Maharashtra Board Class 12 Chemistry Chapter 9 Coordination Compounds Solutions

Coordination Compounds Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 9 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 9 Exercise Solutions

1. Choose The Most Correct Option.

 

Question i. The oxidation state of cobalt ion in the complex [Co(NH3)5Br]SO4 is
a. + 2
b. + 3
c. + 1
d. + 4
Answer: (b) + 3
In simple words: The oxidation state of cobalt is determined by considering the charges of the ligands and the overall charge of the complex. In [Co(NH3)5Br]SO4, the sulfate ion has a -2 charge, NH3 is neutral, and Br is -1, leading to Co having a +3 oxidation state.

🎯 Exam Tip: Remember to consider the charge of the counter-ion and the common charges of ligands (neutral or anionic) to correctly calculate the oxidation state of the central metal atom.

 

Question ii. IUPAC name of the complex [Pt(en)2(SCN)2]2+ is
a. bis (ethylenediamine dithiocyanatoplatinum (IV) ion
b. bis (ethylenediamine) dithiocyantoplatinate (IV) ion
c. dicyanatobis (ethylenediamine) platinate IV ion
d. bis (ethylenediammine)dithiocynato platinate (IV) ion
Answer: (a) bis(ethylenediamine dithiocyanatoplatinum (IV) ion
In simple words: The IUPAC name is derived by naming the ligands alphabetically, followed by the central metal, and then its oxidation state. 'en' (ethylenediamine) is a bidentate ligand, and 'SCN' (thiocyanato) indicates sulfur as the donor atom.

🎯 Exam Tip: Pay close attention to the donor atom in ambidentate ligands (like SCN vs. NCS) and use appropriate prefixes (bis, tris) for polydentate ligands when naming coordination compounds.

 

Question iii. Formula for the compound sodium hexacynoferrate (III) is
a. [NaFe(CN)6]
b. Na2[Fe(CN)6]
c. Na [Fe(CN)6]
d. Na3[Fe(CN)6]
Answer: (d) Na3[Fe(CN)6]
In simple words: "Hexacyanoferrate(III)" indicates a complex anion with iron in the +3 oxidation state and six cyanide ligands. To balance the charge, three sodium cations are needed.

🎯 Exam Tip: When writing formulas from IUPAC names, ensure the charges are balanced between the cation and the complex anion, using subscripts for the number of counter-ions.

 

Question iv. Which of the following complexes exist as cis and trans isomers?
1. [Cr(NH2)2Cl4]®
2. [Co(NH3)5Br]2®
3. [PtCl2Br2]2® (square planar)
4. [FeCl2(NCS)2]2 (tetrahedral)
a. 1 and 3
b. 2 and 3
c. 1 and 3
d. 4 only
Answer: (a) 1 and 3
In simple words: Cis-trans isomerism occurs in square planar and octahedral complexes where there are at least two identical ligands that can occupy adjacent or opposite positions. Complex 1 ([Cr(NH2)2Cl4]®) is octahedral (MA2B4 type) and Complex 3 ([PtCl2Br2]2®) is square planar (MA2B2 type), both capable of exhibiting cis-trans isomerism.

🎯 Exam Tip: Tetrahedral complexes generally do not exhibit geometric isomerism due to their symmetrical structure. Square planar and octahedral complexes with appropriate ligand arrangements are key candidates for cis-trans isomerism.

 

Question v. Which of the following complexes are chiral?
1. [Co(en)2Cl2]®
2. [Pt(en)Cl2]
3. [Cr(C2O4)3]3
4. [Co(NH3)4CI2]®
a. 1 and 3
b. 2 and 3
c. 1 and 4
d. 2 and 4
Answer: (a) 1 and 3
In simple words: Chiral complexes are non-superimposable mirror images (enantiomers). Octahedral complexes with bidentate ligands often exhibit chirality, especially those of the MA2B2 type (cis isomer, like [Co(en)2Cl2]®) and M(AA)3 type (like [Cr(C2O4)3]3).

🎯 Exam Tip: Look for complexes containing bidentate or polydentate ligands that can create a non-superimposable mirror image. Cis isomers of MA2(AA)2 type octahedral complexes and M(AA)3 type octahedral complexes are common examples of chiral coordination compounds.

 

Question vi. On the basis of CFT predict the number of unpaired electrons in [CrF6]3®.
a. 1
b. 2
c. 3
d. 4
Answer: (c) 3
In simple words: In [CrF6]3®, chromium is in the +3 oxidation state (d³ configuration). Fluoride is a weak field ligand, causing high spin, so the three d electrons occupy the three t2g orbitals individually before pairing, resulting in 3 unpaired electrons.

🎯 Exam Tip: To predict the number of unpaired electrons using CFT, first determine the oxidation state and d-electron configuration of the metal. Then, classify the ligands as strong or weak field to decide if the complex is high spin or low spin, which dictates the electron filling order in t2g and eg orbitals.

 

Question vii. When an excess of AgNO3 is added to the complex one mole of AgCl is precipitated. The formula of the complex is
a. [CoCl2(NH3)4]CI
b. [CoCI(NH3)4] Cl2
c. [COCI3(NH3)3]
d. [Co(NH3)4]Cl3
Answer: (a) [COCI3(NH3)4]CI
In simple words: If one mole of AgCl precipitates, it means there is one chloride ion outside the coordination sphere. Option (a) is the only formula where exactly one chloride ion is present as a counter-ion.

🎯 Exam Tip: The number of counter-ions that precipitate with AgNO3 directly corresponds to the number of ionizable halide ions outside the coordination sphere.

 

Question viii. The sum of coordination number and oxidation number of M in [M(en)2C2O4]CI is
a. 6
b. 7
c. 9
d. 8
Answer: (c) 9
In simple words: 'en' (ethylenediamine) is a bidentate ligand, and 'C2O4' (oxalate) is also a bidentate ligand. With two 'en' and one 'C2O4', the coordination number is 2*2 + 1*2 = 6. For the oxidation state, the chloride counterion indicates a +1 complex charge. Since 'en' is neutral and oxalate is -2, if M is +3, then (+3) + 2(0) + 1(-2) = +1. So, M is +3. The sum is 6 + 3 = 9.

🎯 Exam Tip: Always identify the denticity of each ligand to correctly calculate the coordination number. Then, determine the overall charge of the complex ion to deduce the oxidation state of the central metal.

2. Answer The Following In One Or Two Sentences.

 

Question i. Write the formula for tetraammineplatinum (II) chloride.
Answer: Formula of tetraamineplatinum(II) chloride : [Pt(NH3)4]Cl2
In simple words: "Tetraammineplatinum(II)" means platinum in the +2 oxidation state with four ammonia ligands. "Chloride" is the counter-ion, requiring two chloride ions to balance the +2 charge of the complex.

🎯 Exam Tip: Remember that in IUPAC nomenclature, the Roman numeral in parentheses indicates the oxidation state of the central metal, which helps in balancing charges with counter-ions.

Table 9.1 : IUPAC Names Of Anionic And Neutral Ligands

Anionic ligand	IUPAC name	Anionic ligand	IUPAC name
Br-, Bromide	Bromo	CO32-, Carbonate	Carbonato
Cl-, Chloride	Chloro	OH-, Hydroxide	Hydroxo
F-, Fluoride	Fluoro	C2O42-, Oxalate	Oxalato
I- Iodide	Iodo	NO2-, Nitrite	Nitro (For N-bonded ligand)
CN-, Cyanide	Cyano	ONO-, Nitrite	Nitrito (For O-bonded ligand)
SO42-, Sulphate	Sulphato	SCN-, Thiocyanate	Thiocyanato (For ligand donor atom S)
NO3-, Nitro	Nitrato	NCS-, Thiocyanate	Isothiocyanato (For ligand donor atom N)
Neutral ligand	IUPAC name	Neutral ligand	IUPAC names
NH3, Ammonia	Ammine (Note the spelling)	H₂O, water	Aqua
CO, Carbon monoxide	Cabonyl	en, Ethylene diamine	Ethylenediamine

Table 9.2: IUPAC Names Of Anionic Complexes

Metal	Name
A1	Aluminate
Cr	Chromate
Cu	Cuprate
Co	Cobaltate
Au(Gold)	Aurate
Fe	Ferrate
Pb	Plumbate
Mn	Manganate
Mo	Molybdate
Ni	Nickelate
Zn	Zincate
Ag	Argentate
Sn	Stannate

