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Radiation changes the solution's chemistry by radiolysis; it does not turn Am(VI) nuclei into Am(V) nuclei. An emitted alpha particle deposits energy into many water and solute molecules, generating ionized/excited species, hydrated electrons, hydrogen atoms and molecular products including hydrogen peroxide. Reducing products can supply the electron needed forFor example, hydrogen peroxide can be oxidized while reducing two americyl(VI) ions:The $0.3$ mM loss of Am(VI) equals the $0.3$ mM appearance of Am(V), preserving the measured dissolved americium total while transferring $0.3\mathrm{mmol}{\mathrm{L}}^{-1}$ of electron equivalents. Oxidizing radiolysis products are also present, and the net reduction depends on acid, dissolved gases and scavengers; not every radiolytic event is reducing. In nitric acid, primary kinetic work identifies hydrogen peroxide and radiolytically produced nitrous acid as important reductants, with a rate controlled by total americium activity rather than necessarily by the instantaneous Am(VI) concentration.^[ref]

An isolobal analogy compares fragment frontier orbitals, not atomic sizes or total electron counts. The neutral BH fragment has four valence electrons; two make the external B–H bond, leaving two skeletal electrons in three frontier orbitals directed towards possible cluster neighbours. For the neutral $\mathrm{CpCo}$ fragment, $\mathrm{Cp}={\mathrm{C}}_5{\mathrm{H}}_5$ contributes five electrons and cobalt nine, giving 14. Removing the 12 electrons assigned to filled, predominantly non-skeletal fragment orbitals again leaves two electrons in three cluster-directed frontier orbitals.

The corresponding frontier sets have comparable numbers, spatial symmetries and occupations, allowing replacement of a BH vertex by CpCo in a suitable cluster-electron count. This is why the metal fragment can cap an open carborane face without adding an arbitrary new set of skeletal electrons. Exact orbital energies, ligand preferences and bond lengths need not be identical.^[ref]

Neither formula requires a bridging carbonyl to repair an electron deficiency: the terminal-CO donor–acceptor structures in (d) and (e) already give an electron-counting rationale for the metal–metal interaction. A predominantly terminal-carbonyl model is a reasonable starting proposal for A; bridging or semibridging CO is more geometrically plausible for the shorter Fe–Fe contact in B. Larger metal separation makes a carbonyl simultaneously engaging both Mo atoms less favourable, while carbonyl redistribution or tilting can be easier around a compact Fe pair.

This is a qualified structural prediction, not a determination from molecular formula. A bridging CO can change the relative hapticity and local electron allocation, and the two metal centres are inequivalent; a large frequency shift by itself need not prove a symmetric bridge. Infrared carbonyl frequencies combined with diffraction would test terminal versus bridging or semibridging assignments. The given low-temperature preparation can trap a coordination pattern before complete carbonyl redistribution.

Writing $A(\mathbf{r})={\varphi }_{1sA}(\mathbf{r})$ and $B(\mathbf{r})={\varphi }_{1sB}(\mathbf{r})$, the normalized determinant isIts two spin orbitals are orthogonal because $⟨\alpha |\beta ⟩=0$, even if the spatial atomic functions have nonzero overlap. Exchanging both the spatial and spin coordinates of the two electrons reverses the sign, as required for fermionic antisymmetry.

At this separation the orbital overlap and interatomic two-electron exchange integrals are negligible. Each localized electron then contributes essentially the isolated hydrogen energy, while the interatomic electrostatic terms cancel to leading order for neutral fragments. Including nuclear repulsion in the molecular energy givesThe closed-shell RHF determinant instead doubly occupies $g\simeq (A+B)/\sqrt{2}$. It retains ionic configurations with both electrons on the same atom, even at dissociation. Using atomic units and the Hartree energy in the stated unrelaxed hydrogenic 1s basis, the same-atom two-electron Coulomb integral is $J_{AA}=J_{BB}=5E_h/8$. Consequently at infinite separationwhere energies and integrals inside the first expression are in atomic units. It is much too high compared with two neutral atoms. Retaining the leading intersite Coulomb term at $R=10\text{A˚}=18.90a_0$ gives $E_{\mathrm{RHF}}\simeq -11/16-1/(2R)\simeq -0.714E_h$, still about $0.286E_h$ above UHF. The numerical RHF estimates specify the supplied fixed 1s basis; a larger optimized orbital space changes them. If nuclear repulsion is excluded by an electronic-energy convention, subtract the same $1/R$ from both quoted total molecular energies. The restricted/unrestricted comparison and their dissociation behaviour are unchanged.^[ref]

Number the four sites of the cyclobutadienyl dianion cyclically and apply the Hückel method. The orthonormal pi basis givesA convenient real orthonormal eigenbasis isThus one bonding and two degenerate nonbonding orbitals are filled, and the antibonding orbital is empty. There are six pi electrons, including the two added by the dianion charge. Rotating the two occupied nonbonding orbitals within their degenerate subspace changes their individual sketches but not the total density matrix.

