Arrange the following given acids in decreasing order of their Ka values
`FCH_(2)COOH(P), ClCH_(2)COOH(Q), O_(2)NCH_(2)COOH(R ), NC CH_(2)COOH(S)`

Options

A. `R gt P gt S gt Q`
B. `P gt Q gt R gt S`
C. `R gt S gt Q gt P`
D. `R gt S gt P gt Q`

Answer

To arrange the given acids in decreasing order of their Ka values, we need to analyze the electron-withdrawing effects of the substituents attached to the carboxylic acid group (-COOH) in each compound. The acids given are:

1. FCH₂COOH (P)
2. ClCH₂COOH (Q)
3. O₂NCH₂COOH (R)
4. NCCH₂COOH (S)

### Step 1: Identify the Electron-Withdrawing Groups
Each of the substituents (F, Cl, NO₂, CN) is an electron-withdrawing group. The strength of the acid is influenced by how effectively these groups can stabilize the negative charge on the conjugate base formed after the acid donates a proton (H⁺).

### Step 2: Determine the Strength of Electron-Withdrawing Groups
The order of electron-withdrawing ability of the groups is as follows:
- NO₂ (strongest)
- CN
- F
- Cl (weakest)

### Step 3: Analyze the Stability of Conjugate Bases
The stability of the conjugate base increases with the strength of the electron-withdrawing group. Therefore, the acid with the strongest electron-withdrawing group will have the highest Ka value (strongest acid).

### Step 4: Arrange the Acids Based on the Electron-Withdrawing Groups
Based on the analysis:
- R (O₂NCH₂COOH) has the NO₂ group, making it the strongest acid.
- S (NCCH₂COOH) has the CN group, making it the second strongest.
- P (FCH₂COOH) has the F group, making it the third strongest.
- Q (ClCH₂COOH) has the Cl group, making it the weakest.

### Final Order of Acids in Decreasing Order of Ka Values
Thus, the order of the acids in decreasing order of their Ka values is:
1. R (O₂NCH₂COOH)
2. S (NCCH₂COOH)
3. P (FCH₂COOH)
4. Q (ClCH₂COOH)

### Summary of the Solution
The final arrangement of the acids in decreasing order of their Ka values is:
**R > S > P > Q**