Question

Which one of the following group 16 elements does not exist in -2 oxidation state?
a.) S
b.) Se
c.) O
d.) Po
e.) Te

Answer 1

Hint: This element is rare and highly radioactive metal with no stable isotopes. It is also a known carcinogen. This element is widely used in heaters in space probes, antistatic devices, sources of neutrons and alpha particles, and also as a poison.

Complete step by step solution:
Most common oxidation state of group 16 elements is -2 oxidation state . ie, they can easily gain 2 electrons to attain noble gas configuration (stable configuration).
But in the case of Polonium it is an exemption, it’s because of the fact it is having d electrons also.
So, it is clear that option (D) Po is our answer , it cannot show -2 oxidation state.
The filling of the d shell makes the atom smaller because of the poor shielding effect of d electrons and hence the electrons are tightly packed.
Because of this same reason, Selenium cannot acquire the highest oxidation state of $ \left( +VI \right) $.
The oxidation state of elements is defined as the number of electrons gained or lost to form a bond and its sign is the indication of ionic charge on the element.
Oxygen is the most electronegative element after fluorine.
Sulphur, selenium and tellurium usually show + 4 oxidation state in their compounds with oxygen and + 6 with fluorine.
Other elements of the group exhibit + 2, + 4, + 6 oxidation states but + 4 and + 6 are more common.
The stability of + 6 oxidation state decreases down the group and stability of + 4 oxidation state increases due to inert pair effect.
 Bonding in +4 and +6 oxidation states is primarily covalent.
In case of Oxygen and Sulphur , they are having only s and p electrons , so they can readily gain 2 more electrons to gain the stable noble gas configuration and hence are more stable in their -2 oxidation state.
The stability of -2 oxidation state decreases down the group.
Polonium hardly shows –2 oxidation states. Since electronegativity of oxygen is very high, it shows only a negative oxidation state as –2 except in the case of ${ OF }_{ 2 }$ where its oxidation state is + 2.
From the above discussion , let us conclude that (D) Po is the answer.

Additional information:
For making you more clear regarding this oxidation state concept, a table is shown below :

Element	Electronic configuration	Oxidation state
O	$ \left[ He \right] { 2s }^{ 2 }{ 2p }^{ 4 }$	-2, -1, 1, 2
S	$\left[ Ne \right] { 3s }^{ 2 }{ 3p }^{ 4 }$	-2, 2, 4, 6
Se	$ \left[ Ar \right] { { 3d }^{ 10 }4s }^{ 2 }{ 4p }^{ 4 }$	-2, 2, 4, 6
Te	$\left[ Kr \right] { { 4d }^{ 10 }5s }^{ 2 }{ 5p }^{ 4 }$	-2, 2, 4, 6
Po	$\left[ Kr \right] { 4f }^{ 14 }{ { 5d }^{ 10 }6s }^{ 2 }{ 6p }^{ 4 }$	2, 4

Let us now look at inert pair effect, which we mentioned earlier in the solution section:
Inert pair effect is defined as the non-participation of the two s electrons in bonding due to the high energy needed for unpairing them.
The inert pair theory was proposed by Sidgwick. He along with Powell accounted for the shapes of several molecules and correlated the shapes with some of their physical properties.

For Example: The inert pair effect among group 4 and group 5 elements. ${ Sn }^{ 2+ }$ and ${ Pb }^{ 2+ }$ and ${ Sb }^{ 3+ }$ and ${ Bi }^{ 3+ }$ which are the lower oxidation states of the elements are formed because of the inert pair effect. When the s electrons remain paired the oxidation state is lower than the characteristic oxidation state of the group.
The group 16 elements, also known as the chalcogens have 6 valence electrons, and hence they can achieve noble gas configuration either by gaining 2 electrons or by sharing two electrons i.e., by forming ${ M }^{ 2- }$ ions, or forming two covalent bonds.


Note: Along with Po, in case of Te and Se also, -2 oxidation state is most unstable for them. Don’t confuse inert pair effect with shielding effect, both are different concepts. Shielding effect describes the balance between the attraction between the electrons on valence electrons and the repulsion from the inner electrons.