Summer 2025 Paper VI

Based on the examination paper provided, here are detailed answers for the questions. These solutions are structured according to the marks allocated, providing the depth required for an M.Sc. level examination.

Question 1

1. a) Explain electronic spectra and structure of tetrahalocobalt (II) complex. (8 Marks)

Structure: Tetrahalocobalt(II) complexes, such as $[CoCl_4]^{2-}$, typically adopt a tetrahedral ($T_d$) geometry.

The Cobalt ion is in the $+2$ oxidation state ($d^7$ configuration).

Ligands like $Cl^-$, $Br^-$, or $I^-$ are weak field ligands and relatively large, which favors the less sterically hindered tetrahedral geometry over octahedral.

Electronic Configuration:

In a tetrahedral field, the d-orbitals split into two sets: $e$ (lower energy) and $t_2$ (higher energy).

Configuration ($d^7$): $e^4 t_2^3$. This is a high-spin complex.

Electronic Spectra (Transitions):

The ground state term for a $d^7$ tetrahedral ion is ${}^4A_2(F)$.

Three spin-allowed transitions are expected to excited quartet states:

$\nu_1$: ${}^4A_2(F) \rightarrow {}^4T_2(F)$ (Usually in the near-IR region, often not observed in visible).

$\nu_2$: ${}^4A_2(F) \rightarrow {}^4T_1(F)$ (Visible region, often appears as a shoulder due to spin-orbit coupling).

$\nu_3$: ${}^4A_2(F) \rightarrow {}^4T_1(P)$ (Visible region, strong intensity).

Intensity & Color:

These complexes are known for their intense blue color (e.g., $[CoCl_4]^{2-}$).

Why intense? In tetrahedral symmetry, there is no center of inversion ($i$). Therefore, the Laporte selection rule is relaxed due to $d-p$ orbital mixing. This makes the $d-d$ transitions in tetrahedral complexes much more intense ($\epsilon \approx 100-1000$ $L mol^{-1} cm^{-1}$) than in octahedral complexes (which are centrosymmetric).

1. b) Explain the electronic spectra of $d^3$ and $d^7$ metal ion in weak octahedral field with suitable example using Orgel diagram. (8 Marks)

Theory: Orgel diagrams describe the splitting of electronic terms in weak ligand fields. They are useful for high-spin octahedral and tetrahedral complexes.

The $d^3$ Case (e.g., $V^{2+}$, $Cr^{3+}$):

Ground Term: Free ion ground term is ${}^4F$. An excited term ${}^4P$ also exists.

Splitting: In an octahedral field ($O_h$), the ${}^4F$ term splits into ${}^4A_{2g}$, ${}^4T_{2g}$, and ${}^4T_{1g}$. The ${}^4P$ term transforms into ${}^4T_{1g}(P)$.

Transitions:

${}^4A_{2g} \rightarrow {}^4T_{2g}$

${}^4A_{2g} \rightarrow {}^4T_{1g}(F)$

${}^4A_{2g} \rightarrow {}^4T_{1g}(P)$

The $d^7$ Case (High Spin, e.g., $Co^{2+}$):

Hole Formalism: $d^7$ can be treated as a "3-hole" system ($d^{10-3}$), meaning its diagram is the inverse of $d^3$.

Splitting: The ground state is ${}^4T_{1g}(F)$.

Transitions:

${}^4T_{1g}(F) \rightarrow {}^4T_{2g}(F)$

${}^4T_{1g}(F) \rightarrow {}^4A_{2g}(F)$

${}^4T_{1g}(F) \rightarrow {}^4T_{1g}(P)$

Combined Orgel Diagram:

The left side of the diagram typically represents $d^3 (O_h) / d^7 (T_d)$.

The right side represents $d^7 (O_h) / d^3 (T_d)$.

Visual: The lines for $T_{1g}$ and $T_{2g}$ cross, but the $A_{2g}$ line is linear. The non-crossing rule applies to the two $T_{1g}$ states (one from F, one from P), causing curvature.

OR (Alternative Questions for Q1)

1. c) Explain abnormal magnetic properties in octahedral complex. (4 Marks)

Abnormal magnetic properties (deviations from the spin-only formula $\mu_s = \sqrt{n(n+2)}$) arise due to:

Orbital Contribution: If the ground state has orbital degeneracy (e.g., $t_{2g}$ is unsymmetrically occupied), the orbital angular momentum ($L$) contributes to the magnetic moment, making $\mu_{obs} > \mu_{spin-only}$. (Common in high spin $d^1, d^2, d^6, d^7$ octahedral).

