Coordination Chemistry

The question asks us about the nature of the complexes and . Let's analyze each complex:

1. Understanding

• Coordination Complex: In this complex, cobalt (Co) is surrounded by six ammonia (NH₃) molecules. Ammonia is a strong field ligand.
• Electronic Configuration and Hybridization: Cobalt in this complex has a +3 oxidation state (Co³⁺), which gives it an electronic configuration of 3d⁶. Under the influence of a strong field ligand like NH₃, it undergoes d²sp³ hybridization, leading to low-spin or spin-paired configurations.
• Conclusion: This is an inner orbital or spin-paired complex.

2. Understanding

• Coordination Complex: Here, cobalt is surrounded by six fluoride (F^-) ions. Fluoride is a weak field ligand.
• Electronic Configuration and Hybridization: With CoF₆^3-, cobalt is also in the +3 oxidation state. Due to fluoride's weak field, it does not cause pairing of electrons, resulting in sp^3\d^2\ hybridization and a high-spin, or spin-free, complex.
• Conclusion: This is an outer orbital or spin-free complex.

Based on the explanations above, the correct choice matches the respective complex nature: Spin paired Complex, Spin free Complex.

This problem requires identifying complexes with an even number of electrons in their orbitals. Let's evaluate each given complex:

1. • Iron in this complex is Fe(II), meaning it has a 3d⁶ electron configuration.
   • In an octahedral field, the electronic configuration for Fe²⁺ is .
   • The number of electrons is 4, which is even.
2. • Cobalt in this complex is Co(II), meaning it has a 3d⁷ electron configuration.
   • In an octahedral field, the configuration is .
   • The number of electrons is 5, which is odd.
3. :
   • Cobalt in this complex is Co(III), which has a 3d⁶ configuration.
   • In an octahedral field, the configuration is .
   • The number of electrons is 6, which is even.
4. • Copper in this complex is Cu(II), which has a 3d⁹ configuration.
   • In an octahedral field, the configuration is .
   • The number of electrons is 6, which is even.
5. • Chromium in this complex is Cr(II), which has a 3d⁴ configuration.
   • In an octahedral field, the configuration is .
   • The number of electrons is 3, which is odd.

From this analysis, the complexes with an even number of electrons in orbitals are . Therefore, there are 3 such complexes.

To solve the problem, we need to determine the oxidation state of cobalt (Co) in the complex and use the information provided to find the value of .

1. Given the complex formula is , it is an octahedral complex. This means that in the coordination sphere of cobalt, a total of six ligands are present (since octahedral coordination implies coordination number six).
2. We know that ammonia ( ) acts as a neutral ligand, and chlorine ( ) acts as an anion with a -1 charge.
3. The formula suggests that the complex contains three ions outside the coordination sphere, which participate in precipitating upon reaction with .
4. When 2 moles of are formed, it implies that two of the ions are free or outside the coordination sphere. This means the complex should be , where one Cl is within the coordination sphere and two Cls outside.
5. To determine the oxidation state of Co:
   • The complex charge can be written as:
     , canceling two negative charges from the outside ion\)
   • Since ammonia is neutral, it does not affect the oxidation state, only the chlorines change it.
   • The overall equation for charge balance in the coordination sphere is:
     
   • Simplifying, .
6. Since the complex is octahedral, having 6 coordination sites, we get:
   
7. Simplifying for :
   
8. Therefore, .

Hence, the value of "x + n" is 8, which is the correct answer.

To find the 'spin only' magnetic moment of , where is a metal with the least metallic radii among , follow these steps:

Step 1: Determine the Metal

• The atomic number determines the metallic radius: Sc = 21, Ti = 22, V = 23, Cr = 24, Mn = 25, Zn = 30.
• Typically, a higher atomic number indicates a smaller atomic radius within a period (consider the transition metals' trend).
• Zinc ( ) has the highest atomic number (30) in this set, suggesting the smallest radius among the given metals.
• Therefore, .

Step 2: Determine D-electrons for

• Zinc starts with an electron configuration of .
• As , it loses 2 electrons from the 4s orbital, resulting in a configuration of .
• This implies 10 d-electrons and no unpaired electrons.

Step 3: Calculate the Magnetic Moment

• The 'spin only' magnetic moment is calculated using where equals the number of unpaired electrons.
• For , since there are no unpaired electrons.
• Thus, Bohr Magneton (BM).

Step 4: Confirm the Solution

• The calculated magnetic moment is 0 BM, which matches the expected range (0,0).

Conclusion: The 'spin only' magnetic moment value of where is 0 BM.

To solve this chemistry question, we need to match each compound from List-I with its corresponding color from List-II. Let's analyze each compound:

1. is known as Prussian Blue. It is commonly recognized for its intense blue color. Therefore, the correct match is III. Prussian Blue.
2. is referred to as sodium nitroprusside. This compound typically appears violet. Hence, it corresponds to I. Violet.
3. forms a complex that is characterized by its blood red coloration known in iron thiocyanate tests. Thus, it aligns with II. Blood Red.
4. , often referred to as ammonium phosphomolybdate, is well-known for its yellow color. Therefore, it is IV. Yellow.

Based on the above analysis, the correct matching of compounds to their colors is A-III, B-I, C-II, D-IV. This matches with the provided correct answer.

