Why does Beryllium form an sp hybrid orbital?

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Why does Beryllium form an sp hybrid orbital?

I noticed that when $B e$ $Be$ reacts with $H_{2}$ $H_2$ it forms an sp hybrid orbital even though the $B e$ $Be$ electron configuration is $1 s^{2} 2 s^{2}$ $1s^2 2s^2$ and thus the electrons do not occupy a $p$ $p$ orbital. Am I wrong in saying that an element's electrons need to occupy an orbit to it to exhibit a hybrid orbital with that respective orbital?

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Answer 1 · 2018-07-25T20:35:56

In this it has nothing to do with bond angles not being $180^{\circ}$ $180^@$ , nor does it matter that the $2 p$ $2p$ orbitals are not occupied.

The issue here is the orbital phases are incorrect for a bonding molecular orbital.

The $2 s$ $2s$ orbital does not stick out far enough to bond with two atoms at the same time.
The $2 p$ $2p$ orbital is the opposite phase on one side, which would have meant making two DIFFERENT $Be - H$ $"Be"-"H"$ bonds.

Upon hybridization, two IDENTICAL bonds can be made, to give:

![https://chem.libretexts.org/]( useruploads.socratic.org )

instead of:

![https://chem.libretexts.org/]( useruploads.socratic.org )

I assume you are referring to the formation reaction:

$Be (s) + H_{2} (g) \to {BeH}_{2} (g)$ $"Be"(s) + "H"_2(g) -> "BeH"_2(g)$ , $DeltaH_f^@ = "125.52 kJ/mol"$

It does not matter that the $2p$ orbitals are not formally occupied by $"Be"$ atom.

Orbital hybridization is a theory invented by Linus Pauling, and we only use this theory to help describe known molecular geometries around the CENTRAL atom only, that

$(i)$ utilize angles that are not $90^@$ and/or

$(ii)$ form multiple identical bonds even though different instead of identical pure orbitals are available.

In this theory for beryllium, we know that $2p$ orbitals are available, but not occupied in the atom:

$" "" "" "underbrace(ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr)))$
$" "" "" "" "" "" "color(white)(/)2p$

$" "$

$ul(uarr darr)$
$color(white)(/)2s$

Since beryllium needs to form two identical $"Be"-"H"$ bonds, it needs two identical orbitals. The easiest way to do that is to allow the $s$ and $p$ orbitals to mix, in the manner known as hybridization.

![https://courses.lumenlearning.com/](https://useruploads.socratic.org/QlOTQvv1Rj2Bc9Rc6yGI_8pwfx83tbmz3mbad0g1n.jpe)

Pay close attention to the relative orbital energies now, which are shown below:

$" "" "" "" "" "" "underbrace(ul(color(white)(uarr darr))" "ul(color(white)(uarr darr)))$
$" "" "" "" "" "" "" "color(white)(./)2p$
$" "$
$ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))$
$color(white)(/)sp" "" "sp$

$" "$
$ul"2s orbital was previously here in energy!"$

Due to this mixing,

One previously pure $2p$ orbital from above is lowered in energy to form an $sp$ orbital.
One previously pure $2s$ orbital rises in energy a bit to form an $sp$ orbital.
The two electrons previously in the $2s$ atomic orbital of beryllium can now spread out amongst the two hybrid $sp$ orbitals.

And this yielded two $sp$ orbitals, because one $2s$ and one $2p$ orbital is used per hybridized $sp$ orbital, and (as it follows directly from conservation of mass and charge), one must have conservation of orbitals... two in, two out.

They are identical orbitals, which must then make identical bonds.

Bond being made:

$overbrace("H")^(1s) -> larr overbrace("Be")^(sp) -> larr overbrace("H")^(1s)$

Bond made:

$" "" ""H"stackrel(1s-sp)stackrel("bond")stackrel(darr)(—)"Be"stackrel(sp-1s)stackrel("bond")stackrel(darr)(—)"H"$

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Why does Beryllium form an sp hybrid orbital?

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