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Consider the elements : Cs, Ne, I and F (a) Identify the element that exhibits only negative oxidation state. (b) Identify the element that exhibits only postive oxidation state. (c) Identify the element that exhibits both positive and negative oxidation states. (d) Identify the element which exhibits neither the negative nor does the positive oxidation state.

Short Answer

Expert verified
(a) F exhibits only negative oxidation state. (b) Cs exhibits only positive oxidation state. (c) I exhibits both positive and negative. (d) Ne exhibits neither oxidation state.

Step by step solution

01

Understanding Oxidation States

Oxidation states refer to the charge of an atom would have if all bonds were ionic. Elements can exhibit different oxidation states depending on their ability to gain or lose electrons.
02

Analyze Elements for Negative Oxidation State

Among the given elements (Cs, Ne, I, and F), Fluorine (F) is known to exhibit only a negative oxidation state, as it is the most electronegative element and tends to gain electrons.
03

Analyze Elements for Positive Oxidation State

Caesium (Cs), a highly electropositive element, exhibits only positive oxidation states, commonly +1 as it tends to lose one electron.
04

Explore Both Positive and Negative Oxidation States

Iodine (I) can exhibit both positive and negative oxidation states. It commonly shows -1 when it behaves like a halogen and various positive states (e.g., +1, +3, +5, +7) in its oxyanions and in combination with more electronegative elements.
05

Identify Element with No Oxidation States

Neon (Ne) is a noble gas, making it chemically inert under normal conditions. It neither gains nor loses electrons, thus exhibiting neither positive nor negative oxidation states.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Negative Oxidation State
When an element exhibits a negative oxidation state, it means the element has gained electrons. This typically happens with non-metals that are highly electronegative and tend to attract electrons when forming bonds. A classic example is Fluorine (F). Its electronegativity is very high, the highest among all elements, which enables it to almost always acquire a negative oxidation state of -1.
Fluorine's nature makes it very stable with this negative oxidation state because it satisfies the octet rule—having eight electrons in its valence shell. Hence, in most compounds, Fluorine is found carrying a -1 charge due to gaining an electron needed to complete its outer electron shell.
  • Fluorine (F) is the only element that consistently exhibits a negative oxidation state.
  • Gaining one electron gives fluorine its stable -1 oxidation state.
Positive Oxidation State
A positive oxidation state indicates that an element tends to lose electrons. This trait is commonly seen in metals, especially in the alkali group. Cesium (Cs) is an excellent example of this characteristic. Being highly electropositive, cesium loses an electron to achieve a noble gas electron configuration, typically resulting in a +1 oxidation state.
The tendency of cesium to lose an electron rather than gain attributes to its large atomic size and low ionization energy. Therefore, cesium readily gives up that single electron, characteristically promoting it to a stable cation with a +1 charge. This means cesium possesses positive oxidation states in compounds typically involving ionic bonding.
  • Cesium (Cs) often exhibits a positive oxidation state, almost exclusively +1.
  • Loss of one electron allows cesium to reach the electronic structure of noble gases.
Noble Gases Oxidation States
Noble gases, known for their stability, generally exhibit no oxidation states under normal conditions. Neon (Ne) is a classic example of a noble gas that maintains its stability and inertness due to its complete valence electron shell.
Since neon's outer electron shell is full, it does not tend to gain or lose electrons, making its oxidation state zero. This inertness makes neon chemically non-reactive under typical conditions; hence, it does not participate in conventional bonding that would change its oxidation state.
Though noble gases are mostly inert, under extreme conditions, some can form compounds with unusual oxidation states temporarily. However, typically, these states are not stable or ubiquitous, so noble gases remain largely "inactive."
  • Neon (Ne) does not exhibit standard oxidation states under normal conditions.
  • The complete electron shell in neon prevents it from participating in typical chemical reactions.

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Most popular questions from this chapter

The \(\mathrm{Mn}^{3+}\) ion is unstable in solution and undergoes disproportionation to give \(\mathrm{Mn}^{2+}, \mathrm{MnO}_{2}\), and \(\mathrm{H}^{+}\) ion. Write a balanced ionic equation for the reaction.

Justify that the following reactions are redox reactions: (a) \(\mathrm{CuO}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (b) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{CO}(\mathrm{g}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{~g})\) (c) \(4 \mathrm{BCl}_{3}(\mathrm{~g})+\) 3LiAlH \(_{4}(\mathrm{~s}) \rightarrow 2 \mathrm{~B}_{2} \mathrm{H}_{6}(\mathrm{~g})+3 \mathrm{LiCl}(\mathrm{s})+3 \mathrm{AlCl}_{3}(\mathrm{~s})\) (d) \(2 \mathrm{~K}(\mathrm{~s})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{~K}^{+} \mathrm{F}^{-}(\mathrm{s})\) (e) \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

Consider the reactions : \(2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{~s}) \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq})\) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq})+2 \mathrm{Br}_{2}(\mathrm{l})+5 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{SO}_{4}^{2-}(\mathrm{aq})+4 \mathrm{Br}^{-}(\mathrm{aq})+10 \mathrm{H}^{+}(\mathrm{aq})\) Why does the same reductant, thiosulphate react differently with iodine and bromine?

Assign oxidation number to the underlined elements in each of the following species: (a) \(\mathrm{NaH}_{2} \mathrm{PO}_{4}\) (b) \(\mathrm{NaH} \underline{\mathrm{S} O}_{4}\) (c) \(\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}\) (d) \(\mathrm{K}_{2} \underline{\mathrm{Mn}} \mathrm{O}_{4}\) (e) \(\mathrm{Ca} \underline{\mathrm{O}}_{2}\) (f) \(\mathrm{NaB} \mathrm{H}_{4}\) (g) \(\mathrm{H}_{2} \underline{\mathrm{S}}_{2} \mathrm{O}_{7}\) (h) \(\operatorname{KAl}\left(\underline{\mathrm{S}} \mathrm{O}_{4}\right)_{2} \cdot 12 \mathrm{H}_{2} \mathrm{O}\)

In Ostwald's process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with \(10.00 \mathrm{~g}\). of ammonia and \(20.00 \mathrm{~g}\) of oxygen ?

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