Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?
Mn (Z = 25) = 3d5 4s2
Mn+2 is the most stable ion for manganese, the d-orbital can be made to remove 0 to 7 electrons. Compounds of manganese therefore range from Mn(0) as Mn(s), Mn(II) as MnO, Mn(II,III) as Mn3O4, Mn(IV) as MnO2, or manganese dioxide, Mn(VII) in the permanganate ion MnO4-, and so on. Mn has the maximum number of unpaired electrons present in the d-subshell (5 electrons). Hence, Mn exhibits the largest number of oxidation states, ranging from +2 to +7.
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Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 3: Which of the 3d series of the transition metals exhibits the largest number of oxidation states and ....
Comments
But fe has 8 unpaired electron Why they not
Thanks
oxidation number or state is defined as the charge present on an atom or ion. in case of transition metals, there are five orbitals in the d subshell . when the number of unpaired valence electrons increases, the d-orbital increase & the highest oxidation state increases. This is because unpaired valence electrons are unstable and ready to bond with other chemical species,ie, the oxidation states wwill be the highest in the very middle of the transition metal periods due to the presence of the highest number of unpaired valence electrons.this is the reason why Mn has largest number of oxidation state in its period
please write full answers