filled orbitals of the ligand, creating ligand-metal coordinate bonds.
(5) Complexes can be classified as outer orbital (high spin) or inner orbital (low spin) depending on whether the d orbitals from the outer shell or inner shell are utilized for hybridization.
In simple words: Valence bond theory explains how metal ions and ligands form bonds by sharing electron pairs, leading to specific shapes and magnetic properties. It involves the metal's orbitals mixing (hybridization) to accommodate the incoming ligands.

🎯 Exam Tip: When applying VBT, correctly determine the metal's oxidation state, its d-electron configuration, and whether the ligand is strong or weak field (which dictates electron pairing). This helps predict hybridization, geometry, and magnetic properties.

 

Question 16. Draw a figure to show the splitting of d orbitals in an octahedral crystal field?

Answer: The splitting of d orbitals in an octahedral crystal field, as well as the fundamental aspects of Valence Bond Theory and Crystal Field Theory, are detailed below.
Types of hybridisation and geometries of complexes:

Coordination number	Type of hybridisation	Geometry of complex	Examples
4	sp3	Tetrahedral	[Ni(CO)4], [Ni(Cl)4]2-
4	dsp2	Square planar	[Ni(CN)4]2-, [Cu(NH3)4]2+
6	sp3d2	Octahedral	[CoF6]3-
6	d2sp3	Octahedral	[Co(NH3)6]3+, [Fe(CN)6]3-

Ligands can be categorized into two types:
(a) Strong field ligands lead to maximum electron pairing.
(b) Weak field ligands do not cause electron pairing.
\( \text{I}^- < \text{Br}^- < \text{Cl}^- < \text{F}^- < \text{OH}^- < \text{C}_2\text{O}_4^{2-} < \text{H}_2\text{O} < \text{NH}_3 < \text{en} < \text{CN}^- < \text{CO} \)
(weak field \( \implies \) strong field)
Illustration:
(i) Hexaamminecobalt (III) ion, \( \text{[Co(NH}_3\text{)}_6\text{]}^{3+} \)
In this complex, cobalt is in a +3 oxidation state. The outer electronic configuration of the \( \text{Co}^{3+} \) ion is \( 3d^6 \). \( \text{NH}_3 \) is a strong field ligand, forcing the electrons in the \( 3d \) orbitals to pair up against Hund's rule, thus making two \( 3d \) orbitals available.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [Co(NH3)6]3+ कॉम्प्लेक्स में कोबाल्ट (III) आयन के 3d, 4s, और 4p ऑर्बिटल्स के इलेक्ट्रॉन विन्यास और संकरण को दर्शाता है। 3d ऑर्बिटल्स में इलेक्ट्रॉन युग्मन के बाद, दो खाली 3d ऑर्बिटल्स, एक 4s ऑर्बिटल और तीन 4p ऑर्बिटल्स d2sp3 संकरण बनाते हैं। भरे हुए xxxx ब्लॉक लिगेंड से इलेक्ट्रॉन युग्म के दान को दर्शाते हैं।
This diagram illustrates the electron configuration and hybridization for the cobalt(III) ion in the [Co(NH3)6]3+ complex. After electron pairing in the 3d orbitals, two vacant 3d orbitals, one 4s orbital, and three 4p orbitals undergo d2sp3 hybridization. The filled 'xxxx' blocks represent the donation of electron pairs from the ligands.
Two \( 3d \), one \( 4s \) and three \( 4p \) orbitals hybridize to form six vacant \( d^2sp^3 \) hybrid orbitals. These orbitals accept electron pairs from six \( \text{NH}_3 \) ligands, forming \( \text{[Co(NH}_3\text{)}_6\text{]}^{3+} \). The resulting complex is an inner orbital complex, which is low spin and diamagnetic (no unpaired electrons), and has an octahedral shape.
(ii) Hexafluoridocobaltate (III) ion, \( \text{[CoF}_6\text{]}^{3-} \)
In this complex, cobalt is in a +3 oxidation state. The outer electronic configuration of \( \text{Co}^{3+} \) is \( 3d^6 \). Since \( \text{F}^- \) is a weak field ligand, it cannot pair up the unpaired electrons in the \( 3d \) orbitals.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [CoF6]3- कॉम्प्लेक्स में कोबाल्ट (III) आयन के 3d, 4s, 4p, और 4d ऑर्बिटल्स के इलेक्ट्रॉन विन्यास और संकरण को दर्शाता है। चूंकि F- एक दुर्बल क्षेत्र लिगेंड है, 3d इलेक्ट्रॉनों का युग्मन नहीं होता है। परिणामस्वरूप, एक 4s, तीन 4p, और दो 4d ऑर्बिटल्स sp3d2 संकरण बनाते हैं। भरे हुए xxxx ब्लॉक लिगेंड से इलेक्ट्रॉन युग्म के दान को दर्शाते हैं।
This diagram illustrates the electron configuration and hybridization for the cobalt(III) ion in the [CoF6]3- complex across 3d, 4s, 4p, and 4d orbitals. Because F- is a weak field ligand, the 3d electrons do not pair up. Consequently, one 4s, three 4p, and two 4d orbitals undergo sp3d2 hybridization. The filled 'xxxx' blocks represent the donation of electron pairs from the ligands.
The resulting complex is a high spin (paramagnetic - possessing four unpaired \( 3d \) electrons) outer orbital complex with an octahedral shape.
(iii) Tetracyanidonickelate (II) ion, \( \text{[Ni(CN)}_4\text{]}^{2-} \)
In this complex, nickel is in a +2 oxidation state. The electronic configuration of \( \text{Ni}^{2+} \) ion is \( 3d^8 \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [Ni(CN)4]2- कॉम्प्लेक्स में निकल (II) आयन के 3d, 4s, और 4p ऑर्बिटल्स के इलेक्ट्रॉन विन्यास और संकरण को दर्शाता है। CN- एक प्रबल क्षेत्र लिगेंड होने के कारण, दो अयुग्मित 3d इलेक्ट्रॉनों का युग्मन होता है, जिससे एक खाली 3d ऑर्बिटल उपलब्ध होता है। एक 3d, एक 4s, और दो 4p ऑर्बिटल्स dsp2 संकरण बनाते हैं। भरे हुए xxxx ब्लॉक लिगेंड से इलेक्ट्रॉन युग्म के दान को दर्शाते हैं।
This diagram illustrates the electron configuration and hybridization for the nickel(II) ion in the [Ni(CN)4]2- complex across 3d, 4s, and 4p orbitals. Due to CN- being a strong field ligand, the two unpaired 3d electrons pair up, making one 3d orbital vacant. One 3d, one 4s, and two 4p orbitals undergo dsp2 hybridization. The filled 'xxxx' blocks represent the donation of electron pairs from the ligands.
The resulting complex is a low spin (diamagnetic - no unpaired electrons) inner orbital complex with square planar geometry.
(iv) Tetrachloridonickelate (II) ion, \( \text{[NiCl}_4\text{]}^{2-} \)
In this complex, nickel is in a +2 oxidation state. The electronic configuration of \( \text{Ni}^{2+} \) is \( 3d^8 \). Since \( \text{Cl}^- \) is a weak field ligand, it cannot pair up the unpaired electrons in the \( 3d \) orbitals.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [NiCl4]2- कॉम्प्लेक्स में निकल (II) आयन के 3d, 4s, और 4p ऑर्बिटल्स के इलेक्ट्रॉन विन्यास और संकरण को दर्शाता है। चूंकि Cl- एक दुर्बल क्षेत्र लिगेंड है, 3d इलेक्ट्रॉनों का युग्मन नहीं होता है। परिणामस्वरूप, एक 4s और तीन 4p ऑर्बिटल्स sp3 संकरण बनाते हैं। भरे हुए xxxx ब्लॉक लिगेंड से इलेक्ट्रॉन युग्म के दान को दर्शाते हैं।
This diagram illustrates the electron configuration and hybridization for the nickel(II) ion in the [NiCl4]2- complex across 3d, 4s, and 4p orbitals. Because Cl- is a weak field ligand, the 3d electrons do not pair up. Consequently, one 4s and three 4p orbitals undergo sp3 hybridization. The filled 'xxxx' blocks represent the donation of electron pairs from the ligands.
The resulting complex is a high spin (paramagnetic - two unpaired electrons) outer orbital complex with tetrahedral geometry.
(v) Tetracarbonylnickel (0), \( \text{[Ni(CO)}_4\text{]} \)
Here, nickel is in a zero oxidation state. Ni has a \( 3d^84s^2 \) electronic configuration. Since CO is a strong field ligand, the \( 4s \) electrons are forced to pair up against the Aufbau principle, leaving the \( 4s \) orbitals empty.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [Ni(CO)4] कॉम्प्लेक्स में निकल (0) आयन के 3d, 4s, और 4p ऑर्बिटल्स के इलेक्ट्रॉन विन्यास और संकरण को दर्शाता है। CO के प्रबल क्षेत्र लिगेंड होने के कारण, 4s इलेक्ट्रॉनों का युग्मन 3d ऑर्बिटल्स में हो जाता है। परिणामस्वरूप, एक 4s और तीन 4p ऑर्बिटल्स sp3 संकरण बनाते हैं। भरे हुए xxxx ब्लॉक चार CO अणुओं से इलेक्ट्रॉन युग्म के दान को दर्शाते हैं।
This diagram illustrates the electron configuration and hybridization for the nickel(0) ion in the [Ni(CO)4] complex across 3d, 4s, and 4p orbitals. Due to CO being a strong field ligand, the 4s electrons are paired into the 3d orbitals. Consequently, one 4s and three 4p orbitals undergo sp3 hybridization. The filled 'xxxx' blocks represent the donation of four electron pairs from four CO molecules.
Thus, nickel tetracarbonyl is a low spin (diamagnetic - no unpaired electron) outer orbital complex with tetrahedral geometry.
**Crystal Field Theory (CFT):**
The limitations of valence bond theory regarding coordination compounds are largely addressed by the crystal field theory, proposed by H. Bethe and Van Vleck (1930). Crystal field theory views the metal ion as being in an electrostatic field created by the surrounding ligands.
The five d-orbitals in an isolated gaseous metal atom/ion are degenerate, meaning they have the same energy. When ligands approach the central atom, this degeneracy of the d-orbitals is removed. This results in the splitting of the d-orbital energies, a phenomenon known as crystal field splitting. The specific pattern of this splitting depends on the crystal field's nature.
**Splitting in the octahedral field:**
As ligands approach the central atom in an octahedral geometry, the d-orbital energies increase. The electrons in the \( d_{x^2-y^2} \) and \( d_{z^2} \) orbitals (which have lobes along the axes) experience more repulsion from the ligand electron pairs than the electrons in the \( d_{xy} \), \( d_{xz} \), and \( d_{yz} \) orbitals (which have lobes between the axes).
Thus, the five d-orbitals in an octahedral field split into two sets: \( d_{z^2} \) and \( d_{x^2-y^2} \) (designated as 'eg'), and \( d_{xy} \), \( d_{xz} \), and \( d_{yz} \) (designated as 't2g'). The energy difference between the eg and t2g orbitals is called the crystal field splitting energy, denoted by \( \Delta_o \) (where 'o' stands for octahedral).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख अष्टफलकीय क्रिस्टल क्षेत्र में d ऑर्बिटल्स के ऊर्जा विपाटन को दर्शाता है। मुक्त अवस्था में समान ऊर्जा वाले पांच d ऑर्बिटल्स, क्रिस्टल क्षेत्र में औसत ऊर्जा में वृद्धि के बाद दो ऊर्जा स्तरों में विभाजित हो जाते हैं: उच्च ऊर्जा वाले eg (dx2-y2, dz2) ऑर्बिटल्स और निम्न ऊर्जा वाले t2g (dxy, dxz, dyz) ऑर्बिटल्स। इन दोनों स्तरों के बीच का ऊर्जा अंतर क्रिस्टल क्षेत्र विपाटन ऊर्जा (\( \Delta_o \)) कहलाता है।
This diagram illustrates the energy splitting of d orbitals in an octahedral crystal field. The five degenerate d-orbitals in the free state, after an increase in average energy within the crystal field, split into two energy levels: the higher energy eg \( (d_{x^2-y^2}, d_{z^2}) \) orbitals and the lower energy t2g \( (d_{xy}, d_{xz}, d_{yz}) \) orbitals. The energy difference between these two levels is denoted as crystal field splitting energy (\( \Delta_o \)).
In \( d^1 \) coordination complexes, the single d-electron occupies one of the lower t2g orbitals. In \( d^2 \) and \( d^3 \) coordination entities, the d-electrons occupy the t2g orbitals singly, following Hund's rule. For \( d^4 \) ions, two possible electron distribution patterns arise: the electron can either enter the t2g or eg levels. The actual configuration is determined by the relative energies of \( \Delta_o \) and P (pairing energy).
If \( \Delta_o < \text{P} \), we have a weak field, high spin situation where the fourth electron enters one of the eg orbitals.
If \( \Delta_o > \text{P} \), we have a strong field, low spin situation where the pairing occurs in the t2g level.
**Splitting in the tetrahedral field:**
In a tetrahedral field, ligands approach the metal ion through positions located between the axes. Consequently, the d-orbitals along the Cartesian axes \( (d_{x^2-y^2} \) and \( d_{z^2}) \) are less repelled than the \( d_{xy} \), \( d_{xz} \), and \( d_{yz} \) orbitals. Therefore, the eg orbitals have lower energy and t2g orbitals have relatively higher energy.
Thus, in a tetrahedral field, the d-orbital splitting is inverted compared to octahedral splitting. However, the field splitting and splitting energies are smaller than in an octahedral field. The splitting energy \( \Delta_t = \frac{-4}{9} \Delta_o \) ('t' for tetrahedral). In a tetrahedral field, low spin complexes are rarely observed.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख चतुष्फलकीय क्रिस्टल क्षेत्र में d ऑर्बिटल्स के ऊर्जा विपाटन को दर्शाता है। मुक्त अवस्था में समान ऊर्जा वाले पांच d ऑर्बिटल्स, क्रिस्टल क्षेत्र में औसत ऊर्जा में वृद्धि के बाद दो ऊर्जा स्तरों में विभाजित हो जाते हैं: उच्च ऊर्जा वाले t2g (dxy, dxz, dyz) ऑर्बिटल्स और निम्न ऊर्जा वाले eg (dx2-y2, dz2) ऑर्बिटल्स।
This diagram illustrates the energy splitting of d orbitals in a tetrahedral crystal field. The five degenerate d-orbitals in the free state, after an increase in average energy within the crystal field, split into two energy levels: the higher energy t2g \( (d_{xy}, d_{xz}, d_{yz}) \) orbitals and the lower energy eg \( (d_{x^2-y^2}, d_{z^2}) \) orbitals.
In simple words: This question explains how electrons in metal orbitals behave when ligands (other molecules) surround them. Valence bond theory describes orbital mixing and shapes, while crystal field theory explains how ligands cause the metal's d-orbitals to split into different energy levels, affecting properties like colour and magnetism.

