AP EAPCET 22 May 2024 Shift 2 Solved Paper

Correct answer: option D, i.e. $1,1,1$.Interpreting $\mathrm{BrFF}_5$ as the intended $\mathrm{BrF}_5$:For $\mathrm{BrF}_5$, Br has $7$ valence electrons.Five electrons are used in five Br-F bonding pairs, so $7-5=2$ electrons remain as $1$ lone pair on Br.The VSEPR type is $AX_5E$, confirming $1$ lone pair on the central atom.For $\mathrm{XeO}_3$, the total valence electrons are $8+3\times6=26$.A valid Lewis structure has three Xe=O double bonds, one lone pair on Xe, and two lone pairs on each O, giving $12+2+12=26$ electrons.Thus Xe has $1$ lone pair, and the VSEPR type is $AX_3E$.For $\mathrm{SO}_2$, the total valence electrons are $6+2\times6=18$.The Lewis structure is $O=S=O$, with $1$ lone pair on S and two lone pairs on each O: $8+2+8=18$ electrons.Thus S has $1$ lone pair, and the VSEPR type is $AX_2E$.Therefore the required numbers are $1,1,1$, which match option D.