Spin-Orbit Coupling: The interaction between spin and orbital angular momentum can raise or lower the magnetic moment depending on the shell filling (less than half-filled vs. more than half-filled).

Temperature Dependence: Magnetic susceptibility ($\chi_m$) may deviate from the Curie Law due to antiferromagnetic interactions (in clusters) or thermal population of excited states.

1. d) Define charge transfer spectra with suitable example. (4 Marks)

Definition: Electronic transitions that involve the transfer of an electron between orbitals that are predominantly centered on different atoms (e.g., Ligand to Metal, or Metal to Ligand). Unlike $d-d$ transitions, these are Laporte allowed and very intense.

LMCT (Ligand-to-Metal Charge Transfer): Occurs when the metal is in a high oxidation state and the ligand has lone pairs.

Example: Permanganate ion $[MnO_4]^-$. The intense purple color is due to an electron moving from an oxygen $p$-orbital to an empty manganese $d$-orbital ($O \rightarrow Mn^{+7}$).

MLCT (Metal-to-Ligand Charge Transfer): Occurs when the metal is low-valent and the ligand has low-lying empty $\pi^*$ orbitals.

Example: $[Ru(bipy)_3]^{2+}$.

1. e) Explain high spin and low spin crossover. (4 Marks)

Concept: For $d^4$ to $d^7$ octahedral complexes, the difference in energy between the high-spin (HS) and low-spin (LS) states depends on the crystal field splitting energy ($\Delta_o$) versus the pairing energy ($P$).

Crossover: When $\Delta_o \approx P$, the complex sits at the "crossover point" (visible on Tanabe-Sugano diagrams). External factors like temperature or pressure can switch the complex between HS and LS states.

Example: $[Fe(phen)_2(NCS)_2]$. At high temperatures, it is High Spin ($S=2$); at low temperatures, it transitions to Low Spin ($S=0$). This change is often accompanied by a dramatic color change (thermochromism).

1. f) Explain Racah parameter. (4 Marks)

Definition: Racah parameters ($A, B, C$) describe the inter-electronic repulsion within an atom or ion. In complex formation, we mostly focus on B.

Nephelauxetic Effect: When a metal ion forms a complex, the electron cloud expands (delocalizes) over the ligands. This reduces the repulsion between d-electrons compared to the free ion.

Parameter B:

$B_{free}$: Value in free ion.

$B_{complex}$: Value in complex.

The ratio $\beta = B_{complex} / B_{free}$ (Nephelauxetic ratio) is always $< 1$. It indicates the degree of covalency in the metal-ligand bond. Smaller $\beta$ means greater covalency.

Question 2

2. a) What are trans effect? Discuss theories of trans effect with suitable example. (8 Marks)

Definition: The trans effect is the ability of a ligand to direct the substitution of the ligand trans (opposite) to it in a square planar complex (typically Pt(II)).

Trans Effect Series: (Strong) $CN^- \approx CO \approx NO > H^- \approx PR_3 > CH_3^- > I^- > Br^- > Cl^- > NH_3 > H_2O$ (Weak).

Theories:

Polarization Theory (Grinberg): Focuses on the ground state. A strong trans-directing ligand (L) is highly polarizable. The metal induces a dipole in L, which in turn induces a dipole in the metal that repels the negative charge of the ligand trans to L (X). This weakens the M-X bond, making X easier to replace.

$\pi$-Bonding Theory (Chatt & Orgel): Focuses on the transition state. Strong $\pi$-acceptor ligands (like $CO, C_2H_4$) remove electron density from the metal via back-bonding. This stabilizes the trigonal bipyramidal intermediate by lowering the energy of the transition state, facilitating the entry of the incoming nucleophile at the site trans to the $\pi$-acceptor.

Example (Synthesis of Isomers):

Starting with $[PtCl_4]^{2-}$ and adding $NH_3$: Since $Cl^-$ has a stronger trans effect than $NH_3$, the second $NH_3$ will replace a $Cl$ cis to the first $NH_3$ (directed by remaining $Cl$). Result: Cis-platin.

Starting with $[Pt(NH_3)_4]^{2+}$ and adding $Cl^-$: The first $Cl^-$ enters. Since $Cl^-$ has a stronger trans effect than $NH_3$, the next $Cl^-$ replaces the $NH_3$ trans to the first $Cl^-$. Result: Trans-platin.