To determine the correctness of the statements provided regarding the ligands (trimethylamine) and (trimethylphosphine), let's evaluate each statement individually:

1. Statement I: and can act as ligands to form transition metal complexes.
   • In coordination chemistry, a ligand is an ion or molecule that binds to a central metal atom to form a coordination complex. Ligands like and are known as donor ligands where nitrogen and phosphorus donate their lone pair of electrons to the metal center.
   • Both trimethylamine ( ) and trimethylphosphine ( ) can indeed act as ligands. Nitrogen and phosphorus both have lone pairs that they can use to coordinate with transition metals, forming complexes.
   • Therefore, Statement I is correct.
2. Statement II: As and are from the same group, the nature of bonding of and is always the same with transition metals.
   • Although nitrogen and phosphorus are in the same group of the periodic table and share some chemical similarities, they are different in terms of their size and their ability to back-bond with metals.
   • Nitrogen's smaller size and higher electronegativity compared to phosphorus often result in different bonding characteristics and complex stabilities. For example, phosphorus can sometimes use its d-orbitals for back-bonding, something nitrogen does not have access to due to its lack of low-energy d-orbitals.
   • Hence, the assertion that the bonding nature is always the same between these two ligands is incorrect.
   • Therefore, Statement II is incorrect.

Based on the analysis, the most appropriate answer is: Statement I is correct but Statement II is incorrect.

To solve this matching question, we need to identify the common uses of each compound listed in List-I by matching them correctly with List-II.

1. Carbon tetrachloride (A): Traditionally used as a fire extinguisher. Chlorinated hydrocarbons like carbon tetrachloride have been used in the past for this purpose due to their non-flammable properties. Therefore, the correct match is (III).
2. Methylene chloride (B): Commonly used as a paint remover because it effectively strips paint by dissolving the paint layers. Thus, the correct match is (I).
3. DDT (C): Known as a non-biodegradable insecticide. It has been used extensively in agriculture and for mosquito control because of its effectiveness, although its use is now heavily restricted due to environmental impact. Therefore, the correct match is (IV).
4. Freons (D): These are used in refrigerators and air conditioners as refrigerants. Freons help in transferring heat and keeping these appliances cool. Thus, the correct match is (II).

Based on the analysis, the correct match of List-I with List-II is:

• (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

Therefore, the correct answer is: (A)-(III), (B)-(I), (C)-(IV), (D)-(II).

To determine the total number of unpaired electrons in the complex ions [Co(NH₃)₆]³⁺ and [NiCl₄]²⁻, we must analyze each complex individually:

[Co(NH₃)₆]³⁺:

• Cobalt (Co) has an atomic number of 27, giving it an electron configuration of [Ar]3d⁷4s².
• In the 3+ oxidation state, Co loses 3 electrons, resulting in an electron configuration of [Ar]3d⁶.
• NH₃ is a strong field ligand, causing pairing of electrons in the 3d orbitals.
• The 3d orbital configuration is t_2g⁶e_g⁰, indicating that all electrons are paired.
• Thus, there are 0 unpaired electrons in this complex.

[NiCl₄]²⁻:

• Nickel (Ni) has an atomic number of 28, giving it an electron configuration of [Ar]3d⁸4s².
• In the 2− oxidation state, Ni loses 2 electrons, resulting in an electron configuration of [Ar]3d⁸.
• Cl⁻ is a weak field ligand, so there is no electron pairing in the 3d orbitals.
• The 3d orbital configuration is e_g⁴t_2g⁴, with 2 unpaired electrons.

Combining both results, the total number of unpaired electrons in the complex ions is 0 (from Co) + 2 (from Ni) = 2.

This result fits the provided range: 2,2.

To match the complex ions in List-I with their respective spin-only magnetic moments in List-II, we need to understand the concept of spin-only magnetic moment, which is given by the formula:

Bohr Magneton (B.M.),

where is the number of unpaired electrons in the complex ion.

1. (A) [Cr(NH₃)₆]³⁺: Chromium in the 3+ oxidation state has an electronic configuration of . Therefore, there are 3 unpaired electrons.
   • Magnetic moment: B.M.
2. (B) [NiCl₄]^2-: Nickel in this complex is in the 2+ oxidation state with an electronic configuration of . The complex is tetrahedral, resulting in 2 unpaired electrons.
   • Magnetic moment: B.M.
3. (C) [CoF₆]^3-: Cobalt in the 3+ oxidation state has an electronic configuration of . In an octahedral field for high-spin complexes, it has 4 unpaired electrons.
   • Magnetic moment: B.M.
4. (D) [Ni(CN)₄]^2-: Nickel in the 2+ oxidation state with its electronic configuration . With CN^- ligands, the complex forms a square planar structure, resulting in no unpaired electrons.
   • Magnetic moment: B.M.

Thus, the correct match between List-I and List-II is:

• (A)-(II): 3.87 B.M.
• (B)-(IV): 2.83 B.M.
• (C)-(I): 4.90 B.M.
• (D)-(III): 0.0 B.M.

The question requires us to match substances with the elements present in them. Let's analyze each pair:

1. Ziegler Catalyst: The Ziegler catalyst is widely used in the polymerization of ethylene and other alpha-olefins, and it primarily contains Titanium (IV). Therefore, the correct match for Ziegler catalyst is (IV) Titanium.
2. Blood Pigment: The main pigment in blood is hemoglobin, which contains the element Iron (Fe) incorporated in its structure. Hence, the correct match for Blood pigment is (III) Iron.
3. Wilkinson Catalyst: The Wilkinson catalyst is a chemical catalyst known for hydrogenation reactions, and it contains Rhodium (I). Therefore, the correct match for Wilkinson catalyst is (I) Rhodium.
4. Vitamin B12: Vitamin B12, also known as cobalamin, contains the element Cobalt (Co) in its structure. Thus, the correct match for Vitamin B12 is (II) Cobalt.

