Question
The pair of the compounds in which both the metals are in the highest possible oxidation state is
A.
$${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }},{\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}$$
B.
$$Cr{O_2}C{l_{2\,}}\,,\,MnO_4^ - $$
C.
$$Ti{O_3}\,,\,Mn{O_2}$$
D.
$${\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }},\,Mn{O_3}$$
Answer :
$$Cr{O_2}C{l_{2\,}}\,,\,MnO_4^ - $$
Solution :
TIPS/Formulae:
The highest $$O.S.$$ of an element is equal to the number of its valence electrons
$$\left( {\text{i}} \right){\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }},O.N{\text{.}}\,{\text{of}}\,Fe = + 3,$$ $${\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }},O.N\,{\text{of}}\,Co = + 3$$
$$({\text{ii}})\,Cr{O_2}C{l_2},O.N.\,{\text{of}}\,Cr = + 6,$$ $$\left( {{\text{Highest}}\,O.S.\,{\text{of}}\,Cr} \right)$$
$${\left[ {Mn{O_4}} \right]^ - }O.N\,{\text{of}}\,Mn = + 7$$ $$\left( {{\text{Highest}}\;O.S.\,{\text{of}}\,Mn} \right)$$
$$({\text{iii}})\,Ti{O_3},O.N.\,{\text{of}}\,Ti = + 6,$$ $$Mn{O_2}O.N.\,\,{\text{of}}\,\,Mn = + 4$$
$$({\text{iv}})\,{\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }},\,O.N.\,\,{\text{of}}\,\,Co$$ $$ = + 3,Mn{O_3},\,O.N.\,\,{\text{of }}\,Mn = + 6$$