Match the species given in Column I with the properties mentioned in Column II.

 Column I Column II A.  $$\text{BF}^-_4$$ 1. Oxidation state of the central atom is +4 B.  AlCl3 2. Tetrahedral shape C.  SnO 3. Lewis acid D.  PbO2 4. Can be further oxidized

Codes

 A B C D 1. 2 3 4 1 2. 1 2 3 4 3. 1 4 3 2 4. 4 1 3 2
Hint: The hybridization of B in ${\mathrm{BF}}_{4}^{-}$
A. In ${\mathrm{BF}}_{4}^{-}$, boron hybridization is sp3 hybridisation. The shape is tetrahedral.
B. In ${\mathrm{AlCl}}_{3}$ , Al octet is not completed.  Hence, AlCl3 acts as Lewis acid.
C. In $\mathrm{SnO}$, Sn2+ can show a +4 oxidation state. Hence, it can be further oxidised.
D. Oxidation state of Pb in PbO2 is +4. Due to the inert pair effect, Pb4+ is less stable than Pb2+, acts as a strong oxidising agent.