Question

The bond energies of ${\text{H - H , X - X}}$ and ${\text{H - X}}$ are $104\,,\,60$ and $102\,{\text{KCal mo}}{{\text{l}}^{ - 1}}$ respectively. The electronegativity of hydrogen is $2.1$ .The electronegativity of ${\text{X}}$ is :
A.$2.5$
B.$3.5$
C.$3.0$
D.$4.0$

Answer 1

Hint:Electronegativity is defined as the measure of tendency of an atom to attract a bonding pair of electrons. The more the electronegative is an atom then higher the tendency of that atom to attract the bond pair of electrons towards it. The Pauling scale is commonly used to quantify the value of electronegativity of a particular element. Whereas bond energy is the energy that needs to break one mole of covalent bonds between two atoms in the gaseous state. It is represented as
${\text{A - B}}\,\left( g \right)\,\, \to \,{\text{A}}\left( g \right)\,\, + \,{\text{B}}\,\left( g \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\vartriangle {{\text{H}}^0} = \, + {\text{x}}\,{\text{J}}$

Complete answer:
 The Pauling scale of electronegativity used is based on excess of bond energies. From this it is observed that for a heteropolar bond ${\text{A - B}}$ , the energy is generally greater than the average arithmetic mean value of the homopolar bond energies of ${\text{A - A}}$ and ${\text{B - B}}$ bonds. It is given by
${{\text{D}}_{{\text{A - B}}}}\, = \,\dfrac{1}{2}\,\left[ {{{\text{D}}_{{\text{A - A}}}} + {{\text{D}}_{{\text{B - B}}}}} \right] + {\vartriangle _{{\text{AB}}}}$
Where ,
$
  {{\text{D}}_{{\text{A - B}}}}\, - \,{\text{heteropolar bond energy of A - B}} \\
  {{\text{D}}_{{\text{A - A}}}}\, - \hom {\text{opolar bond energy of A - A}} \\
  {{\text{D}}_{{\text{B - B}}}}\, - \,{\text{homopolar bond energy of B - B}} \\
  {\vartriangle _{{\text{AB}}}}\, - \,{\text{excess bond energy}} \\
 $
The excess bond energy ${\vartriangle _{{\text{AB}}}}$is related to the electronegativity of element ${\text{A}}$ and ${\text{B}}$ is given by
$
  {\chi _{{\text{A}}\,}}\, - \,{\text{electronegativity of A}} \\
  {\chi _{\text{B}}}\, - \,{\text{electronegativity of B}} \\
 $
So here by assigning arbitrary values of electronegativity of one element , the electronegativity of the other can be calculated.
So here in the question bond energies of ${\text{H - H , X - X}}$ and ${\text{H - X}}$ is given $104\,,\,60$and $102\,{\text{KCal mo}}{{\text{l}}^{ - 1}}$ . So by using the values we can calculate the excess energy of ${\text{H - X}}$ , using the formula
${{\text{D}}_{{\text{A - B}}}}\, = \,\dfrac{1}{2}\,\left[ {{{\text{D}}_{{\text{A - A}}}} + {{\text{D}}_{{\text{B - B}}}}} \right] + {\vartriangle _{{\text{AB}}}}$
So by putting the values $\therefore \,{\vartriangle _{{\text{AB}}}}\, = \,102\, - \,\dfrac{1}{2}\left[ {104 + 60} \right]\, = \,\,20\,{\text{KCal}}\,{\text{ mol}}{{\text{ }}^{ - 1}}$
Here the electronegativity of one element is given already then we can calculate the electronegativity of ${\text{X}}$ by using the equation
  $
  {\chi _{{\text{A}}\,}}\, - \,{\text{electronegativity of A}} \\
  {\chi _{\text{B}}}\, - \,{\text{electronegativity of B}} \\
 $
${\chi _A} - {\chi _B}\, = \,\,0.208\,\sqrt {20} \, = \,0.93$
The value of electronegativity of an hydrogen is given by $2.1$
$
  {\chi _{\text{A}}} - \,2.1\, = \,0.93 \\
  \therefore \,\,{\chi _{\text{A}}}\, = \,0.93 + 2.1\, = \,3.0 \\
 $
So here the electronegativity of element ${\text{X}}$ is given by $3$.

Hence, the correct answer is option C.

Note:
Some factors regarding the bond energy given below
- The shorter the bond length , the stronger the bond.
-The stronger the bond , the greater will be the bond energy.
- Shorter bond length , the bond energy will be high.