315151
The number of elements from the following which exhibit(s) variable oxidation numbers is ____. \({\mathrm{\qquad}}\) \({\mathrm{\mathrm{Be}, \mathrm{B}, \mathrm{C}, \mathrm{Na}, \mathrm{F}, \mathrm{Cl}, \mathrm{Al}, \mathrm{H}, \mathrm{Li}, \mathrm{O}}}\)
1 6
2 3
3 1
4 4
Explanation:
CHXI08:REDOX REACTIONS
315152
What is the oxidation state of \(\mathrm{\mathrm{Mn}}\) in \(\mathrm{\mathrm{MnO}_{4}^{-}}\)?
1 +5
2 +6
3 +4
4 +7
Explanation:
Let the oxidation state of \(\mathrm{M n}\) be \(\mathrm{x}\). Oxidation state of \(\mathrm{\mathrm{O}=-2}\) \(\mathrm{\therefore \mathrm{x}+4(-2)=-1 \Rightarrow \mathrm{x}=+7}\)
MHTCET - 2022
CHXI08:REDOX REACTIONS
315153
What is the oxidation state of nitrogen in \(\mathrm{\mathrm{NaN}_{3}}\) ?
1 \({\rm{ - 3/1}}\)
2 \(3\)
3 \({\rm{ - 3}}\)
4 \({\rm{ - 1/3}}\)
Explanation:
\(\mathrm{\mathrm{NaN}_{3} \Rightarrow+1+3 \mathrm{x}=0}\) \(\mathrm{\Rightarrow 3 \mathrm{x}=-1 \Rightarrow \mathrm{x}=-\dfrac{1}{3}}\) So, oxidation number of nitrogen in \(\mathrm{\mathrm{NaN}_{3}}\) is \(\mathrm{-1 / 3}\)
CHXI08:REDOX REACTIONS
315154
The oxidation number of \(\mathrm{\mathrm{Cl}}\) in is \(\mathrm{\mathrm{NOClO}_{4}}\) is
1 +11
2 +9
3 +7
4 +5
Explanation:
The compound may be written as \(\mathrm{\mathrm{NO}^{+} \mathrm{ClO}_{4}^{-}}\) For \(\mathrm{\mathrm{ClO}_{4}^{-}}\), Let Ox. No.of \(\mathrm{\mathrm{Cl}=a}\) \({\rm{a + 4x( - 2) = - 1}}\) \({\rm{a = + 7}}\) Hence, the oxidation no. of \(\mathrm{\mathrm{Cl}}\) in \(\mathrm{\mathrm{NOClO}_{4}}\) is +7 .
NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXI08:REDOX REACTIONS
315151
The number of elements from the following which exhibit(s) variable oxidation numbers is ____. \({\mathrm{\qquad}}\) \({\mathrm{\mathrm{Be}, \mathrm{B}, \mathrm{C}, \mathrm{Na}, \mathrm{F}, \mathrm{Cl}, \mathrm{Al}, \mathrm{H}, \mathrm{Li}, \mathrm{O}}}\)
1 6
2 3
3 1
4 4
Explanation:
CHXI08:REDOX REACTIONS
315152
What is the oxidation state of \(\mathrm{\mathrm{Mn}}\) in \(\mathrm{\mathrm{MnO}_{4}^{-}}\)?
1 +5
2 +6
3 +4
4 +7
Explanation:
Let the oxidation state of \(\mathrm{M n}\) be \(\mathrm{x}\). Oxidation state of \(\mathrm{\mathrm{O}=-2}\) \(\mathrm{\therefore \mathrm{x}+4(-2)=-1 \Rightarrow \mathrm{x}=+7}\)
MHTCET - 2022
CHXI08:REDOX REACTIONS
315153
What is the oxidation state of nitrogen in \(\mathrm{\mathrm{NaN}_{3}}\) ?
1 \({\rm{ - 3/1}}\)
2 \(3\)
3 \({\rm{ - 3}}\)
4 \({\rm{ - 1/3}}\)
Explanation:
\(\mathrm{\mathrm{NaN}_{3} \Rightarrow+1+3 \mathrm{x}=0}\) \(\mathrm{\Rightarrow 3 \mathrm{x}=-1 \Rightarrow \mathrm{x}=-\dfrac{1}{3}}\) So, oxidation number of nitrogen in \(\mathrm{\mathrm{NaN}_{3}}\) is \(\mathrm{-1 / 3}\)
CHXI08:REDOX REACTIONS
315154
The oxidation number of \(\mathrm{\mathrm{Cl}}\) in is \(\mathrm{\mathrm{NOClO}_{4}}\) is
1 +11
2 +9
3 +7
4 +5
Explanation:
The compound may be written as \(\mathrm{\mathrm{NO}^{+} \mathrm{ClO}_{4}^{-}}\) For \(\mathrm{\mathrm{ClO}_{4}^{-}}\), Let Ox. No.of \(\mathrm{\mathrm{Cl}=a}\) \({\rm{a + 4x( - 2) = - 1}}\) \({\rm{a = + 7}}\) Hence, the oxidation no. of \(\mathrm{\mathrm{Cl}}\) in \(\mathrm{\mathrm{NOClO}_{4}}\) is +7 .
