The atomic number of an element having the va | vedclass

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(B) The given valence shell electronic configuration is $4s^2 4p^6$. This indicates that the $4th$ shell is completely filled,which corresponds to a n

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$36$

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(B) The given valence shell electronic configuration is $4s^2 4p^6$. This indicates that the $4th$ shell is completely filled,which corresponds to a noble gas configuration.The complete electronic configuration is $[Ar]^{18} 3d^{10} 4s^2 4p^6$.Total number of electrons $= 18 + 10 + 2 + 6 = 36$.Since the number of electrons in a neutral atom equals the number of protons,the atomic number $Z = 36$.

The atomic number of an element having the valence shell electronic configuration $4s^2 4p^6$ is

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