Basic Introduction
When we enter the world of chemistry as learners and students, we are exposed to a variety of chemical reactions, elements and equations. It is intriguing as we learn more about how various substances in our world react and bond with each other to produce different natural and man-made resources that we use for our living.
One such topic of study in Chemistry is the oxidation states of elements. Let us understand what this means.
Oxidation Process of Group 16 Elements
The elements of Group 16, termed as ore-forming elements belong to the p-block of the periodic table because as their last electron enters into the p-orbital. The group 16 elements are Oxygen (O), Sulphur (S), Selenium (Se), Tellurium (Te) and Polonium (Po). Out of these, selenium and tellurium are metalloids, polonium is unstable because it is a radioactive element and oxygen and sulphur are non-metals.
Science tells us that the valency of the electrons of the group 16 elements is 6 which in turn means that 2 more electrons will be required to achieve the octet state.
The electronic configuration of group 16 elements is ns2np4.
Oxidation State of Oxygen (O2)
For oxygen, its oxidation state is determined by how many elements it is able to gain or let go of to attain a noble gas configuration. Depending on how stable the compound formation is, an element can have only one oxidation state. It also depends on electronegativity and the electronic configuration of these elements. Electronegativity depends on the ability of the elements to attract electrons towards itself. The smaller, the greater its electronegativity.
Oxygen has high electronegativity. It displays a -2 oxidation state, which means that it gains 2 electrons in most of its metallic oxides. Because oxygen is small and d-orbitals are absent, oxygen does not have space for unpaired electrons. However exceptions include a +2 oxidation state in OF2, +1 oxidation state in O2F2, and -1 oxidation state in hydrogen peroxide (H2O2 ).
Oxidation State of Sulphur
Sulphur is bigger in size. It has an empty d-orbital 3d because of which it can expand its valency. While it has a -2 oxidation state, sulphur also exhibits +2, +4 and +6 oxidation states, out of which +4 and +6 are common oxidation states.
the oxidation state of sulphur is (+II).
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FAQs on Oxidation States of Group 16 Elements
1. What are the common oxidation states shown by the Group 16 elements?
The elements of Group 16, also known as the chalcogens, exhibit several oxidation states. The most common states are -2, +2, +4, and +6. The stability of these oxidation states varies down the group. Oxygen predominantly shows a -2 state, while other elements like sulfur and selenium commonly display +4 and +6 states in addition to -2.
2. What is the most common oxidation state of oxygen, and are there any important exceptions?
The most common oxidation state of oxygen is -2, due to its high electronegativity. However, there are notable exceptions:
- In peroxides like hydrogen peroxide (H₂O₂), oxygen shows an oxidation state of -1.
- In superoxides such as potassium superoxide (KO₂), it has an oxidation state of -1/2.
- In combination with fluorine, the most electronegative element, as in oxygen difluoride (OF₂), oxygen exhibits a positive oxidation state of +2.
3. How can you calculate the oxidation state of sulfur in a compound like sulfuric acid (H₂SO₄)?
To calculate the oxidation state of sulfur (S) in H₂SO₄, you can use the following method, knowing the overall charge of the molecule is zero:
- The standard oxidation state of Hydrogen (H) is +1.
- The standard oxidation state of Oxygen (O) is -2.
- Let the oxidation state of Sulfur be 'x'.
The equation is: (2 × (+1)) + x + (4 × (-2)) = 0
Solving for x: 2 + x - 8 = 0 → x - 6 = 0 → x = +6. Therefore, the oxidation state of sulfur in sulfuric acid is +6.
4. Why does the stability of the +6 oxidation state decrease down Group 16?
The stability of the +6 oxidation state decreases down the group due to the inert pair effect. As we move from sulfur to polonium, the two electrons in the outermost s-orbital (ns²) become more reluctant to participate in bonding because of poor shielding by the inner d and f-orbitals. This makes it difficult to remove all six valence electrons. Consequently, for heavier elements like Tellurium (Te) and Polonium (Po), the +4 oxidation state (involving only p-electrons) becomes more stable than the +6 state.
5. Why can sulfur exhibit multiple positive oxidation states like +4 and +6, while oxygen generally does not?
The key difference lies in the availability of d-orbitals. Sulfur (and other elements below it in Group 16) has vacant d-orbitals in its valence shell. This allows it to expand its octet by promoting electrons from p-orbitals to d-orbitals, enabling it to form up to six bonds and show +4 and +6 oxidation states. Oxygen, in the second period, lacks vacant d-orbitals and cannot expand its octet, thus limiting its oxidation states, with -2 being the most common.
6. How does electronegativity influence the stability of the -2 oxidation state in Group 16?
Electronegativity decreases as we move down Group 16. Oxygen is the most electronegative element in the group, giving it a strong tendency to accept two electrons to complete its octet. This makes the -2 oxidation state highly stable for oxygen. As we go down to sulfur, selenium, and tellurium, the decreasing electronegativity reduces their ability to attract and hold two extra electrons, causing the stability of the -2 oxidation state to diminish significantly.
7. What is the inert pair effect and how does it specifically affect the oxidation state of Polonium?
The inert pair effect is the phenomenon where the two electrons in the outermost s-orbital (the ns² pair) are reluctant to participate in chemical bonding. This effect is most prominent in heavier p-block elements. For Polonium (Po), the heaviest member of Group 16, this effect is very strong. As a result, only its p-electrons are typically involved in bonding, making the +4 oxidation state its most stable and common oxidation state. The +6 state is highly unstable and rarely observed for Polonium.
8. Provide an example of a compound for both the +4 and +6 oxidation states of a Group 16 element.
Certainly. Here are common examples for sulfur:
- +4 Oxidation State: In Sulfur dioxide (SO₂), sulfur has an oxidation state of +4. Oxygen has a state of -2, so for the molecule to be neutral, S + 2(-2) = 0, which gives S = +4.
- +6 Oxidation State: In Sulfur hexafluoride (SF₆), sulfur shows an oxidation state of +6. Fluorine always has a state of -1, so S + 6(-1) = 0, which gives S = +6.
