A water sample has ppm level concentration of following anions
F$$^{-}$$ = 10; SO$$_{4}^{2-}$$ = 100; NO$$_{3}^{-}$$ = 50
The anion/anions that make/makes the water sample unsuitable for drinking is/are

To decide whether any of the given anions make the sample unfit for drinking, we first recall the generally accepted maximum permissible limits for potable water (as recommended by bodies such as BIS or WHO).

For each anion we have the following limits:

$$\text{Fluoride ion, }F^- : 1.5\ \text{ppm (mg}\,\text{L}^{-1})$$

$$\text{Sulfate ion, }SO_4^{2-} : 250\ \text{ppm}$$

$$\text{Nitrate ion, }NO_3^- : 50\ \text{ppm}$$

Now we compare the given concentrations with these limits one by one.

For fluoride we have

$$\text{Given concentration of }F^- = 10\ \text{ppm}$$

$$\text{Permissible limit} = 1.5\ \text{ppm}$$

Since $$10\ \text{ppm} > 1.5\ \text{ppm},$$ the fluoride level is far above the allowed value, so fluoride definitely makes the water unsuitable.

For sulfate we have

$$\text{Given concentration of }SO_4^{2-} = 100\ \text{ppm}$$

$$\text{Permissible limit} = 250\ \text{ppm}$$

Because $$100\ \text{ppm} < 250\ \text{ppm},$$ the sulfate level lies well within the safe range; hence sulfate is not a cause for rejection.

For nitrate we have

$$\text{Given concentration of }NO_3^- = 50\ \text{ppm}$$

$$\text{Permissible limit} = 50\ \text{ppm}$$

Here $$50\ \text{ppm} = 50\ \text{ppm},$$ i.e. the nitrate content is exactly at the upper permissible boundary. Water containing an ion exactly at its maximum prescribed level is still regarded as acceptable. Therefore nitrate alone does not render the water unfit.

Summarizing these comparisons:

$$F^- : \text{Exceeds limit}$$

$$SO_4^{2-} : \text{Within limit}$$

$$NO_3^- : \text{At limit (acceptable)}$$

Consequently, out of the three anions listed, only fluoride makes the water sample unsuitable for drinking.

Hence, the correct answer is Option B.