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Oxidation state of Group 16 Elements: The Chalcogens

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 M2- ions, or forming two covalent bonds.

Oxidation State
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.

The oxygen of group 16 is the most electronegative element after fluorine. The difference in electronegativity between metals and oxygen is generally very high. We know that ionic bonds are formed when the difference in electronegativity of any two elements is very high. Thus, metals and oxygen combine to form a metal oxide with oxygen attaining O2- form, so the oxidation state of oxygen is mostly (-II).

Electronegativity of chalcogens decreases down the group in the periodic table. Other chalcogen elements like sulphur, selenium, and tellurium react with a more electronegative element of groups 1, 2 and lanthanides i.e., the uppermost elements of groups 1, 2 and lanthanides to form sulphides, selenides, and tellurides. These compounds are the most stable compounds formed by these elements. Compounds formed above can be denoted as S2-, Se2-, and Te2-. The electronegativity difference between the compounds (sulphides, selenides, and tellurides) shows that they are on the borderline of 50% ionic and 50% covalent characters as in the case of PCl5. Phosphorus pentachloride is sometimes covalent in the solid-state but is ionic in the aqueous state.

The chalcogen group can also share their two electrons with another element to form two covalent bonds for example

\(\begin{array}{l}H_2O, F_2O, Cl_2O, H_2S ~and ~SCl_2\end{array} \)
. In the given examples the chalcogen elements have the least electronegativity except for H2O and H2S in which hydrogen is less electronegative in comparison to O and S. As we can see in the case of
\(\begin{array}{l}SCl_2\end{array} \)
(the electronegativity of chlorine = 3.5 and sulphur =2.5) and the oxidation state of sulphur is (+II).

Electronic configuration and Oxidation states (Unstable states are written in parentheses)



Oxidation States

O [He]2s2p4 -II, (-I)
S [Ne]3s3p4 -II, (II), IV, VI
Se [Ar]3d10 4s4p4 (-II), II, IV, VI
Te [Kr]4d10 5s5p4 II, IV, VI
Po [Xe]4f14 5d10 6s6p4 II, IV

Oxygen and sulphur have only s and p electrons, whereas the Se, Te, and Po have d electrons. The filling of the d shell makes the atom smaller and hence the electrons are tightly packed. Because of this reason, Se cannot acquire the highest oxidation state of (+VI). Whenever S is oxidised by

\(\begin{array}{l}HNO_3\end{array} \)
to form
\(\begin{array}{l}H_2SO_4\end{array} \)
(S +VI) but Se is oxidized by
\(\begin{array}{l} HNO_3\end{array} \)
to form
\(\begin{array}{l}H_2SeO_3\end{array} \)
(Se +IV). Some elements of group 16 like S, Se, and Te can attain oxidation states of IV and VI, which is more stable than the +II state. This is just the brief layout of oxidation states of group 16 elements.

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