Which one of the following statements is NOT correct?

We begin by recalling what eutrophication means. Eutrophication is the enrichment of a water body with nutrients, chiefly nitrates ($$\text{NO}_3^-$$) and phosphates ($$\text{PO}_4^{3-}$$). This nutrient loading stimulates excessive algal and plant growth.

Because of the dense algal bloom, sunlight penetration into deeper layers of water is hindered. Aquatic plants in the deeper zone therefore cannot photosynthesise efficiently, so their production of oxygen decreases. Later, when the algal mass dies, its decomposition by bacteria consumes the dissolved oxygen (DO) present in water.

The stoichiometric oxygen demand of the decomposition can be expressed as

$$\text{Organic Matter} + O_2 \longrightarrow \text{CO}_2 + \text{H}_2O + \text{Energy}.$$

The more biomass there is, the more $$O_2$$ is required for complete oxidation. Hence, eutrophication always tends to lower the dissolved oxygen level.

Now we check each statement one by one.

Option A says, “Eutrophication indicates that the water body is polluted.” We have just seen that eutrophication arises from an overload of nutrients, which is indeed a form of water pollution. So this statement is correct.

Option B says, “The dissolved oxygen concentration below 6 ppm inhibits fish growth.” The thumb rule used by environmental engineers is that fish need at least about $$5 \text{-} 6 \text{ mg L}^{-1}$$ (ppm) of DO for normal growth; if $$\text{DO} \lt 6 \text{ ppm},$$ fish become stressed and growth is hampered. Hence Option B is correct.

Option C says, “Eutrophication leads to increase in the oxygen level in water.” We have already explained that decomposition of the excess biomass consumes oxygen according to

$$\text{BOD} = \frac{\text{Mass of } O_2 \text{ consumed}}{\text{Volume of water}},$$

and a high biochemical oxygen demand (BOD) means lower residual $$O_2$$. Therefore, eutrophication does not increase oxygen; it depletes it. Thus Option C is an incorrect statement.

Option D says, “Eutrophication leads to anaerobic conditions.” As the DO falls far below $$1 \text{ ppm},$$ aerobic microorganisms die, and anaerobic bacteria begin to dominate, producing gases like $$\text{H}_2S$$ and $$\text{CH}_4$$. So Option D is a correct statement.

Only Option C contradicts the scientific facts. Hence, the correct answer is Option C.