The bond energies of $\mathrm{H}-\mathrm{H}, \mathrm{X}-\mathrm{X}$ and $\mathrm{H}-\mathrm{X}$ are $104 \mathrm{K}.\mathrm{Cal}$, $38 \mathrm{K}.\mathrm{Cal}$ and $138 \mathrm{K}.\mathrm{Cal}$ respectively the electronegativity of ' $\mathrm{X}$ ' is ___________.  Take $\sqrt{67}=8.18$.

$\mathrm{H}-\mathrm{H}$ bond is  $104 \mathrm{K}.\mathrm{Cal}$,

$\mathrm{X}-\mathrm{X}$ bond is $38 \mathrm{K}.\mathrm{Cal}$

and $\mathrm{H}-\mathrm{X}$ bond is $138 \mathrm{K}.\mathrm{Cal}$

We know that formula of electronegativity of two atoms can be mathematically represented using the formula:

${\mathrm{X}}_{\mathrm{A}}-{\mathrm{X}}_{\mathrm{B}}=0.208\times \sqrt{∆} --------\left(1\right)$

Firstly, calculate the value of$∆$ using the formula as shown below:

$$\begin{gathered} ∆={\mathrm{E}}_{\mathrm{H}-\mathrm{X}}-\frac{1}{2}\left({\mathrm{E}}_{\mathrm{H}-\mathrm{H}}+{\mathrm{E}}_{\mathrm{X}-\mathrm{X}}\right) \\ ∆=138-\frac{1}{2}\left(104+38\right) \\ ∆=138-\frac{1}{2}\left(142\right) \\ ∆=138-71 \\ ∆=67 \end{gathered}$$
 

Substituting the values provided in the question along with the calculated one, we can modify the formula (1) as:

$$\begin{gathered} {\mathrm{X}}_{\mathrm{X}}-{\mathrm{X}}_{\mathrm{H}}=0.208\times \sqrt{67} \\ \Rightarrow {\mathrm{X}}_{\mathrm{X}}=2.1+0.208\times 8.18 \\ \Rightarrow {\mathrm{X}}_{\mathrm{X}}=2.1+1.70144 \\ \Rightarrow {\mathrm{X}}_{\mathrm{X}}\simeq 3.8 \end{gathered}$$
Hence, the required answer is 3.8.