Cages and Metal Clusters Questions and Answers

Answer: B

The carborane is closo with an icosahedral framework.

Count the skeletal electron pairs with Wade's rules. Each B-H vertex contributes 2 skeletal electrons and each C-R vertex 3. For Me₂C₂B₁₀H₁₀, with 12 vertices:

\[ \begin{aligned} 10 \times 2 + 2 \times 3 &= 26\ \text{electrons} \\ &= 13\ \text{pairs} = n + 1 \end{aligned} \]

n + 1 pairs for n vertices is the closo count, a closed polyhedron with every vertex occupied, and the closed 12-vertex deltahedron is the icosahedron. This is the dimethyl derivative of the well-known icosahedral carborane C₂B₁₀H₁₂.

A nido cluster would need n + 2 pairs. The dodecahedron (the 8-vertex deltahedron) is the closo shape for eight vertices, not twelve.

Answer: A

The answer is 1, 2, 1.

In [Mo₂(CH₃CO₂)₄] each Mo(II) is d⁴, so the Mo₂ unit has eight metal d electrons. With the Mo-Mo axis as z, the d orbitals overlap to give one σ bond (d_z2), two π bonds (d_xz, d_yz) and one δ bond (d_xy). The configuration σ²π⁴δ² fills all four bonding orbitals: a quadruple bond.

The eclipsed acetate bridges hold the two MoO₄ units in the alignment the δ overlap needs. '1, 2, 0' describes a triple bond (as in a d³-d³ dimer), and the other options do not count two π components.

Answer: C

The compound with a structure like B₂H₆ is Al₂Cl₆.

Diborane has two boron atoms joined by two bridging hydrogens, with the four terminal hydrogens in a plane at right angles to the B-H-B bridges; each boron is roughly tetrahedral. Al₂Cl₆ (the vapour-phase dimer of AlCl₃) has exactly this arrangement: two edge-sharing AlCl₄ tetrahedra with two bridging chlorines and four terminal chlorines (the bridges there are ordinary electron-pair bonds rather than three-centre bonds).

I₂Cl₆ is also chlorine-bridged but is completely planar, each iodine being square planar. Si₂Cl₆ is ethane-like, with a direct Si-Si bond. Cl₂O₆ is [ClO₂]⁺[ClO₄]^- in the solid, with no bridged dimer of this type. Only one option fits, which is the whole key of this multiple-select question.

Answer: A

The styx number of B₄H₁₀ is 4012.

Lipscomb's code counts four kinds of bond: s = B-H-B bridges, t = three-centre B-B-B bonds, y = two-centre B-B bonds, x = extra terminal H beyond one per boron (BH₂ groups). From the structure: 4 bridging hydrogens, no closed B-B-B bond, 1 B-B bond between the two central borons, and 2 BH₂ groups (the outer borons), so s, t, y, x = 4, 0, 1, 2.

The balance equations for B_pH_p+q, here p = 4 and q = 6, confirm it:

\[ \begin{aligned} s + x &= q:\ 4 + 2 = 6 \\ s + t &= p:\ 4 + 0 = 4 \\ t + y &= p - q/2:\ 0 + 1 = 1 \end{aligned} \]

4102 swaps t and y, claiming a three-centre bond the arachno structure does not have. 2104 and 2014 count only two bridges, so they fail the balance \(s + t = p\), giving 2 + 1 = 3 and 2 + 0 = 2 instead of 4.

Answer: D

[Re₂Cl₈]^2- has four metal-metal bonds between its two rhenium atoms: a quadruple bond.

Each Re(III) is d⁴, so the pair has eight d electrons for the Re-Re bond. Overlap of the d orbitals along the Re-Re axis gives one σ bond (d_z²), two π bonds (d_xz, d_yz) and one δ bond (d_xy), and the eight electrons fill exactly these four bonding orbitals: σ²π⁴δ², a bond order of 4.

The δ bond is why the two ReCl₄ units are eclipsed rather than staggered: rotating them would destroy the d_xy overlap. Cotton's discovery of this ion (1964) established the quadruple bond. One, two or three bonds would leave bonding orbitals empty.

Answer: C

The two structures are closo and arachno.

Wade's rules count skeletal electron pairs: each B-H unit gives 2 electrons, each C-H unit 3, each extra hydrogen 1 and each two-electron ligand 2. For n cage atoms, n+1 pairs means closo, n+2 nido and n+3 arachno.

\[ \begin{aligned} \mathrm{B_{10}C_{2}H_{12}} &: 10(2) + 2(3) = 26\ e \\ &= 13\ \text{pairs} = n+1 \ (n=12) \end{aligned} \]

So the dicarbadodecaborane is a closo icosahedron. For B₁₀H₁₂(SEt₂)₂, the ten B-H units give 20 electrons, the two extra hydrogens 2 and the two SEt₂ donors 4, a total of 26 electrons or 13 pairs for 10 cage atoms, which is n+3: arachno.

Options calling the carborane nido would need 14 pairs, and calling the ligand adduct nido ignores the two electron pairs the SEt₂ groups add to the parent nido B₁₀H₁₄.