🎯 Exam Tip: For CFT, clearly explain the d-orbital splitting patterns for octahedral and tetrahedral geometries. Remember that the magnitude of \( \Delta_o \) or \( \Delta_t \) compared to pairing energy (P) determines whether a complex is high spin or low spin, which is critical for magnetic properties.

 

Question 17. What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand?

Answer: The spectrochemical series is an arrangement of ligands in order of their increasing field strength (or crystal field splitting energy, CFSE).

**Weak field ligands** are those that cause a small crystal field splitting (low CFSE). They do not cause electron pairing in the d-orbitals of the central metal ion. Examples include \( \text{I}^- \), \( \text{Br}^- \), \( \text{Cl}^- \), \( \text{F}^- \), \( \text{OH}^- \), \( \text{C}_2\text{O}_4^{2-} \), and \( \text{H}_2\text{O} \).

**Strong field ligands** are those that cause a large crystal field splitting (high CFSE). They induce electron pairing in the d-orbitals of the central metal ion (if possible), especially in low spin complexes. Examples include \( \text{NH}_3 \), \( \text{en} \) (ethylenediamine), \( \text{CN}^- \), and \( \text{CO} \).
In simple words: The spectrochemical series ranks ligands by how strongly they affect a metal's electron energy levels. Weak field ligands cause a small effect and don't force electrons to pair up, while strong field ligands cause a large effect and often make electrons pair up.

🎯 Exam Tip: Memorize the approximate order of common ligands in the spectrochemical series. Understanding this series is fundamental for predicting the magnetic properties (high spin vs. low spin) and colour of coordination compounds.

 

Question 18. What is crystal field splitting energy? How does the magnitude of \( \Delta_o \) decide the actual configuration of d orbitals in a coordination entity?

