There are two groups of compounds and . Group contains three compounds . Group also contains three compounds . Hybridization of each central atom of group compounds is same as that of iodine in while

Question

There are two groups of compounds $A$ and $B$. Group $A$ contains three compounds $Px_4, Qy_3, Rz_2$. Group $B$ also contains three compounds $Sx_4, Ty_3, Uz_2$. Hybridization of each central atom of group $A$ compounds is same as that of iodine in $IBrCl^-$ while in group $B$ compounds it is same as that of iodine in $IBrCl^+$. Substituents $X, Y$ and $Z$ exhibit covalency of one in ground state. Then find the value of $x/y$. Where, $x$ and $y$ are total number of lone pairs present at central atoms of compounds of group $A$ and $B$ respectively.

Prashant Kotian · Accepted Answer

Correct answer: 2

**SOLUTION:** $IBRCL^-$ HAS $SP^3D$ HYBRIDIZATION. GROUP A COMPOUNDS: $PX_4$ (1 LONE PAIR), $QY_3$ (2 LONE PAIRS), $RZ_2$ (3 LONE PAIRS) $\RIGHTARROW X = \TEXT{TOTAL L.P.} = 6$.
$IBRCL^+$ HAS $SP^3$ HYBRIDIZATION. GROUP B COMPOUNDS: $SX_4$ (0 LONE PAIRS), $TY_3$ (1 LONE PAIR), $UZ_2$ (2 LONE PAIRS) $\RIGHTARROW Y = \TEXT{TOTAL L.P.} = 3$.
$X/Y = 6/3 = 2$.. Step 1: Understand that the potential energy of a bonding system reaches a minimum at the optimum internuclear distance (bond length), balancing attractions and repulsions. Step 2: If the internuclear distance is decreased below this optimum distance, the nuclei and core electron clouds are forced too close together. Step 3: This leads to a massive, steep increase in potential energy due to the sharp increase in electrostatic repulsions between the positively charged nuclei and between the negatively charged electron clouds.

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