The Hückel orbital charge density and Hückel pi bond order giveEach site receives $1/2$ electron from the lowest orbital and one from the occupied nonbonding subspace, so $n_i=3/2$ at every carbon, summing to six. Relative to one neutral pi electron per carbon this is a pi charge of $-e/2$ at each site.

For adjacent sites, only the lowest orbital contributes, giving $p_{12}=p_{23}=p_{34}=p_{41}=1/2$. For opposite sites the nonbonding contribution reverses the sign: $p_{13}=p_{24}=-1/2$. These non-neighbour negative density-matrix entries are not extra sigma bonds or extra resonance integrals in the Hamiltonian. If only skeletal nearest-neighbour pi bond orders are requested, the four values $1/2$ are the relevant ones.

molecular mechanics evaluates a parametrized potential of atomic positions, with terms such as bond stretching, angle bending, torsion and nonbonded interactions. electrons are represented implicitly through the fitted force field, rather than by electronic orbitals, a density or an electronic wavefunction. This example is different from both preceding electronic methods.

Write $H=\alpha I+K$ with the same carbon diagonal parameter throughout, and define $P$ to have entries $+1$ on one sublattice and $-1$ on the other. Bipartite connectivity impliesThus $Kc=ϵc$ implies $K(Pc)=-ϵ(Pc)$. This proves energy pairing about $\alpha$, $E=\alpha \pm ϵ$, and equal site-probability magnitudes in each paired pair, since $|(Pc{)}_i{|}^2=|c_i{|}^2$. This is the Coulson-Rushbrooke pairing property.^[ref]

For the usual neutral half-filled closed-shell occupation, the lower member of each pair is doubly occupied. Combining paired coefficients with completeness of the orthonormal basis gives one pi electron per carbon, so the model predicts no net pi charge separation. The same argument makes off-diagonal pi bond orders between sites on the same sublattice zero. These statements assume a neutral half-filled density and do not apply unchanged to the charged six-electron four-site ion considered separately.

An imbalance of sublattice sizes also guarantees at least $|N_A-N_B|$ nonbonding orbitals at energy $\alpha$, because the rectangular coupling block has insufficient rank to remove all its null space. None of the spectral pairing, equal paired densities, neutral charge uniformity or sublattice bond-order restrictions is required for a non-alternant graph. Long-range same-sublattice couplings or unequal diagonal parameters can likewise destroy the simple proof.

Choose x along O–O. The valence orbital order appropriate to O${}_2$ is approximatelyIts twelve valence electrons occupyThe two degenerate pi-antibonding orbitals are singly occupied with parallel spins according to Hund's rules. Thus $S=1$, multiplicity $2S+1=3$, with three possible $M_S$ values. A triplet is a many-electron spin classification, not three electrons in one orbital. The resulting ground term is ${}^3{\Sigma }_g^{-}$. Exchange stabilization favours this triplet over the singlets formed within the same frontier-orbital manifold.

The active set contains four oxygen pi orbitals and two carbon pi orbitals, with eight electrons in six orbitals. The noninteracting levels, using the parameters inferred below, areEach bonding oxygen orbital has equal-phase coefficients on its two O atoms; its antibonding partner reverses one coefficient. The two pi directions are z in the prospective ring plane and y perpendicular to it. carbon pi and pi-star use z-directed p orbitals with equal and opposite coefficient signs. In this separated-fragment comparison the occupations are those of neutral triplet O${}_2$ plus neutral ethene. Do not refill the combined list solely by energy, which would move an electron between infinitely separated fragments and describe a charge-transfer state instead.

