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3 years ago

Why can hydrogen atoms form covalent bonds but not helium

Chemistry

1 answer:

IgorLugansk [536]3 years ago

4 0

Explanation:

Atomic number of hydrogen is 1 and its electronic configuration is $1s^{1}$ . So, in order to attain stability it needs to gain one more electron from another atom.

As helium cannot donate its one valence electron because if it does so then there will be no electron present in a hydrogen atom which is not possible. Therefore, it needs to share its valence electron with another atom.

A chemical bond formed by sharing of electrons is known as a covalent bond. So, hydrogen atom always shares its valence electron and hence, it always forms a covalent bond.

On the other hand, atomic number of helium is 2 and its electronic configuration is $1s^{2}$ .

Since, helium atom has completely fill orbital so, it is stable in nature. Therefore, it will neither gain or lose electrons under normal conditions.

As a result, helium will not form a covalent bond.

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Explanation:

$PbBr_{2}$ will dissociate into ions as follows.

$PbBr_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^{-}(aq)$

Hence, $K_{sp}$ for this reaction will be as follows.

$K_{sp} = [Pb^{2+}][Br^{-}]^{2}$

We take x as the molar solubility of $PbBr_{2}$ when we dissolve x moles of solution per liter.

Hence, ionic molarities in the saturated solution will be as follows.

$[Pb^{2+}]$ = $[Pb^{2+}]_{o}$ + x

$[Br^{-}]^{2}$ = $[Br^{-}]_{o}$ + 2x

So, equilibrium solubility expression will be as follows.

$K_{sp}$ = $([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}$

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So, $[Br^{-}]_{o} + 2x$ will approximately equals to $[Br^{-}]_{o}$ .

Hence, $K_{sp} = x[Br^{-}]^{2}_{o}$

$4.67 \times 10^{-6} = x \times (0.10)^{2}$

x = $4.67 \times 10^{-4}$ M

Thus, we can conclude that molar solubility of $PbBr_{2}$ is $4.67 \times 10^{-4}$ M.

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