How do you determine which orbitals are used in hybridization?

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How do you determine which orbitals are used in hybridization?

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Answer 1 · 2017-05-30T21:56:16

Yes. For general chemistry, we just count the number of electron groups around the central atom, and assume that the orbitals used are in order of angular momentum $l$ $l$ .

"NORMAL", GENERAL CHEMISTRY WAY

We assume an ordering of

$s, p, p, p, d, d, d, d, d, f, f, f, f, f, f, f$ $s, p, p, p, d, d, d, d, d, f, f, f, f, f, f, f$

or

$s p^{3} d^{5} f^{7}$ $sp^3d^5f^7$ .

For example, in ${NH}_{3}$ $"NH"_3$ , three $N - H$ $"N"-"H"$ bonds are made and one lone pair of electrons is leftover, so we assume that the hybridization is $s + p + p + p \to s p^{3}$ $s+p+p+p -> color(blue)(sp^3)$ .

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(You can think of it as the central atom "anticipating" future bonds it can possibly make.)

Or, in ${BH}_{3}$ $"BH"_3$ , three $B - H$ $"B"-"H"$ bonds are made, and there are no other lone pairs or bonds. So, we assume the hybridization is $s + p + p \to s p^{2}$ $s+p+p -> color(blue)(sp^2)$ .

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In short, count the orbitals from left to right in the list above until you have the same number of orbitals as the number of electron groups.

MORE RIGOROUS WAY

There is a more rigorous way to do it using Group Theory. You do not have to know this for general chemistry, but I am just providing it for the interested reader.

Here it is in a scratchpad:
https://socratic.org/scratchpad/25f1d95969583d1e1c9d

Answer 2 · 2017-05-31T02:27:26

Here's a method using the VSEPR Theory to determine number of hybrid orbitals needed when using the Valence Bond Theory.

Explanation:

Here's a method that uses the VSEPR theory to determine how many hybrid orbitals would be needed for the central element of a given binary compound.
Need:
1. number of bonded electron pairs
2. number of non-bonded electron pairs
3. (NBPr + BPr)/2 => Parent Geometry => Number of Hybrids needed from central element's valence electrons.

Note: The detailing of this procedure will appear to be extensive, however, once the sequence is understood & with a bit of practice, these can almost be determined by inspection. I will finish with several examples of how this is a 'short cut' system.

DETAILED VERSION:
Example: Determine the geometry of the following:
$B e C l_{2}$ $BeCl_2$

(BPr) = Bonded Pr (BPr) = 2 => That is, 2 substrates attached to central element)

(NBPr) = NonBonded Pr (NBPr)
= $\frac{Valence Number (V) - Substrate Number (S)}{2}$ $("Valence""Number"(V) - "Substrate""Number"(S))/2$

Valence Number = Number of Valence Electrons in given compound = Be + 2Cl = 1(2) + 2(7) = 16

Substrate Number = Number of Octet electrons associated with bonded substrates in given compound = 2Cl = 2(8) = 16 (Duet if working with Hydrogen as a substrate)

$(N B P r) = \frac{V - S}{2}$ $(NBPr) = (V - S)/2$ = $\frac{16 - 16}{2} = 0$ $(16-16)/2=0$

Total Electron Pairs Associated with Central Element of Parent Geometry = BPr + NBPr = 2 + 0 = 2 => $A X_{2}$ $AX_2$ Geometry => Linear.

The TTL $e^{-}$ $e^-$ pairs associated with the central element also is the number of hybrid orbital needed for the VB Theory.

That is, 2 sp hybrid orbitals. sp hybrids => linear geometry
![http://www.adichemistry.com/general/chemicalbond/vbt/http://hybridization-illustrations.html](https://useruploads.socratic.org/5Uu7iXOgQOyrxR1lymP0_sp%20hybridization%20BeCl2.png)

ABREVIATED VERSION:
Determine geometry of $S n C l_{2}$ $SnCl_2$ .
(Is this linear of a derived structure?)

