TG EAMCET 4 May 2025 Shift 1 Paper

Step 1: Reaction with Lithium Aluminium HydrideWhen boron trifluoride $(\mathrm{BF}_3)$ reacts with lithium aluminium hydride $(\mathrm{LiAlH}_4)$ in ether, it forms lithium fluoride $(\mathrm{LiF})$, aluminium fluoride $(\mathrm{AlF}_3)$, and diborane $(\mathrm{B}_2\mathrm{H}_6)$ :$4 \mathrm{BF}_3+3 \mathrm{LiAlH}_4 \xrightarrow{\text{ Ether }}3 \mathrm{LiF}+3 \mathrm{AlF}_3+\mathrm{B}_2\mathrm{H}_6$Step 2: Naming the Product XSo, here $X=\mathrm{B}_2\mathrm{H}_6$ (which is called diborane).Step 3: Reaction of $X$ with AmmoniaDiborane $(\mathrm{B}_2\mathrm{H}_6)$ reacts with ammonia $(\mathrm{NH}_3)$ to first form an adduct, which is a combined compound: $[\mathrm{B}_2\mathrm{H}_6 \cdot 2 \mathrm{NH}_3]$.Step 4: Heating to Get ZWhen this adduct is heated, it breaks down to give borazine $(\mathrm{B}_3\mathrm{N}_3\mathrm{H}_6)$, which is compound $Z$, and hydrogen gas $(\mathrm{H}_2)$ :$2 \mathrm{B}_2\mathrm{H}_6+2 \mathrm{NH}_3 \xrightarrow[\mathrm{B}_2\mathrm{H}_6 \cdot 2 \mathrm{NH}_3]{} \rightarrow 2 \mathrm{B}_3\mathrm{N}_3\mathrm{H}_6(Z)+\mathrm{H}_2$Step 5: Counting Sigma $(\sigma)$ and $\Pi(\pi)$ Bonds in $Z$Borazine ( $\mathrm{B}_3\mathrm{N}_3\mathrm{H}_6$ ) is a ring, and its bonds are:There are 6 sigma ( $\sigma$ ) bonds between B and N in the ring.There are 3 sigma ( $\sigma$ ) bonds between B and H .There are 3 sigma $(\sigma)$ bonds between N and H .So, the total number of $\sigma$ bonds $(x)$ is: $6+3+3=12$.There are 3 pi ( $\pi$ ) bonds ( $y$ ) in the ring. So, $x+y=12+3=15$.