In the compound K2C1O4 , what is the correct oxidation number of each of its elements? (K is a Group I element) (The rules for assigning Oxidation States are given as below) 1) The oxidation numbers of the atoms in a neutral molecule must add up to zero. 2) Group I → +1 ; Group II → +2 ; Group I → +3 3) Fluorine → -1 (always) Halogens – -1 (except in compounds with O and other halogens) 4) Hydrogen → +1 (except in metal hydrides, e.g. LiH, Rule 2 and H has -1) 5) Oxygen -2 (except in compounds with F, Rule 3) and compounds with 0-0 bonds (Rules 2 & 4) OA K:+2 , Cr:+2 ,O: -2 O B. K:+1 , Cr:+4 ,O: -2 ocK:+1 , Cr:+2 ,O: -2 O D.K:+2, Cr:+6 ,O: -2 K:+1 , Cr:+6 ,O: -2 O E. None In the compound Kâ‚‚CrO4, what is the correct oxidation number of each of its elements? (K is a Group I element) (The rules for assigning Oxidation States are given as below) 1) The oxidation numbers of the atoms in a neutral molecule must add up to zero. 2) Group I → +1; Group II → +2 ; Group III → +3 3) Fluorine →→-1 (always) Halogens -1 (except in compounds with O and other halogens) 4) Hydrogen +1 (except in metal hydrides, e.g. LiH, Rule 2 and H has -1) 5) Oxygen → -2 (except in compounds with F, Rule 3) and compounds with O-O bonds (Rules 2 & 4) → OAK:+2. Cr:+2.0: -2 B. K:+1, Cr:+4,0:-2 Oc. K:+1, Cr:+2,0:-2 OD. K:+2, Cr:+6,0:-2 K:+1, Cr:+6,0: -2 O E.
The Correct Answer and Explanation is:
The correct answer is:
E. K: +1, Cr: +6, O: –2
Explanation
To determine the oxidation numbers in the compound K₂CrO₄ (potassium chromate), we apply standard oxidation number rules:
1. Total Oxidation Number = 0
Since K₂CrO₄ is a neutral compound, the sum of the oxidation numbers of all atoms must equal zero.
2. Potassium (K):
Potassium is a Group I element, and Rule 2 tells us Group I elements always have an oxidation number of +1 in compounds.
There are two K atoms, so the total contribution from potassium is:
2 × (+1) = +2
3. Oxygen (O):
Oxygen typically has an oxidation number of –2, as per Rule 5.
There are four O atoms, so the total contribution from oxygen is:
4 × (–2) = –8
4. Chromium (Cr):
Let x be the oxidation number of Cr.
Now apply the rule that the sum of oxidation numbers = 0: 2(+1)+x+4(−2)=0⇒2+x−8=0⇒x=+62(+1) + x + 4(-2) = 0 \Rightarrow 2 + x – 8 = 0 \Rightarrow x = +6
So, Cr has an oxidation number of +6.
Summary of Oxidation Numbers in K₂CrO₄:
- K = +1
- Cr = +6
- O = –2
Conclusion:
Therefore, the correct oxidation states are:
K: +1, Cr: +6, O: –2, which matches option E.
This structure is consistent with the known chemistry of chromates, where chromium is commonly in the +6 oxidation state.