Table 9.3: IUPAC Names Of Some Complexes

Complex	IUPAC name
(i) Anionic complexes :
(a) [Ni(CN)]2-	Tetracyanonickelate(II) ion
(b) [Co(C204)3]3-	Trioxalatocobaltate(III) ion
(c) [Fe(CN)6]4-	Hexacyanoferrate(II) ion
(ii) Compounds containing complex anions and metal cations :
(a) Na3[Co(N02)6]	Sodium hexanitrocobaltate(III)
(b) K3[A1(C204)3]	Potassium trioxalatoaluminate(III)
(c) Na3[AlF6]	Sodium hexafluoroaluminate(III)
(iii) Cationic complexes :
(a) [Cu(NH3)4]2+	Tetraamminecopper(II) ion
(b) [Fe(H20)5(NCS)]2+	Pentaaquai sothiocyanatoiron(III)
(c) [Pt(en)2(SCN)2]2+	ionBis(ethylenediamine)dithiocyanatoplatinum(IV)
(iv) Compounds containing complex cation and anion :
(a) [PtBr2(NH3)4]Br2	Tetraamminedibromoplatinum(IV) bromide,
(b) [Co(NH3)5C03]CI	Pentaamminecarbonatocobalt(III) chloride,
(c) [Co(H20)(NH3)5]13	Pentaammineaquacobalt(III) iodide
(v) Neutral complexes :
(a) Co(NO2)3(NH3)3	Triamminetrinitrocobalt(III)
(b) Fe(CO)5	Pentacarbonyliron(0)
(c) Rh(NH3) 3(SCN) 3	Triamminetrithiocyanatorhodium(III)

 

Question ii. Predict whether the [Cr(en)2(H2O)2]³+ complex is chiral. Write structure of its enantiomer.
Answer: (i) Complex is chiral.
(ii) The following are its enantiomers
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक ऑक्टाहेड्रल क्रोमियम (Cr) कॉम्प्लेक्स \( \text{[Cr(en)2(H2O)2]}^{3+} \) के दो नॉन-सुपरइम्पोज़ेबल दर्पण चित्र (d-फॉर्म और l-फॉर्म) को दर्शाता है। केंद्रीय क्रोमियम धातु दो 'en' (एथिलीनडाइमाइन) बाइडेनटेट लिगेंड और दो H2O लिगेंड से घिरी हुई है, जो इस कॉम्प्लेक्स की किरलता (chirality) को प्रदर्शित करती है।
In simple words: The complex [Cr(en)2(H2O)2]³+ is chiral because its cis isomer has a non-superimposable mirror image. This means it can exist as two enantiomers, d-form and l-form.

🎯 Exam Tip: To identify chiral complexes, look for octahedral complexes with bidentate ligands that can form cis isomers, or those with M(AA)3 general formula. These often lack an plane of symmetry or center of inversion, leading to chirality.

 

Question iv. Name the Lewis acids and bases in the complex [PtCl2(NH3)2].
Answer: Lewis acid : Pt2+
Lewis bases : Cl¯ and NF3
In simple words: In the complex, the central metal ion (Pt2+) acts as a Lewis acid by accepting electron pairs, while the ligands (Cl- and NH3) act as Lewis bases by donating electron pairs to the metal.

🎯 Exam Tip: The central metal atom or ion in a coordination complex always functions as a Lewis acid, while the ligands act as Lewis bases due to their ability to donate electron pairs.

 

Question v. What is the shape of a complex in which the coordination number of central metal ion is 4?
Answer: A complex with the coordination number of central metal ion equal to 4 may be tetrahedral or square planar.
In simple words: When a central metal ion is bonded to four ligands, the complex can adopt either a tetrahedral geometry (like sp³ hybridization) or a square planar geometry (like dsp² hybridization).

🎯 Exam Tip: The geometry for coordination number 4 depends on the metal ion, its d-electron configuration, and the nature of the ligands (strong vs. weak field), which influence the hybridization (sp³ for tetrahedral, dsp² for square planar).

 

Question vi. Is the complex [C0F6] cationic or anionic if the oxidation state of cobalt ion is +3?
Answer: In the complex, Co carries + 3 charge while 6F¯ carry – 6 charge. Hence the net charge on the complex is – 3. Therefore it is an anionic complex.
In simple words: With cobalt as +3 and six fluoride ligands each as -1, the total charge of the complex is +3 + 6(-1) = -3, making it an anionic complex.

🎯 Exam Tip: To determine if a complex is cationic or anionic, sum the oxidation state of the central metal ion and the charges of all ligands. A negative total charge indicates an anionic complex.

 

Question vii. Consider the complexes [Cu(NH3)4][PtCl4] and [Pt(NH3)4] [CuCl4]. What type of isomerism these two complexes exhibit?
Answer: Since in these two given complexes, there is an exchange of ligands between cationic and anionic constituents, they exhibit coordination isomerism.
In simple words: Coordination isomerism occurs when there's an exchange of ligands between the cationic and anionic coordination spheres within a complex compound. Here, ammonia and chloride ligands switch places between copper and platinum centers.

🎯 Exam Tip: Look for compounds where both the cation and anion are complex ions, and the ligands are distributed differently between them. This is the hallmark of coordination isomerism.

 

Question viii. Mention two applications of coordination compounds.
Answer: (1) In biology: Several biologically important natural compounds are metal complexes which play an important role in number of processes occurring in plants and animals.
For example, chlorophyll in plants is a complex of Mg2+ ions, haemoglobin in blood is a complex of iron, vitamin B12 is a complex of cobalt.
(2) In medicine : The complexes are used on a large scale in medicine. Many medicines in the complex form are more stable, more effective and can be assimilated easily.
For example, platinum complex [Pt(NH3)2Cl2] known as cisplatin is effectively used in cancer treatment. EDTA is used to treat poisoning by heavy metals like lead.
(3) To estimate hardness of water :
- The hardness of water is due to the presence Mg2+ and Ca2+ ion in water.
- The strong field ligand EDTA forms stable complexes with Mg2+ and Ca2+. Hence these ions can be removed by adding EDTA to hard water.
Similarly these ions can be selectively estimated due to the difference in their stability constants.
(4) Electroplating : This involves deposition of a metal on the other metal. For smooth plating, it is necessary to supply continuously the metal ions in small amounts.
For this purpose, a solution of a coordination compound is used which dissociates to a very less extent. For example, for uniform and thin plating of silver and gold, the complexes K[Ag(CN)2] and K[Au(CN)2] are used.
In simple words: Coordination compounds are vital in biology (e.g., chlorophyll, hemoglobin) and medicine (e.g., cisplatin for cancer, EDTA for heavy metal poisoning). They also find applications in water treatment (removing hardness with EDTA) and electroplating for smooth metal deposition.

🎯 Exam Tip: Focus on understanding diverse applications across different fields. For biology, remember chlorophyll (Mg) and hemoglobin (Fe). For medicine, cisplatin and EDTA are key examples. Electroplating and water hardness are good industrial applications.

3. Answer In Brief.

 

Question i. What are bidentate ligands? Give one example.
Answer: Bidentate ligand : This ligand has two donor atoms in the molecule or ion. For example, ethylenediamine, H2N - (CH2)2 - NH2.
In simple words: Bidentate ligands are molecules or ions that can form two coordinate bonds with a central metal atom because they possess two donor atoms. Ethylenediamine is a common example, with its two nitrogen atoms acting as donors.

🎯 Exam Tip: When identifying bidentate ligands, look for two atoms within the ligand that have lone pairs of electrons and are positioned such that they can simultaneously coordinate to the metal center, forming a stable chelate ring.

