Question

the pauling electronegativity of flourine and chlorine are

Answer 1

HELLO FRIEND HERE IS YOUR ANSWER,,,,,,,,,,

To calculate the required electronegativity of Hydrogen and fluorine are to be found by certain assumptions which can be represented as $X_H$ for Hydrogen and $X_F$ for fluorine respectively.

Now by implying and putting out the Principles of Pauling's formula. Basically it was developed and the relationship between day two of those atoms "A and B" are cordially bonded in together to give the following relationship, it was given by $\bf{Linus \: Pauling}$ who was a American Scientist to find out the difference between electronegativity between two atoms.

$\boxed{\bf{X_A - X_B = 0.1017 \sqrt{\triangle}}}$

Here this $\triangle$ has a separate formula accordingly...

$\boxed{\bf{\triangle = E_{A - B} - \sqrt{E_{A - A} \times E_{B - B}}}}$

Now, all the variables up there, like , $\bf{E_{A - B} \: E_{A - A} \: E_{B - B}}$ are denoting different bind association enthalpies which are expressed by the unit of $\bf{kcal \: mol^{- 1}}$, which basically changes the difference in their respective electronegativities to form a different valued equation altogether of two typical bonded atoms given by this relationship.

$\boxed{\bf{X_A - X_B = 0.208 \sqrt{\triangle}}}$

Whereas when it's representation is done in $\bf{kJ \: mol^{- 1}}$ with the bonds "A and B" or "A--B", "A and A" or "A--A", "B and B" or "B--B" for them respectively. Here, $X_A$ and $X_B$ are denoting their electronegativities of those two elements "A" and "B". If I'd arbitrarily attach a specific numeral value to fluorine as "4.0" which as the highest and most strongest ability to attract stable electrons, all other electronegativities of other elements Lower than that can be calculated.

Here by using that formula of Pauling we've to see the original values of some variables...

$\bf{BE_{H - H} = 104.2 \: kcal \: mol^{- 1}}$

$\bf{BE_{F - F} = 36.6 \: kcal \: mol^{- 1}}$

and,

$\bf{BE_{H - F} = 134.6 \: kcal \: mol^{- 1}}$

These are all bond energies of these elements and atoms of "Hydrogen" and "Fluorine".

Now, using Pauling's formula :

$\bf{| X_H - X_F | = 0.102 \sqrt{\triangle}}$ ; we used this formula because of that unit in $\bf{kJ \: mol^{- 1}}$

Calculate $\triangle$.

$\bf{\triangle = BE (H - F) - \sqrt{BE (H - H) \times BE (F - F)}}$

Substitute the following bond energy values into this equation.

$\bf{= 134.6 - \sqrt{104.2 \times 36.6} \: kcal \: mol^{- 1}}$

$\bf{= 134.6 - \sqrt{3813.72} \: kcal \: mol^{- 1}}$

$\bf{= 134.6 - 61.8 \: kcal \: mol^{- 1}}$

$\bf{= 72.8 \: kcal \: mol^{- 1}}$

Converting $\bf{kcal \: mol^{- 1}}$ to $\bf{kJ \: mol^{- 1}}$.

$\bf{\therefore \: \: = 72.8 \times 4.184 \: kJ \: mol^{- 1}}$

$\bf{= 304.6 \: kJ \: mol^{- 1}}$

Now substitute this to the original equation :

$\bf{\therefore \: \: | X_H - X_F | = 0.012 \sqrt{304.6}}$

$\bf{| X_H - X _F = 0.012 \times 17.4527934727}$

$\bf{\therefore \: \: | X_H - X_F | = 0.102 \sqrt{304.6}}$

$\bf{\therefore \: \: | X_H - X_F | = 1.74}$

Since the Electronegativity of Hydrogen on Pauling's scale is "2.1" or $\bf{X_H = 2.1}$

$\therefore$ As, the value of Electronegativity of Fluorine is higher than that if Hydrogen;

$\bf{X_F = 1.74 + X_H}$

$\bf{\therefore \: \: X_F = 1.74 + 2.1}$

$\boxed{\bf{\therefore \: \: X_F = 3.84 \: or \: 3.8}}$

Likewise, the Electronegativity for Chlorine is $\bf{2.5}$

HOPE THIS HELPS YOU AND SOLVES YOUR DOUBTS FOR CALCULATING ELECTRONEGATIVITY BY USING PAULING'S SCALE!!!!!