Answer : (C) Assertion is true, but Reason is false

Step 1 — Test the Assertion

Bromine water is a mild oxidising agent. It oxidises the $-CHO$ group of glucose into $-COOH$, leaving the primary alcoholic $-CH_2OH$ untouched :

$$\underset{\text{glucose}}{CHO-(CHOH)_4-CH_2OH} \xrightarrow{Br_2\ \text{water}} \underset{\text{gluconic acid}}{COOH-(CHOH)_4-CH_2OH}$$

The product still has six carbon atoms, exactly as the assertion states. The assertion is true.

Step 2 — Test the Reason

The reason claims the carbonyl group is absent from the open chain structure of glucose. This is plainly wrong.

The open chain structure is

$$CHO-(CHOH)_4-CH_2OH$$

and the $-CHO$ at C-1 is a carbonyl group. Its presence is exactly what allows the oxidation to occur.

The reason is false.

Step 3 — State the corrected reason

Glucose contains a free aldehyde group in its open chain structure, and bromine water — being a mild oxidising agent — oxidises only this aldehyde group to a carboxyl group.

Step 4 — Note what the reaction proves

The fact that a mild reagent brings about this oxidation shows that the carbonyl group in glucose is specifically an aldehyde, not a ketone. A ketone would need a far stronger oxidising agent and would break a $C-C$ bond.

This is why bromine water distinguishes glucose from fructose in neutral medium.

Step 5 — Contrast with the strong oxidising agent

Concentrated $HNO_3$ attacks both ends of the chain, giving the dicarboxylic saccharic acid :

$$COOH-(CHOH)_4-COOH$$