# How do you determine if a molecule is linear?

Nov 26, 2017

See below.

#### Explanation:

You would need to take into consideration the Valence Shell Electron Repulsion Theory, or VSEPR Theory.

Here is a sample of a VSEPR Theory table:

The VSEPR Table used by your teacher may look a little different, but it is the same concept (Just keep in mind that "Bent"="Angular").

Based on this, you should be able to find your answer.

I hope that helps!

Nov 26, 2017

Determine if the molecule is linear by looking at the VSEPR theory

#### Explanation:

The Valence Shell Electron Pair Repulsion theory outlines this by its very confusing acronym and name.

Basically, just remember that lone pairs repel each other. Lone pairs are electrons with a negative charge, and since negative and negative repel (while positive charge would attract to negative) thus the reason for the repelling.

In other words, electrons that are not paired up will try to get away from each other as much as possible.

So, if you have a compound that has no lone pairs of electrons, like $B e {H}_{2}$, it will form into a linear compound as the hydrogens will repel each other as much as possible to make a straight line.

Other compounds, $B e {F}_{2}$ do have lone pairs of electrons on the fluorines, 6 electrons each to be exact. This compound is still linear because the equal amount of lone pairs will repel each other so much that the farthest away they can be from each other is when the molecule rests at 180 degrees.

Most compounds like $C {O}_{2}$, $S C {l}_{2}$ are linear (correct me if I am wrong). There are some exceptions.

Another way to look at this is to look at the base. So for example, water. Although ${H}_{2} O$ looks like this compound will be linear, it is not. Since oxygen has 8 electrons it can use up, and the 2 hydrogens only use 4, that means there are 2 pairs of lone electrons (or 4 electrons left). And since electrons repel, the shape of water will not be linear, but a tetrahedral bent shape.

Hope this helps!

Nov 26, 2017

Typically, the determination of molecular geometry involves the 3 dimensional arrangement of substrates attached (bonded) to a 'central' element of a 'binary' structure. The binary consideration for
a linear molecular geometry would be empirically based upon an $A {X}_{2}$ molecular formula with 'A' being the central element and X's the attached substrates. The $A {X}_{n}$ is most commonly used but
the following procedure can be applied to X-A-Y structures where there are different substrates attached to the central element.

For binary systems two items in the structure need be determined ...
(1) The number of bonded pairs of electrons
(2) The number of non-bonded pairs of electrons

The number of bonded pairs is equal to the number of substrates attached to the central element of the binary compound and represents the number of 'Bonded Pairs' of electrons.

For $B e C {l}_{2}$ there are 2 substrates attached to the central element Beryllium. This is the number of 'Bonded Pairs', or

BPrs = 2.

The number of non-bonded pairs is a bit more involved in that one needs to use the expression

$N B P r s$ = $\left(\frac{\text{Valence"."electrons" - "Substrate"."electrons}}{2}\right)$

Valence Electrons = 1Be + 2Cl = 1(2) + 2(7) = 16 valence electrons
Substrate Electrons* = 2Cl = 2(8) = 16 electrons

*when conforming to the octet rule. (Hydrogen fills at 2 electrons)

NBPrs = $\frac{16 - 16}{2}$ = 0 non-bonded pairs on Be

Total electron pairs => Parent structure => BPr + NBPr = 2 + 0 = 2
=> $A {X}_{2}$ => Linear Geometry X - A - X for the $B e C {l}_{2}$ structure.
That is, Cl - Be - Cl is a linear molecule.

To truely understand this concept apply this to $S n C {l}_{2}$ and determine geometry.

Is SnCl_2 Linear or not?

Bonded Pair = 2
NonBonded Pair = 1*
Total electron prs = 2 + 1 = 3 => $A {X}_{3}$ Parent => Trigonal Planar Structure with 2 bonded pair and one non-bonded pair on the central Sn element. The resulting geometry is $A {X}_{2} E$ bent angular and is non-linear.

NBPr = $\left(\frac{V - S}{2}\right)$ = $\left(\frac{18 - 16}{2}\right)$ = $\frac{2}{2}$ = 1 non-bonded pair
V = Sn + 2Cl = 4 + 2(7) = 18
S = 2Cl = 2(8) = 16

This can be applied to many structures to define geometry. The following is a table illustrating the geometries of $C {H}_{4} , N {H}_{3} \mathmr{and} {H}_{2} O$