 

Question ii. What is the coordination number and oxidation state of metal ion in the complex [Pt(NH3)CI5]2?
Answer: Coordination number = 6
Oxidation state of Pt = +4.
In simple words: In [Pt(NH3)Cl5]2-, there is one ammonia and five chloride ligands, totaling six ligands coordinating to platinum, hence the coordination number is 6. With a net charge of -2, and five chlorides (5 * -1 = -5) and one neutral ammonia, platinum must be in the +4 oxidation state to balance the charge (+4 - 5 = -1; wait, the overall charge is -2, so +4 - 5 = -1, if the question meant [Pt(NH3)Cl5]-, then Pt is +4. If the overall complex is [Pt(NH3)Cl5]2-, then Pt + 0 + 5(-1) = -2 => Pt - 5 = -2 => Pt = +3. Based on the provided answer +4, it implies the charge written is a typo and should be [Pt(NH3)Cl5]- or the calculation is based on an incorrect premise for the given formula. Let's assume the given oxidation state +4 is correct for a similar complex, or that the `2` in `[Pt(NH3)CI5]2?` is a typo and should be `-` for oxidation state `+4` to be valid, or is part of a counterion not shown. Given the verbatim rule, I will present the answer as is, acknowledging the potential discrepancy if the full formula with counterions isn't given. However, typical notation `[Complex]2` doesn't mean charge. Let's re-evaluate based on `[Pt(NH3)Cl5]` assuming the `2` is part of a typo. If `[Pt(NH3)Cl5]` is the complex, and if it's overall neutral, Pt would be +5-1 = +4. If it's `[Pt(NH3)Cl5]2-`, then Pt would be +3. The answer states +4. I will go with the provided answer for consistency. The coordination number is 1 (NH3) + 5 (Cl) = 6.
In simple words: The central platinum atom is bonded to one ammonia ligand and five chloride ligands, making the coordination number 6. Given the oxidation state of Pt as +4, and the ammonia being neutral and chlorides being -1 each, the overall charge of the complex is +4 + 0 + 5(-1) = -1. This implies the complex is actually [Pt(NH3)Cl5]-.

🎯 Exam Tip: Carefully count the donor atoms from each ligand to determine the coordination number. To find the oxidation state, remember that neutral ligands have zero charge, and common anionic ligands have known charges (e.g., Cl- is -1). The sum of metal oxidation state and ligand charges must equal the overall complex charge.

 

Question iii. What is the difference between a double salt and a complex? Give an example.
Answer:

Double salt	Coordination compound (complex)
(1) Double salts exist only in the solid state and dissociate into their constituent ions in the aqueous solutions.	(1) Coordination compounds exist in the solid-state as well as in the aqueous or non-aqueous solutions.
(2) Double salts lose their identity in the solution.	(2) They do not lose their identity completely.
(3) The properties of double salts are same as those of their constituents.	(3) The properties of coordination compounds are different from their constituents.
(4) Metal ions in the double salts show their normal valence.	(4) Metal ions in the coordination compounds show two valences namely primary valence and secondary valence satisfied by anions or neutral molecules called ligands.
(5) For example in K2SO4. K2SO4. A12(SO4)3. 24H2O. The ions K+, Al3 + and SO4 show their properties.	(5) In K4[Fe(CN)6], ions K+ and [Fe(CN)6]4′~ ions show their properties.

In simple words: Double salts are stable only in the solid state and completely dissociate into their constituent ions in solution, losing their identity and showing properties of individual ions. Coordination compounds, however, retain their identity in solution as complex ions and exhibit properties distinct from their constituent parts.

🎯 Exam Tip: The key differentiator is stability in solution and retention of identity. Double salts fully ionize, while coordination compounds form stable complex ions that do not fully dissociate into individual metal and ligand ions.

 

Question iv. Classify the following complexes as homoleptic and heteroleptic [Cu(NH3)4]SO4, [Cu(en)2(H2O)CI]3®, [Fe(H2O)5(NCS)]2®, tetraammine zinc (II) nitrate.
Answer: Homoleptic complex :
(a) [Cu(NH3)4]SO4
(d) Tetraaminezinc (II) nitrate : [Zn(NH3)4](NO3)2
Heteroleptic Complex :
(b) [Cu(en)2(H2O)CI]2+
(c) [Fe(H2O)5(NCS)]2+
In simple words: Homoleptic complexes have only one type of ligand coordinated to the central metal atom (e.g., all NH3 ligands). Heteroleptic complexes have two or more different types of ligands coordinated to the central metal atom (e.g., 'en', H2O, and Cl ligands).

🎯 Exam Tip: To distinguish between homoleptic and heteroleptic complexes, simply count the number of different types of ligands attached directly to the central metal atom. One type means homoleptic; multiple types mean heteroleptic.

 

Question v. Write formulae of the following complexes
a. Potassium ammine-tri chloroplatinate (II)
b. Dicyanoaurate (I) ion
Answer: (a) Potassium amminetrichloroplatinate(II) K[Pt(NH3)CI3]
(b) Dicyanoaurate (I) ion [AU(CN)2]¯
In simple words: For "Potassium amminetrichloroplatinate(II)", platinum is +2, with one NH3 (neutral) and three Cl- (-1 each), making the complex ion charge +2 + 0 + 3(-1) = -1. Thus, one potassium ion is needed: K[Pt(NH3)Cl3]. For "Dicyanoaurate(I) ion", gold is +1, with two CN- (-1 each), making the complex ion charge +1 + 2(-1) = -1: [Au(CN)2]-.

🎯 Exam Tip: When converting names to formulas, always start by determining the oxidation state of the central metal and the charges of all ligands. Use this information to find the overall charge of the complex ion and balance it with appropriate counter-ions if necessary.

 

Question vi. What are ionization isomers? Give an example.
Answer: Ionisation isomers : The coordination compounds having same molecular composition but differ in the compositions of coordination (or inner) sphere and outer sphere and produce different ions on ionisation in the solution are called ionisation isomers. For example, Pentaamminesulphatocobalt (III) bromide [Co(NH3)5SO4] Br, Pentaamminebromocobalt(III) sulphate [Co(NH3)5Br] SO4.
In simple words: Ionization isomers are complexes that have the same molecular formula but differ in which ion is inside the coordination sphere and which is outside as a counter-ion. They produce different ions when dissolved in solution.

🎯 Exam Tip: The key characteristic of ionization isomers is that a ligand acts as a counter-ion in one isomer and as a ligand in the other, and vice versa. This leads to the formation of different ions in solution.

 

Question vii. What are the high-spin and low-spin complexes?
Answer: (1) High spin complex (HS) :
- The complex which has greater iwmher of unpaired electrons and hence a higher value of resultant spin and magnetic moment is called high spin (or spin free) or IIS complex.
- It is formed with weak field ligands and the complexes have lower values for crystal field splitting energy (CFSE). Δο
- The paramagnetism of HS complex is larger.
(2) Low spin complex (LS) :
- The complex which has the Icasi number of unpaired electrons or all electrons paired and hence the lowest (or no) resultant spin or magnetic moment is called low spin (or spin paired) or LS complex.
- It is formed with strong tickl ligands and the complexes have higher values of crystal field splitting energy (Δο).
- Low spin complex is diamagnetic or has low paramagnetism.
In simple words: High-spin complexes form with weak field ligands, resulting in a smaller crystal field splitting energy and more unpaired electrons (paramagnetic). Low-spin complexes form with strong field ligands, leading to a larger crystal field splitting energy, causing electron pairing and fewer or no unpaired electrons (diamagnetic or less paramagnetic).

🎯 Exam Tip: The distinction between high-spin and low-spin complexes hinges on the strength of the ligand field (weak vs. strong) and its effect on electron pairing versus occupation of higher energy orbitals. This determines the magnetic properties and color of the complex.

Table 9.5: d-Orbitai Diagrams Fir High Spin And Low Spin Complexes

ℹ️ चित्र व्याख्या (Diagram Explanation): यह तालिका d-इलेक्ट्रॉन विन्यास (d4, d5, d6, d7) के लिए उच्च स्पिन (high spin) और निम्न स्पिन (low spin) कॉम्प्लेक्सों के d-कक्षक विभाजन आरेखों को दर्शाती है। यह दिखाता है कि लिगेंड क्षेत्र की शक्ति के आधार पर इलेक्ट्रॉन eg और t2g कक्षकों को कैसे अलग तरह से भरते हैं, जिससे अयुग्मित इलेक्ट्रॉनों की संख्या प्रभावित होती है।

(Only the electronic configurations c4 to d1 render the high spin and low spin complexes)

 

Question viii. [CoCl4]2® is a tetrahedral complex. Draw its box orbital diagram. State which orbitals participate in hybridization.
Answer: 27Co [Ar] 3d74s²
Oxidation state of Co = +2 Co2+ [Ar] 3d7 4s°
Since Cl¯ is a weak ligand, there is no pairing of electrons. Since C.N. is 4, there is sp³ hybridisation.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक टेट्राहेड्रल कॉम्प्लेक्स में \( \text{Co}^{2+} \) के लिए इलेक्ट्रॉन विन्यास को दर्शाता है। इसमें 3d कक्षकों में 7 इलेक्ट्रॉन (3 युग्मित, \( \text{t}_{2g} \) में 1 अयुग्मित और \( \text{e}_g \) में 2 अयुग्मित) दिखाए गए हैं, क्योंकि एक कमजोर क्षेत्र लिगेंड (Cl-) संकरण से पहले 3d में युग्मन नहीं करता है। इसके बाद, खाली 4s और 4p कक्षक हैं। \( \text{sp}^3 \) संकरण खंड में एक 4s और तीन 4p कक्षकों का मिश्रण दिखाया गया है, जो चार संकर कक्षक बनाते हैं।
In simple words: In [CoCl4]2-, cobalt is in the +2 oxidation state (d7). As chloride is a weak field ligand, no electron pairing occurs in the 3d orbitals. With a coordination number of 4, the hybridization is sp³, involving one 4s and three 4p orbitals, leading to a tetrahedral geometry.

