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. $B{e}^0$

Answer 1

Answer: (D)

Effective nuclear charge $Z_{\text{eff }}=Z-S$
Where, $Z=$ atomic number
and $S=$ screening constant
$=0.35$ per electron for electron in $n$th orbit
$=0.85$ per electron for electron 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 (nth orbit) and two electrons in $1s$-orbital $((n-1)$ th orbit $)$
$\therefore S=0.35+2\times 0.85=2.05$
$\therefore Z_{\text{eff }}=4-2.05=1.95$ (given)
$B{e}^{+}1s^{2}2s^{1}2s$ electron is screened by two electrons in $1s$-orbital $((n-1)$ th orbital $)$
$S=2\times 0.85=1.70$
$\therefore Z_{\text{eff }}=4-1.70=2.30$
$B{e}^{3+}1s^1$ one electron in 1s-orbital (alone exists) is screened by another electron in same orbit. Thus, $S=0.30$
$\therefore Z_{\text{eff }}=4-0.30=3.70$
$B{e}^{3+}1s^1$ no-screening (single electron)
Thus, $Z_{\text{eff }}=4$
Thus, $Z_{\text{effective }}:Be<B{e}^{+}<B{e}^{2+}<B{e}^{3+}$