Answer: **Crystal Field Splitting Energy \( (\Delta_o) \)** is defined as the energy difference between the two sets of d-orbitals (eg and t2g) that results from the interaction of a metal ion's d-orbitals with the electrostatic field generated by the surrounding ligands in an octahedral complex.
The magnitude of \( \Delta_o \) (crystal field splitting energy) in comparison to the pairing energy (P) determines the actual electron configuration of d-orbitals in a coordination entity:
(1) If \( \Delta_o < \text{P} \) (weak field case): The crystal field splitting energy is less than the energy required to pair electrons. In this scenario, the fourth electron (for \( d^4 \) configuration) will enter the higher energy eg orbitals, leading to a **high spin complex**. Electrons will occupy orbitals singly before pairing up in the lower energy t2g orbitals.
(2) If \( \Delta_o > \text{P} \) (strong field case): The crystal field splitting energy is greater than the pairing energy. In this situation, the fourth electron will pair up in the lower energy t2g orbitals, leading to a **low spin complex**. This means electrons will fill the t2g orbitals completely before occupying the eg orbitals.
In simple words: Crystal field splitting energy is the energy gap between d-orbitals when ligands pull them apart. If this gap is small, electrons will spread out (high spin). If the gap is large, electrons will crowd together (low spin) to save energy.

🎯 Exam Tip: Clearly define \( \Delta_o \) and explain its role in determining electron spin state. A thorough understanding of how \( \Delta_o \) compares to pairing energy (P) is crucial for predicting the magnetic moment and spectroscopic properties of complexes.

 

Question 19. [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2- is diamagnetic. Explain why?

Answer:**For [Cr(NH3)6]3+:** The central metal ion is \( \text{Cr}^{3+} \). Chromium (Cr) has an atomic number of 24, with an electronic configuration of \( \text{[Ar]}3d^54s^1 \). For \( \text{Cr}^{3+} \), the electronic configuration becomes \( \text{[Ar]}3d^3 \). The \( \text{NH}_3 \) ligand is a strong field ligand. However, with a \( d^3 \) configuration, there are already three unpaired electrons, one in each of the \( t_{2g} \) orbitals (in an octahedral field), and there is no pairing required or possible to make it low spin. Thus, \( \text{Cr}^{3+} \) undergoes \( d^2sp^3 \) hybridization, forming an inner orbital complex. Due to the presence of three unpaired electrons, \( \text{[Cr(NH}_3\text{)}_6\text{]}^{3+} \) is **paramagnetic**.
**For [Ni(CN)4]2-:** The central metal ion is \( \text{Ni}^{2+} \). Nickel (Ni) has an atomic number of 28, with an electronic configuration of \( \text{[Ar]}3d^84s^2 \). For \( \text{Ni}^{2+} \), the electronic configuration becomes \( \text{[Ar]}3d^8 \). The \( \text{CN}^- \) ligand is a strong field ligand. In the presence of a strong field ligand, the electrons in the \( 3d \) subshell pair up. For a \( d^8 \) system, the eight electrons are arranged such that two orbitals contain paired electrons, and the remaining two electrons also pair up due to the strong ligand field, leaving one \( 3d \) orbital vacant. This allows for \( dsp^2 \) hybridization, resulting in a square planar geometry. Due to the complete pairing of electrons and the absence of any unpaired electrons, \( \text{[Ni(CN)}_4\text{]}^{2-} \) is **diamagnetic**.
In simple words: The chromium complex is paramagnetic because it has unpaired electrons. The nickel complex is diamagnetic because its strong cyanide ligands force all the electrons to pair up, leaving no unpaired electrons.

🎯 Exam Tip: Always determine the oxidation state and d-electron configuration of the central metal ion first. Then, consider the nature of the ligand (strong vs. weak field) to predict electron pairing, hybridization, geometry, and ultimately, the magnetic properties.

 

Question 20. A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2- is colourless. Explain?

Answer: Both complexes contain \( \text{Ni}^{2+} \) as the central metal ion, which has a \( 3d^8 \) electronic configuration. The difference in their colour can be explained by the nature of their ligands and the resulting crystal field splitting:

**For [Ni(H2O)6]2+ (green solution):**
\( \text{H}_2\text{O} \) is a weak field ligand. It causes a relatively small crystal field splitting. In a \( d^8 \) system with a weak field ligand, the electrons in the \( 3d \) orbitals do not pair up completely. This results in the presence of unpaired d electrons. These unpaired electrons can undergo d-d electronic transitions by absorbing specific wavelengths of visible light, and transmitting the complementary colours, making the complex appear green.

**For [Ni(CN)4]2- (colourless solution):**
\( \text{CN}^- \) is a strong field ligand. It causes a large crystal field splitting. In a \( d^8 \) system with a strong field ligand (and square planar geometry), the \( 3d \) electrons are forced to pair up completely, resulting in no unpaired electrons. Without unpaired electrons and with a large energy gap (often extending beyond the visible region), there are no d-d transitions in the visible light range that would absorb specific colours. Therefore, the complex appears colourless.
In simple words: The nickel-water complex is green because its weak water ligands allow electrons to absorb some visible light and show colour. The nickel-cyanide complex is colourless because its strong cyanide ligands force all electrons to pair up, so no visible light is absorbed for colour.

🎯 Exam Tip: Colour in transition metal complexes is primarily due to d-d electronic transitions. The presence of unpaired electrons and the magnitude of crystal field splitting (influenced by ligand strength) determine if a complex will be coloured and what colour it will be. Colourless complexes typically have no d-d transitions in the visible spectrum.

 

Question 21. [Fe(CN)6]4- and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why?

Answer: Both complexes, [Fe(CN)6]4- and [Fe(H2O)6]2+, contain iron in the +2 oxidation state \( (\text{Fe}^{2+}) \), which has a \( 3d^6 \) electronic configuration. The difference in their colours in dilute solutions arises from the varying field strengths of their ligands and the resulting electron configurations and d-d transitions:

**For [Fe(H2O)6]2+:**
\( \text{H}_2\text{O} \) is a weak field ligand. In the presence of \( \text{H}_2\text{O} \), the crystal field splitting energy \( (\Delta_o) \) is smaller than the pairing energy (P). This leads to a high spin complex where the \( 3d^6 \) electrons are arranged as \( t_{2g}^4e_g^2 \), resulting in four unpaired electrons. These unpaired electrons allow for d-d transitions by absorbing light from the visible spectrum, giving the complex a distinct colour.

**For [Fe(CN)6]4-:**
\( \text{CN}^- \) is a strong field ligand. In the presence of \( \text{CN}^- \), the crystal field splitting energy \( (\Delta_o) \) is larger than the pairing energy (P). This results in a low spin complex where the \( 3d^6 \) electrons are completely paired up as \( t_{2g}^6e_g^0 \), meaning there are no unpaired electrons. The absence of unpaired electrons, coupled with a large energy gap, means there are no d-d transitions in the visible region that would absorb specific wavelengths, leading to a different or often colourless appearance compared to high-spin complexes.
In simple words: Both complexes have iron (II), but with water ligands, iron has unpaired electrons and shows colour. With strong cyanide ligands, all electrons pair up, leading to a different colour because of how light is absorbed or not absorbed.

🎯 Exam Tip: Highlight the role of ligand field strength (strong vs. weak) in determining the spin state (high vs. low spin) and consequently the number of unpaired electrons. This directly impacts the d-d transitions and thus the observed colour of the complex.

 

Question 22. Discuss the nature of bonding in metal carbonyls?

Answer: The bonding in metal carbonyls is a unique and important aspect of coordination chemistry, characterized by a synergistic interaction between the metal and the carbonyl (CO) ligand. This bonding involves both \( \sigma \) and \( \pi \) character.

(1) **Sigma \( (\sigma) \) bond formation:** A \( \sigma \) bond is formed by the donation of a lone pair of electrons from the carbon atom of the carbonyl ligand into a vacant d-orbital or hybridized orbital of the central metal atom. This forms a traditional ligand-to-metal coordinate bond.

(2) **Pi \( (\pi) \) bond formation (Back bonding):** A \( \pi \) bond is formed by the back-donation of electrons from a filled d-orbital of the metal into a vacant antibonding pi-molecular orbital \( (\pi^*) \) of the carbon monoxide ligand. This process is known as **synergic bonding** or **back bonding**. This metal-to-ligand back donation strengthens the metal-carbon bond and weakens the carbon-oxygen bond in the CO ligand.

This synergistic bonding mechanism (metal \( \to \) ligand \( \pi \) back bonding along with ligand \( \to \) metal \( \sigma \) donation) leads to highly stable metal carbonyl complexes.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख धातु कार्बोनिल में धातु-कार्बन बंधन के दोहरे स्वभाव को दर्शाता है: सिग्मा (\( \sigma \)) और पाई (\( \pi \)) बंधन। ऊपरी भाग में, CO अणु का कार्बन अपना इलेक्ट्रॉन युग्म धातु के खाली ऑर्बिटल में दान करके एक \( \sigma \) बंधन बनाता है। निचले भाग में, धातु अपने भरे हुए d-ऑर्बिटल से इलेक्ट्रॉन घनत्व को CO अणु के खाली \( \pi^* \) एंटीबॉन्डिंग ऑर्बिटल में वापस दान करके एक \( \pi \) बंधन बनाता है, जिसे बैक बॉन्डिंग कहा जाता है।
This diagram illustrates the dual nature of the metal-carbon bond in metal carbonyls: sigma (\( \sigma \)) and pi (\( \pi \)) bonding. In the upper part, the carbon of the CO molecule donates its electron pair into a vacant metal orbital, forming a \( \sigma \) bond. In the lower part, the metal back-donates electron density from its filled d-orbital into the vacant \( \pi^* \) antibonding orbital of the CO molecule, forming a \( \pi \) bond (back bonding).
In simple words: Bonding in metal carbonyls is special: carbon donates electrons to the metal (sigma bond), and then the metal gives some electrons back to carbon's antibonding orbital (pi bond). This two-way electron sharing makes the bond super strong and stable.

