Question

For $Be$, $Z_{eff}=1.95$ and for $B{e}^{x+},Z_{eff}=2.30$. Hence, ion is:

• A. $B{e}^{+}$
• B. $B{e}^{2+}$
• C. $B{e}^{3+}$
• D. $Be$

Answer 1

Answer: (A)

$B{e}^{+}$

Effective nuclear charge $Z_{eff}=Z-S$
where $Z$ $=$ atomic number
and $S$ $=$ screening constant
$=$ 0.35 per electron for electron in $n^{th}$orbit.
$=$ 0.85 per electron for electrons in$(n-1{)}^{th}$ orbit
$=$ 1.00 per electron for electrons in $(n-2{)}^{th}$, $(n-3{)}^{th}$,$(n-4{)}^{th}$ orbit
$=$ 0.30 per electron in 1s-orbital (when alone)

$Be=$ $1s^{2}2s^2$ one valence-electron in $2s$ is screened by one electron in $2s-$orbital ($n^{th}$ orbit) and two electrons in 1s orbital ($(n-1{)}^{th}$ orbit)
$\therefore S=0.35+2\times 0.85=2.05$
$\therefore Z_{eff}=4-2.05=1.95$ (given)

$B{e}^{+}:1s^{2}2s^1,2s$-electron is screened by two electrons in 1s-orbital ($(n-1{)}^{th}$orbital)

$S=2\times 0.85=1.70$
$\therefore Z_{eff}=4-1.70=2.30$

$B{e}^{2+}:1s^2$ one electron in 1s-orbital (alone exists) is screened by another electron in same orbit. Thus,
$S=0.30$
$\therefore Z_{eff}=4-0.30=3.70$
$B{e}^{3+}:1s^1$ no-screening (single electron)
Thus, $Z_{eff}=4$

Thus, option $A$ is correct.