# How many bonds in "ammonium hydroxide?"

Dec 10, 2016

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

#### Explanation:

$N {H}_{4} O H$ has 5 sigma bonds, $4 \times N - H + 1 \times O - H$, and an ionic bond between $N {H}_{4}^{+}$ and $H {O}^{-}$. Nevertheless, there is really no such thing as $\text{ammonium hydroxide}$. In water, $\text{ammonia}$ is solvated by several water molecules, and ammonia behaves as a base in water:

$N {H}_{3} \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s N {H}_{4}^{+} + H {O}^{-}$

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

Dec 10, 2016

I would say there are 5 bonds.

#### Explanation:

There are 4 covalent bonds in the sf(NH_4^+ ion and 1 covalent bond in $\textsf{O {H}^{-}}$.

$\textsf{N {H}_{4} O H}$ 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:

$\textsf{N {H}_{3} + {H}_{2} O r i g h t \le f t h a r p \infty n s N {H}_{4}^{+} + O {H}^{-}}$

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