The $\mathrm{C}-\mathrm{C}$ single bond length is $1.54 \mathrm{A}°$ and that of $\mathrm{Cl}-\mathrm{Cl}$ is $1.98 \mathrm{A}°$. If the electronegativity of $\mathrm{Cl}$ and $\mathrm{C}$ are $3.0$ and $2.5$ respectively, the $\mathrm{C}-\mathrm{Cl}$ bond-length will be equal to __________ $\mathrm{A}°$.

It is provided in the question single bond lengths are:

$\mathrm{C}-\mathrm{C}$ is  $1.54 \mathrm{A}°$

$\mathrm{Cl}-\mathrm{Cl}$ is $1.98 \mathrm{A}°$

And the electronegativity of $\mathrm{Cl}$ is $3.0$ and that of $\mathrm{C}$ is $2.5$.

The bond length of C-Cl can be calculated using the formula as shown below:

$\mathrm{C}-\mathrm{Cl}=\frac{{\mathrm{r}}_{\mathrm{C}}+{\mathrm{r}}_{\mathrm{Cl}}}{2}-0.09\left({\mathrm{X}}_{\mathrm{Cl}}-{\mathrm{X}}_{\mathrm{C}}\right)$

Substituting the values provided in the question and solving further we get:

$$\begin{gathered} \mathrm{C}-\mathrm{Cl}=\frac{1.54+1.98}{2}-0.09\left(3-2.5\right) \\ \mathrm{C}-\mathrm{Cl}=\frac{3.52}{2}-0.09\left(0.5\right) \\ \mathrm{C}-\mathrm{Cl}=\frac{3.52}{2}-0.045 \\ \mathrm{C}-\mathrm{Cl}=1.76-0.45 \\ \mathrm{C}-\mathrm{Cl}\simeq 1.71 \end{gathered}$$

Hence, the required answer is 1.71.