🎯 Exam Tip: Focus on explaining the synergic bonding in metal carbonyls, highlighting both the sigma donation from CO to metal and the pi back-bonding from metal to CO. This concept is crucial for understanding the stability and properties of these complexes.

 

Question 23. Give the oxidation state, d orbital occupation and coordination number of the central metal ion in the following complexes:

(1) K3[Co(C2O4)3]
(2) cis-[Cr(en)2Cl2]Cl
(3) (NH4)2[CoF4]
(4) [Mn(H2O)6]SO4
Answer: The following table provides the required details for each complex:

Complex	Oxidation state of metal	d-orbital occupation	Coordination number
i. K3[Co(C2O4)3]	+3	3d6 t2g6eg0	6
ii. cis-[Cr(en)2Cl2]Cl	+3	3d3 t2g3eg0	6
iii. (NH4)2[CoF4]	+2	3d7 eg4t2g3	4
iv. [Mn(H2O)6]SO4	+2	3d5 t2g3eg2	6

In simple words: This table summarizes key information for several metal complexes: the charge on the central metal, how its d-electrons are arranged, and how many ligands are directly attached to it.

🎯 Exam Tip: Accurately calculating the oxidation state of the central metal is the first step. Then, determine the d-electron configuration and, based on ligand type and coordination number, predict the d-orbital occupation (t2g/eg) and overall coordination number.

 

Question 24. Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:

(1) K[Cr(H2O)2(C2O4)2].3H2O
(2) [Co(NH3)5Cl]Cl2
(3) [CrCl3(py)3]
Answer:

1. **K[Cr(H2O)2(C2O4)2].3H2O** - **IUPAC Name:** Potassium diaquadioxalatochromate (III) trihydrate - **Oxidation state of Cr:** +3 - **Electronic configuration of Cr3+:** \( 3d^3 = t_{2g}^3e_g^0 \) - **Coordination number:** 6 - **Shape/Stereochemistry:** Octahedral - **Magnetic moment:** \( \sqrt{n(n+2)} \) BM = \( \sqrt{3(3+2)} \) = \( \sqrt{15} \) = 3.87 BM (3 unpaired electrons) 2. **[Co(NH3)5Cl]Cl2** - **IUPAC Name:** Pentaamminechloridocobalt (III) chloride - **Oxidation state of Co:** +3 - **Electronic configuration of Co3+:** \( 3d^6 = t_{2g}^6e_g^0 \) (NH3 is a strong field ligand, causing pairing) - **Coordination number:** 6 - **Shape/Stereochemistry:** Octahedral - **Magnetic moment:** 0 BM (0 unpaired electrons, diamagnetic) 3. **[CrCl3(py)3]** - **IUPAC Name:** Trichloridotripyridine chromium (III) - **Oxidation state of Cr:** +3 - **Electronic configuration of Cr3+:** \( 3d^3 = t_{2g}^3e_g^0 \) - **Coordination number:** 6 - **Shape/Stereochemistry:** Octahedral (fac- and mer- isomers possible) - **Magnetic moment:** \( \sqrt{n(n+2)} \) BM = \( \sqrt{3(3+2)} \) = \( \sqrt{15} \) = 3.87 BM (3 unpaired electrons) 4. **Caesium tetrachloridoferrate (III)** (Note: The original text had Ferrate (II) but Fe oxidation state +3, d5 configuration, and magnetic moment suggest Ferrate (III). Corrected for consistency.) - **IUPAC Name:** Caesium tetrachloridoferrate (III) - **Oxidation state of Fe:** +3 - **Electronic configuration of Fe3+:** \( 3d^5 = e_g^2t_{2g}^3 \) (Cl- is a weak field ligand, high spin) - **Coordination number:** 4 - **Shape/Stereochemistry:** Tetrahedral - **Magnetic moment:** \( \sqrt{n(n+2)} \) BM = \( \sqrt{5(5+2)} \) = \( \sqrt{35} \) = 5.92 BM (5 unpaired electrons) 5. **Potassium hexacyanidomanganate (II)** - **IUPAC Name:** Potassium hexacyanidomanganate (II) - **Oxidation state of Mn:** +2 - **Electronic configuration of Mn2+:** \( 3d^5 = t_{2g}^5e_g^0 \) (CN- is a strong field ligand, low spin) - **Coordination number:** 6 - **Shape/Stereochemistry:** Octahedral - **Magnetic moment:** \( \sqrt{n(n+2)} \) BM = \( \sqrt{1(1+2)} \) = \( \sqrt{3} \) = 1.73 BM (1 unpaired electron)
In simple words: For each complex, we identify its full chemical name, the charge on the central metal, how its electrons are arranged, and how many ligands it has. We also determine its 3D shape and whether it is magnetic (and how much so) based on unpaired electrons.

🎯 Exam Tip: For each complex, systematically determine: 1. Oxidation state of metal. 2. d-electron count. 3. Ligand field strength (strong/weak). 4. Spin state (high/low). 5. Hybridization and geometry. 6. Number of unpaired electrons and magnetic moment. This systematic approach ensures accuracy.

 

Question 25. What is meant by stability of a coordination compound in solution? State the factors which govern stability of complexes?
Answer: The reaction between a metal ion and its ligands can be understood as a Lewis acid-base reaction. If the interaction is strong, the resulting complex will be more stable from a thermodynamic perspective. The equilibrium representing the reaction between a metal ion and its ligands is shown below:
\(Ma^+ + nL^{x-} \rightleftharpoons [ML_n]^{b+}\)
Here, \(a^+\), \(x^-\), and \(b^+\) denote the charges on the metal, ligand, and complex, respectively. The stability constant \(K\) of the complex is defined as:
\[ K = \frac { [ML_{ n }]^{ b+ } }{ [M^{ a+ }][L^{ x- }]^{ n } } \]
The magnitude of this stability constant indicates the complex's stability in solution. For instance, some stability constants are:
\(Ag^+ + 2NH_3 \rightleftharpoons [Ag(NH_3)_2]^+\); \(K = 1.6 \times 10^7\)
\(Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}\); \(K = 4.5 \times 10^{11}\)
\(Cu^{2+} + 4CN^- \rightleftharpoons [Cu(CN)_4]^{2-}\); \(K = 2.0 \times 10^{27}\)
From these values, it's evident that the cyanide ion (\(CN^-\)) is a stronger ligand than ammonia (\(NH_3\)). The stability of a complex depends on several factors:
1. The charge on the central ion: Higher the charge on the metal ion, greater the stability of the complex.
2. The basic nature of the ligand: Greater the basic strength of the ligand, higher the stability of the complex.
3. The presence of chelate rings (Chelate effect): The formation of chelate rings significantly increases the stability, with maximum stability observed for 5- and 6-membered rings.
In simple words: Stability of a coordination compound in solution refers to how strongly the metal ion and ligands bond. It's measured by a stability constant (K). Factors influencing this include the charge on the central metal ion, the basicity of the ligand, and the formation of stable ring structures by chelating ligands.

🎯 Exam Tip: Remember to define the stability constant K and list the three main factors influencing complex stability. Providing an example reaction for each factor or a general equation for K can fetch full marks.

 

Question 26. What is meant by the chelate effect? Give an example?
Answer: A chelating ligand is an atom or group of atoms that binds to a central atom, forming a ring-like structure known as a 'chelate'. These ligands form more stable coordination compounds compared to their monodentate counterparts. The chelate effect refers to this enhanced stability of coordination complexes due to the formation of such rings. For example, the chelate rings formed between a \(Cu^{2+}\) ion and ethylenediamine can be illustrated as follows:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र इथाइलिनेडाइमाइन (en) लिगेंड के साथ एक केंद्रीय कॉपर आयन (\(Cu^{2+}\)) द्वारा बनाए गए चिलेट वलय को दर्शाता है। इथाइलिनेडाइमाइन एक द्विदंती लिगेंड है जो दो नाइट्रोजन परमाणुओं के माध्यम से धातु से जुड़ता है, जिससे एक स्थिर पांच-सदस्यीय वलय बनता है। यह वलय संरचना चिलेट प्रभाव के कारण बनने वाले संकुल की स्थिरता में वृद्धि करती है।
In simple words: The chelate effect describes the increased stability of complexes when a polydentate ligand (a ligand that attaches at multiple points) forms ring structures with the central metal atom. These ring-like complexes, called chelates, are more stable than those formed by monodentate ligands.

