Question

Be is an element which is invariable bivalent and whose oxide is soluble in excess of ____ and its dipostive ion has a noble gas core.

• A. $NaOH$
• B. $HCl$
• C. $NaCl$
• D. $noneofthese$

Answer 1

Answer: (A)

$NaOH$

The electronic configuration of Be is $1s^{2}2s^2$ or $\left[He\right]2s^2$. It loses its two valence electrons to form $B{e}^{2+}$ with electronic configuration $1s^2$ or $\left[He\right]$.

$BeO\left(s\right)+2NaOH\left(aq\right)+H_{2}O\left(l\right)\to N{a}_{2}Be\left(OH{)}_4\right(aq)$

Hence, Be is an element which is invariable bivalent and whose oxide is soluble in excess of NaOH and its dipostive ion has a noble gas core.