MCH-011 Solved Assignment January 2026 | INORGANIC CHEMISTRY I | IGNOU MSCCHEM

Q: Answer the following parts:

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Q1.. Answer the following parts:

(a)) Explain the trends of ionization energy down the groups of the elements in the periodic table. How are the higher values of ionisation energies of the 5d transition elements justified? (300 words)
(b)) What is the trend of the stability of the +6 state for Group 6 elements like tungsten, molybdenum and chromium? Which other group show a similar trend? Justify your answer. (300 words)

Key Points:

Ionization energy generally decreases down a group due to increased atomic radius and shielding.
5d transition elements have anomalously high ionization energies due to lanthanide contraction.
Relativistic effects in heavy elements contract s/p orbitals, increasing ionization energy for 5d elements.
Stability of +6 oxidation state for Group 6 (Cr, Mo, W) increases down the group.

Answer: Ionization energy, a fundamental periodic property, generally decreases down a group due to increasing atomic size and shielding effects. However, 5d transition elements exhibit anomalously high ionization energies, often exceeding those of their 4d counterparts, a phenomenon attributed to lanthanide contraction and relativistic effects. Similarly, the stability of higher oxidation states, such as the +6 state for Group 6 elements (Cr, Mo, W), typically increases down the group, a trend also obs...

Q2.. Answer the following parts:

(a)) Explain the structure of the pentafluorotellurate(IV) anion with the help of VSEPR theory. What is the point group of the symmetry in it? (300 words)
(b)) With suitable illustration explain the theory of Dewar for a bonding in phosphazenes. What is phospham and how it is synthesized? Give suitable reaction. (300 words)

Key Points:

Pentafluorotellurate(IV) anion [TeF₅]⁻ has a square pyramidal geometry.
VSEPR model for [TeF₅]⁻ is AX₅E₁, with one lone pair on Te.
The point group of square pyramidal [TeF₅]⁻ is C₄v.
Dewar's theory explains phosphazene bonding via pπ-dπ interactions.

Answer: The following response addresses the structure and symmetry of the pentafluorotellurate(IV) anion using VSEPR theory, and then explains Dewar's theory for bonding in phosphazenes and the synthesis of phospham, drawing upon general inorganic chemistry principles consistent with the MCH-011 course material. Each sub-question is answered comprehensively within its specified word limit, employing academic terminology and clear, concise explanations.

Q3.. Answer the following parts:

(a)) Give the classification of nonclassically bonded organometallic compounds. (300 words)
(b)) With any two suitable examples show how organanometallic complexes of different classes may be stabilized by bulky ligands. (300 words)

Key Points:

Nonclassically bonded organometallics involve π-electron system donation (alkenes, alkynes, Cp, arenes).
Classes include π-arene, π-cyclopentadienyl, π-alkene/alkyne, and π-allyl/diene complexes.
Bulky ligands provide steric protection, preventing decomposition and aggregation.
Bulky ligands stabilize low-coordinate, reactive metal centers by kinetic hindrance.

Answer: Nonclassically bonded organometallic compounds are characterized by bonding modes that extend beyond simple two-electron, two-center sigma (σ) bonds, often involving the delocalized π-electron systems of organic ligands. These complexes play a pivotal role in catalysis and materials science, exhibiting diverse structures and reactivities. Stabilization of such compounds, especially reactive ones, can be effectively achieved through the strategic use of bulky ligands. These ligands provide steric...

Q4.. Answer the following parts:

(a)) With suitable illustration explain the bent tin-tin double bond in (SnR2)2; where R = (Me3Si)2CH. (300 words)
(b)) With the help of suitable examples and structures, explain why some of the 16-electron organometallic complexes are stable. For a d8 metal ion which form will be more stable, 16-electron square planar complex (d8 and 8 ligand electrons) or a 18-electron complex (d8 and 10 ligand electrons)? Justify your answer. (300 words)

Key Points:

Bent Sn-Sn bond in distannenes like (SnR2)2 contrasts classical C=C due to heavier element bonding.
(SnR2)2 exhibits a trans-bent geometry and significant singlet diradical or 'double bond, no-bond' character.
Bulky R groups ((Me3Si)2CH) kinetically stabilize distannenes, preventing undesirable reactions.
16-electron square planar d⁸ complexes are highly stable due to their favorable d-orbital splitting pattern.