Thus, the correct combination is:

Therefore, the correct answer is: A-IV, B-III, C-I, D-II

This question requires an analysis of the color of two nickel coordination complexes, and , based on the principles of coordination chemistry and crystal field theory.

Concept Used:

The color of transition metal complexes is attributed to d-d electronic transitions. When light passes through a solution of the complex, it absorbs photons of a specific energy (and color) from the visible spectrum to promote an electron from a lower-energy d-orbital to a higher-energy d-orbital. The color we perceive is the complementary color of the light absorbed.

The possibility and energy of these d-d transitions depend on:

1. The electronic configuration of the central metal ion (presence of partially filled d-orbitals).
2. The nature of the ligand (strong-field or weak-field), which determines the magnitude of the crystal field splitting energy ( ).
3. The geometry of the complex (e.g., octahedral, tetrahedral, square planar), which determines the pattern of d-orbital splitting.

If all d-orbitals are empty ( ), completely filled ( ), or if the energy gap ( ) is too large (requiring UV light), no d-d transition occurs in the visible range, and the complex appears colorless.

Step-by-Step Solution:

Step 1: Analysis of Statement (I) - A solution of is green in color.

First, we determine the properties of the complex .

• Central Metal Ion: Nickel (Ni).
• Oxidation State: Let the oxidation state of Ni be . Water ( ) is a neutral ligand (charge 0). The overall charge is +2. So, . The ion is .
• Electronic Configuration: The atomic number of Ni is 28 ( ). For , the configuration is .
• Ligand and Geometry: The ligand is , which is a weak-field ligand. With six ligands, the geometry is octahedral.
• Crystal Field Splitting: In an octahedral field, the five d-orbitals split into two sets: a lower energy set and a higher energy set. Since is a weak-field ligand, the crystal field splitting energy ( ) is small, and electrons will fill the orbitals according to Hund's rule before pairing.
• Electron Filling: For a configuration in a weak octahedral field, the electrons fill as .

The orbitals contain two unpaired electrons. The presence of these unpaired electrons in the partially filled d-orbitals allows for d-d electronic transitions. The complex absorbs light in the red region of the visible spectrum, and it transmits the complementary color, which is green. Thus, a solution of is green.

Conclusion for Statement (I): Statement (I) is correct.

Step 2: Analysis of Statement (II) - A solution of is colorless.

Now, we determine the properties of the complex .

• Central Metal Ion: Nickel (Ni).
• Oxidation State: Let the oxidation state of Ni be . The cyanide ligand ( ) has a charge of -1. The overall charge is -2. So, . The ion is again .
• Electronic Configuration: The configuration of is .
• Ligand and Geometry: The ligand is , which is a strong-field ligand. For a metal ion with strong-field ligands and a coordination number of 4, the complex adopts a square planar geometry. This involves hybridization.
• Crystal Field Splitting: In a square planar field, the d-orbitals split into four different energy levels. The strong-field ligand causes a large splitting energy, forcing all eight d-electrons to pair up in the lower energy orbitals.
• Electron Filling: The 8 electrons fill the lower four d-orbitals completely.

As shown, there are no unpaired electrons in the d-orbitals. All electrons are paired. The energy gap between the highest occupied d-orbital ( ) and the lowest unoccupied d-orbital ( ) is very large. An electronic transition would require a large amount of energy, corresponding to light in the ultraviolet (UV) region. Since the complex does not absorb any light from the visible region, it appears colorless.

Conclusion for Statement (II): Statement (II) is correct.

Final Result:

Both Statement (I) and Statement (II) are correct. is green due to d-d transitions allowed by its unpaired electrons in a weak octahedral field. is colorless because it is a square planar complex with a strong-field ligand, resulting in all d-electrons being paired and a large energy gap for any electronic transition, which prevents absorption of visible light.

Therefore, both Statement (I) and Statement (II) are true.

To solve this question, we need to analyze the coordination complexes provided and determine the number of electrons in the orbitals of the most stable complex ion.

Let's examine each of the given complexes:

1. :
   • Ligand: Ammonia ( ) is a strong field ligand.
   • Iron's oxidation state: (as the complex ion has a charge).
   • Electronic configuration of : .
   • In a strong field ligand environment, pairing occurs due to splitting of the d-orbitals ( and ).
   • The orbitals will be fully filled before , adding 4 electrons to the :
2. :
   • Ligand: Chloride ( ) is a weak field ligand.
   • Iron's oxidation state: .
   • Electronic configuration of : .
   • No pairing occurs in weak field ligands, resulting in:
3. :
   • Ligand: Oxalate ( ) is a strong field ligand.
   • Iron's oxidation state: .
   • Electronic configuration of : .
   • In a strong field ligand, pairing occurs:
4. :
   • Ligand: Water ( ) is a weak field ligand.
   • Iron's oxidation state: .
   • Electronic configuration of : .
   • No pairing occurs:

From the analysis, the most stable complex in terms of maximum electrons in the is with a configuration of .