2. b) What are the types of electron transfer mechanism? Explain outer sphere mechanism. (8 Marks)

Types:

Outer Sphere Mechanism: Electron transfer occurs without a bridging ligand; coordination spheres remain intact.

Inner Sphere Mechanism: Electron transfer occurs via a bridging ligand shared between the oxidant and reductant.

Outer Sphere Mechanism (Detailed):

Process: The oxidant and reductant approach each other, form an outer-sphere complex (ion pair), transfer the electron, and then separate. No bonds are broken or formed during the transfer step.

Requirements:

Franck-Condon Principle: Electron transfer is much faster ($10^{-15}$ s) than nuclear motion ($10^{-13}$ s). Therefore, the nuclear configurations (bond lengths) of both reactants must reorganize to be identical before the electron can jump.

Tunneling: The electron "tunnels" through the energy barrier created by the solvent and ligand shells.

Marcus Theory: Relates the rate of electron transfer to the driving force ($\Delta G^\circ$) and the reorganization energy ($\lambda$).

Example: Self-exchange reaction of $[Fe(CN)_6]^{4-} / [Fe(CN)_6]^{3-}$.
$$[Fe(CN)_6]^{4-} (d^6) + [Fe^*(CN)_6]^{3-} (d^5) \rightarrow [Fe(CN)_6]^{3-} (d^5) + [Fe^*(CN)_6]^{4-} (d^6)$
Since the ligands ($CN^-$) are the same and bond lengths change very little between $Fe(II)$ and $Fe(III)$, this reaction is very fast.

OR (Alternative Questions for Q2)

2. c) Discuss the mechanism of substitution reaction in Pt(II) square planar complex. (4 Marks)

Mechanism: Substitution typically proceeds via an Associative ($A$ or $S_N2$) mechanism.

Pathway:

The incoming ligand ($Y$) approaches the square planar complex $[ML_3X]$.

Formation of a 5-coordinate Trigonal Bipyramidal (TBP) intermediate.

The leaving group ($X$) departs, restoring square planar geometry.

Rate Law: Rate $= k_1[Complex] + k_2[Complex][Y]$. The two terms represent a solvent-assisted path ($k_1$) and the direct nucleophilic attack ($k_2$).

2. d) Discuss the solvent effect, effect of leaving group in nucleophilic substitution reaction in square planar complexes. (4 Marks)

Solvent Effect: Since the mechanism is associative, solvents that can coordinate (act as nucleophiles) can participate. A "solvent path" ($k_s$) often competes with the reagent path. Good coordinating solvents increase the rate.

Leaving Group Effect: The rate of reaction depends on how easily the bond M-X breaks. Generally, weaker bases make better leaving groups.

Order of leaving ability: $NO_3^- > H_2O > Cl^- > Br^- > I^- > CN^-$. Harder ligands generally leave faster than softer ligands from soft Pt(II) centers (inverse of trans effect stability).

2. e) Explain complementary and non-complementary reaction with suitable example. (4 Marks)

Complementary: The oxidant and reductant change their oxidation states by the same number of electrons.

Example: $Fe^{2+} + Ce^{4+} \rightarrow Fe^{3+} + Ce^{3+}$ (Both are 1-electron changes). These are usually fast.

Non-Complementary: The oxidant and reductant change oxidation states by different numbers of electrons.

Example: $2Fe^{2+} + Tl^{3+} \rightarrow 2Fe^{3+} + Tl^{+}$. This requires a trimolecular collision (unlikely) or a multi-step mechanism with unstable intermediates (e.g., $Tl^{2+}$), making these reactions generally slower.

2. f) Discuss the bridge activated complex mechanism for electron transfer reaction. (4 Marks)

This is the Inner Sphere Mechanism.

Steps:

Bridging: The reductant (usually labile) forms a bridge with the oxidant (usually inert) via a ligand (e.g., $Cl^-$).

Electron Transfer: The electron moves from Reductant $\rightarrow$ Bridge $\rightarrow$ Oxidant.

Dissociation: The bridge breaks, usually transferring the bridging ligand to the new oxidized species.

Classic Example (Taube's Experiment):
$[Co(NH_3)_5Cl]^{2+} + [Cr(H_2O)_6]^{2+} \rightarrow [Co(H_2O)_6]^{2+} + [Cr(H_2O)_5Cl]^{2+}$
Here, the Chlorine atom is transferred from Cobalt to Chromium, proving the bridge existed.

