Disconnect the internal Z alkene by a non-stabilized Wittig reaction:The aldehyde carbon becomes the right-hand internal alkene carbon; the ylide carbon becomes the left-hand one. Both the methyl ester and the remote terminal alkene survive the olefination.

Make the aldehyde from pentanal by Stork enamine reaction: form its pyrrolidine enamine, react at carbon with allyl bromide, then hydrolyse to $\mathrm{OHCCH}(\mathrm{nPr})\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$. The original alpha carbon of pentanal now bears allyl and n-propyl substituents. Control the alkylation to avoid multiple substitutions.

The other branch needs a three-carbon extension of the supplied four-carbon bromo alcohol. Protect the OH of $\mathrm{HO}(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{Br}$ as a tetrahydropyranyl ether. Convert the bromide into an organocuprate and add it conjugatively to methyl acrylate:Here the organocopper formula is shorthand for a suitable cuprate preparation, not a claim that an isolated monomer is necessary. The acrylate contributes two methylenes and the ester carbonyl. Remove the THP group, replace the resulting terminal OH by Br, and react with $\mathrm{PP}{\mathrm{h}}_3$ to form $[\mathrm{P}{\mathrm{h}}_3\mathrm{PC}{\mathrm{H}}_2(\mathrm{C}{\mathrm{H}}_2{)}_5\mathrm{C}{\mathrm{O}}_2\mathrm{Me}{]}^{+}\mathrm{B}{\mathrm{r}}^{-}$. Deprotonate adjacent to phosphorus to generate the non-stabilized ylide. This proton is much more acidic than the ester alpha protons, but controlled base stoichiometry and low temperature remain important.

Add the allylated aldehyde under conditions favouring the usual kinetic Z-selective Wittig olefination, and separate any E isomer. The result isThe four-carbon starting alcohol plus the three-carbon acrylate backbone supplies the seven-carbon acid-derived chain; the ester Me gives an eighth carbon. Pentanal plus allyl gives the other eight. The total is sixteen carbons, including the terminal alkene. A two-carbon extension would produce the wrong chain length.

Disconnect the amide into 2-propylhexanoic acid and N-methyl-2-(4-hexylphenyl)ethylamine. This keeps the nitrogen substituents distinct: one is Me and the other is the two-carbon aryl side chain. The acid's alpha carbon must bear both n-butyl and n-propyl groups.

For the acid fragment, esterify the supplied hexanoic acid, form its ester enolate with LDA, and alkylate with 1-iodopropane:hydrolysis gives 2-propylhexanoic acid; conversion with $\mathrm{SOC}{\mathrm{l}}_2$ gives the acyl chloride for the final coupling. Limit the alkylating reagent and isolate the monoalkylated ester to avoid overalkylation.

For the aryl fragment, use another portion of hexanoic acid to make hexanoyl chloride. A Friedel-Crafts acylation of bromobenzene with this reagent and $\mathrm{AlC}{\mathrm{l}}_3$ gives ortho and para products; isolate the para bromoaryl ketone. A Wolff-Kishner reduction converts $\mathrm{ArCO}(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{Me}$ into $\mathrm{ArC}{\mathrm{H}}_2(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{Me}$ without deliberately hydrogenolyzing the aryl bromide. Acylation followed by carbonyl reduction installs an unrearranged n-hexyl group; direct Friedel–Crafts alkylation is a less reliable way to do that.

Convert the retained aryl bromide into its grignard reagent, open ethylene oxide, and work up to obtain $\mathrm{p}-\mathrm{Hex}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$. Oxidize the primary alcohol to $\mathrm{p}-\mathrm{Hex}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$ under conditions that stop at the aldehyde, then perform reductive amination with methylamine and sodium triacetoxyborohydride. Excess methylamine helps favour the desired secondary amine over further alkylation. Finally acylate that secondary amine with the isolated acyl chloride, using a separate base to capture HCl.

The final amide has an N-methyl group, the para n-hexyl phenethyl group, and the required n-butyl/n-propyl acyl branching. The carbon check is $9$ in the acyl fragment plus $1+2+6+6=15$ in the amine fragment, giving $24$ overall. No absolute stereochemistry is specified at the acid alpha carbon.

The bipyrimidine ligand has two distinct N,N-chelating pockets. It can use one pocket as a terminal bidentate ligand, or use both to bridge two Eu centres. Take all three nitrates per Eu to be bidentate O,O donors.

A consistent structure for B is $\mathrm{Eu}(\mathrm{N}{\mathrm{O}}_3{)}_3(\mathrm{bpm}{)}_2$, with two terminal, singly chelating bpm ligands. Six nitrate oxygens plus four bpm nitrogens give coordination number ten. The unused pocket of each terminal bpm points away from that Eu centre.

For C, use a dimer $\mathrm{E}{\mathrm{u}}_2(\mathrm{N}{\mathrm{O}}_3{)}_6(\mathrm{bpm}{)}_3$. Place one bpm between the metals, chelating each Eu through a different N,N pocket. Add one terminal bidentate bpm to each Eu and three bidentate nitrates to each Eu. Each metal again has $2+2+6=10$ donor atoms, while the overall Eu:bpm ratio is $2:3$. In this model all four nitrogens of the bridging bpm are used, but only two of each terminal bpm are used.

Both structures have their accessible coordination sites occupied by multidentate ligands, and the chelate effect favours retention of those donors over adding or exchanging a solvent molecule. This accounts for the absence of coordinated solvent while allowing a higher coordination number than the thiocyanate example. Stoichiometry alone does not exclude polymeric alternatives or different nitrate binding modes; the monomer/dimer pair is an explicit internally consistent prediction.

Strong ligand-to-metal pi donation is favoured by an occupied ligand p orbital and a suitable empty metal d orbital of matching symmetry, with reasonably similar energies and good spatial overlap. Small donor atoms such as O or N can supply short bonds and efficient overlap; oxo, imido and nitrido ligands are familiar examples. Early transition metals in high oxidation states, often $d^0$, are particularly favourable because the metal acceptor orbitals are vacant and relatively low in energy. More spatially extended $4d$ and $5d$ orbitals can strengthen the overlap.

Metal-to-ligand pi backbonding is the complementary case: electron-rich, low-oxidation-state metals donate occupied d orbitals into low-lying ligand antibonding orbitals, as for CO. In both directions, filling a metal–ligand pi-antibonding orbital offsets the multiple-bond stabilization. The relevant prediction is orbital occupancy and overlap, rather than the presence of a formal double-bond line in a drawing.

Sulfur substituting for phosphorus acts as a donor impurity, giving n-type semiconductor behaviour and usually increasing electrical conductivity. A P site normally supplies five valence electrons; S supplies six. After forming the surrounding bonds, the extra electron can occupy a donor level below the conduction band and be thermally ionized into that band. The donor impurity then carries positive charge.

This assignment is supported experimentally: infrared transmission measurements on S-doped gallium phosphide resolved bound-donor transitions and determined a sulfur donor optical ionization energy of about $0.104\mathrm{eV}$, far smaller than the host band gap.^[ref]

The effect depends on substitution on P sites, donor ionization, compensation by acceptors and charge carrier mobility. A dilute active donor population increases the electron term in $\sigma =e(n{\mu }_n+p{\mu }_p)$; not every dopant must donate a mobile electron at room temperature. Sulfur's larger Allen electronegativity does not overturn this electron-counting argument or make a P-site sulfur atom an acceptor.

A $2.24\mathrm{eV}$ band gap is consistent with a semiconductor. In pure gallium phosphide, the valence band is essentially full and the conduction band essentially empty at low temperature; thermal or optical excitation can produce mobile electrons and electron holes.

At room temperature, $k_{B}T\approx 0.026\mathrm{eV}$ is much smaller than the band gap, so intrinsic excitation is strongly suppressed. Pure material consequently has a low intrinsic carrier density, while semiconductor doping can markedly increase its electrical conductivity.

Gallium phosphide forms a tetrahedral network of polar covalent bonds, with electron density displaced toward phosphorus. The alternating diamond-related structure is the zinc blende structure: each Ga has four P neighbours and each P has four Ga neighbours. A local orbital hybridization picture combines outward-pointing s,p-derived orbitals into four approximately $s{p}^3$ bond directions; in the crystal these interactions form extended occupied bonding bands.

Since ${\chi }_{\mathrm{Allen}}(\mathrm{P})-{\chi }_{\mathrm{Allen}}(\mathrm{Ga})=2.25-1.76=0.49>0$, P attracts shared electron density more strongly. Write ${\mathrm{Ga}}^{\delta +}-{\mathrm{P}}^{\delta -}$ to indicate an ionic contribution superposed on covalent bonds. Formal oxidation states Ga($+3$) and P($-3$) are electron-assignment labels, not proof of completely transferred charges or a purely ionic crystal.

Neither atom requires $3d$ hybrids for these four bonds. The Ga $3d^{10}$ shell is filled and relatively core-like; the relevant Ga valence orbitals are $4s,4p$. P uses principally $3s,3p$; its empty $3d$ orbitals are too high in energy to be a necessary bonding set. Detailed band calculations may include d-state mixing or core–valence effects, but that does not turn the tetrahedral bonding into an expanded-valence d-hybrid construction.

Allen electronegativity is proportional to the average binding energy of the free ground-state atom's valence electrons. For an s,p main-group atom, use positive, spectroscopically inferred one-electron binding energies ${\epsilon }_s,{\epsilon }_p$ and their occupancies $n_s,n_p$:Here $C>0$ fixes the conventional scale; the one-electron energies are obtained from appropriate atomic ionization/term energies, not merely from the first ionization energy. For group 15 the weighted average is $(2{\epsilon }_s+3{\epsilon }_p)/5$.^[ref]

The approximate connection to effective nuclear charge is ${\epsilon }_{n\ell }\sim R_{\infty }[Z_{\mathrm{eff}}(n\ell )/n_{\mathrm{eff}}{]}^2$. At a fixed shell, stronger effective nuclear attraction generally raises the valence binding energies and therefore Allen electronegativity. Down a group, increasing shell size and changes in penetration offset that attraction. There is no universal relation ${\chi }_{\mathrm{Allen}}\propto Z_{\mathrm{eff}}$ across different shells, nor can one substitute a core-electron $Z_{\mathrm{eff}}$ for the valence average.