🎯 Exam Tip: Clearly define the chelate effect and provide a common example like ethylenediamine with a metal ion. Highlighting the formation of stable ring structures is key.

 

Question 27. Discuss briefly giving an example in each case the role of coordination compounds in:
Answer: The ability of metal ions to form complexes with various molecular species, leading to different physico-chemical properties, is utilized in numerous ways. Some key roles include:
1. **In analytical chemistry:**
(a) The multidentate ligand EDTA (Ethylenediamine Tetraacetic Acid) forms highly stable complexes with metal ions like \(Ca^{2+}\) and \(Mg^{2+}\). This property is leveraged in a simple titration method to estimate water hardness.
(b) A confirmatory test for detecting copper (II) involves the formation of a deep-blue colored complex, \([Cu(NH_3)_4]^{2+}\), which appears when ammonia solution is added to a copper (II) salt solution.
\[Cu^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+} \text{ (deep blue)}\]
(c) The separation of Group I precipitates (in qualitative inorganic analysis), such as \(AgCl\), \(Hg_2Cl_2\), and \(PbCl_2\), involves adding aqueous ammonia solution to the precipitate. Only silver chloride dissolves, forming the complex ion \([Ag(NH_3)_2]^+\), while \(Hg_2Cl_2\) and \(PbCl_2\) do not form such complex ions and thus remain undissolved.
\[AgCl + 2NH_3 \rightarrow [Ag(NH_3)_2]^+ + Cl^-\]
(d) A confirmatory test for nickel involves adding a dimethyl glyoxime solution, which forms a scarlet red chelate precipitate.
2. **In extraction of metals:**
(a) Metals like silver and gold are extracted by converting them into soluble cyanide complexes. For example:
\[4Au + 8CN^- + 2H_2O + O_2 \rightarrow 4[Au(CN)_2]^- + 4OH^- \text{ (soluble)}\]
The solution containing the cyanide complex is then treated with zinc, which precipitates gold:
\[2[Au(CN)_2]^- + Zn \rightarrow [Zn(CN)_4]^{2-} + 2Au\]
(b) Coordination compounds of silver and gold are used in electroplating baths for the controlled delivery of \(Ag^+\) and \(Au^+\) ions during electrorefining processes.
(c) The formation of volatile \([Ni(CO)_4]\) is utilized in the purification of nickel through the Mond process.
3. **In medicines:**
Many medications are complexes. For instance, Vitamin B12, a cobalt complex, is vital for preventing anemia. The coordination compound cis-\([PtCl_2(NH_3)_2]\), commonly known as cisplatin, is used in chemotherapy to treat cancer. Complexing agents are also employed to remove metal poisoning. For example, D-penicillamine treats copper poisoning, while EDTA is used for lead poisoning. In cases of exposure to radioactive materials like Plutonium (Pu), chelating agents are used to remove these metals from the body (chelate therapy).
4. **In photographs:**
Unexposed silver bromide (\(AgBr\)) in photographic film is dissolved using a sodium thiosulphate solution. This dissolution forms a soluble silver thiosulphato complex, a process known as 'fixing'.
\[AgBr + 2Na_2S_2O_3 \rightarrow Na_3[Ag(S_2O_3)_2] + NaBr\]
5. **In biological systems:**
Chlorophyll, a coordination compound of magnesium (II), plays a catalytic role in photosynthesis, producing glucose from \(CO_2\) and \(H_2O\). Hemoglobin, an iron (II) coordination compound, is responsible for oxygen transport. Metalloenzymes are also coordination compounds, such as carboxy peptidase A and carbonic anhydrase.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र विभिन्न जैविक और औषधीय प्रणालियों में समन्वय यौगिकों की भूमिका को दर्शाता है। इसमें कैंसर-रोधी एजेंट सिस-प्लैटिन की संरचना, एंजाइम कार्बोक्सीपेप्टिडेस-ए में जिंक आयन का समन्वय, हीमोग्लोबिन में हीम समूह की संरचना, और क्लोरोफिल ए और बी की संरचनाएं शामिल हैं। ये सभी जटिल संरचनाएं धातु आयनों की महत्वपूर्ण जैविक भूमिकाओं को उजागर करती हैं।
6. **In industrial processes:**
(a) Zeigler-Natta catalysts, such as \([(C_2H_5)_3Al \text{ and } TiCl_4]\), are employed as heterogeneous catalysts for the polymerization of alkenes.
(b) Wilkinson's catalyst, \([(Ph_3P)_3RhCl]\), is used as a catalyst for the hydrogenation of alkenes.
In simple words: Coordination compounds are used extensively in various fields like chemical analysis (e.g., EDTA for water hardness, copper tests), metal extraction (e.g., cyanide complexes for gold), medicine (e.g., cisplatin for cancer, chelate therapy for metal poisoning), photography (fixing with thiosulphate), and biological systems (e.g., chlorophyll in plants, hemoglobin for oxygen transport). They also play a role in industrial catalysis.

🎯 Exam Tip: When discussing the role of coordination compounds, provide at least one clear example for each category (analytical, medicinal, biological, industrial). Ensure chemical equations are balanced and correct, and state the function of each compound.

 

Question 28. How many ions are produced from the complex \([Co(NH_3)_6]Cl_2\) in solution?
Answer: In solution, the complex \([Co(NH_3)_6]Cl_2\) ionizes. Since the coordination number of cobalt is 6, the complex dissociates into one complex ion \([Co(NH_3)_6]^{2+}\) and two chloride ions (\(2Cl^-\)). Therefore, a total of three ions are produced.
In simple words: When the complex \([Co(NH_3)_6]Cl_2\) dissolves, it splits into one large complex ion and two chloride ions, resulting in a total of three ions.

🎯 Exam Tip: To determine the number of ions, identify the complex ion (inside the square bracket) and the counter ions (outside the bracket). Each counter ion dissociates, and the entire complex ion acts as a single unit.

 

Question 29. Amongst the following ions which one has the highest magnetic moment value?
1. \([Cr(H_2O)_6]^{3+}\)
2. \([Fe(H_2O)_6]^{2+}\)
3. \([Zn(H_2O)_6]^{2+}\)
Answer: To determine the highest magnetic moment, we need to find the number of unpaired electrons in each complex. The magnetic moment is directly proportional to the number of unpaired electrons.

Complex	Metal ion	Electronic Configuration (d)	Unpaired electrons
\([Cr(H_2O)_6]^{3+}\)	\(Cr^{3+}\)	\(3d^3\)	3
\([Fe(H_2O)_6]^{2+}\)	\(Fe^{2+}\)	\(3d^6\)	4
\([Zn(H_2O)_6]^{2+}\)	\(Zn^{2+}\)	\(3d^{10}\)	0

Since the number of unpaired electrons increases, the magnetic moment also increases. The complex \([Fe(H_2O)_6]^{2+}\) has the highest number of unpaired electrons (4), and thus exhibits the highest magnetic moment.
In simple words: The magnetic moment of a complex depends on the number of unpaired electrons. By calculating the number of unpaired electrons for each metal ion in the given complexes, we find that \([Fe(H_2O)_6]^{2+}\) has the most unpaired electrons (4), giving it the highest magnetic moment.

🎯 Exam Tip: To compare magnetic moments, first determine the oxidation state of the central metal, then its d-electron configuration. For weak field ligands like \(H_2O\), electrons populate orbitals singly before pairing up (high spin). Count the unpaired electrons to identify the complex with the highest magnetic moment.

 

Question 30. The oxidation number of cobalt in K[Co(CO)4 is
1. +1
2. +3
3. -1
4. -3
Answer: 3. -1
In simple words: In the compound K[Co(CO)4], potassium has a +1 charge. Carbonyl (CO) is a neutral ligand, so it has no charge. To balance the overall charge of the complex ion [Co(CO)4]-, cobalt must have an oxidation state of -1.

🎯 Exam Tip: Remember that carbonyl (CO) is a neutral ligand. Assign known oxidation states (e.g., K is +1) and the ligand charge to find the unknown oxidation state of the central metal.

