Chemistry, 18.09.2019 18:00, kaziyahf2006

How do valence electrons relate to cation formation?

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"A pair of oxygen atoms can form an O2 molecule in which each atom has a total of eight valence electrons by sharing two pairs of electrons. The term covalent bond is used to describe the bonds in compounds that result from the sharing of one or more pairs of electrons."

answered: davidoj13

The valence electrons are the ones that are lost to form the cations

Explanation:

The cations have to do with the valence electrons since in most cases the valence electrons are those that are outside the full-layer configuration.

The full layer configuration corresponds to the configuration of the noble gases which is the most stable.

Atoms of different elements tend to gain (anions) or lose electrons (cations) to reach this more stable configuration.

Therefore, to reach this configuration, the electrons that are "lost" or "won" are the so-called valence electrons.

answered: Guest

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

answered: Guest

remember that hf is only the energy of one photon. not two. not three. one. 1) 5.12 * 10^3 photons find the energy of one photon, then determine how many photons are needed to reach the sensitivity. e = hf = (6.626 * 10^-34 js)(5.95 * 10^19 hz) = 3.94247 * 10^-14 j per photon (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons round to three sig figs. 2) 5.77 * 10^-17 j energy of pulse = energy per photon * number of photons = hfn = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250) = 5.774559 * 10^-17 j round to three sig figs. 3) 4.07 * 10^8 kj/mol 10^12 pm = 1 m c = fλ f = c/λ e = hf = hc/λ = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol) = 40702571.4 kj/mol round to three sig figs. 4) 9.94 * 10^-1 m e = hc/λ λ = hc / e = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j) = 0.9939 m round to three sig figs. 5) 9.06 * 10^3 photons energy of pulse = energy per photon * number of photons number of photons = energy of pulse / energy per photon = energy of pulse / (hf) = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)] = 9055.23695 photons round to three sig figs. 6) 2.78 * 10^-15 j this is problem 1, except in reverse. sensitivity = minumum number of photons * energy per photon = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz) = 2.78070758 × 10^-15 j round to three sig figs.

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