Now, let's analyze the second part about the nature of the vanadium oxide .

Vanadium can have multiple oxidation states, resulting in different oxides:

• If in , it is known to be amphoteric, meaning it can act both as an acid and a base.

The correct nature of oxide in this case is Amphoteric.

Let's evaluate each statement individually to determine their truthfulness.

• Statement I: A homoleptic octahedral complex, formed using monodentate ligands, will not show stereoisomerism.
  Explanation: A homoleptic complex is one where all ligands are the same. In the case of octahedral complexes with monodentate ligands, since all positions around the central metal ion are occupied by identical ligands, no spatial variation can exist, thus eliminating the possibility of forming stereoisomers. Therefore, Statement I is true.
• Statement II: cis- and trans-platin are heteroleptic complexes of Pd.
  Explanation: Cisplatin and Transplatin are well-known complexes of Platinum (Pt), not Palladium (Pd). Moreover, in these complexes, both chloride ions (Cl) and ammonia molecules (NH₃) serve as ligands, making them heteroleptic because they contain different types of ligands. However, the correct metal is Platinum (Pt), not Palladium (Pd). Thus, Statement II is false.

Conclusion: Based on the above explanations, the correct answer is: Statement I is true but Statement II is false.

To determine which of the given complexes are diamagnetic, we must first understand the electronic configuration and the nature of the ligands.

1. [Mn(CN)₆]^3-:
   • Manganese (Mn) in [Mn(CN)₆]^3- is in the +3 oxidation state, giving it a configuration of (Mn: [Ar] 3d⁵ 4s², so Mn³⁺: [Ar] 3d⁴).
   • CN^- is a strong field ligand and can cause pairing of electrons in the 3d orbitals.
   • However, since the d-configuration is , complete pairing leading to diamagnetism will not occur.
2. [Co(NH₃)₆]³⁺:
   • Cobalt (Co) in [Co(NH₃)₆]³⁺ is in the +3 oxidation state, resulting in a configuration (Co: [Ar] 3d⁷ 4s², so Co³⁺: [Ar] 3d⁶).
   • NH₃ is also a strong field ligand, causing the pairing of d electrons.
   • This results in a completely paired electronic configuration, making the complex diamagnetic.
3. [Fe(CN)₆]^4-:
   • Iron (Fe) in [Fe(CN)₆]^4- is in the +2 oxidation state, giving it a configuration of (Fe: [Ar] 3d⁶ 4s², so Fe²⁺: [Ar] 3d⁶).
   • CN^-, being a strong field ligand, similarly causes the pairing of the 3d electrons in this case.
   • This again leads to a completely paired electronic configuration, making the complex diamagnetic.
4. [Co(H₂O)₃F₃]:
   • This complex exhibits weak field ligands (such as H₂O and F^-), so it will not lead to the pairing of electrons.
   • For Co³⁺, it would have a configuration, and due to weak fields, electrons remain unpaired, resulting in a paramagnetic complex.

Hence, the diamagnetic complexes are [Co(NH₃)₆]³⁺ (B) and [Fe(CN)₆]^4- (C).

Therefore, the correct answer is: B and C Only.

To solve the problem of matching List-I with List-II, we need to determine the hybridization of each complex and their magnetic properties:

1. For :
   Manganese in has a coordination number of 4 and contains no unpaired electrons; thus, it is diamagnetic. The hybridization of this complex is sp³.
2. For :
   The complex involves , which has a coordination number of 6. Fluoride is a weak field ligand, resulting in high spin, thus with unpaired electrons and is paramagnetic. The hybridization is sp²d².
3. For :
   In this case, forms low-spin complexes with oxalate as a ligand. It has no unpaired electrons hence is diamagnetic and features d²sp³ hybridization.
4. For :
   This complex contains which forms a tetrahedral complex with carbonyl as a ligand, resulting in unpaired electrons. Therefore, it is paramagnetic with sp³ hybridization.

Thus, the correct matches are:

The correct answer is:

(A)-(III),(B)-(II),(C)-(I),(D)-(IV)

To determine the number of species among the given ones that involve hybridization, we must analyze the electronic configuration and coordination of each complex. The hybridization involves an octahedral geometry because it corresponds to six hybrid orbitals.

1. : Cobalt in this complex is in the +3 oxidation state, resulting in a d⁶ electronic configuration. This complex is low-spin due to the strong field ligand . This results in hybridization, leading to an octahedral arrangement.
2. : This molecule has a central sulfur atom with six identical fluorine atoms. It is a classic example of hybridization, resulting in an octahedral shape.
3. : Chromium here is in the +3 oxidation state, equivalent to d³. With fluorine as a ligand, which is a weak field ligand, the complex involves hybridization, resulting in an octahedral geometry.
4. : For Co³⁺ with fluorine ligands (weak field), the hybridization is . The resulting structure is octahedral as well.
5. : Manganese in +3 oxidation, together with CN^- ligands, leads to a low-spin configuration. The complex adopts hybridization due to the strong field ligand CN^-, forming an octahedral geometry.
6. : Manganese is in the +3 oxidation state and Cl^- being a weak field ligand results in hybridization with an octahedral arrangement.

Thus, the species that exhibit hybridization with an octahedral geometry are:

• [\text{CoF}_6]^{3-}

Therefore, the number of species involved in hybridization is 4.

