- A$\mathrm{Cr}$
- ✓$\mathrm{Mn}$
- C$\mathrm{V}$
- D$\mathrm{Fe}$
$\mathrm{V}: 2833\ \mathrm{KJ} / \mathrm{mol}$
$\mathrm{Cr}: 2990\ \mathrm{KJ} / \mathrm{mol}$
$\mathrm{Mn}: 3260\ \mathrm{KJ} / \mathrm{mol}$
$\mathrm{Fe}: 2962\ \mathrm{KJ} / \mathrm{mol}$
alternative
$Mn$ : $3 d^5 4 s^2$
$\mathrm{Fe}: 3 \mathrm{~d}^6 4 \mathrm{~s}^2$
$\mathrm{Cr}: 3 \mathrm{~d}^5 4 \mathrm{~s}^1$
$V: 3 d^3 4 s^2$
So Mn has highest $3rd$ IE among all the given elements due to $d^5$ configuration.
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Assertion $A :$ Flourine forms one oxoacid.
Reason $R$ : Flourine has smallest size amongst all halogens and is highly electronegative
In the light of the above statements, choose the most appropriate answer from the options given below.

