What is the oxidation number for an element with the atomic number 16? *

[He]2s2 2p

Explanation:

group 16 elements, also known as the chalcogens have 6 valence electrons, and hence they can achieve noble gas configuration either by gaining 2 electrons or by sharing two electrons i.e., by forming M2- ions, or forming two covalent bonds. The oxidation state of elements is defined as the number of electrons gained or lost to form a bond and its sign is the indication of ionic charge on the element. Oxygen is the most electronegative element after fluorine. The difference in electronegativity between metals and oxygen is generally very high. We know that ionic bond is formed when the difference in electronegativity of any two elements is very high. Thus, metals and oxygen combine to form a metal oxide with oxygen attaining O2- form, so the oxidation state of oxygen is mostly (-II).

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.

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