Answer:
C.
Explanation:
The electronic configuration of N (7 electrons): 1s² 2s² 2p³.
The orbital 1s is filled with two electrons and their spinning direction is opposite and also electrons of 2s.
3p contains (3 electrons) should fill the 3 orbitals firstly. Every orbital contains 1 electron and be in the same spin direction.
So, the right choice is c.
A is wrong because 2 electrons of 3p are paired in the first orbital before filling every orbital.
B is wrong because the 2 electrons of 1s and 2s are in the same direction and also 2 electrons of 3p are paired in the first orbital before filling every orbital.
D is also wrong the 2 electrons of 1s and 2s are in the same direction and the electron in the second orbital of 3p are in opposite direction of the other 2 electrons.
1. Solid
2. Liquid
3. Gas
4. Plasma
H2SO4 + 2RbOH -> Rb2SO4 + 2H2O
If you want an explanation, keep reading.
In the first portion, there are two hydrogen ions and four sulfate ions.
The second portion has one rubidium ions and one hydroxide ion.
On the other side of the equation, in order to keep those two rubidiums balanced, you'll need to add a two at the beginning of the second portion, but in that process you are giving a second hydroxide value.
Back to the right side, there is there is water (H2O).
On the first portion, there were two hydrogen ions. The second portion also has two hydroxides because of the value change (adding the two to the front).
So on the fourth portion, you'd have to add another two so you could balance the four hydrogen ions (H2 and 2OH) and the two oxygen ions (2OH).
I hope this was easy to understand.
Answer:
D
Explanation:
We must study the reaction pictured in the question closely before we begin to attempt to answer the question.
Now, the reaction is a free radical reaction. This implies that only one electron is transferred. The transfer of one electron is shown using a half arrow rather than a full arrow. The both species are radicals (odd electron species) and contribute one electron each.
Hence we must show electron movements in both species using a half arrow.