🎯 Exam Tip: For tetrahedral complexes with weak field ligands, predict sp³ hybridization. Remember that weak field ligands do not cause pairing of electrons in the d-orbitals, which influences the number of unpaired electrons and the magnetic properties.

 

Question ix. What are strong field and weak field ligands? Give one example of each.
Answer: The ligands are then classified as (a) strong field and (b) weak field ligands. Strong field ligands are those in which donor atoms are C,N or P. Thus CN¯, NC¯, CO, HN3, EDTA, en (ethylenediammine) are considered to be strong ligands. They cause larger splitting of d orbitals and pairing of electrons is favoured. These ligands tend to form low spin complexes. Weak field ligands are those in which donor atoms are halogens, oxygen or sulphur. For example, F¯, CI¯, Br¯, I¯, SCN¯, C2O42-. In case of these ligands the A0 parameter is smaller compared to the energy required for the pairing of electrons, which is called as electron pairing energy. The ligands then can be arranged in order of their increasing field strength as I¯ < Br¯ < CI¯ < S2- < F¯ < OH¯ < C2O42- < H2O < NCS¯ < EDTA < NH3 < en < CN- < CO.
In simple words: Strong field ligands (like CN-, CO, NH3) cause a large splitting of d-orbitals, promoting electron pairing and forming low-spin complexes. Weak field ligands (like F-, Cl-, H2O) cause a small splitting, resulting in electrons occupying higher energy orbitals before pairing, forming high-spin complexes.

🎯 Exam Tip: Memorize the spectrochemical series to quickly identify strong and weak field ligands. Strong field ligands are typically those with C, N, P donor atoms, while weak field ligands often have O, S, or halogen donor atoms.

 

Question x. With the help of a crystal field energy-level diagram explain why the complex [Cr(en)3]3® is coloured?
Answer: s. Cr+3 = [Ar]3d34s°
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक ऑक्टाहेड्रल कॉम्प्लेक्स में \( \text{Cr}^{3+} \) के लिए d-कक्षकों के क्रिस्टल क्षेत्र विभाजन को दर्शाता है। \( \text{Cr}^{3+} \) (3d³) से तीन इलेक्ट्रॉन प्रारंभ में कम ऊर्जा वाले \( \text{t}_{2g} \) कक्षकों में समानांतर स्पिन के साथ होते हैं। आरेख आगे उच्च ऊर्जा स्तर पर \( \text{e}_g \) कक्षकों को दिखाता है, जो d-कक्षकों के दो सेटों में विभाजन को दर्शाता है।
Since (en) is a strong field ligand there is pairing of electrons. The electrons occupy the t2g orbitals of lower energy. It has one unpaired electron. Due to d-d transition, it is coloured.
In simple words: The [Cr(en)3]3® complex is colored because chromium(III) has a d³ configuration. The strong field ligand 'en' causes crystal field splitting of the d-orbitals, allowing electrons to absorb specific wavelengths of light and transition from lower energy t2g orbitals to higher energy eg orbitals (d-d transition), thus showing complementary color.

🎯 Exam Tip: The d-d transition is the primary reason for the color of transition metal complexes. For a complex to be colored, it must have partially filled d-orbitals and the ability for electrons to transition between split d-orbitals upon absorption of visible light.

4. Answer The Following Questions.

 

Question i. Give valence bond description for the hybrid orbitals are used by the metal? State the number of unpaired electrons.
Answer: Since Cl¯ is a weak ligand, there is no pairing of electrons.
Number of unpaired electrons = 2
Type of hybridisation = sp³
Geometry of complex ion = Tetrahedral
The complex ion is paramagnetic.
In simple words: For a complex with a weak field ligand and sp³ hybridization, there is no electron pairing, leading to unpaired electrons and a tetrahedral geometry. The presence of unpaired electrons makes the complex paramagnetic.

🎯 Exam Tip: In Valence Bond Theory (VBT), weak field ligands usually lead to outer orbital complexes (sp³d² or sp³), while strong field ligands typically form inner orbital complexes (d²sp² or d²sp³). The number of unpaired electrons determines paramagnetism (unpaired) or diamagnetism (all paired).

 

Question ii. Draw a qualitatively energy-level diagram showing d-orbital splitting in the octahedral environment. Predict the number of unpaired electrons in the complex [Fe(CN)6]4. Is the complex diamagnetic or paramagnetic? Is it coloured? Explain.
Answer: (A) r-orbital splitting in the octahedral environment :
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख ऑक्टाहेड्रल वातावरण में d-कक्षकों के गुणात्मक ऊर्जा-स्तर विभाजन को दर्शाता है। इसमें उच्च ऊर्जा \( \text{e}_g \) कक्षक और निम्न ऊर्जा \( \text{t}_{2g} \) कक्षक दिखाए गए हैं, जिनमें इलेक्ट्रॉन इन विभाजित स्तरों को भरते हैं, जो आमतौर पर एक मजबूत क्षेत्र लिगेंड परिदृश्य को दर्शाता है जहाँ \( \text{t}_{2g} \) कक्षक पहले भरे जाते हैं, जिससे युग्मन होता है।
(B) [Fe (CN)6]4- is an octahedral complex.
(C) Since CN is a strong ligand, there is pairing of electrons and the complex is diamagnetic.
(D) The complex exists as lemon yellow crystals.
(In the complex all electrons in t2g are paired and requires high radiation energy for excitation.)
In simple words: In [Fe(CN)6]4-, iron is Fe(II) (d6). Cyanide (CN-) is a strong field ligand, causing large crystal field splitting and forcing all six d-electrons to pair up in the t2g orbitals, making the complex diamagnetic with zero unpaired electrons. It appears lemon yellow because it absorbs blue-violet light, with its color originating from ligand-to-metal charge transfer (LMCT) rather than d-d transitions, which are not allowed due to all paired electrons.

🎯 Exam Tip: For d6 complexes with strong field ligands, electrons will pair up in the t2g orbitals, leading to a low-spin, diamagnetic complex. If d-d transitions are suppressed, color may arise from charge-transfer bands.

 

Question iii. Draw isomers in each of the following
a. [Pt(NH3)2CINO2]
b. [Ru(NH3)4CI2]
c. [Cr(en2)Br2]®
Answer: (a) [Pt(NH3)2CINO2]
Geometric isomers :
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख कॉम्प्लेक्स \( \text{[Pt(NH3)2CINO2]} \) के लिए ज्यामितीय आइसोमर (cis और trans) को दर्शाता है। केंद्रीय प्लेटिनम परमाणु वर्ग समतल है, जिसमें cis आइसोमर में दो समान लिगेंड एक-दूसरे के निकट होते हैं और trans आइसोमर में वे विपरीत दिशा में होते हैं।
(b) [RU(NH3)4CI2]
Geometric isomers :
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख ऑक्टाहेड्रल कॉम्प्लेक्स \( \text{[Ru(NH3)4Cl2]} \) के लिए ज्यामितीय आइसोमर (cis और trans) को दर्शाता है। केंद्रीय रूथेनियम परमाणु चार अमोनिया और दो क्लोराइड लिगेंड से बंधा हुआ है, जिसमें cis रूप में क्लोराइड लिगेंड निकट होते हैं और trans रूप में वे विपरीत होते हैं।

🎯 Exam Tip: For square planar MA2B2 type complexes, cis and trans isomers are possible. For octahedral MA4B2 type complexes, cis and trans isomers are also possible. For chiral complexes like those with bidentate ligands, enantiomers (optical isomers) are key.

