JEE MAIN - Chemistry (2020 - 2nd September Evening Slot - No. 8)
The oxidation states of transition metal atoms
in K2Cr2O7, KMnO4 and K2FeO4, respectively,
are x, y and z. The sum of x, y and z is _______.
Answer
19
Explanation
K2Cr2O7
Let oxidation state of Ce
2(+1) + 2x + 7(–2) = 0
$$ \Rightarrow $$ x = +6
In K2Cr2O7, Transition metal (Cr) present in +6 oxidation state.
KMnO4
(+1) + y + 4(–2) = 0
$$ \Rightarrow $$ x = +7
In KMnO4, transition metal (Mn) present in +7 oxidation state.
K2FeO4
2(+1) + z + 4(–2) = 0
$$ \Rightarrow $$x = +6
In K2FeO4, transition metal (Fe) present in +6 oxidation state.
$$ \therefore $$ x + y + z = 6 + 7 + 6 = 19
Let oxidation state of Ce
2(+1) + 2x + 7(–2) = 0
$$ \Rightarrow $$ x = +6
In K2Cr2O7, Transition metal (Cr) present in +6 oxidation state.
KMnO4
(+1) + y + 4(–2) = 0
$$ \Rightarrow $$ x = +7
In KMnO4, transition metal (Mn) present in +7 oxidation state.
K2FeO4
2(+1) + z + 4(–2) = 0
$$ \Rightarrow $$x = +6
In K2FeO4, transition metal (Fe) present in +6 oxidation state.
$$ \therefore $$ x + y + z = 6 + 7 + 6 = 19
Comments (0)