 

Question 31. Amongst the following, the most stable complex is
Answer: The most stable complex is \([Fe(C_2O_4)_3]^{3-}\). Oxalate (\(C_2O_4^{2-}\)) is a bidentate chelating ligand. It forms stable chelate rings with the central metal ion, which significantly enhances the complex's stability due to the chelate effect. Therefore, complexes with chelating ligands are generally more stable than those with monodentate ligands.
In simple words: The complex \([Fe(C_2O_4)_3]^{3-}\) is the most stable because oxalate is a chelating ligand, meaning it binds at two points and forms a ring. This "chelate effect" makes the complex much more stable.

🎯 Exam Tip: Look for chelating ligands in the options. Complexes formed with chelating ligands are typically more stable due to the chelate effect, which involves the formation of stable ring structures.

 

Question 32. What will be the correct order for the wavelengths of absorption in the visible region for the following?
\([Ni(NO_2)_6]^{4-}\), \([Ni(NH_3)_6]^{2+}\), \([Ni(H_2O)_6]^{2+}\)
Answer: The order of ligands in the spectrochemical series, from weak field to strong field, is:
\(H_2O < NH_3 < NO_2^-\)
A stronger ligand causes a larger crystal field splitting energy (\(\Delta_o\)), which means it absorbs light of higher energy and thus shorter wavelength. The relationship between energy and wavelength is given by \(E = \frac{hc}{\lambda}\). Therefore, a larger energy corresponds to a shorter wavelength.
Based on the spectrochemical series, the crystal field splitting energy (\(\Delta_o\)) will be in the order:
\([Ni(H_2O)_6]^{2+} < [Ni(NH_3)_6]^{2+} < [Ni(NO_2)_6]^{4-}\)
Consequently, the absorbed wavelengths will be in the opposite order:
\(\lambda([Ni(H_2O)_6]^{2+}) > \lambda([Ni(NH_3)_6]^{2+}) > \lambda([Ni(NO_2)_6]^{4-})\)
In simple words: Stronger ligands cause higher energy light absorption and shorter wavelengths. Since \(NO_2^-\) is the strongest ligand and \(H_2O\) is the weakest among the given, the complex with \(NO_2^-\) will absorb the shortest wavelength, and the complex with \(H_2O\) will absorb the longest wavelength.

🎯 Exam Tip: To solve this, recall the spectrochemical series for ligands. Remember that strong field ligands cause a large crystal field splitting, leading to absorption of high-energy light and thus shorter wavelengths. The order of absorption wavelength will be inverse to the order of ligand field strength.

Gseb Class 12 Chemistry Coordination Compounds Additional Important Questions And Answers

 

Question 1. (I) Write the postulates of Werner's coordination theory. (II) Write an example of a polydentate ligand. (III) Write the formulae of the following co-ordination compounds.
(a) hexamminecobalt (III) chloride
(b) dichloridodiammineplatinum (II)
Answer: Alfred Werner's coordination theory, proposed in 1893, provided a comprehensive explanation for the characteristics of coordination compounds. The fundamental postulates are:
(a) Metals exhibit two types of valencies: primary and secondary. The primary valency is ionizable, while the secondary valency is non-ionizable.
(b) Every metal atom or ion has a fixed number of secondary valencies, which corresponds to its coordination number.
(c) Primary valencies are satisfied by negative ions, whereas secondary valencies can be satisfied by negative ions or neutral groups (ligands).
(d) Ligands satisfying the secondary valencies are oriented in specific fixed positions in space, giving the complex a definite geometry. Primary valencies are non-directional.
For example, hexaamminecobalt (III) chloride, \([Co(NH_3)_6]Cl_3\), can be represented with three dotted lines indicating primary valencies (oxidation state of cobalt) and six thick lines indicating secondary valencies (coordination number).

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र \([Co(NH_3)_6]Cl_3\) की संरचना को दर्शाता है, जिसमें कोबाल्ट केंद्र में है और छह अमोनिया लिगेंड इससे सीधे जुड़े हुए हैं। तीन बिंदीदार रेखाएं कोबाल्ट की प्राथमिक संयोजकता (ऑक्सीकरण अवस्था) को दर्शाती हैं, जबकि छह मोटी रेखाएं इसकी द्वितीयक संयोजकता (समन्वय संख्या) को दर्शाती हैं, जो एक निश्चित ज्यामिति का निर्माण करती हैं।
(II) A polydentate ligand possesses multiple donor sites. An example is Ethylenediaminetetraacetic acid (EDTA), which is a hexadentate ligand.
(III) The formulae for the given coordination compounds are:
(a) Hexaamminecobalt (III) chloride: \([Co(NH_3)_6]Cl_3\)
(b) Dichloridodiammineplatinum (II): \([PtCl_2(NH_3)_2]\)
In simple words: Werner's theory explains how metal ions bond with other molecules (ligands) to form complexes, using primary (ionizable, for charge) and secondary (non-ionizable, for coordination number and geometry) valencies. A polydentate ligand, like EDTA, has multiple attachment points. For example, hexaamminecobalt (III) chloride is \([Co(NH_3)_6]Cl_3\) and dichloridodiammineplatinum (II) is \([PtCl_2(NH_3)_2]\).

🎯 Exam Tip: Be sure to clearly state all four postulates of Werner's theory. Provide a suitable example for a polydentate ligand and write the correct chemical formulae for the coordination compounds, paying attention to oxidation states and ligand names.

 

Question 2. When \(CuSO_4\) is mixed with \(NH_3\) a blue coloured solution is obtained?
1. Identify the compound formed?
2. Write the IUPAC name of the compound?
3. What do you understand by the terms co-ordination number and ligand?
4. Explain the structure of tetracarbonyl nickel with the help of valence bond theory?
Answer:
1. When copper sulfate (\(CuSO_4\)) is mixed with ammonia (\(NH_3\)), a deep blue solution is formed due to the creation of the cuprammonium complex.
\[CuSO_4 + 4NH_3 \rightarrow [Cu(NH_3)_4]SO_4\]
2. The IUPAC name of the compound formed is Tetraamminecopper (II) sulfate.
3. **Coordination number:** This refers to the total number of ligands directly attached to the central metal atom or ion via coordinate bonds within the coordination sphere. For instance, in \([Cr(NH_3)_6]Cl_3\), the \(Cr^{3+}\) ion has a coordination number of six.
**Ligand:** A ligand is an atom or group of atoms that binds to the central metal atom or ion by donating a lone pair of electrons. Ligands can be neutral molecules or ions, and they must possess at least one donor atom capable of forming a coordinate bond. Examples include \(NH_3\), \(H_2O\), \(CO\), \(Cl^-\), \(F^-\), \(OH^-\), etc.
4. **Structure of tetracarbonyl nickel (\([Ni(CO)_4]\)) using Valence Bond Theory:**
In \([Ni(CO)_4]\), nickel is in a zero oxidation state. Its electronic configuration is \(3d^84s^2\). Since carbon monoxide (CO) is a strong field ligand, it causes the \(4s\) electrons to pair up with the \(3d\) electrons, leaving the \(4s\) and \(4p\) orbitals vacant.
Nickel (ground state): \[Ni: [Ar] 3d^8 4s^2 \Rightarrow \boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow} \quad \boxed{\uparrow\downarrow} \quad \boxed{\phantom{\uparrow}\phantom{\downarrow}}\boxed{\phantom{\uparrow}\phantom{\downarrow}}\boxed{\phantom{\uparrow}\phantom{\downarrow}}\]
Nickel (after rearrangement due to strong field ligand CO): \[Ni: [Ar] 3d^{10} 4s^0 \Rightarrow \boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow} \quad \boxed{\phantom{\uparrow}\phantom{\downarrow}} \quad \boxed{\phantom{\uparrow}\phantom{\downarrow}}\boxed{\phantom{\uparrow}\phantom{\downarrow}}\boxed{\phantom{\uparrow}\phantom{\downarrow}}\]
The empty \(4s\) and three \(4p\) orbitals then undergo \(sp^3\) hybridization to form four equivalent \(sp^3\) hybrid orbitals. These hybrid orbitals accept electron pairs from four CO molecules, forming \([Ni(CO)_4]\).
Hybridization: \(sp^3\)
\([Ni(CO)_4]\): \[ \boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow} \quad \boxed{\text{XX}} \quad \boxed{\text{XX}}\boxed{\text{XX}}\boxed{\text{XX}} \] (where XX represents electron pairs from CO ligands)
This results in a low spin (diamagnetic) outer orbital complex with a tetrahedral geometry.
In simple words: When copper sulfate mixes with ammonia, it forms a deep blue compound called Tetraamminecopper (II) sulfate. A coordination number tells us how many ligands are attached to the central metal, and a ligand is a molecule or ion that donates electrons to the metal. For tetracarbonyl nickel, nickel's electrons pair up due to strong CO ligands, leading to \(sp^3\) hybridization and a tetrahedral, diamagnetic structure.

