Answer
c) 17.7 is the correct answer
Work Step by Step
Observed from the Table 8.5 ,
Experimental Dipole moment = $1.08 D$
Theoretical Dipole moment = $4.8\times bond length$ = $4.8 \times 1.27$ = $6.096 D$
Percentage ionic character = $μ _{Experimental}$/$μ _{Theoretical}$ X 100
Percentage ionic character = ($\frac{1.08}{6.096}$)$\times100$
Percentage ionic character = 17.7165 $\approx$ 17.7
So, Percentage ionic character of HCl is 17.7