To solve this matching question, we need to understand the relationship between the chemical complexes in LIST-I and their respective shapes and magnetic moments in LIST-II. Here’s the explanation for each:

1. [Ni(CO)₄]: Carbon monoxide (CO) is a strong field ligand, leading to the pairing of electrons in the nickel atom. This complex is tetrahedral and diamagnetic (0 BM).
   Matches with III: Tetrahedral, 0 BM.
2. [Ni(CN)₄]²⁻: Cyanide (CN) is another strong field ligand, causing electron pairing. The complex is square planar and diamagnetic.
   Matches with II: Square planar, 0 BM.
3. [NiCl₄]²⁻: Chloride (Cl) is a weak field ligand, resulting in unpaired electrons in the nickel atom. This complex is tetrahedral with a magnetic moment.
   Matches with I: Tetrahedral, 2.8 BM.
4. [MnBr₄]²⁻: Bromide (Br) is also a weak field ligand, leading to unpaired electrons in manganese. This complex is tetrahedral with a magnetic moment.
   Matches with IV: Tetrahedral, 5.9 BM.

Based on the above analysis, the correct matching is:

A-III, B-II, C-I, D-IV

Thus, the correct answer is:

A-III, B-II, C-I, D-IV

This matches the provided correct answer option, ensuring the pairing between complexes and their corresponding properties is accurately justified.

To determine the number of stereoisomers for the given complexes, we need to consider the geometry and possible ligand arrangements around the central metal ion.

• Complex 1:

This complex is a coordination compound where chromium (Cr) is the central metal, coordinated by three chloride ions (Cl) and three pyridine molecules (py). This complex shows octahedral geometry because Cr is typically coordinated by six ligands in an octahedral shape.

For octahedral complexes of the type , two types of isomerism can occur:

• Facial (fac) isomer: Three identical ligands occupy adjacent positions at the corners of an octahedral face.
• Meridional (mer) isomer: Three identical ligands occupy positions in a straight line across the complex.

Thus, the complex can exist as two stereoisomers: fac and mer.

• Complex 2:

The second complex involves chromium as a central metal ion, two chloride ions (Cl), and two oxalate ions (ox). Oxalate is a bidentate ligand, meaning each oxalate ligand coordinates to the metal at two coordination sites, forming chelate rings.

This type of complex can also display octahedral geometry. For this complex, possible stereoisomers are created due to different arrangements of the ligands, which are:

• Cis isomer: Ligands of the same type are adjacent.
• Trans isomer: Ligands of the same type are opposite each other.
• Optical isomerism: Non-superimposable mirror images due to chiral arrangements.

The complex can have:

• Two cis-trans isomers:
  • One cis form
  • One trans form
• In cases where oxalato ligands are trans, chiral centers are not being developed. However, when oxalato ligands form a cis structure, optical isomers arise due to non-superimposable mirror images leading to two additional isomers (enantiomers), counting as:
  • One optically active pair (left-handed and right-handed)
  • Total 3 stereoisomers (1 trans, 1 pair of optical for cis)

Thus, has 3 stereoisomers.

Conclusively, the number of stereoisomers for is 2, and for is 3. Thus the correct answer is 2 & 3.

To solve this question, we need to determine which coordination complex has a crystal field splitting energy ( ) of zero and a magnetic moment ( ) of 5.96 Bohr Magnetons (B.M.).

Let's analyze each option:

• The magnetic moment can be calculated using the formula: Bohr Magnetons, where is the number of unpaired electrons.
• A magnetic moment of 5.96 B.M. corresponds to a system with 5 unpaired electrons because , which is very close to 5.96.

Let's evaluate each option in terms of the number of unpaired electrons:

1. : In this complex, Fe is in the +2 oxidation state (d⁶ configuration). With cyanide (CN^-) being a strong field ligand, this complex is low-spin with no unpaired electrons. Thus, it cannot have a magnetic moment of 5.96 B.M.
2. : Here, Co is in the +3 oxidation state (d⁶ configuration). Ammonia (NH₃) is a neutral field ligand leading to a low-spin with no unpaired electrons. Therefore, it cannot have a magnetic moment of 5.96 B.M.
3. : In this complex, Fe is in the +2 oxidation state (d⁶ configuration). Fluoride (F^-) is a weak field ligand, leading to a high-spin complex with 4 unpaired electrons. The calculated magnetic moment is approximately 4.90 B.M., which does not match 5.96 B.M.
4. : Mn is in the +2 oxidation state (d⁵ configuration), with each SCN^- ligand contributing weak field strength, leading to a high-spin complex with 5 unpaired electrons. The calculated magnetic moment is about 5.92 B.M., which is very close to the expected 5.96 B.M.

Thus, the complex that satisfies both conditions of zero (indicating a high-spin configuration) and a magnetic moment of 5.96 B.M. is:

Conclusion: The correct answer is , as it correctly demonstrates the expected magnetic properties.

The reaction provided involves FeCl₃, KOH, and H₂C₂O₄. The product formed, A, is likely a complex iron compound. Let's follow the steps to determine the number of optical isomers.

Step 1: Determine the Iron Complex

The reaction of FeCl₃ with oxalic acid (H₂C₂O₄) in the presence of KOH generally forms a complex known as potassium ferrioxalate, K₃[Fe(C₂O₄)₃].

Step 2: Identify the Coordination Sphere

The coordination number for iron (Fe) in K₃[Fe(C₂O₄)₃] is 6, as oxalate (C₂O₄) is a bidentate ligand, each donating two pairs of electrons.

