Answer:
![AU^{3+} : [Rn] 5f^3](https://tex.z-dn.net/?f=AU%5E%7B3%2B%7D%20%3A%20%5BRn%5D%205f%5E3)
Explanation:
Writing electronic configuration of any element you should know atomic number of that element ,
and also electrons are filling according to their energy level and first electron is filled in the lower energy orbital
and it follows n+1 rule if n+1 is same for two orbital electron will go first in the lowest value of n.
writing electronic configuration of ion can be done like first for their neutral atom and then add or remove electron it will make things easy because there are also some eception case their you may do wrong.
![AU : [Rn] 5f^3 6d^1 7s^2](https://tex.z-dn.net/?f=AU%20%3A%20%5BRn%5D%205f%5E3%206d%5E1%207s%5E2)
remove three electron from outer most shell of AU
A) For balanced chemical equation: 2HgO(s) → 2Hg(l) + O₂(g).
1) Mole ratio 1: n(HgO) : n(Hg) = 2 : 2 (1 : 1).
2) Mole ratio 2: n(HgO) : n(O₂) = 2 : 1.
3) Mole ratio 3: n(Hg) : n(O₂) = 2 : 1.
B) Balanced chemical equation: 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l).
1) Mole ratio 1: n(NH₃) : n(NO) = 4 : 6 (2 : 3).
2) Mole ratio 2: n(NH₃) : n(N₂) = 4 : 5.
3) Mole ratio 3: n(NH₃) : n(H₂O) = 4 : 6 (2 : 3).
4) Mole ratio 4: n(NO) : n(N₂) = 6 : 5.
5) Mole ratio 5: n(NO) : n(H₂O) = 6 : 6 (1 :1).
6) Mole ratio 6: n(N₂) : n(H₂O) = 5 : 6.
Polar molecules exhibit an unequal balance of charges between the individual elements of the compound. This is brought about by the large difference in their electronegativities. The H atom has the least amount of electronegativity. Then, it is a known periodic trend, that as you go downwards in a group, electronegativity decreases, and increase as you go from left to right. Thus, you can deduce that the most electronegative elements are found in the upper right corner which includes O, N and F atoms. Any bond created between Hydrogen and any of O, N and F atoms is a polar bond.
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