Which of the following represents the correct order of the acidity...

Which of the following represents the correct order of the acidity in the given compounds?

Options

A $\mathrm{FCH}_2 \mathrm{COOH}>\mathrm{CH}_2 \mathrm{COOH}> \mathrm{BrCH}_2 \mathrm{COOH}>\mathrm{ClCH}_2 \mathrm{COOH}$
B $\mathrm{BrCH}_2 \mathrm{COOH}>\mathrm{ClCH}_2 \mathrm{COOH}> \mathrm{FCH}_2 \mathrm{COOH}>\mathrm{CH}_3 \mathrm{COOH}$
C $\mathrm{FCH}_2 \mathrm{COOH}>\mathrm{ClCH}_2 \mathrm{COOH}> \mathrm{BrCH}_2^2 \mathrm{COOH}>\mathrm{CH}_3 \mathrm{COOH}$
D $\mathrm{CH}_3 \mathrm{COOH}>\mathrm{BrCH}_2 \mathrm{COOH}> \mathrm{ClCH}_2 \mathrm{COOH}>\mathrm{FCH}_2 \mathrm{COOH}$

A. $\mathrm{FCH}_2 \mathrm{COOH}>\mathrm{CH}_2 \mathrm{COOH}> \mathrm{BrCH}_2 \mathrm{COOH}>\mathrm{ClCH}_2 \mathrm{COOH}$

Electron withdrawing substituent increases the acidity by increasing the ionic character of

-\mathrm{OH}

by inductive effect. Electronegativity decreases in the order.

\mathrm{F}>\mathrm{Cl}>\mathrm{Br}

-I effect also decreases in the same order. Thus, the correct sequence is

\begin{aligned} & \mathrm{FCH}_2 \mathrm{COOH}>\mathrm{ClCH}_2 \mathrm{COOH}> \\ & \mathrm{BrCH}_2 \mathrm{COOH}>\mathrm{CH}_3 \mathrm{COOH} \end{aligned}

Related Theory I effect is a permanent effect & generally represented by an arrow on the bond.