The atom with the electron configuration
1s\(^{2}\)2s\(^{2}\)2p63s\(^{2}\)3p\(^{6}\)3d\(^{10}\)4s\(^{2}\)4p\(^{4}\) is in
a
period 4, p- block
b
period 3, p- block
c
period 4, d- block
d
period 3, d- block
Explanation
Correct Option
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DrSponsor
1 year ago
Just remember:
The highest number (n) = Period
Last orbital type (s, p, d, f) = Block
The highest value of 𝙣 in this configuration is 𝟒 (from 4s and 4p).
So, the element is in 𝙋𝙚𝙧𝙞𝙤𝙙 𝟒.
The last orbital here is 𝟒𝐩.
Elements in which the 𝙡𝙖𝙨𝙩 electrons enter the p-orbital belong to the 𝙥-𝙗𝙡𝙤𝙘𝙠.
Period 4 → 
p-block →
Thus, the correct answer is:
A. Period 4, p-block
