Isolobal Analogy, Metal Metal Bonds and Cluster Compounds: Complete Notes on Boranes and Carbonyl Clusters

 

📘 Chapter 15: Parallels between Main Group & Organometallic Chemistry

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⚡ 15.1 — Main Group Parallels with Binary Carbonyl Complexes

🔑 Core Concept: Species needing the same number of electrons to reach a filled-shell (octet or 18e⁻) config are called electronically equivalent species.

📊 Key Table 1 — Electronically Equivalent Species

Electrons Short of Filled Shell	Main Group	Metal Carbonyl
1	Cl, Br, I	Mn(CO)₅, Co(CO)₄
2	S	Fe(CO)₄, Os(CO)₄
3	P	Co(CO)₃, Ir(CO)₃

💡 Trick to Remember:
Halogen (7e⁻ main group) ↔ 17e⁻ metal carbonyl
S group-16 (6e⁻) ↔ 16e⁻ metal carbonyl
P group-15 (5e⁻) ↔ 15e⁻ metal carbonyl

📊 Table 15.1 — Parallels: Cl vs Co(CO)₄

Property	Halogen (Cl)	Co(CO)₄
1– ion	Cl⁻	[Co(CO)₄]⁻
Neutral dimer	Cl₂	[Co(CO)₄]₂
Hydrohalic acid (strong)	HCl	HCo(CO)₄
Interhalogen-type	BrCl from Br₂+Cl₂	ICo(CO)₄ from I₂+[Co(CO)₄]₂
Heavy metal precipitate	AgCl	AgCo(CO)₄
Addition to alkene	Cl₂ + H₂C=CH₂ → ClCH₂CH₂Cl	[Co(CO)₄]₂ + F₂C=CF₂ → product
Disproportionation by Lewis base	Cl₂ + N(CH₃)₃ → [ClN(CH₃)₃]Cl	[Co(CO)₄]₂ + piperidine → ionic product

🔴 Exam Highlight: Co(CO)₄ and other 17e⁻ binary carbonyls are called PSEUDOHALOGENS — this term is directly asked in CSIR-NET & GATE!

📊 Table 15.2 — Parallels: S vs Fe(CO)₄

Property	Sulfur	Fe(CO)₄
2– ion	S²⁻	[Fe(CO)₄]²⁻
Neutral compound	S₈	Fe₂(CO)₉, [Fe(CO)₄]₃
Hydride	H₂S (pKa₁=7.24, pKa₂=14.92)	H₂Fe(CO)₄ (pKa₁=4.44, pKa₂=14)
Phosphine adduct	Ph₃PS	Ph₃PFe(CO)₄
Ethylene complex	Ethylene sulfide (3-membered ring)	Fe(CO)₄–ethylene π-complex

💡 Note: P and Ir(CO)₃ both form tetrahedral tetramers — P₄ and [Ir(CO)₃]₄ — a beautiful analogy!
Also: Co(CO)₃ can replace P atoms in P₄ tetrahedron. So P₄, P₃Co(CO)₃, etc. are all known.

⚠️ Limitation of Electronic Equivalency: Organometallic analogs of IF₇, XeF₄ (expanded octet) don't exist. Also, CO loss is very common in metal carbonyls but NOT in main group — so chemistry can differ significantly.

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🧩 15.2 — The Isolobal Analogy (⭐ Most Important for All Exams)

📖 Hoffmann's Definition (Nobel 1981):
Two fragments are ISOLOBAL if they have the same:

• Number of frontier orbitals
• Symmetry properties of frontier orbitals
• Approximate energy of frontier orbitals
• Number of electrons in those orbitals

Symbol used: ⟺ (double-headed arrow with lobe)

🧱 Building Isolobal Fragments from CH₄ & ML₆

• CH₄ (sp³) & ML₆ (d²sp³) — both 0 vertices missing, 0e⁻ short → parent closed-shell
• Remove 1 H / 1 L → CH₃ (7e⁻) ↔ ML₅ e.g. Mn(CO)₅ (17e⁻)
• Remove 2 → CH₂ (6e⁻) ↔ ML₄ e.g. Fe(CO)₄ (16e⁻)
• Remove 3 → CH (5e⁻) ↔ ML₃ e.g. Co(CO)₃ (15e⁻)

📊 Table 15.3 — Isolobal Fragment Summary (Most Asked!)