Question iv. Draw geometric isomers and enantiomers of the following complexes.
a. [Pt(en)3]4+
b. [Pt(en)2ClBr]2+
Answer:
The complex [Pt(en)3]4+ has two optical isomers.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [Pt(en)3]4+ के दो प्रकाशीय समावयवों, d-फॉर्म और l-फॉर्म को दर्शाता है। दोनों संरचनाएं ऑक्टाहेड्रल ज्यामिति में केंद्रीय प्लैटिनम आयन (Pt) से जुड़े तीन एथिलीनडाइमाइन (en) लिगेंड को दर्शाती हैं, जहां एक फॉर्म दूसरे का गैर-अध्यारोपण योग्य दर्पण प्रतिबिंब है, जो चिरलता को दर्शाता है।
In simple words: [Pt(en)3]4+ exists as two optical isomers, d-form and l-form, which are non-superimposable mirror images of each other. These isomers have the same chemical formula but differ in their spatial arrangement, leading to distinct optical properties.

🎯 Exam Tip: Understanding optical isomerism and the ability to draw enantiomers for complexes like [Pt(en)3]4+ is crucial. Pay attention to the non-superimposable mirror image relationship and the octahedral geometry of the complex for full marks.

Question v. What are ligands? What are their types? Give one example of each type.
Answer:
Ligands: The neutral molecules or negatively charged anions (or rarely positive ions) which are bonded by coordinate bonds to the central metal atom or metal ion in a coordination compound are called ligands or donor groups. For example in [Cu(CN)4]2-, four CN- ions are ligands coordinated to central metal ion Cu2+.
Ligands can be classified on the basis of number of electron donor atoms in the ligand i.e. denticity.
(1) Monodentate or unidentate ligand : A ligand molecule or an ion which has only one donor atom with a lone pair of electrons forming only one coordinate bond with metal atom or ion in the complex is called monodentate or unidentate ligand. For example NH3, Cl-, OH-, H2O, etc.
(2) Polydentate or multidentate ligand : A ligand molecule or an ion which has two or more donor atoms with the lone pairs of electrons forming two or more coordinate bonds with the central metal atom or ion in the complex is called polydentate or multidentate ligand. For example, ethylene diamine, H2N - (CH2)2 - NH2.
According to the number of donor atoms they are classified as follows :
• Bidentate ligand: This ligand has two donor atoms in the molecule or ion. For example, ethylenediamine, H2N - (CH2)2 - NH2.
• Tridentate ligand : This ligand molecule has three donor atoms or three sites of attachment.
E.g. Diethelene triamine, H2N - CH2 - CH2 - NH - CH2 - CH2 - NH2. This has three N donor atoms.
• Tetradentate (or quadridentate) ligand : This ligand molecule has four donor atoms.
Eg. Triethylene tetraamine which has four N donor atoms.
• Hexdentate ligand : This ligand molecule has six donor atoms. E.g. Ethylenediamine tetracetato.
(3) Ambidentate ligand : A ligand molecule or an ion which has two or more donor atoms, however in the formation of a complex, only one donor atom is attached to the metal atom or an ion is called ambidentate ligand. For example, NO2- which has two donor atoms N and O forming a coordinate bond, M ← ONO (nitrito) or M ← NO2 (nitro).
(4) Bridging ligand: A monodentate ligand having more than one lone pairs of electrons, hence can attach to two or more metal atoms or ions and hence acts as a bridge between different metal atoms is called bridging ligand. For example : OH-, F-, SO4-2, etc.
In simple words: Ligands are ions or molecules that bind to a central metal atom in a coordination compound, forming coordinate bonds. They are classified based on the number of donor atoms (denticity), like monodentate (one donor atom) or polydentate (multiple donor atoms), and can also be ambidentate (multiple donor sites, but only one binds) or bridging (connects two metal atoms).

🎯 Exam Tip: A clear definition of ligands and their classification based on denticity, with specific examples for each type, is essential. Understanding the concept of ambidentate and bridging ligands also scores well.

Question vi. What are cationic, anionic and neutral complexes? Give one example of each.
Answer:
(1) Cationic sphere complexes : A positively charged coordination sphere or a coordination compound having a positively charged coordination sphere is called cationic sphere complex.
For example : [Zn(NH3)4]2+ and [Co(NH3)5Cl] SO4 are cationic complexes. The latter has coordination sphere [Co(NH3)5Cl]2+, the anion SO42- makes it electrically neutral.
(2) Anionic sphere complexes : A negatively charged coordination sphere or a coordination compound having negatively charged coordination sphere is called anionic sphere complex. For example, [Ni(CN)4]2- and K3[Fe(CN)6] have anionic coordination sphere; [Fe(CN)6]3- and three K+ ions make the latter electrically neutral.
(3) Neutral sphere complexes : A neutral coordination complex does not possess cationic or anionic sphere.
[Pt(NH3)2Cl2] or [Ni(CO)4] are neither cation nor anion but are neutral sphere complexes.
In simple words: Cationic complexes have a positive charge, anionic complexes have a negative charge, and neutral complexes have no overall charge. The charge of a complex depends on the central metal ion's oxidation state and the charges of its ligands.

🎯 Exam Tip: Provide clear definitions for each type of complex (cationic, anionic, neutral) and ensure the examples given correctly demonstrate the charge distribution. Emphasize how the overall charge is determined from the metal and ligand contributions.

Question vii. How stability of the coordination compounds can be explained in terms of equilibrium constants?
Answer:
Stability of the coordination compounds : The stability of coordination compounds can be explained on the basis of their stability constants. The stability of coordination compounds depends on metal-ligand interactions. In the complex, metal serves as electron-pair acceptor (Lewis acid) while the ligand as Lewis base (since it is electron donor). The metal-ligand interaction can be realized as the Lewis acid-Lewis base interaction. Stronger the interaction greater is stability of the complex.
Consider the equilibrium for the metal-ligand interaction :
Ma+ + nLx- \( \rightleftharpoons \) [MLn]a+(-nx)
where a, x, [a + (-nx)] denote the charge on the metal, ligand and the complex, respectively. Now, the equilibrium constant K is given by
\[K = \frac{[ML_n]^{a+(-nx)}}{[M^{a+}][L^{x-}]^n}\]
Stability of the complex can be explained in terms of K. Higher the value of K larger is the thermodynamic stability of the complex hence K is called stability constant, and denoted by Kstab. The equilibria for the complex formation with the corresponding K values are given below.
Ag+ + 2CN- \( \rightleftharpoons \) [Ag(CN)2]- K = 5.5 x 10¹⁸
Cu2+ + 4CN- \( \rightleftharpoons \) [Cu(CN)4]2- K = 2.0 x 10²⁷
Co3+ + 6NH3 \( \rightleftharpoons \) [Co(NH3)6]3+ K = 5.0 x 10³³
From the above data, the stability of the complexes is [Co(NH3)6]3+ > [Cu(CN)4]2- > [Ag(CN)2]-.
In simple words: The stability of a coordination compound is measured by its stability constant (K). A higher K value indicates a more stable complex because it signifies a stronger interaction between the central metal ion (Lewis acid) and the ligands (Lewis base).

🎯 Exam Tip: Clearly define the stability constant (K) and its relationship to the thermodynamic stability of a complex. Including the general equilibrium equation and specific examples with their K values strengthens the answer.

Question viii. Name the factors governing the equilibrium constants of the coordination compounds.
Answer:
The equilibrium constant of the complex depends on the following factors :
(a) Charge to size ratio of the metal ion : Higher the ratio greater is the stability. For the divalent metal ion complexes their stability shows the trend : Cu2+ > Ni2+ > Co2+ > Fe2+ > Mn2+ > Cd2+. The above stability order is called the Irving-William order. In the above list both Cu and Cd have the charge + 2, however, the ionic radius of Cu2 + is 69 pm and that of Cd2 + is 97 pm. The charge to size ratio of Cu2+ is greater than that of Cd2+. Therefore the Cu2+ forms stable complexes than Cd2+.
(b) Nature of the ligand : A second factor that governs stability of the complexes is related to how easily the ligand can donate its lone pair of electrons to the central metal ion that is, the basicity of the ligand. The ligands those are stronger bases tend to form more stable complexes.
In simple words: The stability of coordination compounds is mainly influenced by the charge-to-size ratio of the metal ion (higher ratio means more stable) and the nature of the ligand, particularly its basicity (stronger bases form more stable complexes).

🎯 Exam Tip: When discussing factors affecting stability, always mention the charge-to-size ratio of the metal ion and the basicity of the ligand. Illustrate with an example like the Irving-William order to show application.

Activity:

1. The reaction of chromium metal with H 2SO4 in the absence of air gives blue solution of chromium ion.
Cr(s) + 2H+(aq) → Cr2+(aq) + H2(s)
Cr2+ forms octahedral complex with H2O ligands.
a. Write formula of the complex
b. Describe bonding in the complex using CFT and VBT.
Draw crystal field splitting and valence bond orbital diagrams.
Answer:
In simple words: This activity asks to determine the formula and bonding description of an octahedral chromium(II) complex with water ligands, considering its formation from chromium metal and sulfuric acid in the absence of air.

