Match the species given in column $I$ with the properties mentioned in column $I I$

$\begin{matrix} Column I & Column II (A) & B F_{4}^{-} & (i) & oxidation state of central atom is + 4 (B) & A l C l_{3} & (i i) & strong oxidising agent (C) & S n O & (i i i) & Lewis acid (D) & P b O_{2} & (i v) & can be further oxidised (v) & Tetrahedral shape \end{matrix}$

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Solution

(A) $B F_{4}^{-}$ has boron in $s p^{3}$ hybridisation due to its formation of $4 σ$ bonds with fluorine.

$∴$ Thus, $B F_{4}^{-}$ has tetrahedral shape.

(B) $A l C l_{3}$ is electron deficient species due to incomplete octet of aluminium. Hence, it acts as lewis acid and have tendency to accept lone pairs of electrons from any lewis base.

(C) In $S n O$ , $S n$ is having $+ 2$ oxidation state. It is more stable in $+ 4$ oxidation state and hence it can be further oxidized.

(D) In $P b O_{2}$ , $P b$ is having $+ 4$ oxidation state. Due to the inert pair effect, it is more stable in $+ 2$ oxidation state.

Hence, $P b O_{2}$ can act as a strong oxidizing agent by itself getting reduced from $+ 4$ to $+ 2$ oxidation state and oxidising the other species.

Hence, the correct match will be

$\begin{matrix} Column I & Column II (A) & B F_{4}^{-} & (v) & Tetrahedral shape (B) & A l C l_{3} & (i i i) & Lewis acid (C) & S n O & (i v) & can be further oxidised (D) & P b O_{2} & (i) & oxidation state of central atom is + 4 (i i) & strong oxidising agent \end{matrix}$

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Octet Rule

Standard XII Chemistry

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Match the species given in column I with the properties mentioned in column II Column IColumn II(A)BF−4(i)oxidation state of central atom is+4(B)AlCl3(ii)strong oxidising agent(C)SnO(iii)Lewis acid(D)PbO2(iv)can be further oxidised(v)Tetrahedral shape