Cu2+與過量的氨水反應形成深藍色溶液的離子方程式?

Cu(OH)2(s) + 4NH3(aq) → [Cu(NH3)4]2+(aq) + 2OH-(aq)

左邊4個NH3係要平衡方程, 因為原子不會無原無故地出現或消失.

右邊有4個NH3 :[Cu(NH3)4], 自然左邊也要有, 不然的話就不平衡了.


2009-07-23 11:01:53 補充：
一條化學方程裡, 左右的原子數一定要一樣
即左邊有1個Cu, 右邊也只可以有一個Cu
右邊有(NH3)4, 即4*NH3, 那左邊也要有4個NH3

如果只列出參與的化學元數而不列出數目, 就如這樣:
N2+H2→NH3
人家就會質疑: 另一個N去了那?為何多了一個H, 從哪來的?
(正確為: 2N2+3H2→2NH3)

if you are a CE level student, forgive me to say
that there is nothing to understand. It is a matter of fact.
So, you can only remember it.

If you are a very good CE or a AL student :
First of all, there are NH3 molecules in ammonia water
i.e. NH3 (aq)

Then, Cu is a transition metal and there are many more vacant orbitals in which the lone pair electrons of NH3 can form a dative bond
i.e. bind to Cu2+

if you want to know why there are FOUR.
try to write the s, p, d, f box diagram of Cu2+
and you will see.