🎯 Exam Tip: For complex formation, remember the key reaction and naming conventions. For coordination number and ligand definitions, be precise. For VBT explanations, correctly determine the central metal's oxidation state, electron configuration, and the effect of the ligand (strong/weak field) on electron pairing and hybridization. Clearly state the geometry and magnetic property.

 

Question 3. What is the co-ordination number? Give any two examples.
Answer: The coordination number is defined as the total number of ligands directly attached to the central metal atom or ion through coordinate bonds within the coordination sphere. It dictates the complex's geometry.
**Examples:**
(i) In \([Co(NH_3)_6]Cl_3\), the \(Co^{3+}\) ion has a coordination number of 6, as there are six ammonia ligands directly bonded to it.
(ii) In \([Ag(CN)_2]^-\), the \(Ag^+\) ion has a coordination number of 2, as there are two cyanide ligands directly bonded to it.
In simple words: The coordination number is simply how many other molecules or ions (ligands) are directly attached to the central metal atom in a complex. For instance, in \([Co(NH_3)_6]Cl_3\), cobalt has 6 ligands attached, so its coordination number is 6. In \([Ag(CN)_2]^-\), silver has 2 ligands, so its coordination number is 2.

🎯 Exam Tip: Define coordination number accurately. When providing examples, clearly identify the central metal ion and the number of ligands bonded directly to it.

 

Question 4. Cisplatin is used for the treatment of cancer.
1. What is cisplatin?
2. Write the IUPAC name of the following coordination compounds
• K3[Fe(C2O4)3]
• [Pt(NH3)6]Cl4
3. Using an example write about ionization isomerism shown by co-ordination compounds.
4. Explain with the help of electronic configuration why \([Fe (CN)_6]^{3-}\) is paramagnetic.
Answer:
1. Cisplatin is the common name for cis-\([PtCl_2(NH_3)_2]\). It is a coordination compound used as an anticancer drug.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सिस-प्लैटिन (\([Pt(NH_3)_2Cl_2]\)) की वर्ग समतलीय ज्यामिति को दर्शाता है। केंद्रीय प्लैटिनम आयन से दो अमोनिया (\(NH_3\)) लिगेंड और दो क्लोराइड (\(Cl\)) लिगेंड जुड़े हुए हैं। सिस-रूप में, समान लिगेंड एक-दूसरे के निकट (90 डिग्री पर) स्थित होते हैं, जो इसे एक प्रभावी एंटी-कैंसर दवा बनाता है।
2. The IUPAC names of the given coordination compounds are:
• K3[Fe(C2O4)3]: Potassium trioxalatoferrate (III)
• [Pt(NH3)6]Cl4: Hexaammineplatinum (IV) chloride
3. **Ionization isomerism:** This type of isomerism occurs when coordination compounds with the same molecular formula yield different ions in solution. It involves the interchange of groups between the coordination sphere of the metal ion and ions existing outside the coordination sphere. Consequently, the isomers produce different ions when dissolved in water.
**Example:** Violet-colored \([Co(NH_3)_5Br]SO_4\) is an ionization isomer of red-colored \([Co(NH_3)_5SO_4]Br\). When dissolved, the first complex releases sulfate ions (\(SO_4^{2-}\)), while the second releases bromide ions (\(Br^-\)).
4. **Paramagnetism of \([Fe(CN)_6]^{3-}\):**
In \([Fe(CN)_6]^{3-}\), iron is in the +3 oxidation state. The electronic configuration of \(Fe^{3+}\) is \(3d^5\). Cyanide (\(CN^-\)) is a strong field ligand, which causes the \(3d\) electrons to pair up.
\(Fe^{3+}\) ion configuration (free state): \[ \boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow} \quad \boxed{\phantom{\uparrow}\phantom{\downarrow}}\boxed{\phantom{\uparrow}\phantom{\downarrow}}\boxed{\phantom{\uparrow}\phantom{\downarrow}}\]
\(Fe^{3+}\) ion in \([Fe(CN)_6]^{3-}\) (strong field): \[ \boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}\boxed{\uparrow} \quad \boxed{\phantom{\uparrow}\phantom{\downarrow}}\boxed{\phantom{\uparrow}\phantom{\downarrow}}\boxed{\phantom{\uparrow}\phantom{\downarrow}}\]
After pairing, only one unpaired electron remains in the \(3d\) orbitals. The hybridization involved is \(d^2sp^3\). Due to the presence of this single unpaired electron, \([Fe(CN)_6]^{3-}\) is slightly paramagnetic.
In simple words: Cisplatin is a platinum compound used in cancer treatment. Ionization isomerism is when two complexes with the same formula release different ions in solution (e.g., one releases sulfate, the other bromide). In \([Fe(CN)_6]^{3-}\), iron is Fe(III) with five 3d electrons. Since cyanide is a strong ligand, four electrons pair up, leaving one unpaired electron, which makes the complex paramagnetic.

🎯 Exam Tip: For cisplatin, remember its common name and formula. For IUPAC naming, correctly identify oxidation states and ligand names. For ionization isomerism, explain the concept with a clear example. For magnetic properties, correctly determine the metal's oxidation state, electron configuration, and apply the strong/weak field ligand effect on electron pairing.

 

Question 5. "Strength of the ligand can affect the magnetic property of the co-ordination complex"?
1. Can you agree with this statement?
2. Illustrate the statement by taking two complexes?
Answer:
1. Yes, the statement is correct. The strength of the ligand significantly affects the magnetic properties of a coordination complex by influencing electron pairing in the metal's d-orbitals.
2. **Illustration with two complexes:**
(i) **Tetracyanonickelate (II) ion, \([Ni(CN)_4]^{2-}\):**
In this complex, nickel is in the +2 oxidation state, and its electronic configuration is \(3d^8\). Cyanide (\(CN^-\)) is a strong field ligand. It causes the two unpaired electrons in the \(3d\) orbitals to pair up, making a vacant \(3d\) orbital available for hybridization.
\(Ni^{2+}\) ion (free state): \[ \boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}\boxed{\uparrow}\boxed{\uparrow} \quad \boxed{\phantom{\uparrow}\phantom{\downarrow}}\boxed{\phantom{\uparrow}\phantom{\downarrow}}\boxed{\phantom{\uparrow}\phantom{\downarrow}}\]
\(Ni^{2+}\) in \([Ni(CN)_4]^{2-}\) (strong field): \[ \boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}\boxed{\text{XX}} \quad \boxed{\text{XX}}\boxed{\text{XX}}\boxed{\text{XX}}\]
The empty \(3d\), \(4s\), and two \(4p\) orbitals undergo \(dsp^2\) hybridization to form four hybrid orbitals that accept electron pairs from four \(CN^-\) ligands. Due to the complete pairing of electrons, this complex is diamagnetic.
(ii) **Hexafluorocobaltate (III) ion, \([CoF_6]^{3-}\):**
In this complex, cobalt is in the +3 oxidation state, and its electronic configuration is \(3d^6\). Fluoride (\(F^-\)) is a weak field ligand. It is unable to force the unpaired electrons in the \(3d\) orbitals to pair up.
\(Co^{3+}\) ion (free state): \[ \boxed{\uparrow\downarrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow} \quad \boxed{\phantom{\uparrow}\phantom{\downarrow}}\boxed{\phantom{\uparrow}\phantom{\downarrow}}\boxed{\phantom{\uparrow}\phantom{\downarrow}}\]
\(Co^{3+}\) in \([CoF_6]^{3-}\) (weak field): \[ \boxed{\uparrow\downarrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow} \quad \boxed{\text{XX}} \quad \boxed{\text{XX}}\boxed{\text{XX}}\boxed{\text{XX}} \quad \boxed{\text{XX}}\boxed{\text{XX}}\]
The \(4s\), three \(4p\), and two \(4d\) orbitals hybridize to form \(sp^3d^2\) hybrid orbitals, accepting electron pairs from six \(F^-\) ligands. Because there are four unpaired electrons, this complex is paramagnetic.
These examples clearly show that strong field ligands (like \(CN^-\)) cause electron pairing, leading to diamagnetism, while weak field ligands (like \(F^-\)) do not, resulting in paramagnetism.
In simple words: Yes, ligand strength affects magnetic properties. Strong ligands (like cyanide) force electrons to pair up, making the complex diamagnetic (no unpaired electrons). Weak ligands (like fluoride) don't force pairing, leaving unpaired electrons and making the complex paramagnetic.

🎯 Exam Tip: To illustrate this concept, choose one complex with a strong field ligand and another with a weak field ligand. Clearly show the d-electron configuration before and after ligand interaction, indicating electron pairing or lack thereof, to explain the resulting magnetic properties (diamagnetic or paramagnetic).

