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Asked Nov 20, 2019
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8. What is the oxidation state of the vanadium atom in V02+ (aq)?
(a) -2
(b) 0
(c) +2
(d) +4
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8. What is the oxidation state of the vanadium atom in V02+ (aq)? (a) -2 (b) 0 (c) +2 (d) +4

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Expert Answer

Step 1

Oxidation is the process that involved the loss of electrons and oxidized to cation whereas reduction is the process that involved the gain of electrons and reduced to anion.

Hence when a molecule loses an electron, that molecule is said to be oxidized whereas when a molecule gains an electron, that molecule is said to be reduced.  

The oxidized substances is called as reducing agent whereas the reduced substance is called as oxidizing agent.

Step 2

To calculate the oxidation state or number of any atom in the given molecule or ion, the rule of oxidation number must be followed. It states that the sum of oxidation number or state of all bonded atoms is always zero for neutral molecule or equal to charge of ion given.

Step 3

In the given ion VO2+; charge of O is -2 and whole charge on the ion is...

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vo2 assume charge on V as x x+-2 x 1)2 x=+2 +2+4

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Chemical bonding

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