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 BF4-
A. In BF4-, boron hybridization is sp3 hybridisation. The shape is tetrahedral.
B. In AlCl3 , Al octet is not completed.  Hence, AlCl3 acts as Lewis acid.
C. In 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.