Type	CH₄	CH₃	CH₂	CH	C
Neutral MCs	Cr(CO)₆	Mn(CO)₅	Fe(CO)₄	Co(CO)₃	Ni(CO)₂
Cationic MCs	—	[Mn(CO)₆]⁺	[Fe(CO)₅]⁺	[Co(CO)₄]⁺	[Ni(CO)₃]⁺
Cp Complexes	—	CpMn(CO)₃	CpFe(CO)₂	CpCo(CO)	CpNi
Anionic (from CH₃⁻)	—	Fe(CO)₅	Co(CO)₄	Ni(CO)₃	—
Cationic (from CH₃⁺)	V(CO)₆	Cr(CO)₅	Mn(CO)₄	Fe(CO)₃	—

📐 Step-by-step to find isolobal organometallic fragment:
1. Count how many e⁻ short & how many vertices missing from CH₄
2. Same for ML₆ parent — both must match
3. Adjust charge: +1 charge ↔ 1 extra d-electron, –1 ↔ 1 fewer d-electron

🌟 Extensions of the Isolobal Analogy

• Extension 1: Isoelectronic fragments with same coordination number are isolobal. e.g. Mn(CO)₅ ↔ Re(CO)₅ ↔ [Fe(CO)₅]⁺ ↔ [Cr(CO)₅]⁻ — all isolobal with CH₃
• Extension 2: Gain/loss of e⁻ gives isolobal fragments. e.g. Cr(CO)₅, Mo(CO)₅, W(CO)₅ all isolobal with CH₃⁺
• Extension 3: PR₃, NCR, Cl⁻ — all 2e⁻ donors — treated same as CO. e.g. Mn(CO)₅ ↔ Mn(PR₃)₅ ↔ [MnCl₅]⁵⁻
• Extension 4: η⁵-C₅H₅ and η⁶-C₆H₆ = 6e⁻ donors occupying 3 coord sites. So (η⁵-Cp)Fe(CO)₂ ↔ Mn(CO)₅ (both 17e⁻)
• Extension 5: Octahedral MLₙ (dˣ) ↔ Square-planar MLₙ₋₂ (dˣ⁺²). e.g. Cr(CO)₅ (d⁶ oct) ↔ [PtCl₃]⁻ (d⁸ sq-pl)

✅ Solved Example 15.1 — Find isolobal organometallic for CH₂⁺

Q: Propose organometallic fragments isolobal with CH₂⁺

Solution:
• CH₂⁺ = 2 ligands short + 3e⁻ short (6e⁻ – 3 = fragment with 5e⁻, cation)
• Octahedral fragment: ML₄ (2 vertices short), 15 electrons = Mn(CO)₄ (d⁷ Mn + 4×2CO = 7+8=15 ✓)
• Vary charge: d⁸ → [Fe(CO)₄]⁺ ; d⁶ → [Cr(CO)₄]⁻

Answer: Mn(CO)₄, [Fe(CO)₄]⁺, [Cr(CO)₄]⁻

📊 Table 15.4 — Isolobal Fragments from Different Polyhedra

Organic Fragment	CN=5 parent	CN=6 parent	CN=7 parent	Valence e⁻
CH₃	d⁹-ML₄	d⁷-ML₅	d⁵-ML₆	17
CH₂	d¹⁰-ML₃	d⁸-ML₄	d⁶-ML₅	16
CH	d⁹-ML₃	d⁷-ML₄	d⁵-ML₅	15

💡 Real Applications of Isolobal:

• CH≡CH (two CH) → can replace by Co(CO)₃ + Co(CO)₃ → (OC)₃Co-Co(CO)₃
• C₃H₆ (cyclopropane) ↔ Os₃(CO)₁₂ (triangle of Os)
• AuPPh₃ behaves like H — both bridge triosmium clusters identically!
• P₅⁻ is isolobal with C₅H₅⁻ → metallocenes with P₅ rings exist! e.g. (η⁵-C₅Me₅)Fe(η⁵-P₅)
• Carbon-free metallocene: [(η⁵-P₅)₂Ti]²⁻ prepared from Ti(naphthalene)₂²⁻ + P₄