$B P r = 2$ $BPr = 2$

$N B P r = \frac{V - S}{2} = \frac{18 - 16}{2} = \frac{2}{2} = 1$ $NBPr = (V-S)/2 = (18-16)/2 = 2/2 = 1$

$V = 1 S n + 2 C l = 1 (4) + 2 (7) = 18$ $V = 1Sn + 2Cl = 1(4) + 2(7) = 18$
$S = 2 C l = 2 (8) = 16$ $S = 2Cl = 2(8) = 16$

Parent = BPr + NBPr = 2 + 1 = 3 => Trigonal Planar Parent $(A X_{3})$ $(AX_3)$

=> Derived Geometry $(A X_{2} E)$ $(AX_2E)$ => Bent Angular Geometry with 2 bonded pair and 2 nonbonded pair.

=> This is also the number of hybrid orbitals needed. => $3 (s p^{2})$ $3(sp^2)$ hybrids (2 bonded pr and 1 with nonbonded pr)
![http://www.adichemistry.com/general/chemicalbond/vbt/http://hybridization-illustrations.html](https://useruploads.socratic.org/bNtLTiUYTKiID7roYAIj_sp2%20hybridization%20SnCl2.jpg)

Determine the Geometry of $N H_{3}$ $NH_3$ .
(Is this Trigonal Planar or a Derived Structure?)

$B P r = 3$ $BPr = 3$ (number substrates attached to central element)
$N B P r = \frac{V - S}{2} = \frac{8 - 6}{2} = \frac{2}{2} = 1$ $NBPr = (V-S)/2 = (8-6)/2 = 2/2 = 1$

$V = 1 N + 3 H = 1 (5) + 3 (1) = 8$ $V = 1N + 3H = 1(5) + 3(1) = 8$
$S = 3 H = 3 (2) = 6$ $S = 3H = 3(2) = 6$

BPr + NBPr = 3 + 1 = 4 => Tetrahedral Parent $(A X_{4})$ $(AX_4)$ => Pyrimidal Geometry $(A X_{3} E)$ $(AX_3E)$ => 4 $(s p^{3})$ $(sp^3)$ Hybrid Orbitals ( 3 with bonded pairs & 1 with nonbonded pair).
![http://www.adichemistry.com/general/chemicalbond/vbt/http://hybridization-illustrations.html](https://useruploads.socratic.org/atHpn9kVSn6eR7pQ4OcP_ammonia%20hybridization.gif)

Determine the Geometry of $I C l_{4}^{-}$ $ICl_4^-$ .
(Is this a Tetrahedral Geometry or a Derived Structure?)

$B P r = 4$ $BPr = 4$
$N B P r = \frac{V - S}{2} = \frac{36 - 32}{2} = \frac{4}{2} = 2$ $NBPr = (V-S)/2 = (36-32)/2 = 4/2 = 2$

$V = 1 I + 4 C l + 1 e^{- =} 1 (7) + 4 (7) + 1 = 36$ $V = 1I + 4Cl + 1e^- = 1(7) + 4(7) + 1 = 36$
$S = 4 C l = 4 (8) = 32$ $S = 4Cl = 4(8) = 32$

BPr + NBPr = 4 + 2 = 6 => Octahedral Parent $(A X_{6})$ $(AX_6)$ => Square Planar Geometry $(A X_{4} E_{2})$ $(AX_4E_2)$ => 6 $(s p^{3} d^{2})$ $(sp^3d^2)$ Hybrid Orbitals ( 4 with bonded pairs & 2 with nonbonded pair).
![http://www.ilectureonline.com/lectures/subject/CHEMISTRY/12/92]( useruploads.socratic.org )

STURCTURES WITH MULTIPLE BONDS
Determine the Geometry of $C H_{2} = O$ $CH_2=O$ .
(The Lewis Structure shows this to be $H_{2} C = O$ $H_2C=O$ . Carbon is the central element and assume the double bond to be a single bond for defining geometry.)

BPr = 3
NBPr = 0
BPr + NBPr = 3 => Trigonal Planar Geometry about Carbon. This means that the structure is composed of $3 s p^{2}$ $3sp^2$ hybrid orbitals and 1 unhybridized $p$ $p$ -orbital at the carbon valence level.

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How do you determine which orbitals are used in hybridization?

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