Question 3

3. a) What are metal carbonyls? Explain structure and bonding in $[Fe_2(CO)_9]$ and $[Fe_3(CO)_{12}]$. (8 Marks)

Metal Carbonyls: Compounds containing metal atoms bonded to carbon monoxide (CO) ligands.

$[Fe_2(CO)_9]$ (Di-iron nonacarbonyl):

Structure:

Contains 3 bridging CO groups and 6 terminal CO groups (3 on each Fe).

There is an Fe-Fe bond to satisfy the 18-electron rule.

Fe atoms are in an octahedral-like environment (sharing a face).

Formula Check: $2 \times Fe (8e) + 9 \times CO (2e) = 16 + 18 = 34e$. For two metals, $34e$ corresponds to 1 M-M bond ($18 \times 2 - 2 = 34$).

$[Fe_3(CO)_{12}]$ (Tri-iron dodecacarbonyl):

Structure:

Triangular arrangement of 3 Iron atoms.

Two Fe atoms are bridged by 2 bridging CO groups.

The remaining 10 COs are terminal.

One Fe atom has no bridging COs attached directly to the other Fe's in a bridging mode (it is unique).

Contains Fe-Fe bonds (Total 3 M-M bonds in the triangle).

3. b) i) Give an account of four important chemical reactions of metal carbonyls. (8 Marks)

Substitution Reactions: CO groups can be replaced by other ligands like $PPh_3$, $NO$, or alkenes.

$Ni(CO)_4 + PPh_3 \rightarrow Ni(CO)_3(PPh_3) + CO$

Reduction (Formation of Carbonylates): Reacting with active metals (Na/Hg) produces anionic carbonyls.

$Co_2(CO)_8 + 2Na/Hg \rightarrow 2Na[Co(CO)_4]$

Reaction with Halogens (Oxidative Cleavage): Breaking M-M bonds.

$Mn_2(CO)_{10} + Br_2 \rightarrow 2Mn(CO)_5Br$

Disproportionation: In basic media.

$3Fe(CO)_5 + 4OH^- \rightarrow [Fe_3(CO)_{11}]^{2-} + CO_3^{2-} + 2H_2O$ (Example pathway). Or simple disproportionation:

$Co_2(CO)_8 \xrightarrow{\text{base}} [Co(CO)_4]^- + Co^{2+}$ salts.

ii) Explain vibrational spectra of metal carbonyl. (Included in 3b marks generally, or treated as separate part)

Principle: The C-O stretching frequency ($\nu_{CO}$) is very sensitive to the electronic environment. Free CO absorbs at $2143 cm^{-1}$.

Back-bonding Effect: Metal $d\pi \rightarrow CO(\pi^*)$ donation adds electron density to the antibonding orbital of CO.

Result: The C-O bond order decreases, the bond weakens, and $\nu_{CO}$ decreases.

Range:

Terminal CO: $2125 - 1850 cm^{-1}$.

Bridging CO ($\mu_2$): $1850 - 1750 cm^{-1}$ (Lower frequency due to sharing).

Triply bridging ($\mu_3$): $1675 - 1600 cm^{-1}$.

Charge Effect: Positive charge on metal $\rightarrow$ Less back-bonding $\rightarrow$ Higher $\nu_{CO}$ (e.g., $[Mn(CO)_6]^+ > 2090$). Negative charge $\rightarrow$ More back-bonding $\rightarrow$ Lower $\nu_{CO}$ (e.g., $[V(CO)_6]^- \approx 1860$).

OR (Alternative Questions for Q3)

3. c) Explain Synergic bonding in metal carbonyls. (4 Marks)

Mechanism:

$\sigma$-Donation: The filled HOMO (carbon lone pair) of CO donates electrons to an empty metal d-orbital (formation of $\sigma$-bond).

$\pi$-Back Donation: A filled metal $d\pi$ orbital overlaps with the empty LUMO ($\pi^*$ antibonding) of CO. The metal donates electron density back to the ligand.

Synergy: The $\sigma$-donation makes the metal more electron-rich, enhancing its ability to $\pi$-back donate. The $\pi$-back donation removes electron density from the metal, making it a better $\sigma$-acceptor. The two effects reinforce each other.

