Calculate the atomic number and number of p-electrons of an atom whose valence shell electronic configuration is 4s^2
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Question asked by Heena
Atomic structure
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Oldest comments (4)
The atom would be in Period 4 Group 2 which is Ca.
The variance shell electronic configuration is 4s^2.
The principal quantum number (4) gives you the Period number. So, starting from element ‘H’ in period 1 and counting down to the period 4 (or fourth row). That gets us to the element ‘K’. Now, start counting to the right through the s-block. 2 elements including the K, which will land us to ‘Ca’. This is the element we’re looking for with 4s^2.
Here’s the snap from atomic table for a little bit more clarity:
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Answered by chimini_chand
Atomic num=20??
Num of p-electrons=??
Didn’t get the idea…😕😕
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Reply posted by Heena
You never mentioned the number of p-electrons. A common equation, we often encounter is 4s^2 4p^5 for which the explanation is same for upto 4s^2 but will continue further to get the full result.
As it needs us to get an element from the p-block, we need to get through the d-block upto ‘Zn’. Once there, count five more atoms for 4p^5 configuration which will land you at ‘Br’. This is the desired element. It being, atom in period 4, group 17 - ‘Br’.
Don’t go on atomic number here, the simple way to calculate would be to look at the atomic table and just count the way around. As it’s 4 s, you just count upto 4th row down and then move to the right for s-block and then past the d-block as it needs you to find the element in p-block. Within p-block as it’s powered by 5, just count upto the 5th element there in that row which is ‘Br’ in our case here.
Hope that clarifies :slight_smile:
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Reply added by chimini_chand
Thank you…
:slight_smile::slight_smile::slight_smile::slight_smile::slight_smile:
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Reply posted by Heena