Chapter 3.5, Problem 10LTS

Interpretation Introduction

Interpretation:

The position of equilibrium in the given reaction has to be predicted.

Concept Introduction:

Predicting position of equilibrium: By comparing the conjugate bases using their structures, the position of equilibrium can be predicted.  The equilibrium favors towards the most stabilized negative charge of conjugate base.

For example: $\text{H}-\text{A}\text{+}{\text{B}}^{\text{-}}\rightleftharpoons {\text{A}}^{\text{-}}\text{+}\text{H}-\text{B}$

In the above reaction if ${\text{B}}^{\text{-}}$ conjugate base is more stable, then equilibrium will favors formation of ${\text{B}}^{\text{-}}$. If ${\text{A}}^{\text{-}}$ is more stable, then equilibrium will favor for formation of ${\text{A}}^{\text{-}}$ Hence, the position of equilibrium can be predicted by comparing the stability of ${\text{A}}^{\text{-}}$ and ${\text{B}}^{\text{-}}$.

Conjugate base stability: As the acid deprotonates, the stability of conjugate base formed is analyzed on the basis of electronegativity.

Factors affecting the stability of Negative charges:

• Electronegativity: Compare the atoms bearing the negative charge. If the negative charge is on high electronegative atom, more the conjugate base is stabilized and the compound readily donates proton.
• Size: Compare the atoms in the same column. If the negative charge is in the same column of periodic table, no more electronegativity will be the dominant effect. Instead dominant effect is SIZE.  Larger the size of the atom, better stabilize a negative charge by that atom.
• Orbitals: Look at the hybridization states of the orbitals that accommodate the negative charge of the conjugate base. The pair of electrons in a $\text{sp-}$ hybridized orbital is held closer to the nucleus than a pair of electrons in a ${\text{sp}}^2{\text{- or sp}}^3-$ hybridized orbital. As a result, electrons residing in $\text{sp-}$ orbital are stabilized by being close to the positively charged nucleus; is more stable.