# How can we know that which element is more electronegative than other ? I've never come across any such formula or trick for this,except learning it by practice. Please tell if you know any.

## Eg: in heterolytic fission of CO, we know that oxygen will take both the electrons but what if there was some other element whose relative electronegativity I dont know, then how can I solve it.

Mar 27, 2017

Electronegativity INCREASES across a Period, from left to right as we face the Table...........

#### Explanation:

And excluding the $\text{Noble Gases}$, electronegatvity INCREASES across the Period from left to right, BUT DECREASES down the Group.

Why should this be so? Electronegativity tends to be a function of NUCLEAR charge (and thus dependent on $Z$, the atomic number), and SHIELDING by other electrons. These work in opposition. Incomplete electronic shells tend to shield nuclear charge VERY ineffectively. And thus across a Period (from RIGHT to LEFT as we face the Table), $\text{fluorine}$ is more electronegative than $\text{oxygen}$, which is more electronegative than $\text{nitrogen}$, which is more electronegative than $\text{carbon}$, which is more electronegative than $\text{boron}$. Can you see the Periodic trend?

Down a Group, the $\text{nucular charge}$ is shielded by inner closed shells that are complete. The valence electrons are also farther removed from the nuclear core, and thus, even tho the nuclear charge has increased, the attraction for the valence electrons should be DECREASED, and this is manifested by the well-known decrease in ionization enthalpies as we go down a Group, a column of the Periodic Table. And thus down the Group, $\text{electronegativity}$ DECREASES.

And so let's have a gander at the Pauling Scale, and examine the electronegativities:

Is this representation roughly consistent with what I have argued? The most electronegative element should thus be fluorine, and indeed it is. On the other hand, the LEAST electronegative elements should be the alkali, and alkaline-earth metals. Is this true?

This is one of the most important trends a chemistry undergraduate can learn. If you are unsatisfied with this answer, or seek clarification, voice your objection, and someone will address the issue.

And thus in your example, for $C \equiv O$, and also for $O = C = O$, electron density will clearly be polarized toward the more electronegative atom, and we may represent this polarization by ""^(+delta)C-=O^(delta+), and also for ""^(-delta)O=stackrel(delta^+)C=O^(delta-),

Mar 28, 2017