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 (integer)
19
Solution
<b>K<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub></b>
<br><br>Let oxidation state of Ce
<br><br>2(+1) + 2x + 7(–2) = 0
<br><br>$\Rightarrow$ x = +6
<br><br>In K<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub>, Transition metal (Cr) present in +6 oxidation state.
<br><br><b>KMnO<sub>4</sub></b>
<br><br>(+1) + y + 4(–2) = 0
<br>$\Rightarrow$ x = +7
<br><br>In KMnO<sub>4</sub>, transition metal (Mn) present in +7
oxidation state.
<br><br><b>K<sub>2</sub>FeO<sub>4</sub></b>
<br><br>2(+1) + z + 4(–2) = 0
<br>$\Rightarrow$x = +6
<br><br>In K<sub>2</sub>FeO<sub>4</sub>, transition metal (Fe) present in +6
oxidation state.
<br><br>$\therefore$ x + y + z = 6 + 7 + 6 = 19
About this question
Subject: Chemistry · Chapter: Redox Reactions · Topic: Oxidation States
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