🎯 Exam Tip: For problems involving reactions and complex formation, ensure you can determine the oxidation state of the metal, the type of ligands, and the geometry to correctly write the formula and discuss bonding theories.

2. Reaction of complex [Co(NH3)3(NO2)3 with HCl gives a complex [Co(NH3)3H2OCl2]+ in which two chloride ligands are trans to one another.
a. Draw possible stereoisomers of starting material
b. Assuming that NH3 groups remain in place, which of two starting isomers would give the observed product?
Answer:
In simple words: This activity requires drawing stereoisomers of [Co(NH3)3(NO2)3] and identifying which isomer would yield the product [Co(NH3)3H2OCl2]+ with trans chloride ligands, assuming ammonia ligands remain unchanged.

🎯 Exam Tip: For isomerism questions, practice drawing all possible geometric and optical isomers. When a product is specified, work backward from its structure to identify the most plausible starting isomer, especially with constraints like "NH3 groups remain in place."

12th Chemistry Digest Chapter 9 Coordination Compounds Intext Questions And Answers

Use Your Brain Power ...... (Textbook page 192)

Question 1. Draw Lewis structures of the following ligands and identify the donor atom in them : NH3, H2O.
Answer:

Ligand	Lewis dot structure	Donor atom
NH3	H - ..N - H | H	N
H2O	H - ..O - H	O

In simple words: The Lewis structures show the valence electrons and bonding. For ammonia (NH3), Nitrogen is the donor atom due to its lone pair, and for water (H2O), Oxygen is the donor atom because it has lone pairs.

🎯 Exam Tip: Accurately drawing Lewis structures and identifying donor atoms is fundamental for understanding coordination chemistry. Pay attention to lone pairs of electrons on potential donor atoms.

Try This ...... (Textbook page 193)

Question 1. Can you write ionisation of [Ni (NH3)6] Cl2?
Answer:
[Ni (NH3)6] Cl2 → [Ni(NH3)6]2+ + 2Cl-
In simple words: The complex [Ni(NH3)6]Cl2 ionizes in solution to form a hexamminenickel(II) cation, [Ni(NH3)6]2+, and two chloride anions, 2Cl-.

🎯 Exam Tip: For ionization reactions of coordination compounds, correctly identify the coordination sphere (which stays intact) and the counter ions (which dissociate). The charge on the complex ion must be balanced by the counter ions.

Question 2. Identify coordination sphere and counter ions.
Answer:
Coordination sphere : [Ni(NH3)6]2+
Counter ions : Cl-
In simple words: In [Ni(NH3)6]Cl2, the coordination sphere is the [Ni(NH3)6]2+ part, which includes the central metal and its directly bonded ligands, while the counter ions are the Cl- ions, which balance the charge of the sphere.

🎯 Exam Tip: Distinguishing between the coordination sphere (the complex ion) and counter ions is crucial for writing correct formulas and understanding the properties of coordination compounds. The coordination sphere is enclosed in square brackets.

Can You Tell? (Textbook page 193)

Question 1. A complex is made of Co (III) and consists of four NH3 molecules and two Cl- ions as ligands. What is the charge number and formula of complexion?
Answer:
The complex ion has formula, [Co(NH3)4Cl2]+.
The charge number is + 1.
In simple words: For a complex with Co(III) as the central metal, four neutral NH3 ligands, and two Cl- ligands, the overall charge is +1, resulting in the formula [Co(NH3)4Cl2]+.

🎯 Exam Tip: To determine the charge and formula of a complex, calculate the sum of the oxidation state of the central metal ion and the charges of all ligands. Remember that neutral ligands do not contribute to the charge.

Use Your Brain Power ...... (Textbook page 193)

Question 1. Coordination number used in coordination of compounds is somewhat different than that used in solid state. Explain.
Answer:
• In a coordination compound the coordination number is the number of donor atoms of ligands directly attached to metal atom or ion.
• In a solid state, the number of closest constituent atoms or ions in contact with a particular atom in the crystal lattice is called coordination number.
• In a coordination compound, coordination number depends upon nature of metal atom or ion, and its electronic configuration.
• In a solid state, the coordination number depends upon the crystalline structure of the unit cell.
In simple words: In coordination chemistry, coordination number refers to the number of ligand donor atoms directly bonded to the central metal. In solid-state chemistry, it refers to the number of nearest neighbors around an atom in a crystal lattice. These definitions differ because one focuses on bonding in a discrete complex, while the other focuses on packing in a continuous solid structure.

🎯 Exam Tip: Clearly differentiate between the definition of coordination number in coordination compounds (donor atoms) and in solid-state chemistry (nearest neighbors). Highlighting the dependence on electronic configuration for complexes and crystal structure for solids will earn full credit.

Can You Tell? (Textbook page 194)

Question 1. What is the coordination number of
(a) Co in [CoCl2(en)2]+ = 6
(b) Ir in [Ir(C2O4)2Cl2]3- and
(c) Pt in [Pt(NO2)2(NH3)2] ?
Answer:
(a) Coordination number of Co in [CoCl2(en)2]+ = 6
(b) Coordination number of Ir in [Ir(C2O4)2Cl2]3- = 6
(c) Coordination number of Pt in [Pt(NO2)2(NH3)2] = 4
In simple words: The coordination number is determined by counting the number of donor atoms attached to the central metal. For 'en' (ethylenediamine) and 'C2O4' (oxalate), each acts as a bidentate ligand, contributing two donor atoms.

🎯 Exam Tip: Always remember that bidentate ligands like 'en' (ethylenediamine) and 'C2O4' (oxalate) contribute 2 to the coordination number each. Monodentate ligands like Cl- and NH3 contribute 1 each.

Use Your Brain Power (Textbook page 195)

Question 1. Classify the complexes as homoleptic and heteroleptic:
(a) [Co (NH3)5Cl]SO4,
(b) [CO(ONO)(NH3)5]Cl2,
(c) [CoCl(NH3)(en)2]2+ and
(d) [Cu(C2O4)3]3-
Answer:
Homoleptic Complexes : (d) [Cu(C2O4)3]3-
Heteroleptic Complexes: (a) [CO(NH3)5Cl]SO4
(b) [CO(ONO)(NH3)5]Cl2,
(C) [CoCl(NH3)(en)2]2+
In simple words: Homoleptic complexes have only one type of ligand, while heteroleptic complexes have two or more different types of ligands attached to the central metal ion.

🎯 Exam Tip: To classify complexes as homoleptic or heteroleptic, simply check if all the ligands coordinated to the central metal atom are identical (homoleptic) or if there are multiple types of ligands present (heteroleptic).

Use Your Brain Power (Textbook page 195)

Question 1. Classify the complexes as cationic, anionic or Cr(H2O)2(C2O4)23-, PtCl2(en)2 and Cr(CO)6.
Answer:
Cationic complexes : [CO(NH3)6]Cl2
Anionic complexes : Na4[Fe(CN)6], [Cr(H2O)2 (C2O4)2]3-
Neutral complexes : Cr(CO)6, Pt Cl2(en)2
In simple words: Complexes are classified as cationic, anionic, or neutral based on their overall charge. To determine this, sum the oxidation state of the metal and the charges of all ligands.

🎯 Exam Tip: Remember to consider the charge of both the central metal ion and each ligand when determining the overall charge of the complex. Bidentate ligands like oxalate (C2O4)2- contribute a -2 charge.

Try This (Textbook page 197)

Question 1. Write the representation of the following :
(i) Tricarbonatocobaltate(III) ion.
(ii) Sodium hexacyanoferrate(III).
(iii) Potassium hexacyafioferrate(II).
(iv) Aquachlorobis(ethylenediamine)cobalt(III).
(v) Tetraaquadichlorochromium(III) chloride.
(vi) Diamminedichloroplatinum(II).
Answer:
(i) [Co(CO3)3]3-
(ii) Na3[Fe(CN)6]
(iii) K4[Fe(CN)6]
(iv) Co(en)2(H2O)(Cl)
(v) [Cr(H2O)4Cl2]Cl
(vi) Pt(NH3)2Cl2
In simple words: This question requires translating IUPAC names of coordination compounds into their chemical formulas, ensuring correct notation for the central metal, ligands, and overall charge or counterions.

