Illustrate by taking examples of transition elements and non-transition elements that oxidation states of elements are largely based on electronic configuration.
(N/A) The oxidation states of elements are determined by their electronic configuration,specifically the number of electrons in the outermost shell and the stability of the resulting configuration.
$1$. Non-transition elements: Consider Nitrogen $(Z=7)$ with configuration $[He] 2s^{2} 2p^{3}$. It can show oxidation states like $-3$ (by gaining $3$ electrons to complete the octet) or $+5$ (by losing $5$ valence electrons).
$2$. Transition elements: Consider Manganese $(Z=25)$ with configuration $[Ar] 3d^{5} 4s^{2}$. Due to the availability of both $4s$ and $3d$ electrons,it exhibits variable oxidation states ranging from $+2$ to $+7$. The $+2$ state arises from the loss of $4s^{2}$ electrons,while higher states involve the progressive loss of $3d$ electrons,reaching stability at $+7$ where the $3d$ subshell becomes empty.
$1$. Non-transition elements: Consider Nitrogen $(Z=7)$ with configuration $[He] 2s^{2} 2p^{3}$. It can show oxidation states like $-3$ (by gaining $3$ electrons to complete the octet) or $+5$ (by losing $5$ valence electrons).
$2$. Transition elements: Consider Manganese $(Z=25)$ with configuration $[Ar] 3d^{5} 4s^{2}$. Due to the availability of both $4s$ and $3d$ electrons,it exhibits variable oxidation states ranging from $+2$ to $+7$. The $+2$ state arises from the loss of $4s^{2}$ electrons,while higher states involve the progressive loss of $3d$ electrons,reaching stability at $+7$ where the $3d$ subshell becomes empty.
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Similar Questions
Which of the following elements in their respective oxidation states develops the highest spin-only magnetic moment?
Determine whether the following statements are True $(T)$ or False $(F)$:
$(a)$ The atomic radii of the elements of the second transition series $(4d)$ are larger than those of the first transition series $(3d)$.
$(b)$ $Cr^{+}$,$Mn^{2+}$,and $Fe^{3+}$ have the same electronic configuration.
$(c)$ $Cr^{+}$,$Mn^{2+}$,and $Fe^{3+}$ have the same $d^5$ electronic configuration.
$(a)$ The atomic radii of the elements of the second transition series $(4d)$ are larger than those of the first transition series $(3d)$.
$(b)$ $Cr^{+}$,$Mn^{2+}$,and $Fe^{3+}$ have the same electronic configuration.
$(c)$ $Cr^{+}$,$Mn^{2+}$,and $Fe^{3+}$ have the same $d^5$ electronic configuration.
Difficult
View SolutionIron $(Fe)$ is classified as:
Easy
View SolutionAlthough $Cr^{3+}$ and $Co^{2+}$ ions have the same number of unpaired electrons,the magnetic moment of $Cr^{3+}$ is $3.87 \ BM$ and that of $Co^{2+}$ is $4.87 \ BM$. Why?
Medium
View SolutionAssertion : The free gaseous $Cr$ atom has six unpaired electrons.
Reason : Half-filled $s$ orbital has greater stability.
Reason : Half-filled $s$ orbital has greater stability.
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