Total number of non- bonding valancy electrons present in \(F_{2}\) is 'x'.The maximum oxidation state that can be exhibited by lodine in its second exited state is 'y' then "x/y" value?

Maximum number of oxidation states which nitrogen can show in its compound is ....

The maximum number of unpaired electrons can present in p_(x) orbital is

What will be the maximum value of Sigma(n+1+m) for all unpaired electrons of chlorine is its second excited state?

" In lanthanides highest oxidation state is " "x" ' and actinides highest oxidation state is "y" .Then "x+y" is "

The most common oxidation state of an element is -2. the number of electrons present in its outer most shell is-

In \(ICl_{3}\) the total number of lone pairs is x. In \(ClO_{2}^{-}\) lone pairs on Cl is y. The oxidation state of lodine in IBr is z. then (x+z)/y is

The oxidation state of fluorine in F_(2) is x. Find value of |x|.

The transition element ( with few exceptions ) show a large number of oxidation states . The various oxidation states are related to the electronic configuration of their atoms. The variable oxidation states of a transition metal is due to the involvement of (n-1)d and outer ns electrons . For the first five elements of 3d transition series , the minimum oxidation state is equal to the number of electrons in 4s shell and the maximum oxidation state is equal to the sum of 4s and 3d electrons. The relative stability of various oxidation states of a given element can be explained on the basis of stability of d^(0),d^(5) and d^(10) configuration . In 3d series, the maximum oxidation state is shown by