🎯 Exam Tip: Practice converting IUPAC names to chemical formulas and vice-versa. Pay close attention to the oxidation state of the metal, charges of ligands, and the rules for naming/writing formulas for complex ions and overall compounds.

Try This (Textbook page 196)

Question 1. Find out the EAN of
(a) [Zn(NH3)4]2+
(b) [Fe(CN)6]4-
Answer:
(a) For the complex ion, [Zn(NH3)4]2+ :
Atomic number of Zn = Z = 30
Charge on metal ion = + 2
Therefore Number of electrons lost by Zn atom = X = 2 Total number of electrons donated by 4NH3 ligands = Y = 2 x 4 = 8
EAN = Z - X + Y
= 30 - 2 + 8
= 36
(Note: This is atomic number of the nearest inert element 36Kr.)
(b) For the complex ion, [Fe(CN)6]4- :
For Fe, Z = 26 (Atomic number)
X = 2 (Due to + 2 charge on Fe)
Y = 12 (Due to 6 CN- ligands)
Therefore EAN = Z - X + Y
= 26 - 2 + 12
= 36
In simple words: The Effective Atomic Number (EAN) is calculated by adding the atomic number of the metal, subtracting its oxidation state, and adding twice the coordination number (representing electrons donated by ligands). For both given complexes, the EAN is 36, matching Krypton's atomic number, suggesting stability.

🎯 Exam Tip: Mastering the EAN rule calculation (Z - oxidation state + 2 * coordination number) is vital. Remember to correctly identify the atomic number (Z), the metal's oxidation state (X), and the total electrons donated by ligands (Y).

Use Your Brain Power (Textbook page 197)

Question 1. Do the following complexes follow the EAN rule
(a) Cr(CO)4,
(b) Ni(CO)4,
(c) Mn(CO)5,
(d) Fe(CO)5?
Answer:
(a) Cr(CO)4 : EAN = Z - X + Y
= 24 - 0 + 8
= 32
(b) Ni(CO)4 : EAN = Z - X + Y
= 28 - 0 + 8
= 36
(c) Mn(CO)5 : EAN = Z - X + Y
= 25 - 0 + 10
= 35
(d) Fe(CO)5 : EAN = Z - X + Y
= 26 - 0 + 10
= 36
Conclusion:
(a) Cr(CO)4 and (c) Mn(CO)5 do not follow EAN Rule.
In simple words: By calculating the EAN for each complex, it's found that Ni(CO)4 and Fe(CO)5 follow the rule (EAN=36), while Cr(CO)4 (EAN=32) and Mn(CO)5 (EAN=35) do not.

🎯 Exam Tip: Be meticulous in calculating the oxidation state of the metal (often 0 for carbonyls) and the number of electrons donated by each ligand (CO is a 2-electron donor). Compare the calculated EAN with the atomic number of the nearest noble gas.

Try This (Textbook page 199)

Question 1. Draw structures of ci,c and trans isomers of [Fe(NH3)2(CN)4]
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [Fe(NH3)2(CN)4] के सिस- और ट्रांस-समावयवों को दर्शाता है। सिस-समावयव में, समान लिगेंड (NH3 और CN) एक-दूसरे के निकट (90° पर) होते हैं, जबकि ट्रांस-समावयव में, समान लिगेंड एक-दूसरे के विपरीत (180° पर) होते हैं, जो इन ऑक्टाहेड्रल परिसरों की ज्यामितीय भिन्नता को स्पष्ट करता है।
In simple words: The cis isomer of [Fe(NH3)2(CN)4] has the two ammonia ligands adjacent to each other, while the trans isomer has them opposite to each other. These are geometric isomers, differing in the spatial arrangement of ligands around the central iron atom.

🎯 Exam Tip: When drawing cis and trans isomers for octahedral complexes, remember that cis isomers have identical ligands at 90° to each other, and trans isomers have identical ligands at 180° to each other. Accurate placement of ligands is key.

Remember ..... (Textbook page 199)

Our hands are non-superimposable mirror images. When you hold your left hand up to a mirror the image looks like right hand.

Try This (Textbook page 199)

Question 1. Draw enantiomers of [Cr(OX)2]3- where OX = C2O4 :
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [Cr(C2O4)3]3- के दो एनैन्शियोमर, d-फॉर्म और l-फॉर्म को दर्शाता है। दोनों संरचनाएं ऑक्टाहेड्रल ज्यामिति में केंद्रीय क्रोमियम आयन (Cr) से जुड़े तीन ऑक्सलेट (ox) लिगेंड को दर्शाती हैं। l-फॉर्म d-फॉर्म का गैर-अध्यारोपण योग्य दर्पण प्रतिबिंब है, जो चिरलता को प्रदर्शित करता है।
In simple words: The enantiomers of [Cr(C2O4)3]3- are d-form and l-form, which are non-superimposable mirror images. This complex is chiral because the three bidentate oxalate ligands create an asymmetric arrangement around the central chromium ion.

🎯 Exam Tip: For complexes with bidentate ligands like oxalate, pay attention to how the ligands chelate to the metal. Chiral complexes (enantiomers) will have non-superimposable mirror images, often identifiable by a propeller-like arrangement.

Question 2. Draw (A) enantiomers and (B) cis and trans isomers of [Cr(H2O)2(OX)2] :
Answer:
(A) Enantiomers :
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [Cr(H2O)2(C2O4)2] के एनैन्शियोमर को दर्शाता है। केंद्रीय क्रोमियम आयन (Cr) से दो पानी (H2O) और दो ऑक्सलेट (ox) लिगेंड जुड़े हुए हैं। d-फॉर्म और l-फॉर्म एक-दूसरे के दर्पण प्रतिबिंब हैं जो अध्यारोपण योग्य नहीं हैं, जो इस जटिल में चिरलता का संकेत देते हैं।
(B) as and trans isomers :
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [Cr(H2O)2(C2O4)2] के सिस- और ट्रांस-समावयवों को दर्शाता है। सिस-फॉर्म में, समान लिगेंड (H2O) एक-दूसरे के निकट होते हैं, जबकि ट्रांस-फॉर्म में, वे एक-दूसरे के विपरीत होते हैं। ऑक्सलेट लिगेंड बाइडेनटेट रूप से बंधे होते हैं, जिससे ऑक्टाहेड्रल ज्यामिति बनती है।
In simple words: For [Cr(H2O)2(OX)2], the enantiomers are non-superimposable mirror images, and the cis and trans isomers differ in the relative positions of the H2O ligands. Cis has H2O ligands adjacent, while trans has them opposite.

🎯 Exam Tip: When a complex can exhibit both geometric and optical isomerism, draw all possible geometric isomers first. Then, identify which of these geometric isomers (if any) are chiral and draw their enantiomeric pairs. This systematic approach ensures no isomers are missed.

Can You Tell? (Textbook page 200)

Question 1. Can you write IUPAC names of isomers (I) [Co(NH3)5SO4]Br and (II) [Co(NH3)5Br]SO4?
Answer:

Coordination compound	IUPAC name
(I) [Co(NH3)5SO4]Br	Pentaamminesulphatocobalt(III) bromide
(II) [Co(NH3)5Br]SO4	Pentaamminebromocobalt(III) sulphate

In simple words: The two compounds are ionization isomers because they differ in the ions outside and inside the coordination sphere. The first is Pentaamminesulphatocobalt(III) bromide, and the second is Pentaamminebromocobalt(III) sulphate.

🎯 Exam Tip: For ionization isomers, correctly identify which ligand acts as the counter ion in each case. The ligand inside the coordination sphere becomes part of the complex name, while the one outside is named separately.

Question 2. Write linkage isomers of [Fe(H2O)5SCN]+. Write their IUPAC names.
Answer:

Linkage isomers	IUPAC name
(a) [Fe(H2O)5SCN]+	Pentaaquathiocyanatoiron(II)
(b) [Fe(H2O)5NCS]+	Pentaaquaisothiocyanatoiron(II)

In simple words: Linkage isomers occur when an ambidentate ligand, like SCN-, can bind to the central metal through different donor atoms. Here, SCN- can bind via Sulfur (thiocyanato) or Nitrogen (isothiocyanato), creating two distinct isomers.

🎯 Exam Tip: To identify linkage isomers, look for ambidentate ligands (e.g., SCN-, NO2-). Ensure the IUPAC names reflect the correct donor atom using prefixes like "thiocyanato-" (S-bonded) and "isothiocyanato-" (N-bonded).

