
Which has the most stable +2 oxidation state?
(a)- Cs
(b)- Cl
(c)- Pb
(d)- Tl
Answer
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Hint: Here,+2 oxidation state means there is the removal of two electrons from its electronic configuration in the ground state. The atomic number of Cs is 55, the atomic number of Cl is 17, the atomic number of Pb is 82, and the atomic number of Tl is 81.
Complete answer:
Each element is found in many forms because the elements are not in stable form, so it combines with other elements to gain stability. So, when they combine with other elements, there is an oxidation state on the element.
With the help of the oxidation number of the element, we can find the removal and gain of electrons in the element. So, +2 oxidation state means there is the removal of two electrons from its electronic configuration in the group state.
So, the given elements in the options are:
Cs is Cesium and it is the element of alkali metals. The atomic number of cesium is 55 so, its electronic configuration is:
$Cs=[Xe]6{{s}^{1}}$
So, its stable oxidation state is +2
Cl is chlorine and it is the element of halogen. The atomic number of chlorine is 17 so, its electronic configuration is:
$Cl=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{5}}$
So, its stable oxidation state is -1
Pb is lead and it is the element of the carbon family. The atomic number of lead is 82 so, its electronic configuration is:
$[Xe]4{{f}^{14}}5{{d}^{10}}6{{s}^{2}}6{{p}^{2}}$
So, there are two oxidation states of lead, i.e., +2 and +4. But due to the inert pair effect, the +2 oxidation state is more stable.
Tl is Thallium and it is the element of the Boron family. The atomic number of thallium is 81 so, its electronic configuration is:
$[Xe]4{{f}^{14}}5{{d}^{10}}6{{s}^{2}}6{{p}^{1}}$
So, there are two oxidation states of thallium, i.e., +1 and +3. But due to the inert pair effect, the +1 oxidation state is more stable.
Therefore, the correct answer is an option (c).
Note:
The propensity of valence shell s-electrons to engage in bond-forming decreases as we pass down the group. The Inert Pair Effect refers to the s-unwillingness electron's to engage in bond forming.
Complete answer:
Each element is found in many forms because the elements are not in stable form, so it combines with other elements to gain stability. So, when they combine with other elements, there is an oxidation state on the element.
With the help of the oxidation number of the element, we can find the removal and gain of electrons in the element. So, +2 oxidation state means there is the removal of two electrons from its electronic configuration in the group state.
So, the given elements in the options are:
Cs is Cesium and it is the element of alkali metals. The atomic number of cesium is 55 so, its electronic configuration is:
$Cs=[Xe]6{{s}^{1}}$
So, its stable oxidation state is +2
Cl is chlorine and it is the element of halogen. The atomic number of chlorine is 17 so, its electronic configuration is:
$Cl=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{5}}$
So, its stable oxidation state is -1
Pb is lead and it is the element of the carbon family. The atomic number of lead is 82 so, its electronic configuration is:
$[Xe]4{{f}^{14}}5{{d}^{10}}6{{s}^{2}}6{{p}^{2}}$
So, there are two oxidation states of lead, i.e., +2 and +4. But due to the inert pair effect, the +2 oxidation state is more stable.
Tl is Thallium and it is the element of the Boron family. The atomic number of thallium is 81 so, its electronic configuration is:
$[Xe]4{{f}^{14}}5{{d}^{10}}6{{s}^{2}}6{{p}^{1}}$
So, there are two oxidation states of thallium, i.e., +1 and +3. But due to the inert pair effect, the +1 oxidation state is more stable.
Therefore, the correct answer is an option (c).
Note:
The propensity of valence shell s-electrons to engage in bond-forming decreases as we pass down the group. The Inert Pair Effect refers to the s-unwillingness electron's to engage in bond forming.
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