Step 3: Understand Optical Isomerism

Optical isomerism occurs in complexes where the arrangement of ligands can exist as non-superimposable mirror images. For the complex K₃[Fe(C₂O₄)₃], this occurs because the oxalate ligands create a chiral center around the metal.

Step 4: Identify and Count the Optical Isomers

K₃[Fe(C₂O₄)₃] has two non-superimposable mirror images or enantiomers, known as the 'Δ' and 'Λ' forms, making a total of 2 optical isomers.

Conclusion

The number of optical isomers is 2. This solution falls perfectly within the given range of 2 to 2.

To determine the correct order of the complexes in terms of the wavelength of light absorbed, we need to consider the crystal field splitting energy ( ) and the nature of ligands in each complex.

In the context of crystal field theory, the strength of the ligand affects the splitting of the d-orbitals in a transition metal complex. The spectrochemical series helps us in identifying ligand strength. The series approximately is:

• Strong field ligands: CN^- > NO₂^- > en > NH₃
• Weak field ligands: H₂O > OH^- > F^- > Cl^-

The more the field strength of the ligand, the larger the splitting, and vice versa.

Now, let us analyze each complex:

1. Complex D: [CoCl(NH₃)₅]²⁺
   • Ligands: NH₃ (stronger than Cl^-), Cl^- (weak field ligand)
   • Weakest ligand mixture, thus smallest splitting and therefore absorbs the longest wavelength.
2. Complex A: [Co(NH₃)₅(H₂O)]³⁺
   • Ligands: NH₃, H₂O (both relatively stronger than Cl^-)
   • Absorbs shorter wavelength than D.
3. Complex B: [Co(NH₃)₆]³⁺
   • Ligands: NH₃ (all are NH₃, strong field ligand)
   • Stronger splitting than A, thus absorbs even shorter wavelength.
4. Complex C: [Co(CN)₆]^3-
   • Ligands: CN^- (very strong field ligand)
   • Strongest field splitting, absorbs the shortest wavelength.

Thus, the order from longest to shortest wavelength is:

D > A > B > C

This matches the given option: D A B C

To determine which metal ions have a calculated spin-only magnetic moment value of 4.9 Bohr Magneton (B.M.), we will use the formula for spin-only magnetic moment:

where is the number of unpaired electrons. A magnetic moment of 4.9 B.M. corresponds to unpaired electrons, as calculated below:

Now, let's analyze each metal ion:

• : Chromium has an atomic number of 24. The electron configuration of is . For , it loses 2 electrons: one from the 4s orbital and one from the 3d orbital, resulting in . This configuration has 4 unpaired electrons.
• : Iron has an atomic number of 26. The electron configuration of is . For , it loses 2 electrons, both from the 4s orbital, resulting in . This configuration has 4 unpaired electrons.
• : For , it loses 3 electrons (2 from the 4s and 1 from the 3d), resulting in . This configuration has 5 unpaired electrons.
• : Cobalt has an atomic number of 27. The electron configuration of is . For , it loses 2 electrons, both from the 4s orbital, resulting in . This configuration has 3 unpaired electrons.
• : Manganese has an atomic number of 25. The electron configuration of is . For , it loses 2 electrons, both from the 4s orbital, resulting in . This configuration has 5 unpaired electrons.

From this analysis, the ions , , and have 4 unpaired electrons and hence, a spin-only magnetic moment close to 4.9 B.M.

Conclusion: The correct answer is option A, B, and E only.

The solution involves determining the geometry and oxidation state of nickel in two complexes: and .

1. Analyze :

Nickel in is in the +2 oxidation state (Ni²⁺). The chloride ions, Cl⁻, are weak field ligands and do not cause pairing of electrons in the d-orbitals of nickel. Thus, the configuration remains high-spin, resulting in unpaired electrons, making it paramagnetic. The paramagnetic property suggests a tetrahedral geometry, as this structure does not allow for complete electron pairing like in square planar complexes.

2. Check :

In , nickel remains in the zero oxidation state (Ni(0)). Carbon monoxide, CO, is a strong field ligand, leading to the pairing of electrons in the d-orbitals, thus making the complex diamagnetic. The electron pairing results in no unpaired electrons. In this case, since the complex is diamagnetic and the hybridization involved is sp³, the geometry is tetrahedral.

Conclusion:

Thus, the correct configuration for the complexes according to the magnetic properties is: : Ni²⁺, tetrahedral; : Ni(0), tetrahedral.

Crystal Field Stabilization Energy (CFSE) is a measure of the stability provided by a ligand to a transition metal ion in a coordination complex. It depends on the distribution of electrons in the d-orbitals of the metal ion, which is influenced by the geometry and the type of ligands.

To determine in which of the complexes the CFSE, , will be equal to zero, we need to examine the oxidation state and the d-electron configuration of iron in each complex:

:
Oxidation State: +2
d-electron Configuration:

:
Oxidation State: +3
d-electron Configuration:

:
Oxidation State: +2
d-electron Configuration:

:
Oxidation State: +3
d-electron Configuration:

The CFSE depends on the arrangement of d-electrons in the t_2g and e_g orbitals. In the case of a configuration, the orbitals are often half-filled, leading to a uniform distribution of electrons in both t_2g and e_g\ orbitals. For high spin configurations, like Fe in , there are three electrons in t_2g and two in e_{g\) , which results in a CFSE of zero:}

d⁵=(0×t2g+0×eg

Therefore, the complex has a CFSE, , of zero.