3. d) Calculate EAN of the metal in following metal carbonyl. (4 Marks)

Formula: EAN = (Atomic No. of Metal) - (Oxidation State) + (2 $\times$ Coordination Number). Or for clusters: Total valence electrons.

i) $Fe_3(CO)_{12}$:

Valence e- count: $3 \times Fe(8) + 12 \times CO(2) = 24 + 24 = 48e$.

Per metal atom: $48/3 = 16e$. Wait, EAN usually refers to the 18e rule check. The cluster is stable with 48e because of 3 M-M bonds ($18 \times 3 - 2 \times 3 = 48$). The "Effective Atomic Number" for one Fe is theoretically Kr (36).

ii) $Ru_2(CO)_9$:

$2 \times Ru(8) + 9 \times CO(2) = 16 + 18 = 34e$. Matches single M-M bond.

iii) $Co_4(CO)_{12}$ (Assuming correction from $CO_4$):

$4 \times Co(9) + 12 \times CO(2) = 36 + 24 = 60e$. Matches tetrahedral cluster ($18 \times 4 - 6 \text{ bonds} \times 2 = 60$).

iv) $Os_2(CO)_9$:

$2 \times Os(8) + 9 \times CO(2) = 34e$. Similar to Iron analog.

3. e) What are metal carbonyl cluster? Give their classification with suitable example. (4 Marks)

Definition: Compounds containing M-M bonds forming a cluster.

Classification:

LNCC (Low Nuclearity Carbonyl Clusters): Small clusters, typically 2-4 metal atoms.

Example: $Mn_2(CO)_{10}$, $Fe_3(CO)_{12}$.

HNCC (High Nuclearity Carbonyl Clusters): Large clusters, 5+ metal atoms.

Example: $Rh_6(CO)_{16}$, $[Os_{10}C(CO)_{24}]^{2-}$.

3. f) Explain $\pi(Pi)$ -back bonding in metal carbonyl. (4 Marks)

(Similar to Synergic bonding but focuses on the back-donation).

The overlap involves the filled $t_{2g}$ orbitals of the metal and the empty $\pi^*$ orbitals of Carbon Monoxide.

Significance: This interaction stabilizes low oxidation states of metals (even zero or negative) by delocalizing the accumulated negative charge from the metal onto the ligands. It shortens the M-C bond length (partial double bond character).

Question 4

4. a) Explain different type of bonding by nitrosyl in metal nitrosyl complex with Example. (8 Marks)

Nitrosyl ($NO$) is a versatile ligand.

1. Linear Nitrosyl ($NO^+$):

NO acts as a 3-electron donor (ionic model: $NO^+$ is isoelectronic with CO, 2e donor).

Bond angle M-N-O is near 180°.

Example: Sodium Nitroprusside $Na_2[Fe(CN)_5NO]$.

2. Bent Nitrosyl ($NO^-$):

NO acts as a 1-electron donor (ionic model: $NO^-$ is 2e donor).

Bond angle M-N-O is near 120°.

Example: $[Co(NH_3)_5NO]^{2+}$ (specifically the black isomer).

3. Bridging Nitrosyl: Like CO, NO can bridge two metals.

4. b) Discuss the structure and bonding in metal dinitrogen and dioxygen complex. (8 Marks)

Dinitrogen ($N_2$):

Isoelectronic with CO but a poorer donor and acceptor.

Bonding Modes: usually End-on ($\eta^1$).

Example: $[Ru(NH_3)_5(N_2)]^{2+}$ (First isolated $N_2$ complex).

Bonding: $\sigma$-donation from N lone pair, weak $\pi$-back donation into $N_2$ $\pi^*$. Weakens N-N bond slightly.

Dioxygen ($O_2$):

Crucial for biological transport (Hemoglobin).

Bonding Modes:

Superoxo ($\eta^1$, end-on): $M-O-O$ (bent). Like $O_2^-$.

Peroxo ($\eta^2$, side-on): Metal bonds to both oxygens. Like $O_2^{2-}$.

Example: Vaska's complex adduct $[IrCl(CO)(PPh_3)_2(O_2)]$ (Peroxo type).

OR (Alternative Questions for Q4)

4. c) Discuss the nitrosylating agent for the synthesis of metal nitrosyls. (4 Marks)

Common agents used to introduce NO into complexes:

Nitric Oxide Gas ($NO$): Direct replacement of CO. ($Fe(CO)_5 + 2NO \rightarrow Fe(CO)_2(NO)_2 + 3CO$).

Nitrosyl Halides ($NOCl$): Used to introduce NO and halogen.

