315291
Which of the following is the most powerful oxidising agent?
1 \(\mathrm{\mathrm{F}_{2}}\)
2 \(\mathrm{\mathrm{O}_{2}}\)
3 \(\mathrm{\mathrm{Br}_{2}}\)
4 \(\mathrm{\mathrm{I}_{2}}\)
Explanation:
\(\mathrm{\mathrm{F}_{2}}\), being most electronegative, is the most powerful oxidising agent with standard oxidation potential \(\mathrm{2.87 \mathrm{~V}}\).
CHXI08:REDOX REACTIONS
315169
The oxidation state of sulphur in \(\mathrm{\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}}\) is
1 \({\rm{ + 4}}\)
2 \({\rm{ + 6}}\)
3 \({\rm{ + 5}}\)
4 \({\rm{ + 7}}\)
Explanation:
Let, the oxidation state of sulphur in \(\mathrm{\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}}\) be \(\mathrm{\mathrm{x}}\). \(\mathrm{\therefore 2(+1)+2(\mathrm{x})+7(-2)=0}\) \({\rm{2 + 2x - 14 = 0}}\) \({\rm{2x = 12x = + 6}}\)
MHTCET - 2019
CHXI08:REDOX REACTIONS
315170
Maximum oxidation state shown by Os, Ru and Xe in their compounds is
1 +8
2 +6
3 +10
4 +4
Explanation:
Oxidation state of an atom in its elemental state is zero.
CHXI08:REDOX REACTIONS
315197
The oxidation number of carbon in \(\mathrm{\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}}\) is
315291
Which of the following is the most powerful oxidising agent?
1 \(\mathrm{\mathrm{F}_{2}}\)
2 \(\mathrm{\mathrm{O}_{2}}\)
3 \(\mathrm{\mathrm{Br}_{2}}\)
4 \(\mathrm{\mathrm{I}_{2}}\)
Explanation:
\(\mathrm{\mathrm{F}_{2}}\), being most electronegative, is the most powerful oxidising agent with standard oxidation potential \(\mathrm{2.87 \mathrm{~V}}\).
CHXI08:REDOX REACTIONS
315169
The oxidation state of sulphur in \(\mathrm{\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}}\) is
1 \({\rm{ + 4}}\)
2 \({\rm{ + 6}}\)
3 \({\rm{ + 5}}\)
4 \({\rm{ + 7}}\)
Explanation:
Let, the oxidation state of sulphur in \(\mathrm{\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}}\) be \(\mathrm{\mathrm{x}}\). \(\mathrm{\therefore 2(+1)+2(\mathrm{x})+7(-2)=0}\) \({\rm{2 + 2x - 14 = 0}}\) \({\rm{2x = 12x = + 6}}\)
MHTCET - 2019
CHXI08:REDOX REACTIONS
315170
Maximum oxidation state shown by Os, Ru and Xe in their compounds is
1 +8
2 +6
3 +10
4 +4
Explanation:
Oxidation state of an atom in its elemental state is zero.
CHXI08:REDOX REACTIONS
315197
The oxidation number of carbon in \(\mathrm{\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}}\) is
315291
Which of the following is the most powerful oxidising agent?
1 \(\mathrm{\mathrm{F}_{2}}\)
2 \(\mathrm{\mathrm{O}_{2}}\)
3 \(\mathrm{\mathrm{Br}_{2}}\)
4 \(\mathrm{\mathrm{I}_{2}}\)
Explanation:
\(\mathrm{\mathrm{F}_{2}}\), being most electronegative, is the most powerful oxidising agent with standard oxidation potential \(\mathrm{2.87 \mathrm{~V}}\).
CHXI08:REDOX REACTIONS
315169
The oxidation state of sulphur in \(\mathrm{\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}}\) is
1 \({\rm{ + 4}}\)
2 \({\rm{ + 6}}\)
3 \({\rm{ + 5}}\)
4 \({\rm{ + 7}}\)
Explanation:
Let, the oxidation state of sulphur in \(\mathrm{\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}}\) be \(\mathrm{\mathrm{x}}\). \(\mathrm{\therefore 2(+1)+2(\mathrm{x})+7(-2)=0}\) \({\rm{2 + 2x - 14 = 0}}\) \({\rm{2x = 12x = + 6}}\)
MHTCET - 2019
CHXI08:REDOX REACTIONS
315170
Maximum oxidation state shown by Os, Ru and Xe in their compounds is
1 +8
2 +6
3 +10
4 +4
Explanation:
Oxidation state of an atom in its elemental state is zero.
CHXI08:REDOX REACTIONS
315197
The oxidation number of carbon in \(\mathrm{\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}}\) is
315291
Which of the following is the most powerful oxidising agent?
1 \(\mathrm{\mathrm{F}_{2}}\)
2 \(\mathrm{\mathrm{O}_{2}}\)
3 \(\mathrm{\mathrm{Br}_{2}}\)
4 \(\mathrm{\mathrm{I}_{2}}\)
Explanation:
\(\mathrm{\mathrm{F}_{2}}\), being most electronegative, is the most powerful oxidising agent with standard oxidation potential \(\mathrm{2.87 \mathrm{~V}}\).
CHXI08:REDOX REACTIONS
315169
The oxidation state of sulphur in \(\mathrm{\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}}\) is
1 \({\rm{ + 4}}\)
2 \({\rm{ + 6}}\)
3 \({\rm{ + 5}}\)
4 \({\rm{ + 7}}\)
Explanation:
Let, the oxidation state of sulphur in \(\mathrm{\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}}\) be \(\mathrm{\mathrm{x}}\). \(\mathrm{\therefore 2(+1)+2(\mathrm{x})+7(-2)=0}\) \({\rm{2 + 2x - 14 = 0}}\) \({\rm{2x = 12x = + 6}}\)
MHTCET - 2019
CHXI08:REDOX REACTIONS
315170
Maximum oxidation state shown by Os, Ru and Xe in their compounds is
1 +8
2 +6
3 +10
4 +4
Explanation:
Oxidation state of an atom in its elemental state is zero.
CHXI08:REDOX REACTIONS
315197
The oxidation number of carbon in \(\mathrm{\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}}\) is