Use Your Brain Power .(Textbook page 201)

Question 1. The stability constant K of the [Ag(CN)2]- is 5.5 x 10 while that for the corresponding [Ag(NH3)2]+ is 1.6 x 107. Explain why [Ag(CN)2]2- is more stable.
Answer:
Stability constant of [Ag(CN)2]2- is larger than that of [Ag(NH3)2]+ and hence [Ag(CN)2]2- is more stable. Also, CN is a stronger ligand than NH3.
In simple words: [Ag(CN)2]- is more stable than [Ag(NH3)2]+ because its stability constant is significantly higher. This indicates that cyanide (CN-) is a much stronger ligand than ammonia (NH3), forming a more robust bond with the silver ion.

🎯 Exam Tip: When comparing stability, always refer to the stability constant (K). A larger K value directly indicates greater stability. Also, be able to explain the reason in terms of ligand strength.

Remember ...... (Textbook Page 202)

Question 1. Complete the missing entries.
Answer:

Coordination number	Geometry of complex	Hybridisation
2	Linear	sp
4	Tetrahedral	sp3
4	Square planar	dsp2
6	Octahedral	d2sp3/sp3d2

(Note: The missing entries are underlined.)
In simple words: This table maps the relationship between a complex's coordination number, its resulting geometry, and the type of hybridization involved. For example, a coordination number of 4 can lead to either tetrahedral (sp3) or square planar (dsp2) geometry.

🎯 Exam Tip: Memorize the correlation between coordination number, geometry, and hybridization, especially for common coordination numbers 2, 4, and 6. This knowledge is fundamental for VBT explanations.

Table 9.3: Type of hybridisation and geometry of a complex

Type of hybridisation	sp	sp2	sp3	dsp2	dsp3
Geometry	Linear	Triangular plane	Tetrahedral	Square planar	Trigonal bipyramidal
Type of hybridisation	d2sp3, sp3d2	d3sp3
Geometry	Octahedral	Pentagonal bipyramidal

Try This (Textbook page 204)

Question 1. Based on the VBT predict structure and magnetic behaviour of the [Ni(NH3)6]
Answer:
28Ni [Ar] 3d⁸ 4s²
Ni3+ [Ar] 3d⁷ 4s⁰
Hybridisation : sp³d²
Geometry: Octahedral
Magnetic property: Paramagnetic
In simple words: For [Ni(NH3)6]3+, the Ni3+ ion has a d7 configuration. Ammonia is a strong field ligand, but for Ni3+ in an octahedral field, it undergoes sp3d2 hybridization, resulting in a paramagnetic complex with an octahedral geometry due to the presence of unpaired electrons.

🎯 Exam Tip: For VBT questions, always determine the oxidation state of the metal, its electron configuration, and whether the ligands are strong or weak field. This helps predict hybridization, geometry, and magnetic properties accurately.

Try This (Textbook page 202)

Question 1. Give VBT description of bonding in each of following complexes. Predict their magnetic behaviour.
(a) [ZnCl4]2+
(b) [CO(H2O)6]2- (high spin)
(c) [Pt(CN)4]2- (square planar)
(d) [CoCl4]2- (tetrahedral)
(e) [Cr(NH3)6]3+
Answer:
In simple words: This question requires applying Valence Bond Theory to predict the hybridization, geometry, and magnetic properties (paramagnetic or diamagnetic) for various coordination complexes by considering the central metal's oxidation state and electron configuration, and the nature of its ligands.

🎯 Exam Tip: Practice VBT for different geometries (tetrahedral, square planar, octahedral). Remember to consider strong vs. weak field ligands for electron pairing and the resulting hybridization (e.g., dsp2 for square planar, sp3 for tetrahedral, d2sp3/sp3d2 for octahedral).

Try This ....... (Textbook page 206)

Question 1. Sketch qualitatively crystal field d orbital energy level diagrams for each of the following complexes :
(a) [Ni(en)3]2+
(b) [Mn(CN)6]3-
(c) [Fe(H2O)6]2+
Predict whether each of the complexes is diamagnetic or paramagnetic.
Answer:
(a) The complex ion, [Ni(en)3]2+ is octahedral.
28Ni [Ar] 3d⁸ 4s²
Ni2+ [Ar] 3d⁸ 4s⁰.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [Ni(en)3]2+ के लिए d-ऑर्बिटल ऊर्जा स्तर विखंडन को दर्शाता है। केंद्रीय Ni2+ आयन (3d8) की अष्टफलकीय ज्यामिति में, तीन उच्च ऊर्जा e_g ऑर्बिटलों में से एक में और तीन निम्न ऊर्जा t_2g ऑर्बिटलों में से दो में इलेक्ट्रॉनों का युग्मन होता है। चूंकि 'en' एक मजबूत क्षेत्र लिगेंड है, इसलिए यह इलेक्ट्रॉनों के युग्मन को बढ़ावा देता है, जिसके परिणामस्वरूप t2g में छह इलेक्ट्रॉन और eg में दो इलेक्ट्रॉन होते हैं। चूंकि दो अयुग्मित इलेक्ट्रॉन हैं, जटिल अनुचुंबकीय है।
Since en is a strong ligand there is pairing of electrons.
Number of unpaired electrons = n = 2 in t2g, orbitals
Magnetic moment = μ = \( \sqrt{n(n + 2)} \)
= \( \sqrt{2(2+2)} \) = 2.83 B. M.
The complex ion is paramagnetic.
(b) The complex ion [Mn(CN)6]3- is octahedral.
25Mn [Ar] 3d⁵ 4s²
Mn3+ [Ar] 3d⁴ 4s⁰
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [Mn(CN)6]3- के लिए d-ऑर्बिटल ऊर्जा स्तर विखंडन को दर्शाता है। केंद्रीय Mn3+ आयन (3d4) की अष्टफलकीय ज्यामिति में, CN- एक मजबूत क्षेत्र लिगेंड है, जो इलेक्ट्रॉनों के युग्मन को बढ़ावा देता है। इससे सभी चार इलेक्ट्रॉन निम्न ऊर्जा t_2g ऑर्बिटलों में युग्मित हो जाते हैं। परिणामस्वरूप, e_g ऑर्बिटलों में कोई इलेक्ट्रॉन नहीं होता है, और t_2g ऑर्बिटलों में एक अयुग्मित इलेक्ट्रॉन होता है, जिससे यह अनुचुंबकीय हो जाता है।
Since CN- is a strong ligand there is pairing of electrons.
Number of unpaired electrons = n = 2 in t2g, orbitals
Magnetic moment = μ = \( \sqrt{n(n + 2)} \)
= \( \sqrt{2(2 + 2)} \) = 2.83 B. M.
The complex ion is paramagnetic.
(c) The complex ion [Fe(H2O)6]2+ is octahedral.
26Fe [Ar] 3d⁶ 4s²
Fe2+ [Ar] 3d⁶ 4s⁰
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख [Fe(H2O)6]2+ के लिए d-ऑर्बिटल ऊर्जा स्तर विखंडन को दर्शाता है। केंद्रीय Fe2+ आयन (3d6) की अष्टफलकीय ज्यामिति में, H2O एक कमजोर क्षेत्र लिगेंड है, जो इलेक्ट्रॉनों के युग्मन को बढ़ावा नहीं देता है। इससे इलेक्ट्रॉन t_2g और e_g ऑर्बिटलों में पहले भरे जाते हैं, जिसके परिणामस्वरूप t_2g में चार और e_g में दो इलेक्ट्रॉन होते हैं, जिसमें चार अयुग्मित इलेक्ट्रॉन होते हैं। इस कारण, जटिल अनुचुंबकीय होता है।
Since H2O is a weak ligand, there is no pairing of electrons.
Number of unpaired electrons = n = 4 in t2g and eg orbitals.
Magnetic moment = \( \sqrt{n(n + 2)} \)
= \( \sqrt{4(4+2)} \)
= 4.90 B. M.
The complex ion is paramagnetic.
In simple words: For [Ni(en)3]2+, it's paramagnetic with 2 unpaired electrons (en is strong field). For [Mn(CN)6]3-, it's paramagnetic with 1 unpaired electron (CN- is strong field causing pairing). For [Fe(H2O)6]2+, it's paramagnetic with 4 unpaired electrons (H2O is weak field, no pairing).

🎯 Exam Tip: For CFT diagrams, correctly identify the central metal's oxidation state, d-electron count, and whether the ligand is strong or weak field. This determines the splitting pattern (t2g/eg) and electron distribution, which in turn predicts paramagnetism (unpaired electrons) or diamagnetism (all paired electrons). Calculating magnetic moment using the spin-only formula is also essential.

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