Question
Write about oxidation states of transition series ?

Answer

$\rightarrow$ One of the notable features of a transition element is the great variety of oxidation states it may show in its compounds.
$Sc$ $Ti$ $V$ $Cr$ $Mn$ $Fe$ $Co$ $Ni$ $Cu$ $Zn$
  $+2$ $+2$ $+2$ $+2$ $+2$ $+2$ $+2$ $+1$ $+2$
$+3$ $+3$ $+3$ $+3$ $+3$ $+3$ $+3$ $+3$ $+2$  
  $+4$ $+4$ $+4$ $+4$ $+4$ $+4$ $+4$    
    $+5$ $+5$ $+5$          
      $+6$ $+6$ $+6$        
        $+7$          
Oxidation States of the first row Transition Metals
$($the most common ones are in bold types$)$
$\rightarrow$ The elements which give the greatest number of oxidation states occur in or near the middle of the series. Manganese, for example, exhibits all the oxidation states from $+2$ to $+7.$
$\rightarrow$ In the starting of series, very less number of $d-$electrons are available for chemical bonding. Hence, less number of oxidation states are shown by elements present at the starting of series.
e.g.: $Sc^{+3}, Ti^{+2}, Ti^{+3}, Ti^{+4}$
$\rightarrow$ At the end of the series there are too many $d-$electrons and $d-$orbitals are completely occupied. Hence, these elements show very less number of oxidation states.
$\rightarrow$ Down the group the stability of elements in higher oxidation states increases because removal of electrons from $d-$orbitals become easy.
$\rightarrow$ For example, in group $6,$ $Mo(VI)$ and $W(VI)$ are found to be more stable than $Cr(VI).$ Thus, $Cr(VI)$ in the form of dichromate in acidic medium is a strong oxidising agent, whereas $MoO_3$ and $WO_3$ are not.
$\rightarrow$ Low oxidation states are found when a complex compound has ligands capable of $\pi -$acceptor character in addition to the $o-$bonding. For example, in $Ni(CO)_4$ and $Fe(CO)_5,$ the oxidation state of nickel and iron is zero.

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