Why oxides of metals are less dense than pure metals?

Question

I want to ask that why densitis of most metal oxides is lesser than their metals . and i want to ask a second question that what is electronic confugeration of oxygen atoms and other relative element(for example sodium and potassiam in NaO2 and KO2 respectivly) in super oxides.

Answer 1

because the density of oxygen is lesser than metals

Answer 2

I know the second question. Oxides tend to be ionic compounds. The electron configuration of oxygen is in heaven once it matches that of the nearest noble gas (neon), which means it has a stable octet (8 valence electrons)

Oxygen by itself (neutral) is: 1s2 2s2 2p4
Neon is 1s2 2s2 2p6

The valence electrons are the level 2 ones
Oxygen therefore 2 more electrons to be happy. If I give neutral oxygen 2 more electrons, it gains a -2 charge, one for each, and becomes O^2-^, which has configuration 1s2 2s2 2p6 just like neon.

The metals in column 1 of the periodic table (Alkali metals) need to go the other way to get to the heavenly configuration of the noble gas before them. For sodium, it's again neon.
Sodium is 1s2 2s2 2p6 3s1 for a total of its atomic number 11
To get to neon's 10, sodium needs to LOSE 1 electron. It becomes charged as +1, because it lost 1 negative charge.

Na^+^ is 1s2 2s2 2p6, again like neon.

Potassium is in column 1, number 19 on the table. Its nearest noble gas is argon at 18. So Potassium wants to be like Argon.

K is normally 1s2 2s2 2p6 3s2 3p6 4s1
It, like sodium, needs to lose 1 electron to get where it wants to be:

K^+^ is 1s2 2s2 2p6 3s2 3p6
In this case, the valence is level 3, but it still contains 8 (octet) valence electrons.

Sodium oxide is Na2O, not NaO2. The charges need to be balanced for a stable oxide, and 2 Na's give 2(+1) and 1 O gives 1(-2) for a total of 0 (neutral). Potassium Oxide works much the same.