 

Question 6. Coordination compounds are used for the extraction of metals?
1. Write one application each of the coordination compounds in analytical chemistry and in the extraction of metals.
2. Write the formulae of,
• tetracyanonickelate (II)
• hexaamminecobalt (III) chloride
3. \([Co(NH_3)_6]^{3+}\) is diamagnetic, whereas \([CoF_6]^{3-}\) is paramagnetic. Why?
Answer:
1. **Application in analytical chemistry:** In qualitative analysis, \(Cu^{2+}\) ions are detected by the formation of a deep blue colored complex \([Cu(NH_3)_4]^{2+}\) upon adding ammonia solution.
\[Cu^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+} \text{ (deep blue)}\]
**Application in extraction of metals:** Metals like silver and gold are extracted by converting them into soluble cyanide complexes, from which the pure metal can then be recovered.
\[4Au + 8CN^- + 2H_2O + O_2 \rightarrow 4[Au(CN)_2]^- + 4OH^- \text{ (soluble)}\]
2. **Formulae:**
• Tetracyanonickelate (II): \([Ni(CN)_4]^{2-}\)
• Hexaamminecobalt (III) chloride: \([Co(NH_3)_6]Cl_3\)
3. **Explanation of magnetic properties:**
In both \([Co(NH_3)_6]^{3+}\) and \([CoF_6]^{3-}\), cobalt is in the +3 oxidation state, meaning its electronic configuration is \(3d^6\).
**For \([Co(NH_3)_6]^{3+}\):** Ammonia (\(NH_3\)) is a strong field ligand. When \(NH_3\) approaches the \(Co^{3+}\) ion, it causes the \(3d\) electrons to pair up. This results in all six \(3d\) electrons being paired, leaving no unpaired electrons. Therefore, \([Co(NH_3)_6]^{3+}\) is diamagnetic.
\(Co^{3+}\) in \([Co(NH_3)_6]^{3+}\) (strong field): \[ \boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow}\boxed{\uparrow\downarrow} \quad \boxed{\text{XX}} \quad \boxed{\text{XX}}\boxed{\text{XX}}\boxed{\text{XX}}\]
**For \([CoF_6]^{3-}\):** Fluoride (\(F^-\)) is a weak field ligand. It is unable to force the \(3d\) electrons to pair up. This leaves four unpaired electrons in the \(3d\) orbitals. Therefore, \([CoF_6]^{3-}\) is paramagnetic.
\(Co^{3+}\) in \([CoF_6]^{3-}\) (weak field): \[ \boxed{\uparrow\downarrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow} \quad \boxed{\text{XX}} \quad \boxed{\text{XX}}\boxed{\text{XX}}\boxed{\text{XX}} \quad \boxed{\text{XX}}\boxed{\text{XX}}\]
In simple words: Coordination compounds help in detecting metals (like copper with ammonia) and extracting them (like gold using cyanide). For \([Co(NH_3)_6]^{3+}\), ammonia is a strong ligand, forcing electrons to pair, making it diamagnetic. For \([CoF_6]^{3-}\), fluoride is a weak ligand, so electrons remain unpaired, making it paramagnetic.

🎯 Exam Tip: For applications, provide specific examples with chemical reactions where relevant. For formulae, ensure correct oxidation states and ligand representations. For magnetic properties, clearly distinguish between strong and weak field ligands and their impact on electron pairing and d-orbital occupation.

 

Question 6 (b). What are the important uses of coordination compounds?
Answer: The ability of metal ions to form complexes with various molecular species, leading to different physico-chemical properties, is utilized in numerous ways. Some key roles include:
(I) **In analytical chemistry:**
(a) The multidentate ligand EDTA (Ethylenediamine Tetraacetic Acid) forms highly stable complexes with metal ions like \(Ca^{2+}\) and \(Mg^{2+}\). This property is leveraged in a simple titration method to estimate water hardness.
(b) A confirmatory test for detecting copper (II) involves the formation of a deep-blue colored complex, \([Cu(NH_3)_4]^{2+}\), which appears when ammonia solution is added to a copper (II) salt solution.
\[Cu^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+} \text{ (deep blue)}\]
(c) The separation of Group I precipitates (in qualitative inorganic analysis), such as \(AgCl\), \(Hg_2Cl_2\), and \(PbCl_2\), involves adding aqueous ammonia solution to the precipitate. Only silver chloride dissolves, forming the complex ion \([Ag(NH_3)_2]^+\), while \(Hg_2Cl_2\) and \(PbCl_2\) do not form such complex ions and thus remain undissolved.
\[AgCl + 2NH_3 \rightarrow [Ag(NH_3)_2]^+ + Cl^-\]
(d) A confirmatory test for nickel involves adding a dimethyl glyoxime solution, which forms a scarlet red chelate precipitate.
(II) **In extraction of metals:**
(a) Metals like silver and gold are extracted by converting them into soluble cyanide complexes, from which the pure metal can then be recovered.
\[4Au + 8CN^- + 2H_2O + O_2 \rightarrow 4[Au(CN)_2]^- + 4OH^- \text{ (soluble)}\]
The solution containing the cyanide complex is then treated with zinc, which precipitates gold:
\[2[Au(CN)_2]^- + Zn \rightarrow [Zn(CN)_4]^{2-} + 2Au\]
(b) Coordination compounds of silver and gold are used in electroplating baths for the controlled delivery of \(Ag^+\) and \(Au^+\) ions during electrorefining processes.
(c) The formation of volatile \([Ni(CO)_4]\) is utilized in the purification of nickel through the Mond process.
(III) **In medicines:**
Many medications are complexes. For instance, Vitamin B12, a cobalt complex, is vital for preventing anemia. The coordination compound cis-\([PtCl_2(NH_3)_2]\), commonly known as cisplatin, is used in chemotherapy to treat cancer. Complexing agents are also employed to remove metal poisoning. For example, D-penicillamine treats copper poisoning, while EDTA is used for lead poisoning. In cases of exposure to radioactive materials like Plutonium (Pu), chelating agents are used to remove these metals from the body (chelate therapy).
(IV) **In photography:**
Unexposed silver bromide (\(AgBr\)) in photographic film is dissolved using a sodium thiosulphate solution. This dissolution forms a soluble silver thiosulphato complex, a process known as 'fixing'.
\[AgBr + 2Na_2S_2O_3 \rightarrow Na_3[Ag(S_2O_3)_2] + NaBr\]
(V) **In biological systems:**
Chlorophyll, a coordination compound of magnesium (II), plays a catalytic role in photosynthesis, producing glucose from \(CO_2\) and \(H_2O\). Hemoglobin, an iron (II) coordination compound, is responsible for oxygen transport. Metalloenzymes are also coordination compounds, such as carboxy peptidase A and carbonic anhydrase.
In simple words: Coordination compounds are vital in many areas: they are used in chemical tests to identify substances, extract metals from their ores, and treat diseases like cancer or metal poisoning. They are also essential in photography for processing film and play crucial roles in biological processes like photosynthesis and oxygen transport in living organisms.

🎯 Exam Tip: When asked for uses, provide diverse examples from different fields. Focus on one or two clear applications for each category (e.g., EDTA for hardness, cisplatin for cancer, chlorophyll for photosynthesis) to demonstrate broad understanding.

 

Question 8.
Match the following

A	B
i. Ionisation isomerism	a. \( [\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+ \)
ii. Geometrical isomerism	b. \( [\text{Co}(\text{NH}_3)_5\text{NO}_2]^{2+} \)
iii. Optical isomerism	c. \( [\text{Co}(\text{NH}_3)_6][\text{Cr}(\text{CN})_6] \)
iv. Linkage isomerism	d. \( [\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4 \)
v. Co-ordinate isomerism	e. \( [\text{Co}(\text{en})_3]^{3-} \)

Answer:
The correct pairings for the given types of isomerism are as follows:
(i) Ionisation isomerism is matched with (d) \( [\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4 \).
(ii) Geometrical isomerism corresponds to (a) \( [\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+ \).
(iii) Optical isomerism is correctly identified with (e) \( [\text{Co}(\text{en})_3]^{3-} \).
(iv) Linkage isomerism is paired with (b) \( [\text{Co}(\text{NH}_3)_5\text{NO}_2]^{2+} \).
(v) Co-ordinate isomerism is linked to (c) \( [\text{Co}(\text{NH}_3)_6][\text{Cr}(\text{CN})_6] \).
In simple words: Coordination compounds exhibit various types of isomerism, where molecules have the same chemical formula but different arrangements of atoms. This matching exercise tests the ability to correctly associate each isomerism type (ionisation, geometrical, optical, linkage, and co-ordinate) with a representative chemical complex.

🎯 Exam Tip: For matching questions on isomerism, it is crucial to understand the fundamental definition of each isomer type and be able to recognize its characteristics in complex chemical formulas to ensure accurate pairings and maximize scores.