Question

(a) Explain the Pauling scale for the determination of electronegativity. Give the disadvantages of Pauling scale.
(b) How does Fluorine differ from other halogens?

Answer 1

Hint:Pauling scale gives a numerical value of electronegativity based on the bond energy calculations. Electronegativity is the property of the bonded atoms by which it attracts the shared electron pairs.

Complete answer:
a)Linus Pauling gave a method to determine the electronegativity of the bonded atoms. It demonstrates electronegativity as the power of an atom to attract the bonded electrons towards it. The Pauling scale is expressed as an empirical relation between the energy of a bond and electronegativity.
In order to determine the relation let us consider a bond made of two dissimilar atoms $$A$$ and $$B$$ as $$A-B$$ .
$A-A+B-B\to 2A-B$
The corresponding bond energies of the molecules $$A-A$$ , $$B-B$$ and $$A-B$$ bonds are $$E_{A-B}$$ ​, $$E_{B-B}$$ ​ and $$E_{A-B}$$ respectively. The bond dissociation energies calculation indicates that the bond energy of $$A-B$$ is higher than the geometric mean of the bond dissociation energies of individual $$A-A$$ and $$B-B$$ bonds.
$$E_{A-B}>\sqrt{E_{A-A}\times E_{B-B}}$$ ​.
If $X_A$​ and $X_B$​ are the electronegativities of atoms A and B, the expression of electronegativity is given as the difference of the bond dissociation energies as shown below:
$$\mathrm{\Delta }=E_{A-B}-\sqrt{E_{A-A}\times E_{B-B}}={\left({\chi }_A-{\chi }_B\right)}^2$$
$${\chi }_A-{\chi }_B=0.208\sqrt{E_{A-B}-\sqrt{E_{A-A}\times E_{B-B}}}$$
where $${\chi }_A$$ ​ and $${\chi }_B$$ ​ are the electronegativities of A and B respectively.

b)The disadvantages of Pauling's scale are as follows:
 i) The accurate bond energies are unknown for solid elements so the calculation of the electronegativity is erroneous.
ii) The electronegativity of hydrogen is estimated as $$2.1$$ and it was taken as the reference for obtaining the electronegativity calculation for other elements.

(b) The difference of fluorine and other halogen atoms are as follows:
i) Fluorine is small and the most electronegative compared to other halogens. The difference in electronegativity results in hydrogen bond formation between hydrogen and fluorine atoms.
ii) The association of fluorine in hydrogen bonding makes $$HF$$ a weaker acid than other hydrohalide acids.
iii) The bond dissociation energy of $$F-F$$ bond is less than other elements. So fluorine is the more reactive element than other halogens.
(iv) The silver salt of fluoride anion i.e.$$AgF$$ is water soluble while other silver halides are insoluble in water.
(vi) Fluorine has negative oxidation state due to strong electronegativity character but other halogens have both negative and positive oxidation states.
(viii) Fluorine does not form polyhalide like chlorine, bromine or iodine. The reason lies in the unavailability of empty $$d-$$ orbitals.
Therefore, the correct option is C.

Note:
Fluorine is the smallest halide and has no $$d-$$ orbitals. So the acceptance of more electrons is not possible for fluorine. Besides Pauling scale Mulliken and Allred–Rochow electronegativity calculation are mostly used for determining the electronegativity of the atoms.