✅ Solved Example 15.2

CH₃ is isolobal with 17e⁻ Zn(η⁵-C₅Me₅) (Extension 4)
CH₂ is isolobal with 16e⁻ Ir(PPh₃)₂(CS)Cl (start with 16e⁻ [Ir(CO)₄]⁺, substitute PPh₃, Cl⁻, CS for CO)

→ These isolobal pairs are used to synthesize real compounds (Figure 15.6 in textbook):

• Pt[Ir(PPh₃)₂(CS)Cl]₂ — analog of cyclopropane
• Zn₃ clusters — analog of CH₃ trimers

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🔗 15.3 — Metal–Metal Bonds (⭐ Very Important for GATE/CSIR-NET/IIT-JAM)

Historical Timeline:

• 1935: K₃W₂Cl₉ — first direct M–M bond evidence (W–W = 240 pm vs 275 pm in W metal)
• 1963: [Re₃Cl₁₂]³⁻ crystal structure — Re–Re = 248 pm
• 1964: [Re₂Cl₈]²⁻ — first quadruple bond (Re–Re = 224 pm)
• 2005: Power's dimeric Cr(I) complex — first quintuple bond (Cr–Cr = 183.5 pm)
• 2009: Shortest ever M–M bond: Cr–Cr = 172.9 pm (guanidinate ligand)

⚛️ How Quadruple Bond Forms

d-orbital interactions in M₂ (z-axis = internuclear axis):
1. dz² + dz² → σ bond (strongest overlap)
2. dxz + dxz & dyz + dyz → 2 π bonds
3. dxy + dxy → δ bond (weakest, 4-region overlap)
4. dx²-y² + dx²-y² → used in σ bonding with ligands (in M₂L₈ complexes)

Bond order = σ + 2π + δ = 1+2+1 = 4 (quadruple!)
MO energy order: σ < π < δ < δ* < π* < σ*

[Re₂Cl₈]²⁻ Analysis:
• Re(III) → d⁴ each → 8 d-electrons total
• Fill: σ(2e) + π(4e) + δ(2e) = bond order 4
• Color: Royal Blue (δ→δ* transition in visible region)
• Geometry: eclipsed D₄ₕ — maintained by weak δ bond
• Compare: [Mo₂Cl₈]⁴⁻ = Bright Red

💡 Why are quadruply bonded complexes COLORED?
δ–δ* energy gap = visible light range → absorption → color!
Main group triple bonds (N₂, CO) are colorless because π–π* gap = UV range.

📊 Table 15.5 — Effect of Oxidation on Re–Re Bond Distance

Complex	d-electrons	Bond Order	Re Ox. State	Re–Re (pm)
Re₂Cl₄(PMe₂Ph)₄	10	3	+2	224.1
[Re₂Cl₄(PMe₂Ph)₄]⁺	9	3.5	+2.5	221.8
[Re₂Cl₄(PMe₂Ph)₄]²⁺	8	4	+3	221.5

⚠️ Key Point: Oxidation removes δ* electrons → expect shorter bond, BUT d-orbital contraction at higher oxidation offsets this → very small change in distance (224→221.5 pm only!)

📏 Formal Shortness Ratio (FSR) — Very Important for CSIR-NET

Bond	Multiple/Single ratio	Bond	Ratio
C≡C	0.783	Cr≡≡≡≡Cr (quadruple)	0.767
N≡N	0.786	Mo≡≡≡≡Mo	0.807
Re≡≡≡≡Re	0.848	Cr (quintuple, 2009)	0.729

Smaller FSR = shorter relative bond = stronger multiple bond. Cr–Cr quadruple bonds have SMALLEST FSR known!