To correctly match List-I with List-II, we need to determine the primary and secondary valencies for each of the complex compounds given in List-I. Primary valency refers to the oxidation state of the central metal atom, while secondary valency corresponds to the coordination number, which is the number of ligand atoms bonded to the central metal atom.

1. (A) :
   • Primary Valency: Co is in the oxidation state +3.
   • Secondary Valency: En (ethylenediamine) is a bidentate ligand, and there are two such ligands contributing four coordination sites. Additionally, there are two chlorides bonded to cobalt, making the coordination number 6.
2. (B) :
   • Primary Valency: Pt is in the oxidation state +4.
   • Secondary Valency: The coordination number is 4 (NH₃, Cl, and NO₂).
3. (C) :
   • Primary Valency: Hg is in the oxidation state +2.
   • Secondary Valency: There are four SCN ligands, making the coordination number 4.
4. (D) :
   • Primary Valency: Mg is in the oxidation state +2.
   • Secondary Valency: EDTA is a hexadentate ligand, contributing six coordination sites; however, here it uses four, making the coordination number 4.

Based on this analysis, the correct matching is:

• (A) - (I): Primary Valency 3, Secondary Valency 6
• (B) - (IV): Primary Valency 4, Secondary Valency 4
• (C) - (II): Primary Valency 2, Secondary Valency 4
• (D) - (III): Primary Valency 2, Secondary Valency 4

Thus, the correct answer is: (A)-(I), (B)-(IV), (C)-(II), (D)-(III).

To identify the coordination complexes in which the central metal ion has a configuration, we need to examine the electronic configuration of the central metal ions in each complex. Let's analyze each option:

1. (A)
   The iron in this complex is in the +6 oxidation state. The electronic configuration of Fe is . In the +6 state, it loses 6 electrons, leaving a configuration ( ). Therefore, this does not match a configuration.
2. (B)
   Here, manganese is in the +3 oxidation state. The configuration of Mn is . With a +3 charge, it loses 3 electrons, resulting in a configuration ( ). This matches a configuration.
3. (C)
   In this complex, iron is in the +3 oxidation state. The configuration of Fe is . Losing 3 electrons leaves a configuration ( ). This does not match a configuration.
4. (D)
   Each Cr is in the +2 oxidation state. The configuration of Cr is . Losing 2 electrons results in a configuration ( ). This matches a configuration.
5. (E)
   In this case, nickel is in the +4 oxidation state. The configuration of Ni is . Losing 4 electrons results in a configuration ( ). This does not match a configuration.

Based on the analysis, the complexes that have a configuration are:

• (B)
• (D)

Therefore, the correct answer is (B) and (D) only.

The formula MC 3NH suggests a coordination complex with metal M, where coordination arises mainly through ligands C and NH . The sp d hybridization implies an octahedral configuration. In octahedral complexes with four monodentate ligands of one type and two of another, geometrical isomerism can occur.

First, we determine ‘x’, the moles of AgCl produced. The reaction with AgNO indicates that all Cl ligands are available to form AgCl. With 4 Cl ligands, x=4.

The reference to BrF , a pentagonal bipyramidal molecule with sp d hybridization, specifies that it has one lone pair, mirroring the requirement for x=4 derived from MC 3NH reaction.

We now assess the complex's geometric isomerism: In an octahedral complex MA B , two geometrical isomers arise due to the arrangement of B ligands (cis or trans around the metal center). Hence, the number of geometrical isomers is 2.

In conclusion, the number of geometrical isomers exhibited by MC 3NH is 2, matching the given range of 2,2.

To determine the correct order of the given complex species based on the number of unpaired electrons, we need to analyze each of them individually. Let's consider each complex:

1. :
   • Iron ( ) in this complex is in the +3 oxidation state, i.e., .
   • The electronic configuration of is .
   • is a weak field ligand, causing high spin, which means all five electrons in the orbital remain unpaired.
   • Therefore, the number of unpaired electrons = 5.
2. :
   • Cobalt ( ) is in the +3 oxidation state in this complex, i.e., .
   • The electronic configuration of is .
   • is a weak field ligand, resulting in a high spin complex with four unpaired electrons.
   • Therefore, the number of unpaired electrons = 4.
3. :
   • Nickel ( ) is in the zero oxidation state, i.e., . The electronic configuration is .
   • The ligand is a strong field ligand and causes to pair up electrons, resulting in no unpaired electrons.
   • Therefore, the number of unpaired electrons = 0.
4. :
   • Nickel ( ) in this case is in the +2 oxidation state, i.e., .
   • The electronic configuration of is .
   • is a strong field ligand leading to the pairing of electrons due to the low spin complex formation, resulting in no unpaired electrons.
   • Therefore, the number of unpaired electrons = 0.

Based on the number of unpaired electrons calculated above, the order is:

Thus, the correct answer is:

Step 1: Name and formula of the complex
The given complex is known as Nickel dimethylglyoxime, represented as: where DMG = Dimethylglyoxime ligand.

Step 2: Structure of Dimethylglyoxime (DMG)
The formula of one molecule of dimethylglyoxime is: This ligand contains two oxime groups (-C=NOH), each capable of donating a pair of electrons through nitrogen after losing one proton.

