The phenomenon described involves the deflection of a flowing stream of a liquid when a charged comb is brought near it. This deflection occurs for polar liquids. A charged comb generates a non-uniform electric field in its vicinity. Polar molecules, which have a permanent net dipole moment, will align themselves with this electric field. Due to the non-uniformity of the field, the force on the end of the dipole closer to the comb is stronger than the force on the end farther away, resulting in a net attraction towards the comb, causing the stream to deflect.

Nonpolar molecules do not have a permanent dipole moment and are not significantly affected by the static electric field. Therefore, a stream of a nonpolar liquid will not show any noticeable deflection.

To answer the question, we need to determine which of the given compounds are polar (i.e., have a non-zero net dipole moment, $\mu \neq 0$).

1. $O_2$ (Oxygen): This is a homonuclear diatomic molecule. The bond between the two oxygen atoms is nonpolar, and the molecule itself is nonpolar ($\mu = 0$). It will not be deflected.

2. $HF$ (Hydrogen Fluoride): This is a heteronuclear diatomic molecule. Due to the large difference in electronegativity between fluorine and hydrogen, the $H-F$ bond is highly polar, and the molecule has a significant dipole moment ($\mu \neq 0$). It will be deflected.

3. $H_2O$ (Water): The water molecule has a bent (V-shaped) geometry due to the two lone pairs on the central oxygen atom. The $O-H$ bonds are polar. Because of the bent shape, the bond dipoles do not cancel out, resulting in a net dipole moment for the molecule ($\mu \neq 0$). It will be deflected.

4. $NH_3$ (Ammonia): Ammonia has a trigonal pyramidal structure with a lone pair on the nitrogen atom. The $N-H$ bonds are polar. The geometry is not planar, so the bond dipoles do not cancel, leading to a net dipole moment ($\mu \neq 0$). It will be deflected.

5. $H_2O_2$ (Hydrogen Peroxide): This molecule has a non-planar, open-book structure. The $O-H$ bonds are polar. Due to its asymmetrical structure, the bond dipoles do not cancel each other out, resulting in a net dipole moment ($\mu \neq 0$). It will be deflected.

6. $CCl_4$ (Carbon Tetrachloride): This molecule has a symmetrical tetrahedral geometry. Although the individual $C-Cl$ bonds are polar, their vector sum is zero due to the symmetry of the molecule. Thus, $CCl_4$ is nonpolar ($\mu = 0$). It will not be deflected.

7. $CHCl_3$ (Chloroform): This molecule has a tetrahedral geometry, but it is not symmetrical because the four atoms attached to the central carbon are not identical (one H and three Cl). The dipole moments of the three $C-Cl$ bonds and one $C-H$ bond do not cancel out. Therefore, chloroform is a polar molecule ($\mu \neq 0$). It will be deflected.

8. $C_6H_6$ (Benzene): Benzene is a planar, highly symmetrical hexagonal molecule. Any small polarity in the $C-H$ bonds is cancelled out by the symmetry of the structure. Benzene is nonpolar ($\mu = 0$). It will not be deflected.

9. $C_6H_5Cl$ (Chlorobenzene): The substitution of a chlorine atom for a hydrogen atom on the benzene ring breaks its symmetry. The $C-Cl$ bond is polar, which gives the molecule a net dipole moment ($\mu \neq 0$). It will be deflected.

The compounds that will show deflection are the polar ones: $HF, H_2O, NH_3, H_2O_2, CHCl_3, C_6H_5Cl$.

Counting these compounds, we find there are 6 of them.