🔵 Quintuple Bonds

• 2005 (Power): Dimeric Cr(I) with bulky terphenyl ligands → Cr–Cr = 183.5 pm → FSR = 0.774
• d¹ Cr(I) → 5 d-electrons → σ + 2π + 2δ potentially occupied
• 2007: Cr–Cr reduced to 180.3 pm
• 2008 (Tsai): 174.0 pm
• 2009 (Kempe): 172.9 pm (guanidinate ligand) — shortest M–M bond ever! FSR = 0.729
• Mo–Mo quintuple bonds also known: 201.9 & 201.6 pm

🔴 Controversy: Some calculations give bond order as low as 3.3 for "quintuple" Cr complexes. Trans-bent geometry in Power's complex reduces orbital overlap efficiency. Tsai & Kempe complexes lack trans-bent geometry → more direct d-orbital interactions.

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🧊 15.4 — Cluster Compounds (⭐⭐ Most Numerically Asked in IIT-JAM/GATE)

📐 15.4.1 — Boranes (closo/nido/arachno)

Closo Boranes [BₙHₙ]²⁻: closed polyhedral cages, all triangular faces
• MO theory gives 2n+1 bonding MOs (n B–H + n+1 framework)
• B₆H₆²⁻ (Oh symmetry) = 7 framework bonding pairs = 26 valence electrons
• Framework orbitals: A₁g (central) + T₁u + T₂g types
• Described as spherically aromatic (3D delocalization!)

📊 Table 15.6 — Framework Bonding Pairs in Closo Boranes

Formula	Total Valence e⁻ pairs	A₁ framework pair	Other sym pairs	B–H pairs
B₆H₆²⁻	13	1	6	6
B₇H₇²⁻	15	1	7	7
BₙHₙ²⁻	2n+1	1	n	n

📊 Table 15.7 — Wade's Rules (Classification)

Structure Type	Corners Occupied	Framework Bond Pairs	Empty Corners
Closo	n of n	n+1	0
Nido	n–1 of n	n+1	1
Arachno	n–2 of n	n+1	2
Hypho	n–3 of n	n+1	3
Klado	n–4 of n	n+1	4

💡 All types have SAME n+1 framework pairs for the same parent! Only empty corners differ.

🔢 Electron Counting Method for Boranes

Step 1: Each BH fragment → contributes 2 electrons to framework
Step 2: Each extra H (beyond BH) → contributes 1 electron to framework
Step 3: Charge → add its electrons
Step 4: Total framework electrons ÷ 2 = framework pairs → parent polyhedron has (pairs – 1) corners
Step 5: Compare boron atoms to parent corners → classify as closo/nido/arachno

✅ Solved Example 15.3 — Classify B₁₁H₁₃²⁻

Total valence e⁻ = 33+13+2 = 48
Step 1: 11 BH → 11×2 = 22 e⁻
Step 2: Extra H = 13–11 = 2 → 2 e⁻ (total: 24)
Step 3: charge = –2 → +2 e⁻ (total: 26 e⁻ = 13 pairs)
Step 4: Parent polyhedron corners = 13–1 = 12 (icosahedron)
Step 5: Only 11 B atoms → 1 corner missing → NIDO

🚀 Quick Classification Trick (Easier Method!)

Subtract H⁺ ions to make (# H) = (# B), then check the charge:
• Charge = 2– → Closo
• Charge = 4– → Nido
• Charge = 6– → Arachno
• Charge = 8– → Hypho

✅ Solved Example 15.4

B₁₀H₁₄: B₁₀H₁₄ – 4H⁺ = B₁₀H₁₀⁴⁻ → Nido
B₂H₇⁻: B₂H₇⁻ – 5H⁺ = B₂H₂⁶⁻ → Arachno
B₈H₁₆: B₈H₁₆ – 8H⁺ = B₈H₈⁸⁻ → Hypho

📊 Table 15.8 — Electron Count & Classification Examples

Parent vertices	Type	Boron atoms	Valence e⁻	Framework pairs	Example
6	Closo	6	26	7	B₆H₆²⁻
6	Nido	5	24	7	B₅H₉
6	Arachno	4	22	7	B₄H₁₀
7	Closo	7	30	8	B₇H₇²⁻
12	Closo	12	50	13	B₁₂H₁₂²⁻
12	Nido	11	48	13	B₁₁H₁₃²⁻

🧪 15.4.2 — Carboranes (Heteroboranes)

Key rule: CH⁺ is isoelectronic with BH
So replace C by BH → treat as borane → classify!

