First find the oxidation states of the various atoms:
<span>in Cr2O2 2- Cr @ +1; In NH3 N @ +3; in CrO3 Cr @ +3, N2 N @ 0 </span>
<span>Note that N gained electrons, ie, was reduced; Cr was oxidized </span>
<span>Now there is a problem, because B has NH4+ which the problem did not, and is not balanced, showing e- in/out </span>
<span>B.NH4+ → N2 </span>
<span>Which of the following is an oxidation half-reaction? </span>
<span>A.Sn 2+ →Sn 4+ + 2e- </span>
<span>Sn lost electrons so it got oxidized</span>
Pure air is a mixture of several gases that are invisible and odorless. Consists of 78% nitrogen, 21% oxygen, and less than 1% of argon, carbon dioxide.
Answer:
The configuration of the atom would be 2-8-2.
Explanation:
Any atom of an element combines with other element to complete its octet and become stable.
The electron configuration of the given atom is 2-8-6. That means the atom has 6 electrons in its outermost shell. To become stable the atom should have 8 electrons in its outermost shell. The given atom has 6 electrons so it either lose 6 electrons or gain 2 electrons to complete its octet.
But we know the atom having 5,6,7 electrons in its outermost shell they do not lose, they gain either 3 or 2 or 1 electrons to complete its octet.
So we say that atom with the electron configuration 2-8-6 bond with the atom having electron configuration 2-8-2.
