wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

It is because of the inability of ns2 electrons of the valence shell to participate in bonding that


A

Sn2+ is oxidizing while Pb2+ is reducing

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

Sn2+ and Pb2+ are both oxidizing and reducing

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

Sn4+ is reducing while Pb4+is oxidizing

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

Sn2+ is reducing while Pb4+is oxidizing

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C

Sn4+ is reducing while Pb4+is oxidizing


As we go down the group, the outermost s electrons of the ns2 np2 configuration tend not to participate in bond formation. Hence, the stability if +4 oxidation state decreases and the stability of the +2 oxidation state increases (as we descend). This is called inert pair effect.

This implies that Pb4+ would want to grab two electrons and become the more stable Pb2+ and itself get reduced. Hence Pb4+ is oxidizing. Likewise, for tin, Sn4+ is more stable than Sn2+ and hence, the latter ( as in - Sn2+) is reducing.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Oxides of Phosphorus
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon