Why is more Fe-56 than Ni-62 produced by fusion in massive stars? Suppose we create an Fe-56 nucleus and an Ni-62 nucleus, each from individual protons and neutrons. In the case of Ni-62, more mass per nucleon is converted to binding energy. Thus we could argue the Ni-62 nucleus to be more strongly bound than the Fe-56 nucleus, if I'm correct so far.

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*Why is Fe-56 mentioned in many astrophysics texts as the most strongly bound of all nuclei?

Fe-56 is commonly mentioned as the dominant end product of fusion reactions in the core of massive stars. If I'm correct, fusion reactions beyond Si-28 are accompanied by partial disintegrations, resulting in a cocktail of fragments, not exclusively multiples of He-4 (nuclear statistical equilibrium).
2. Why is much more Fe-56 than Ni-62 produced in the core of a massive star, although Ni-62 is more tightly bound than Fe-56? What determines the share of each nuclide in the resulting iron group?
 A: The final stage of nucleosynthesis at the core of a massive star involves the production
of iron-peak elements, mostly determined by competition between alpha capture and photodisintegration. The starting material is mostly Si28 and weak processes are unable to significantly alter the n/p ratio from unity on short enough timescales. Thus the expected outcome of these quasi-equilibrium reactions should be nuclei with $Z \simeq N$.
Subject to that constraint, then the most stable nucleus resulting from alpha captures onto Si is Ni56.
To produce heavier nuclei (e.g. Zn60) by alpha capture requires higher temperatures (because of the higher Coulomb barrier) and at these higher temperatures photodisintegration drives the equilibrium back towards smaller nuclei.
So where does all the confusion arise? Most of the iron-peak material ejected in a supernova is formed slightly further out from the core in explosive Si burning. The major product is Ni56, as above, and this then undergoes weak decays to Co56 and then Fe56 with half lives of 6 days and 77 days respectively. Thus the most common iron-peak product that ends up in the interstellar medium is Fe56 (also from alpha capture in type Ia supernovae).
A: A very nice question about a common misconception in books on astrophysics (I've made the same mistake in a comment here). According to M.P. Fewell, the origin of this misconception lies in the theory of stellar nucleosynthesis and the abundance of the elements. While other nuclei have higher binding energy per nucleon, $^{56}\mathrm{Fe}$ is more abundant

because the competition between photodisintegration and charged-particle capture starts to favor photodisintegration at iron.

Once the chain of reactions reaches $\mathrm{Fe}$, there is no reason to examine heavier or more stable nuclei, because the conditions are such that they are barely produced.
