
The Bond energies of H-H, X-X, H-X are 104, 60, 102 $\dfrac{{KCal}}{{mol}}$ respectively. The electronegativity of hydrogen is 2.1. The electronegativity of X is :
Answer
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Hint:In this question we will be learning about the electronegativities of particles. How the bond energies are used to calculate the electronegativities of individual particles.
Complete step by step answer:
First let us write down the parameters given to us,
${E_{H - H}}$ = 104 $\dfrac{{KCal}}{{mol}}$
${E_{X - X}}$ = 60 $\dfrac{{KCal}}{{mol}}$
${E_{H - X}}$ = 102 $\dfrac{{KCal}}{{mol}}$
Now, we know that the
Difference in electronegativity of substance X and electro negativity of substance H is given as:
${E_{NX}} - {E_{NH}}$ = $0.208\sqrt \Delta $.
First, let’s calculate the value of delta, to do so let’s take the delta to the left hand side of the above equation and all others to the right hand side to solve it quick and easy:
$\Delta $ = ${E_{H - X}}$ - $\dfrac{1}{2}$ $[{E_{H - H}} + {E_{X - X}}]$
Now, let’s substitute the values we know and we were given in the question,
$ \Rightarrow \Delta = 20$
As we got the value of delta, let’s substitute that value in the first equation, where we calculated the differences in electronegativities. The equation is as follows:
Let’s consider that,
${X_A}$ is electronegativity of the substance that we need to find out, and
${X_B}$ is the electronegativity of the given substance Hydrogen,
$ \Rightarrow {X_A} - {X_B} = 0.208\sqrt {20} $
$ \Rightarrow {X_A} - {X_B} = 0.93$
As we were given the value of electronegativity of Hydrogen let’s substitute that value in th =e above equation,
$ \Rightarrow {X_A} - 2.1 = 0.93$
Finally, we get the electronegativity of ${X_A}$ (substance X) as:
${X_A} = 0.91 + 2.1$
$ \Rightarrow {X_A} = 3.03 \simeq 3.0$
Note:
The tendency of an ion in an atom to pull in the shared pair of electrons towards itself is known as electronegativity. It is a dimensionless property since it is just a tendency. It essentially demonstrates the net results of the tendency of atoms in various elements to draw in the bond forming electron pairs.
Complete step by step answer:
First let us write down the parameters given to us,
${E_{H - H}}$ = 104 $\dfrac{{KCal}}{{mol}}$
${E_{X - X}}$ = 60 $\dfrac{{KCal}}{{mol}}$
${E_{H - X}}$ = 102 $\dfrac{{KCal}}{{mol}}$
Now, we know that the
Difference in electronegativity of substance X and electro negativity of substance H is given as:
${E_{NX}} - {E_{NH}}$ = $0.208\sqrt \Delta $.
First, let’s calculate the value of delta, to do so let’s take the delta to the left hand side of the above equation and all others to the right hand side to solve it quick and easy:
$\Delta $ = ${E_{H - X}}$ - $\dfrac{1}{2}$ $[{E_{H - H}} + {E_{X - X}}]$
Now, let’s substitute the values we know and we were given in the question,
$ \Rightarrow \Delta = 20$
As we got the value of delta, let’s substitute that value in the first equation, where we calculated the differences in electronegativities. The equation is as follows:
Let’s consider that,
${X_A}$ is electronegativity of the substance that we need to find out, and
${X_B}$ is the electronegativity of the given substance Hydrogen,
$ \Rightarrow {X_A} - {X_B} = 0.208\sqrt {20} $
$ \Rightarrow {X_A} - {X_B} = 0.93$
As we were given the value of electronegativity of Hydrogen let’s substitute that value in th =e above equation,
$ \Rightarrow {X_A} - 2.1 = 0.93$
Finally, we get the electronegativity of ${X_A}$ (substance X) as:
${X_A} = 0.91 + 2.1$
$ \Rightarrow {X_A} = 3.03 \simeq 3.0$
Note:
The tendency of an ion in an atom to pull in the shared pair of electrons towards itself is known as electronegativity. It is a dimensionless property since it is just a tendency. It essentially demonstrates the net results of the tendency of atoms in various elements to draw in the bond forming electron pairs.
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