a maximum of how many electrons having a given spin can occupy any given molecular orbital?

Question

1.) 0
2.)4
3.)2
4.)+ or - 1/2
5.) 1

Answer 1

One.

This is due to "Pauli's exclusion principle."

This states that no electrons in an atom can have *identical* quantum numbers.

Since there are only two possibilities for the "spin quantum number" of an electron: +1(spin-up), and -1(spin-down), a maximum of two electrons can occupy any particular orbital.

~WOMBAT

Answer 2

each molecular orbital only contains a maximum of two electrons with anti-parallel spins


In chemistry, a molecular orbital is a region in which an electron may be found in a molecule. A molecular orbital specifies the spatial distribution and energy of one (or one pair of) electron(s). Most commonly a MO is represented as a linear combination of atomic orbitals (the LCAO-MO method), especially in qualitative or very approximate usage.
A molecular orbital describes the behavior of one electron in the electric field generated by the nuclei and some average distribution of the other electrons. In the case of two electrons occupying the same orbital, the Pauli principle demands that they have opposite spin.

The Pauli exclusion principle is a quantum mechanical principle formulated by Wolfgang Pauli in 1925. It states that no two identical fermions (in your case the fermions are the electrons) may occupy the same quantum state simultaneously. A more rigorous statement of this principle is that, for two identical fermions, the total wave function is anti-symmetric. For electrons in a single atom, it states that no two electrons can have the same four quantum numbers, that is, if n, l, and ml are the same, ms must be different such that the electrons have opposite spins.

Answer 3

Whoah, it quite is no longer continually 8. think of roughly hydrogen, it quite is 2. And for heavier aspects with d orbitals, it has a tendency to be better than 8. however the levels of orbitals are like this, you have probable seen: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 (etc) after which you pass diagonally precise-proper to backside-left. (1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, etc...) reckoning on the style of atom, you %. the orbitals till you run out of electrons. Say carbon is atommic#6, with 6 electrons. so 1s2 is interior the 1st shell. So then, 2s2 2p2, provides the the rest 4 electrons in C's 6. yet, would not or no longer it quite is extreme-high quality if there have been 4 greater electrons to end the shell, and for this reason make it good. by fact of this C exists with 4 bonds. each and every bond has 2 electrons, a million of each and every is shared with C. that can provide 4 electrons to end the valence 8 shell of point 2: 2s2 2p6. (those of course hybridize). And as I pronounced, as you pass to heavier aspects, the valence will advance by fact now you have d orbitals which choose greater electrons to end the valence shell.

Answer 4

5.)1

Answer 5

only 2, one positive and one negative. (It's right there in your text book)