The correct options are
A ICl2− B I3− C N3−(i) ICl−22 Total valence electrons around I = 7+2+1
= 10 e− = 5 pairs
No. of bond pairs = 2
No. of lone pairs = 3
∴ It's linear in shape.
(ii) I−3
Total valence electrons around central I = 7+2+1= = 10 e− = 5 pairs
No. of bond pairs = 2
No. of lone pairs = 3
∴ It's linear in shape.
In the above case, the side I atoms can only act as monovalent because other high oxidation states would be possible only if it denotes an e− pair which is not possible here since the central I does not have empty orbital. the same thing cannot be done for N−3
(iii) N−3
Total valance electrons = 16
Resonating structures of IN−3
In all the structures, the central N is sp hybridized
∴ N−3 resonance hybrid is linear in shape.
(iv) ClO2
Total valence electrons around central I = 7+4= 11 e− = 5 pairs + 1 e−
There are two double bonds.
Revised TVEP = 3 pairs + 1 e−
∴ ClO2 is not linear as it contain 2 bond pairs 1 lone pair & 1 odd e− . It is bent in shape