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What how?how? what's Ne? isn't it the symbol for Neon?
Yes it is the symbol for Neon.
[Ne]3s23p6, here [Ne] means electronic config of Neon.
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What how?how? what's Ne? isn't it the symbol for Neon?
Sorry I'm new to this.. can you please explain why are we using Neon while working for Argon?What how?
Yes it is the symbol for Neon.
[Ne]3s23p6, here [Ne] means electronic config of Neon.
Read chapter 3 from Roger Norris book. Page 40Sorry I'm new to this.. can you please explain why are we using Neon while working for Argon?
Read chapter 3 from Roger Norris book. Page 40
Else,
We can do this for any element, BUT, we must use only noble gases in the brackets. I call this thenoble gas simplification. In this method of writing electron configurations, the last noble gas before we get to the element of interest is the noble gas we put into the brackets. For instance, for the element aluminum we write
Sulfur [Ne]3s23p4
Chlorine [Ne]3s23p5
Argon[Ne]3s23p6
Potassium[Ar]4s1
Calcium[Ar]4s2
We may NOT use any element in the brackets, only noble gases.
Yeah, not next element, but the next period until other Noble gas.Actually i don't have that book I'll buy it soon In sha Allah.
Can you please specify a reason for using the last noble gas?
is that true that last noble gas electronic configuration will be similar to the initial electronic comfiguration of the next element therefore we use them in brackets? I'm just guessing, correct me if I'm wrong..
JazakAllah.
for the entire period?Yeah, not next element, but the next period until other Noble gas.
Didn't you got that explanation?
Good luck.for the entire period?
I got it now.
The structure already shows CH3 and H.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w13_qp_43.pdf
guys, how to do Q6 (f) (iii) ?
The structure already shows CH3 and H.
You need to complete it using the following:
H2N
HO2C
H
OH
Hey can I get an A* in chem if I got A in P1 and P2 but D in practical? and how plzzzzz???
What's the reason for Sodium and group 1 elements having a low melting point?
When you go down the group, there will be more orbital shells. Thus, the outer-most (valence) electron shell will be subject to less electrostatic force attraction from the nucleus. Therefore, the valence electrons would be more free to move about. Metallic bonding is the force of attraction between the nucleus and the 'sea' of electrons, so the increased mobility of the valence electrons as it descends down the group will result in a weaker force of attraction. Thus, less energy will be required to overcome the forces of attraction.
Enna tu Post-CorrectorDoesn't answer her question.
Enna tu Post-Corrector
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