$A : 1s^2\, 2s^2\, 2p^6$
$B : 1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^3$
$C : 1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^5$
$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}$
(b) Fourth period.
(c) The electronic configuration of element A is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}$
The electronic configuration of element $C$ is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$
From the electronic configuration of the elements A and C, the atomic size of A is greater than the atomic size of C. Hence, the distance between the outermost electron of element A and its
nucleus is more than the distance between the outermost electron and nucleus of element C. As
the atomic size increases, the effective nuclear attraction experienced by the outermost electron decreases. Therefore, among the atoms A and C, the attraction of the nucleus towards the outermost electron is greater in case of element $C$.
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