Answer: ### (a) Bent tin-tin double bond in (SnR2)2 The presence of a bent tin-tin double bond in compounds like (SnR2)2, where R = (Me3Si)2CH, is a fascinating aspect of main group element chemistry. Unlike the planar, sp² hybridized carbon-carbon double bond, heavier Group 14 elements such as tin (Sn) exhibit a unique structural motif for their dimetallene counterparts. This deviation stems from the different bonding characteristics of heavier p-block elements compared to their lighter congeners. Th...

Q5.. Answer the following parts:

(a)) With suitable diagrams explain the electronic structure of NO and its binding to a metal fragment on the covalent model. (300 words)
(b)) Compare the characteristics of electron-rich phosphines with those of electron poor phosphines regarding their o donor and π acceptor capacities. What happens to these characteristics when the average electronegativity of the three groups attached to P increases? (300 words)

Key Points:

NO is a 15-electron paramagnetic molecule with one unpaired electron in a π* antibonding orbital.
NO binds as a linear 2e⁻ donor (NO⁺-like) or a bent 3e⁻ donor (neutral NO-like).
Electron-rich phosphines (e.g., PMe₃) are strong σ donors and weak π acceptors.
Electron-poor phosphines (e.g., PF₃) are weak σ donors and strong π acceptors.

Answer:

Q6.. Answer the following parts:

(a)) Explain the π system of the allyl group with the help of molecular orbital theory. Illustrate your answer. (300 words)
(b)) Give the structure of titanocene with its IUPAC nomenclature. How is the Hund's rule obeyed in manganocene? (300 words)

Key Points:

Allyl π system involves three p-orbitals combining to form three molecular orbitals (ψ₁, ψ₂, ψ₃).
ψ₁ is bonding (zero nodes), ψ₂ is non-bonding (one node), and ψ₃ is anti-bonding (two nodes).
Electron filling: Allyl cation (2πe⁻ in ψ₁), radical (3πe⁻ in ψ₁, ψ₂), anion (4πe⁻ in ψ₁, ψ₂).
Titanocene: Bis(η⁵-cyclopentadienyl)titanium(II) for Cp₂Ti; Cp₂TiCl₂ is dichlorobis(η⁵-cyclopentadienyl)titanium(IV).

Answer: The following response provides a comprehensive explanation of the π system of the allyl group using molecular orbital theory and details the structure and IUPAC nomenclature of titanocene, along with how Hund's rule applies to manganocene, drawing upon concepts from inorganic chemistry as typically covered in MCH-011.

Q7.. Answer the following parts:

(a)) Give the reactions for preparation of decaborane(14) from B5H8 anion. Which is the only neutral dodecaborane and how is it prepared? Give suitable equation. (300 words)
(b)) Explain the structure of MoCl²⁻. Which are the two main classes of metal atom cluster compounds? (300 words)

Key Points:

Decaborane(14) (B₁₀H₁₄) forms from B₅H₈⁻ via oxidative coupling with I₂.
B₁₂H₁₂ is the unstable neutral dodecaborane, derived from stable B₁₂H₁₂²⁻ dianion.
B₁₂H₁₂²⁻ is synthesized by reacting B₁₀H₁₄ with Et₃N and BH₄⁻, then protonated.
The [Mo₆Cl₈]⁴⁺ cluster has an Mo₆ octahedron with eight μ₃-bridging chloride atoms.