For classification: C in carborane → convert each C to BH, then apply borane rules.

📊 Table 15.9 — Borane vs Carborane Formulas

Type	Borane	Carborane
Closo	BₙHₙ²⁻	C₂Bₙ₋₂Hₙ (e.g. C₂B₁₀H₁₂)
Nido	BₙHₙ₊₄	C₂Bₙ₋₂Hₙ₊₂
Arachno	BₙHₙ₊₆	C₂Bₙ₋₂Hₙ₊₄

✅ Solved Example 15.5

C₂B₉H₁₂⁻: → B₁₁H₁₄⁻ → B₁₁H₁₄⁻ – 3H⁺ = B₁₁H₁₁⁴⁻ → Nido
C₂B₇H₁₃: → B₉H₁₅ → B₉H₁₅ – 6H⁺ = B₉H₉⁶⁻ → Arachno
C₄B₂H₆: → B₆H₁₀ → B₆H₁₀ – 4H⁺ = B₆H₆⁴⁻ → Nido

🔬 Heteroborane Conversion Rules

Heteroatom	Replace with
C, Si, Ge, Sn	BH
N, P, As	BH₂
S, Se	BH₃

✅ Solved Example 15.6

SB₉H₁₁: S→BH₃ → B₁₀H₁₄ → B₁₀H₁₄ – 4H⁺ = B₁₀H₁₀⁴⁻ → Nido
CPB₁₀H₁₁: C→BH, P→BH₂ → PB₁₁H₁₂ → B₁₂H₁₄ → B₁₂H₁₄ – 2H⁺ = B₁₂H₁₂²⁻ → Closo

⚙️ 15.4.3 — Metallaboranes (M replaces BH)

Organometallic Fragment	Valence e⁻	Equivalent BHₓ
Mn(CO)₃	13	B (boron alone)
Co(η⁵-Cp)	14	BH
Co(CO)₃	15	BH₂
Fe(CO)₄	16	BH₃

✅ Solved Example 15.7

B₄H₆(CoCp)₂: Each CoCp(14e⁻) → BH → B₄H₆(BH)₂ = B₆H₈ → B₆H₈ – 2H⁺ = B₆H₆²⁻ → Closo

B₃H₇[Fe(CO)₃]₂: Each Fe(CO)₃(16e⁻) → BH₃? No — wait, BH uses 4e⁻ toward 8, so Fe(CO)₃ is 2e⁻ short → BH. So: B₃H₇(BH)₂ = B₅H₉ → B₅H₉ – 4H⁺ = B₅H₅⁴⁻ → Nido

📊 Table 15.10 — Closo Metallaborane/Metallacarborane Structures

Skeletal Atoms	Shape	Example 1	Example 2
6	Octahedron	B₄H₆(CoCp)₂	C₂B₃H₅Fe(CO)₃
7	Pentagonal bipyramid	C₂B₄H₆Ni(PPh₃)₂	C₂B₃H₅(CoCp)₂
8	Dodecahedron	C₂B₄H₄[(CH₃)₂Sn]CoCp	—
9	Capped sq. antiprism	C₂B₆H₈Pt(PMe₃)₂	C₂B₅H₇(CoCp)₂
10	Bicapped sq. antiprism	[B₉H₉NiCp]⁻	CB₇H₈(CoCp)(NiCp)
12	Icosahedron	C₂B₇H₉(CoCp)₃	C₂B₉H₁₁Ru(CO)₃

💡 Carborane-ferrocene analog: C₂B₉H₁₁²⁻ (nido) has p-orbital lobe at missing vertex → exactly like Cp⁻ ring! → [Fe(η⁵-C₂B₉H₁₁)₂]²⁻ = carboranyl ferrocene

⚗️ 15.4.4 — Carbonyl Clusters & Wade's Rules (Transition Metals)

Key Rule: Transition metal has 5 more valence orbitals than B → needs 10 more electrons per atom for framework equivalency.

