-1
$\begingroup$

Why would the number of covalent bounds typically formed by an element equal to 8 minus the group number?

E.g for C, we have 8-4 bonds, for Cl, 8-7 bonds

$\endgroup$
1
  • $\begingroup$ It isn't always. Boron compounds, sulfur forming three or four delocalized bomds in SF4 and SF6, etc. $\endgroup$ Feb 1 '20 at 19:27
1
$\begingroup$

First of all, one has to adopt an outdated concept of group number for the concept mentioned in the question to work. The group number of carbon, silicon, germanium ... is 14, not 4, per IUPAC.

Reverting to the old definition (sometimes called main group), one notices that the old group number is equal to the valence electrons for the members of the group. In order to fulfill the octet rule$^{1}$, for example a nitrogen atom could engage in 3 bonds, following $$ 8 \quad \text{(due to octet rule)} \quad - \quad 5 \quad \text{(valence electrons = old group number)} \quad = \quad 3 \quad \text{(covalent bonds)} $$

However, there are many, many exceptions to this concept: $\ce{ClO4-}$, $\ce{SF6}$, $\ce{PCl5}$, both polymeric and monomeric $\ce{BeCl2}$, almost all borohydrides beginning with $\ce{B2H6}$, "sub-valent" but stable carbenes (compounds in which at least one carbon has a free electron pair), and more.

So, overall, the concept from the question is a bit of a crutch$^2$ for the more useful valence electrons.


$^{1}$ which follows from quantum mechanics.

$^{2}$ though not a bad observation per se.

$\endgroup$
1
$\begingroup$

In a very general aspect we could think of the following, if $X$ was the more electronegative element, say group 14 - group 17 and $M$ was the less electronegative element, then if the compound $M_mX_x$ forms, $M$ would give its electrons to $X$ so $X$ can form an octett which is the most stable electron configuration. To get this octett at $X$ you will require $8x$ electrons.

$M*e(M)+x*e(X)=8x$

Now let's asume there were covalent bonds between $M$ and $M$. Those would include electrons that cannot be shared with $X$, hence, we need to subtract $b(MM)$ for every bond. Same goes for lone-pairs at $M$ $e(M)$. For $X$ we need to increase the amount of electrons for every $b(XX)$ and $e(X)$. Therefore we get

$8x = m[e(M)-b(MM)-E]+x[e(X)+b(XX)]$

or

$\frac{m*e(M)+x*e(X)}{x}=8+\frac{(m[b(MM)+E]-x*b(XX))}{x}$

The part on the left side is what we call the valence electron concentration (VEC), a useful property to understand many intermetallic phases and structures. Form this we can say

$b(XX)=8-VEC(X)+\frac{m}{x}[b(MM)+E]$

Now if we assume a pure element like sulphur for example:

$m=0$

$VEC(X)=e(X)=N$

$b(XX)=8-VEC(X)=8-N$ so for sulphur (N=6) we get $b(SS)=8-6=2$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.