Phosphorus has electron configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^3$, with $Z=15$. For a $1s$ electron, the only shielding contribution in Slater's rules is $0.30$ from the other $1s$ electron:For a $3p$ electron, the other four electrons in the $3s,3p$ group contribute $4(0.35)$, the eight $n=2$ electrons contribute $8(0.85)$, and the two $1s$ electrons contribute $2(1.00)$:The compact core electron experiences nearly the full nuclear charge, while the valence electron is strongly shielded by the core. The two numbers describe different orbitals of the same atom. Outer electrons do not shield the core in this Slater approximation, and the selected electron never counts toward its own shielding.

Photoelectron spectroscopy measures orbital-dependent binding energies from which an effective charge can be inferred using a stated model. Ultraviolet photoelectron spectroscopy accesses outer-shell ionizations; X-ray photoelectron spectroscopy accesses deeper core states. For an isolated gas-phase atom and a calibrated threshold, measure emitted-electron kinetic energy and use $I=h\nu -E_{\mathrm{kin}}$, with the appropriate energy reference. Optical atomic term spectra and series limits also constrain valence ionization energies.

A hydrogenic inference would set $I\simeq R_{\infty }{Z}_{\mathrm{eff}}^2/n_{\mathrm{eff}}^2$, hence $Z_{\mathrm{eff}}\simeq n_{\mathrm{eff}}\sqrt{I/R_{\infty }}$, with $R_{\infty }$ here an energy of approximately $13.6\mathrm{eV}$ and $n_{\mathrm{eff}}$ a chosen effective shell parameter. The measured binding energies are experimental; the extracted $Z_{\mathrm{eff}}$ values remain model dependent. They need not equal Slater screening parameters fitted to radial functions.

Unlike the equal $ns,np$ values in Slater's rules, spectroscopic values distinguish s from p because their penetration and shielding differ; the more tightly bound s states usually imply larger binding-energy-derived effective charges at a common shell parameter. Subshell, atomic-term and, especially for heavier atoms, spin-orbit coupling structure give further distinctions. Electron relaxation following ionization and electron correlation also matter. Thus real binding energies vary across the heavier group-15 atoms even where the simple Slater outer-shell estimates plateau; no universal constant correction to the Slater values follows.^[ref]

For the outer $ns,np$ valence electrons, Slater's rules predict a modest initial rise down group 15, followed by approximately constant values. Write the effective nuclear charge as $Z_{\mathrm{eff}}=Z-S$, where $S$ is the shielding constant. Each atom has $n{s}^{2}n{p}^3$, so the other four electrons in the outer group contribute $4(0.35)=1.40$. Electrons in shell $n-1$ contribute $0.85$ each, and those deeper contribute $1.00$ each.^[ref]

Applying those weights gives the following original calculations:The increasing nuclear electric charge is largely offset by additional inner-electron shielding; once the preceding shell contains $18$ electrons, these simple weights give a plateau. The $n-1$ shell includes its d electrons. This trend concerns outer-shell values, not every orbital: a $1s$ electron is shielded only by its partner, so $Z_{\mathrm{eff}}(1s)=Z-0.30$ rises strongly down the group. The larger principal shell also means nearly constant outer $Z_{\mathrm{eff}}$ does not imply nearly constant atomic size or binding energy.

For the cell reaction normalized as in 4(b)(i), six moles of electrons are transferred per mole of reaction. ThereforeThis is the Gibbs energy of reaction at the stated composition and gas pressure, not the standard Gibbs energy of reaction. Its negative sign confirms thermodynamic favourability in the written direction under those conditions.

At pH $3$, $[{\mathrm{H}}^{+}]=10^{-3}\mathrm{M}$, so the nitrate concentration is $0.001+2(0.1)=0.201\mathrm{M}$. With ideal thermodynamic activities,HenceThe activity of nitric oxide is $0.1$ because the supplied pressure is $0.1\mathrm{bar}$ and the gas standard is $1\mathrm{bar}$. This numerical result additionally approximates zinc and nitrate thermodynamic activities by their standard-concentration ratios; without their activity coefficients, a more precise nonideal value is not specified.

Bulk electroneutrality gives $\boxed{[\mathrm{N}{\mathrm{O}}_3^{-}]=[{\mathrm{H}}^{+}]+2[\mathrm{Z}{\mathrm{n}}^{2+}]}$. Each dissolved zinc ion contributes two positive elementary charges, whereas each nitrate contributes one negative charge.

This assumes complete nitric acid dissociation, only ${\mathrm{H}}^{+}$ and $\mathrm{Z}{\mathrm{n}}^{2+}$ as significant cations, and only nitrate as a significant anion. It neglects $\mathrm{O}{\mathrm{H}}^{-}$ at acidic pH, other dissolved salts, zinc complexes and additional ionic nitrogen products. It is a bulk charge balance; a thin charged electrode interface does not invalidate bulk electroneutrality.