So: B₆H₆²⁻ has 26 valence e⁻ (closo) → Co₆ analog needs 26 + 6×10 = 86 e⁻
→ Co₆(CO)₁₆ = 86e⁻ → closo octahedral!

📊 Table 15.11 — Electron Counting for Main Group vs Transition Metal Clusters

Structure Type	Main Group	Transition Metal
Closo	4n+2	14n+2
Nido	4n+4	14n+4
Arachno	4n+6	14n+6
Hypho	4n+8	14n+8

📊 Table 15.12 — Borane & TM Cluster Parallels

n	Parent vertices	FBP	Closo B	Nido B	Arachno B	Closo TM	Nido TM	Examples
5	6	7	22	24	26	72	74	Os₅(CO)₁₆(closo) / Os₅C(CO)₁₅(nido)
6	7	8	26	28	30	86	88	Co₆(CO)₁₆(closo) / Os₆(CO)₁₇[P(OMe)₃]₃(nido)
4	—	5	—	—	22	—	60	Co₄(CO)₁₂(nido by count, but actually tetrahedral — exception!)

🔴 Exception to Wade's Rules: M₄(CO)₁₂ (M = Co, Rh, Ir) — 60 e⁻ (= 14n+4 → predicted nido = trigonal bipyramid missing 1 vertex), but X-ray shows tetrahedral metal core! Wade's rules not always correct for TM clusters.

📊 Table 15.13 — Clusters with 7 Metal–Metal Framework Bond Pairs

Framework atoms	Type	Shape	Example
7	Capped closo	Capped octahedron	[Rh₇(CO)₁₆]³⁻, Os₇(CO)₂₁
6	Closo	Octahedron	Rh₆(CO)₁₆, Ru₆C(CO)₁₇
5	Nido	Square pyramid	Ru₅C(CO)₁₅
4	Arachno	Butterfly	[Fe₄(CO)₁₃H]⁻

🧊 15.4.5 — Carbon-Centered (Carbide) Clusters

Encapsulated C contributes its 4 valence e⁻ to the cluster count.
Example: Ru₆C(CO)₁₇ — C contributes 4e⁻ → total = 86e⁻ → closo (octahedral Ru₆ core)

Symmetry analysis (Oh):
• C(2s) = A₁g symmetry
• C(2p) = T₁u symmetry
• Match with Ru₆ framework orbitals of A₁g & T₁u → 4 C–Ru bonding pairs formed
• T₂g Ru orbitals → Ru–Ru bonding only (no C interaction)

📐 mno Rule — Extended Wade's Rules

For closed cluster: Total framework pairs = m + n + o (+ p for open)
• m = number of condensed (linked) polyhedra
• n = total number of vertices
• o = single-atom bridges between polyhedra
• p = missing vertices (1 for nido, 2 for arachno)

✅ Solved Example 15.9

B₁₂H₁₂²⁻: m=1, n=12, o=0, p=0 → m+n+o = 13 pairs ✓

[(η⁵-C₂B₉H₁₁)₂Fe]²⁻: m=2, n=23 (all C,B,Fe), o=1 (Fe bridges), p=0 → 26 pairs

Ferrocene (η⁵-Cp)₂Fe: m=2, n=11, o=1, p=2 (nido each ring) → 16 pairs

🌍 Zintl Ions (Main Group Clusters)

Ionic clusters of main group elements — classified by same 4n+2, 4n+4, 4n+6 rules:

• Pb₅²⁻: 5×4+2 = 22 = 4×5+2 → Closo
• Sn₉⁴⁻: 9×4+4 = 40 = 4×9+4 → Nido
• Sb₄²⁻: 5×4+2=22=4×4+6 → Arachno (square, octahedron –2 vertices)

✅ Solved Example 15.8

a. Pb₅²⁻: 5×4+2 = 22; 4n+2 with n=5 → Closo
b. Sn₉⁴⁻: 9×4+4 = 40; 4n+4 with n=9 → Nido (1 missing vertex)
c. Sb₄²⁻: 5×4+2=22; 4n+6 with n=4 → Arachno (square = octahedron –2)

