1. Principle involved

For substituted acetic acids of type $XCH_2COOH$, acid strength depends mainly on the electron-withdrawing inductive effect ($-I$ effect) of substituent $X$.

• Stronger $-I$ effect $\Rightarrow$ better stabilization of the conjugate base $XCH_2COO^-$
• Better stabilization $\Rightarrow$ stronger acid

So we need the decreasing order of $-I$ effect among:

$NO_2,\ CN(=NC),\ F,\ Cl$

2. Compare the electron-withdrawing groups

For inductive effect at the $\alpha$-carbon:

• $NO_2$ has very strong $-I$ effect
• $CN$ also has strong $-I$ effect, but less than $NO_2$
• Among halogens, $F$ has stronger $-I$ effect than $Cl$

Thus,

$NO_2 > CN > F > Cl \quad \text{(in }-I\text{ effect)}$

3. Hence acid strength order

Therefore,

$NO_2CH_2COOH > NCCH_2COOH > FCH_2COOH > ClCH_2COOH$

4. Match with options

• A: $NO_2CH_2COOH > FCH_2COOH > CNCH_2COOH > ClCH_2COOH$ ❌

  Here $F$ is placed stronger than $CN$, which is incorrect.

• B: $FCH_2COOH > NCCH_2COOH > NO_2CH_2COOH > ClCH_2COOH$ ❌

  This incorrectly places $F$ above both $CN$ and $NO_2$.

• C: $CNCH_2COOH > O_2NCH_2COOH > FCH_2COOH > ClCH_2COOH$ ❌

  This incorrectly places $CN$ above $NO_2$.

• D: $NO_2CH_2COOH > NCCH_2COOH > FCH_2COOH > ClCH_2COOH$ ✅

5. Final answer

The correct decreasing order of acid strength is:

$NO_2CH_2COOH > NCCH_2COOH > FCH_2COOH > ClCH_2COOH$

So, the correct option is D.