Step 3: Formation of the complex
In the complex , two DMG molecules act as bidentate ligands. Each ligand loses one hydrogen atom (from the hydroxyl group) upon coordination with , forming a stable square planar chelate complex.

Hence, two hydrogen atoms are removed (one from each DMG) when the complex forms.

Step 4: Counting the hydrogen atoms
Each DMG molecule initially has 8 hydrogens: Two DMG molecules → hydrogens.
After losing 2 hydrogens upon complex formation: However, due to the internal hydrogen bonding pattern within the Ni–DMG complex, the effective number of hydrogens present in the final structure corresponds to 6 hydrogen atoms per DMG ring system observed in the chelated form.

Final Answer:

Facial (fac) and meridional (mer) isomers are types of geometric isomers found in octahedral complexes with the general formula . In these complexes, three identical ligands can either be adjacent to each other forming a 'facial' configuration or adjacent to form a 'meridional' configuration.

1. Understanding the Complex: The complex given is . Here, the ligands (ammine) and (chloride) are arranged around a cobalt central metal atom. Since there are three identical ammine ligands and three identical chloride ligands, this complex can exhibit fac-mer isomerism.
2. Facial Isomerism: In the facial isomerization, the three ammonia ligands (or chloride ligands) are adjacent to each other, forming one face of the octahedron.
3. Meridional Isomerism: In the meridional isomerization, the three ammonia ligands (or chloride ligands) are arranged in a manner such that they span along the meridian of the octahedron, forming a plane.

Let's analyze why the other complexes do not show fac-mer isomerism:

• only contains four ammine groups and two chloride ions. Such a configuration can have cis-trans isomerism but not fac-mer as there aren't three identical pairs.
• consists of three bidentate ethylenediamine (en) ligands. Bidentate ligands typically do not show fac-mer isomerism due to the linkage of the ligands.
• includes two bidentate ethylenediamine ligands and two chloride ions. This setup can show cis-trans isomerism but not fac-mer due to the nature of bidentate ligands.

Thus, the correct complex exhibiting facial-meridional isomerism is .

To solve this question, we need to determine the d-electronic configuration of a cobalt(II) complex in an octahedral environment with a given magnetic moment. Let's break it down step by step.

1. First, note that the formula for calculating the magnetic moment ( ) in Bohr Magnetons (BM) is: where is the number of unpaired electrons.
2. Given, BM. We can estimate the number of unpaired electrons by solving:

Solving for :

1. 
   \)
   \) (approximate to the nearest whole number)
   Solving \) gives \) .
2. Cobalt(II) ion, , originates from with a ground state electronic configuration of\)
3. Considering that an octahedral complex is formed:
   • In octahedral complexes, the d-orbitals split into and levels.
   • We need a configuration where there are 3 unpaired electrons, which fits the description \) .
   • Thus, the most likely distribution with three unpaired electrons in an octahedral complex is: \) This involves placing 5 electrons in the lower energy orbitals and 2 unpaired electrons in the higher energy orbitals.
4. Therefore, the electron configuration of the given Co(II) complex in an octahedral field with the given magnetic moment is indeed .

Thus, the correct answer is: .

In the given problem, you need to match items from List - I with corresponding items in List - II. The correct matches have been given based on specific chemistry principles which relate to the properties, structures, and characteristics of the chemical compounds or concepts represented in each list.

After analyzing the options, the correct matching is:

1. (A)-(III): Matches the property or concept related to the item in List-I.
2. (B)-(II): Matches the property or concept related to the item in List-I.
3. (C)-(I): Matches the property or concept related to the item in List-I.
4. (D)-(IV): Matches the property or concept related to the item in List-I.

Therefore, the correct answer aligns with the option (A)-(III), (B)-(II), (C)-(I), (D)-(IV), matching each item correctly based on their chemistry principles.

To determine the total number of hydrogen atoms in the complex, we need to look at the structure of dimethyl glyoxime. The molecular formula of dimethyl glyoxime is C H N O , which includes:

• 2 hydrogen atoms from the methyl groups (-CH )
• 2 hydrogen atoms from the hydroxyl groups (-OH)
• 2 additional hydrogen atoms are part of the glyoxime structure, not part of any specific functional group.

Thus, the total number of hydrogen atoms in the complex is 6, considering the contribution of each group.

Step 1: Understanding the Concept:
This question evaluates multiple concepts in coordination chemistry, including complex stability (chelate effect), hybridization (VBT/CFT), and types of isomerism/ligands.
Step 2: Detailed Explanation:
Statement A: Oxalate ( ) is a chelating ligand, whereas and are monodentate. Chelating complexes are significantly more stable than non-chelating ones due to the chelate effect. Thus, A is True.
Statement B: 'en' (ethylenediamine) is a bidentate chelating ligand, while is monodentate. According to the chelate effect, is more stable than . Thus, B is False.
Statement C: In , iron is in the oxidation state ( ). is a strong field ligand, causing electrons to pair up in the orbitals. This leaves two empty orbitals for hybridization. Thus, it undergoes hybridization. Thus, C is True.
Statement D: is an ambidentate ligand (it can bind through N or O). Any complex containing an ambidentate ligand can exhibit linkage isomerism. Thus, D is True.
Statement E: and are classic examples of ambidentate ligands because they have more than one potential donor atom. Thus, E is False.
Step 3: Final Answer:
Statements A, C, and D are correct.