Question

IE (in kJ ${moI}^{-1}$) of $Li$ and $Be$ are given below:

	Z	${\left(IE\right)}_1$	${\left(IE\right)}_2$	${\left(IE\right)}_3$
$Li$	3	520.1	$X\left(1\right)$	11810
$Be$	4	899.3	$Y\left(2\right)$	14810

From above table $X\left(1\right)$ is higher than $Y\left(2\right)$ .

Answer 1

The given statement is true.
X(1) is higher than Y(2).
The electronic configuration of Li is $1s^{2}2s^1$. It loses one electron to obtain stable electronic configuration of $1s^2$ in which dutet is complete. It is electronic configuration of noble gas He. It is difficult to remove second electron from lithium as it will break the stable electronic configuration. Hence, X(1) has high value.
The electronic configuration of Be is $1s^{2}2s^2$. It loses one electron to obtain electronic configuration of $1s^{2}2s^1$. It is easy to remove second electron from beryllium as it will form the stable electronic configuration of noble gas He in which dutet is complete . Hence, Y(2) has low value.