Bond distance in $HF$ is $9.17 \times 10^{-11} \ m$. Dipole moment of $HF$ is $6.104 \times 10^{-30} \ Cm$. The percentage ionic character in $HF$ will be : .............. $\%$
(electron charge $= 1.60 \times 10^{-19} \ C$)

Which of the following dissolves in ionic solvents?

The compound that will have a permanent dipole moment among the following is

The correct set of species with zero dipole moment is:
$(i) \, CO_2, \, (ii) \, COCl_2, \, (iii) \, CH_2Cl_2, \, (iv) \, BCl_3$

Pick out the molecule which has zero dipole moment.

Given below are two statements:
Statement-$I$: Since fluorine is more electronegative than nitrogen,the net dipole moment of $NF_3$ is greater than $NH_3$.
Statement-$II$: In $NH_3$,the orbital dipole due to lone pair and the dipole moment of $N-H$ bonds are in opposite direction,but in $NF_3$ the orbital dipole due to lone pair and dipole moments of $N-F$ bonds are in same direction.
In the light of the above statements,choose the most appropriate from the options given below.