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Q.

+ 2 oxidation state of lead is more stable than + 4, because of

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a

presence of vacant orbitals

b

octet configuration

c

inert pair effect

d

penetration power

answer is C.

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Detailed Solution

IV A group elements
Common oxidation state is : "+4"
C shows :- +4 -4
Si:- +4 [Stable]

Ge, Sn;-  +2,+1,+4(stable)

Pb:-    +2, +4(stable)
→ For Ge, Sn +4 is stable oxidation state hence in +2 oxidation state they acts as reducing agents
→ Where as for Pb +2 oxidation state is state [due to inert pair effect] hence in +4 oxidation state Pb acts as oxidising agent
→ Higher oxidation state is due to involvement of ns and np electron
i.e., ns2 np2    +4
Lower oxidation state is due to involvement of only np electron
i.e., ns2 np(2) ⇒ +2
Higher oxidation stable stability dececreases from top to bottom
 

\large \\\\G{e^{ + 4}}\; > \;S{n^{ + 4}}\; > \;P{b^{ + 4}}\left[ {Stability\;order} \right]\\\\G{e^{ + 4}}\; < \;S{n^{ + 4}}\; < \;P{b^4}\;\left[ {Oxidisin g\;power\;order} \right]


Lower oxidation stability increases from top to bottom
Ge+2 < Sn+2 < Pb+2 Stability order

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