The third shell contanis one s orbital,three p orbitals and five d orbitals.Each orbital hold 2 electron so the total number of electron which can be accommodate in third shell are 18 electrons
Answer:
We are considering an Allene molecule here, CH2CCH2. To answer your question, NO, they don't have to lie on the same plane. The spatial arrangement between them is that the center carbon that forms these pi bind in the left and right are PERPENDICULAR to each other.
Explanation:
We see here that The terminal carbons are sp2 hybridized, and form three σ-bonds each which means that each terminal carbon has one unhybridized p-orbital. The central carbon atom is sp hybridized, and forms two σ-bonds which means it has two unhybridized p-orbitals. For better understanding, let's call these two orbitals px and py. Summarily, These orbitals are perpendicular to each other
You fine fr you need to stop playing games and hml so i can hit it from the back ma!!!!!
Answer:
Your cation is Pb2+
Explanation:
This is the explanation by chemical reactions
HCl (l) ----> H+(aq) + Cl-(aq)
Pb2+(aq) + 2Cl-(aq) ---> PbCl2 (s) ↓
H2SO4 (l) ----> 2H+ (aq) + SO4-2(aq)
Pb2+(aq) + SO4-2(aq) ---> PbSO4 (s) ↓
NaOH (l) ---> Na+(aq) + OH-(aq)
Pb2+(aq) + 2OH-(aq) ---> Pb(OH)2 (s) ↓
If the reaction takes place in a strong alkaline medium, lead hydroxide dissolves in excess of base
Answer:
(S)-3-methoxy-3-methylbutan-2-ol
Explanation:
In this case, we have an <u>epoxide opening in acid medium</u>. The first step then is the <u>protonation of the oxygen</u>. Then the epoxide is broken to generate the most <u>stable carbocation</u>. The nucleophile (
) will attack the carbocation generating a new bond. Finally, the oxygen is <u>deprotonated</u> to obtain an ether functional group and we will obtain the molecule <u>(S)-3-methoxy-3-methylbutan-2-ol</u>.
See figure 1
I hope it helps!