Let ${\sigma }_L,{\sigma }_R$ be localized bonding orbitals of the two new C–O bonds, and let ${\sigma }_L^{\ast },{\sigma }_R^{\ast }$ be their antibonding partners. Their symmetry combinations areThe two remaining active orbitals are out-of-ring oxygen lone-pair combinations, $n_{\pm }=(p_{y,L}\pm p_{y,R})/\sqrt{2}$. The opposite-phase combination $n_{-}$ is the HOMO in the supplied ordering and $n_{+}$ is HOMO−1. Both are filled, as are the two C–O bonding combinations, givingThe original active-orbital sketches show these phase relations, while the correlation diagram gives a qualitative energy ordering. Precise splittings between unrelated symmetry blocks are not determined by the fragment ionization energies. The retained O–O sigma bond, the C–C sigma bond and the other framework orbitals lie outside this six-orbital active comparison; the eight-electron count does not describe all valence electrons of the product.

In the electronic state correlation diagram, the singly occupied reactant orbitals for triplet oxygen have symmetries $B_1$ and $A_2$, so their spatial product is $B_2$. The reactant state is therefore a triplet ${}^3{B}_2$, whereas the closed-shell product is ${}^1{A}_1$. A strictly spin-conserving path cannot correlate triplet reactants directly to the ground-state product. It correlates to an excited triplet product configuration or requires intersystem crossing; the symmetric direct reaction consequently has a substantial electronic obstacle.

Simply changing the oxygen spin does not repair the occupied carbon-orbital correlation. The lowest oxygen singlet manifold has spatial components of $A_1$ and $B_2$ symmetry in this subgroup. The $A_1$ singlet can have an avoided crossing with another $A_1$ configuration and reach the product singlet; the $B_2$ singlet cannot mix directly with the product ${}^1{A}_1$ while $C_{2v}$ is maintained. Following either closed-shell oxygen singlet component in the orbital diagram still leaves the carbon-pi electron pair in the rising ${\sigma }_{+}^{\ast }$ orbital. Its redistribution into the final occupied orbitals requires configuration mixing at an avoided crossing. A barrier is therefore expected even for the symmetry-constrained concerted singlet addition; singlet does not imply barrierless. The state curves in the original diagram are schematic, illustrating this avoided-crossing argument rather than computed energies.^[ref]

The real singlet reaction can lower this obstacle through asymmetric or stepwise pathways that are absent from the stipulated direct approach. A primary multireference study explicitly distinguishes singlet and triplet addition through biradical intermediates, supporting that limitation of a one-coordinate symmetry argument.^[ref] Neither correlation alone nor a spin label supplies a numerical activation energy.

Two contributions must be separated: the underlying contraction of the metal ion across the series and the configuration-dependent crystal field stabilization energy. Increasing effective nuclear charge across the relevant divalent ions generally contracts the metal orbitals and favours shorter metal–chlorine bonds. Superimposed on this, a weak-field tetrahedral chloride environment places the $e$ set below the $t_2$ set, at $-0.6{\Delta }_t$ and $+0.4{\Delta }_t$ relative to the barycentre. The high-spin occupancies and corresponding crystal field stabilization energies areBecause the tetrahedral crystal field splitting increases on bringing the ligands inward, the stabilization is especially effective at shortening the $d^7$ bond. Adding the next electron to the upper $t_2$ set reduces that stabilization; in a simple sigma-bonding description this set also has more metal–ligand antibonding character. That change explains the upward deviation from $d^7$ to $d^8$ despite the general contraction trend. By $d^{10}$ the crystal field stabilization energy is zero, but the background contraction can still produce the shortest bond in the plotted set. Thus the $d^7$ shortening and $d^8$ rebound reflect orbital filling, while the overall decrease reflects contraction. A zero crystal field stabilization energy does not mean no chemical bonding. Absolute distances also involve covalent bonds, repulsion and ligand interactions; this qualitative model does not determine them numerically. No inference about the missing $d^9$ distance is required.

The molecular orbital diagram has one level at $\alpha +2\beta$ labelled $A_{1g}$, two degenerate levels at $\alpha$ labelled $E_u$, and one level at $\alpha -2\beta$ labelled $B_{1g}$, in increasing energy for $\beta <0$. Each spatial molecular orbital admits two opposite-spin electrons. The diagram below also shows how one representative distortion lifts the middle degeneracy; the distorted scale is illustrative, not a fitted molecular spectrum.