Answer: Inorganic chemistry, particularly the study of boranes and metal clusters, involves complex bonding and structural principles. Boranes are electron-deficient compounds that exhibit unique polyhedral structures, with their synthesis often involving controlled reactions of smaller borane units. Metal atom clusters, on the other hand, are characterized by direct metal-metal bonding, leading to diverse architectures and distinct classifications based on their electronic properties and oxidation stat...

Q8.. Answer the following parts:

(a)) Give the electronic configuration of d⁸ and d⁷complexes in both weak and strong fields. Also calculate their CFSE. (300 words)
(b)) Explain the structure of normal spinels with a suitable example. Give suitable illustration. (300 words)

Key Points:

d⁸ octahedral complexes always have t₂g⁶ eg² configuration, irrespective of field strength.
CFSE for d⁸ octahedral is -1.2Δ₀ with 0P (no additional pairing energy).
d⁷ weak field (high spin) octahedral configuration is t₂g⁵ eg² with 3 unpaired electrons.
CFSE for d⁷ weak field is -0.8Δ₀ with 0P.

Answer: Crystal Field Theory (CFT) explains the electronic configurations and stability of transition metal complexes by considering the interaction between metal d-orbitals and ligand electrons. This interaction causes the d-orbitals to split into different energy levels, with the energy difference denoted as Δ₀ for octahedral complexes. The arrangement of electrons in these split orbitals, governed by the strength of the ligand field and the pairing energy, determines the Crystal Field Stabilization E...

Q9.. Answer the following parts:

(a)) Give suitable equations and explain the orbital and spin contributions towards paramagnetism in a complex. (300 words)
(b)) What is the expected magnetic moment of the complex MnF3? What sort of stereochemistry is seen in it? What is the electron configuration in it based on crystal field theory. (300 words)

Key Points:

Paramagnetism arises from unpaired electron spins and orbital angular momentum.
Spin-only magnetic moment: μ_s = √(n(n+2)) BM, 'n' is unpaired electrons.
Orbital angular momentum contribution is often 'quenched' in first-row transition metal complexes.
MnF₃ contains Mn³⁺ (d⁴ ion) and weak-field F⁻ ligands.

Answer: Paramagnetism in coordination complexes arises primarily from the presence of unpaired electrons, contributing both spin and, in some cases, orbital angular momentum to the overall magnetic moment. The spin contribution is generally dominant and calculable using the spin-only formula, μ_s = √(n(n+2)) BM, where 'n' is the number of unpaired electrons. Orbital angular momentum also generates a magnetic moment, but its contribution is often 'quenched' in many complexes due to the ligand field, espe...

Q10.. Answer the following parts:

(a)) Write the Russell-Saunders terms symbols for states with the angular momentum quantum numbers (L,S) (a) (1, 1/2) (b) (2,1) (c) (0, 3/2) (d) (3, 0). (300 words)
(b)) Explain how the R- and S- labels are used for enantiomers. Explain the λ and δ conformations of methyl axial and methyl equatorial respectively. Give suitable illustrations for all of these. (300 words)

Key Points:

Russell-Saunders term symbol: ²ˢ⁺¹L, representing spin multiplicity (2S+1) and total orbital angular momentum (L as S, P, D, F).
R/S enantiomer labels: CIP rules prioritize groups; clockwise path 1-2-3 is R, counter-clockwise is S (with lowest priority group away).
λ conformation: Left-handed helical twist of a five-membered chelate ring, tracing a counter-clockwise path.
δ conformation: Right-handed helical twist of a five-membered chelate ring, tracing a clockwise path.

Answer: This response comprehensively addresses the requested aspects of Russell-Saunders term symbols, R- and S- configurations for enantiomers, and the λ/δ and axial/equatorial conformations, drawing upon fundamental principles of inorganic and organic chemistry. Each sub-question is treated distinctly, adhering to specified word limits and formatting requirements.

MCH-011 Solved Assignment — January 2026 / July 2025

INORGANIC CHEMISTRY I | IGNOU Master of Science in Chemistry Programme (MSCCHEM)

MCH-011—January 2026 / July 2025

INORGANIC CHEMISTRY I

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