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🎯 Exam Tips & Must-Remember Points

🔴 TOP PICKS for GATE/CSIR-NET/IIT-JAM:

• Table 15.3 (Isolobal fragments) — memorize completely!
• Quadruple bond in [Re₂Cl₈]²⁻ — mechanism, color, geometry
• Wade's rules Table 15.11 — 4n+2/14n+2 formulas
• Heteroatom conversion for classification (S→BH₃, N→BH₂, C→BH)
• mno rule for polyhedral clusters

⚡ Quick Formula Sheet:

• Closo BₙHₙ²⁻ → total e⁻ = 4n+2
• Nido BₙHₙ₊₄ → total e⁻ = 4n+4
• Arachno BₙHₙ₊₆ → total e⁻ = 4n+6
• TM closo: 14n+2 | nido: 14n+4 | arachno: 14n+6
• Replace each TM → 10 extra electrons vs B in cluster
• FSR < 1 means multiple bond; smaller = stronger bond

⚠️ Common MCQ Traps:

• HCo(CO)₄ is a strong acid like HCl — but slightly soluble in water
• M₄(CO)₁₂ predicted nido BUT actually tetrahedral — exception!
• Fe(CO)₅ is NOT isolobal with CH₄ — both filled shells but different geometry
• [Os₂Cl₈]²⁻ has 10d electrons → triple bond → staggered geometry (not eclipsed like quadruple)
• δ bond is WEAKEST — that's why complexes are colored (small δ→δ* gap)
• Quintuple bond controversy: bond order calculations give 3.3 to 5 range

📝 Practice MCQs (Based on This Chapter)

Q1. Which of the following is called a "pseudohalogen"?
(A) Fe(CO)₄   (B) Co(CO)₄   (C) Mn(CO)₅   (D) Ni(CO)₄
Ans: (B) — 17e⁻ species, one short of 18e⁻, just like halogen is one short of octet

Q2. CH₂ is isolobal with which organometallic complex?
(A) Mn(CO)₅   (B) Fe(CO)₄   (C) Co(CO)₃   (D) Cr(CO)₆
Ans: (B) — Both 2 vertices short; CH₂ (6e⁻) ↔ Fe(CO)₄ (16e⁻)

Q3. What is the total valence electron count for closo Os₅(CO)₁₆?
(A) 60   (B) 72   (C) 74   (D) 86
Ans: (B) — 14n+2 = 14×5+2 = 72

Q4. [Re₂Cl₈]²⁻ is eclipsed due to:
(A) σ bond   (B) π bond   (C) δ bond   (D) steric repulsion
Ans: (C) — δ bond locks the eclipsed D₄ₕ geometry

Q5. Classify B₁₀H₁₄ by Wade's rules.
(A) Closo   (B) Nido   (C) Arachno   (D) Hypho
Ans: (B) — B₁₀H₁₄ – 4H⁺ = B₁₀H₁₀⁴⁻ → Nido

Q6. Which molecule is the carbon-free metallocene?
(A) [Fe(C₂B₉H₁₁)₂]²⁻   (B) [(P₅)₂Ti]²⁻   (C) (Cp)₂Fe   (D) [Ir(CO)₃]₄
Ans: (B) — Contains only P₅ rings, no carbon ligands

Q7. The shortest metal-metal bond ever recorded is:
(A) Re–Re in [Re₂Cl₈]²⁻   (B) Mo–Mo quintuple   (C) Cr–Cr = 172.9 pm (2009)   (D) W–W in [W₂Cl₉]³⁻
Ans: (C) — Guanidinate Cr(I) complex, FSR = 0.729

Q8. mno rule: For ferrocene (Cp)₂Fe, number of framework e⁻ pairs =?
(A) 13   (B) 14   (C) 16   (D) 26
Ans: (C) — m=2, n=11, o=1, p=2 → 16 pairs

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Chapter 15 Notes · Inorganic Chemistry · All competitive exams · Made with ❤️

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