For two electrons, $n+1+D<4$ whenever $0<n<1$: both sides of $D<3-n$ are positive and squaring reduces the inequality to $n<1$. Since $\beta <0$, the trapezium then has a higher total electronic energy than the square. The two-electron model favours the square. For four electrons, $D>2$, so $2\beta D<4\beta$: the four-electron model favours the distortion toward a trapezium. The lower split middle orbital is occupied while the upper one is empty.

The distinction is especially clear near the square. Write $\delta =1-n>0$; then $D=2+{\delta }^2/4+O({\delta }^4)$ andIn particular the four-electron lowering in this specified one-edge model is quadratic near the square; a first-order lowering of the entire occupied sum should not be asserted merely from the middle-level splitting. This Hückel square-to-trapezium distortion is an electronic comparison under the stated fixed-$\alpha$, fixed-short-edge-$\beta$ assumptions. Nuclear repulsion, other bonding contributions and the structural cost of distortion are omitted, so an actual equilibrium geometry or optimum $n$ is not determined by this calculation alone.

Put $D=\sqrt{n^2-2n+5}=\sqrt{(n-1{)}^2+4}$. For $0<n<1$, $D>n+1$, since $D^2-(n+1{)}^2=4(1-n)>0$. Remembering $\beta <0$, the two lowest trapezium molecular orbital energies areThe remaining levels lie at $2\alpha -{ϵ}_2$ and $2\alpha -{ϵ}_1$. Filling the lowest spatial orbitals with opposite-spin pairs givesFor the square four-electron case, two electrons occupy the lowest orbital and two occupy the degenerate $E_u$ set. Their summed one-electron energy is fixed here, but this Hückel method calculation alone does not decide the spin or correlation within that open shell.

Ni(II) is $d^8$. A square-planar complex can keep its strongly antibonding in-plane orbital empty and has a paired, $S=0$ configuration. A tetrahedral $d^8$ arrangement has two unpaired electrons, $S=1$.

The important change here is the substituent on the imine nitrogen: Me, Et, then tert-butyl. The phenolic ortho-tert-butyl substituent is common to all three ligands. The small N-Me groups permit coplanar N${}_2$O${}_2$ coordination in $\mathrm{Ni}({\mathrm{L}}_{\mathrm{a}}{)}_2$. N-tert-butyl crowds that arrangement and favors twisting the two chelate planes toward a distorted tetrahedron in $\mathrm{Ni}({\mathrm{L}}_{\mathrm{c}}{)}_2$. N-Et gives a closer energetic competition, so $\mathrm{Ni}({\mathrm{L}}_{\mathrm{b}}{)}_2$ accesses both structures. The observed equilibrium reflects ligand-field stabilization, pairing, steric strain and temperature-dependent entropy; it is not a change in nickel oxidation state.

The free end of N-bound $\mathrm{Co}-\mathrm{NCS}$ is sulfur. Cr(II) binds there, giving the remote bridge $\mathrm{C}{\mathrm{o}}^{\mathrm{III}}-\mathrm{N}-\mathrm{C}-\mathrm{S}-\mathrm{C}{\mathrm{r}}^{\mathrm{II}}$. Electron transfer and cleavage of the Co-N bond leave S-bound Cr(III), $[\mathrm{Cr}(\mathrm{SCN})(\mathrm{O}{\mathrm{H}}_2{)}_5{]}^{2+}$; labile Co(II) becomes hexaaquacobalt(II).

The much smaller rate constant shows that the efficiency of bridge formation/electron transmission is strongly sensitive to linkage orientation. Efficient donation and coupling in a Co-S-bound precursor need not be reproduced when cobalt is N-bound. Its reduction is still consistent with inner-sphere electron transfer; a slow rate alone is not a criterion for outer-sphere transfer.

Phosphine binding breaks the hydroxo-bridged dimer into mononuclear Rh-OH units. Hydroxide/boronic-acid association forms a Rh-O-B linkage; transfer of the aryl group from boron to rhodium, with Rh-O bond cleavage, gives the Rh-aryl initiator. This is transmetalation, with rhodium remaining formally Rh(I).

Writing $\mathrm{Ar}=p-{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{Me}$ and $L=\mathrm{PP}{\mathrm{h}}_3$, the balanced net activation isThe stoichiometric byproduct is boric acid, $\mathrm{B}(\mathrm{OH}{)}_3$. Each monomeric Rh(I) product is a 16-electron complex: $d^8$ Rh(I), four electrons from COD, two from phosphine and two from the aryl X ligand. Temporary phosphine dissociation makes room for alkyne binding during propagation.