I dun get it ! I tot Be has to lose 2 e- to gain a stable duplet
structure?
If stable octet it has to gain 6 e- ! How is it possible
here?
Then
does NaBeH3 exist?
Thanksss ultima!!!
To understand, you must draw out the mechanism and structures of
the reactants and products.
Can exist, but won't be stable, coz Be will lack a stable octet
and still kena a uninegative formal charge. Either BeH2 (lack
stable octet, but at least no formal charge), or Na2BeH4 (has
stable octet, but kena dinegative formal charge) are both more
stable than your NaBeH3, which kena the worst of both worlds (in
terms of stabilities).
But of course, you still won't fully understand, you still have
many questions, but that's enough for the forum, go ask your school
teacher or private tutor.
