Question

How many bonds in `"ammonium hydroxide?"`

Answer 1

AS written in the formula there are six......

Explanation:

`NH_4OH` has 5 sigma bonds, `4xxN-H+1xxO-H`, and an ionic bond between `NH_4^+` and `HO^-`. Nevertheless, there is really no such thing as `"ammonium hydroxide"`. In water, `"ammonia"` is solvated by several water molecules, and ammonia behaves as a base in water:

`NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^(+) + HO^-`

The equilibrium lies strongly to the left, but it does operate.

Answer 2

I would say there are 5 bonds.

Explanation:

There are 4 covalent bonds in the `sf(NH_4^+` ion and 1 covalent bond in `sf(OH^-)`.

`sf(NH_(4)OH)` does not exist as a discrete substance, you cannot isolate it .

It is best described as "aqueous ammonia" which is a system at equilibrium which you get by dissolving ammonia in water:

`sf(NH_(3)+H_2OrightleftharpoonsNH_4^(+)+OH^-)`

Since the `"pKa"` of ammonia is about `36`, but that of ammonium is about `9.4`, the equilibrium is pretty much completely towards ammonia (the weaker "acid").