Nitrite Salts ($NaNO_2$): In acidic solution.

Nitrosonium Salts ($NO^+BF_4^-$): For cationic complexes.

4. d) Explain Wilkinson's catalyst. (4 Marks)

Formula: $RhCl(PPh_3)_3$ (Chlorotris(triphenylphosphine)rhodium(I)).

Structure: Square planar, 16-electron complex. $d^8$ Rh(I).

Use: Homogeneous hydrogenation of alkenes.

Mechanism Cycle:

Oxidative addition of $H_2$ (Rh(I) $\rightarrow$ Rh(III)).

Ligand dissociation/Alkene coordination.

Migratory insertion (Hydride moves to alkene).

Reductive elimination (Release of alkane, return to Rh(I)).

4. e) Explain how vibrational spectra is use in the study of structure and bonding in metal nitrosyls. (4 Marks)

IR spectroscopy distinguishes between linear ($NO^+$) and bent ($NO^-$) coordination.

Linear ($NO^+$): $\nu_{NO} = 1650 - 1900 cm^{-1}$. (Strong back-bonding lowers it from free $NO^+$ at 2200, but it remains high).

Bent ($NO^-$): $\nu_{NO} = 1525 - 1690 cm^{-1}$. The lower frequency indicates the reduced nature of the NO ligand (N-O double bond character rather than triple).

4. f) Give important reaction of metal nitrosyls. (4 Marks)

Nucleophilic Attack at N: Since linear NO has character of $NO^+$, nucleophiles ($OH^-$) attack the Nitrogen.

$[Fe(CN)_5NO]^{2-} + 2OH^- \rightarrow [Fe(CN)_5(NO_2)]^{4-} + H_2O$.

Substitution: NO can replace CO in carbonyls.

Reductive Nitrosylation: Metal is reduced while NO coordinates.

Question 5 (2 Marks Each)

5. a) What are the term symbol for $d^2$-configuration.

The ground state term symbol for a free $d^2$ ion (like $V^{3+}$) is ${}^3F_2$.

Other terms include ${}^3P, {}^1G, {}^1D, {}^1S$.

5. b) Explain Hole formulation with example.

A $d^{10-n}$ configuration (holes in the d-shell) behaves electronically like a $d^n$ configuration (positrons), but the sign of the energy splitting is reversed.

Example: The Orgel diagram for $d^8$ (2 holes) is the inverse of $d^2$ (2 electrons).

5. c) Arrange the following ligand in order to their trans effect: NO, $PR_3$, $CH_3$, Br, Cl, $H_2O$.

Order: $NO > PR_3 > CH_3 > Br > Cl > H_2O$.

(Note: NO is a very strong $\pi$-acid, placed very high).

5. d) Explain cross reaction with example.

A cross reaction is an electron transfer between two different metal complexes.

Example: $[Fe(CN)_6]^{4-} + [IrCl_6]^{2-} \rightarrow [Fe(CN)_6]^{3-} + [IrCl_6]^{3-}$. (Calculated using Marcus Cross Relation).

5. e) Draw the structure of i) $Ru_3(CO)_{12}$.

Same as $Fe_3(CO)_{12}$ but often without bridging carbonyls depending on the isomer, though the standard reference for group 8 is the $Fe_3$ type. However, strictly speaking, $Ru_3(CO)_{12}$ is different from Iron: It has a triangle of Ru atoms with only terminal COs ($D_{3h}$ symmetry), unlike Fe which has bridging COs.

5. f) Write the method of preparation of metal carbonyl.

Direct Combination:

$Ni + 4CO \xrightarrow{25^\circ C} Ni(CO)_4$

$Fe + 5CO \xrightarrow{High P, T} Fe(CO)_5$

5. g) Write a note on Vaska's compound.

Formula: $trans-[IrCl(CO)(PPh_3)_2]$.

It is a square planar, 16-electron $d^8$ complex of Iridium(I).

It is famous for its ability to undergo oxidative addition reversibly with small molecules like $O_2, H_2, Cl_2$.

5. h) Give the IUPAC name of $[Rh~Cl(PPh_{3})_{4}]$.

Note: The standard Wilkinson's catalyst is tris ($PPh_3)_3$. However, based on the formula printed ($P_4$):

Name: Chlorotetrakis(triphenylphosphine)rhodium(I).

(If assuming standard typo: Chlorotris(triphenylphosphine)rhodium(I)).