315151
The number of elements from the following which exhibit(s) variable oxidation numbers is ____. \({\mathrm{\qquad}}\) \({\mathrm{\mathrm{Be}, \mathrm{B}, \mathrm{C}, \mathrm{Na}, \mathrm{F}, \mathrm{Cl}, \mathrm{Al}, \mathrm{H}, \mathrm{Li}, \mathrm{O}}}\)
1 6
2 3
3 1
4 4
Explanation:
CHXI08:REDOX REACTIONS
315152
What is the oxidation state of \(\mathrm{\mathrm{Mn}}\) in \(\mathrm{\mathrm{MnO}_{4}^{-}}\)?
1 +5
2 +6
3 +4
4 +7
Explanation:
Let the oxidation state of \(\mathrm{M n}\) be \(\mathrm{x}\). Oxidation state of \(\mathrm{\mathrm{O}=-2}\) \(\mathrm{\therefore \mathrm{x}+4(-2)=-1 \Rightarrow \mathrm{x}=+7}\)
MHTCET - 2022
CHXI08:REDOX REACTIONS
315153
What is the oxidation state of nitrogen in \(\mathrm{\mathrm{NaN}_{3}}\) ?
1 \({\rm{ - 3/1}}\)
2 \(3\)
3 \({\rm{ - 3}}\)
4 \({\rm{ - 1/3}}\)
Explanation:
\(\mathrm{\mathrm{NaN}_{3} \Rightarrow+1+3 \mathrm{x}=0}\) \(\mathrm{\Rightarrow 3 \mathrm{x}=-1 \Rightarrow \mathrm{x}=-\dfrac{1}{3}}\) So, oxidation number of nitrogen in \(\mathrm{\mathrm{NaN}_{3}}\) is \(\mathrm{-1 / 3}\)
CHXI08:REDOX REACTIONS
315154
The oxidation number of \(\mathrm{\mathrm{Cl}}\) in is \(\mathrm{\mathrm{NOClO}_{4}}\) is
1 +11
2 +9
3 +7
4 +5
Explanation:
The compound may be written as \(\mathrm{\mathrm{NO}^{+} \mathrm{ClO}_{4}^{-}}\) For \(\mathrm{\mathrm{ClO}_{4}^{-}}\), Let Ox. No.of \(\mathrm{\mathrm{Cl}=a}\) \({\rm{a + 4x( - 2) = - 1}}\) \({\rm{a = + 7}}\) Hence, the oxidation no. of \(\mathrm{\mathrm{Cl}}\) in \(\mathrm{\mathrm{NOClO}_{4}}\) is +7 .
315151
The number of elements from the following which exhibit(s) variable oxidation numbers is ____. \({\mathrm{\qquad}}\) \({\mathrm{\mathrm{Be}, \mathrm{B}, \mathrm{C}, \mathrm{Na}, \mathrm{F}, \mathrm{Cl}, \mathrm{Al}, \mathrm{H}, \mathrm{Li}, \mathrm{O}}}\)
1 6
2 3
3 1
4 4
Explanation:
CHXI08:REDOX REACTIONS
315152
What is the oxidation state of \(\mathrm{\mathrm{Mn}}\) in \(\mathrm{\mathrm{MnO}_{4}^{-}}\)?
1 +5
2 +6
3 +4
4 +7
Explanation:
Let the oxidation state of \(\mathrm{M n}\) be \(\mathrm{x}\). Oxidation state of \(\mathrm{\mathrm{O}=-2}\) \(\mathrm{\therefore \mathrm{x}+4(-2)=-1 \Rightarrow \mathrm{x}=+7}\)
MHTCET - 2022
CHXI08:REDOX REACTIONS
315153
What is the oxidation state of nitrogen in \(\mathrm{\mathrm{NaN}_{3}}\) ?
1 \({\rm{ - 3/1}}\)
2 \(3\)
3 \({\rm{ - 3}}\)
4 \({\rm{ - 1/3}}\)
Explanation:
\(\mathrm{\mathrm{NaN}_{3} \Rightarrow+1+3 \mathrm{x}=0}\) \(\mathrm{\Rightarrow 3 \mathrm{x}=-1 \Rightarrow \mathrm{x}=-\dfrac{1}{3}}\) So, oxidation number of nitrogen in \(\mathrm{\mathrm{NaN}_{3}}\) is \(\mathrm{-1 / 3}\)
CHXI08:REDOX REACTIONS
315154
The oxidation number of \(\mathrm{\mathrm{Cl}}\) in is \(\mathrm{\mathrm{NOClO}_{4}}\) is
1 +11
2 +9
3 +7
4 +5
Explanation:
The compound may be written as \(\mathrm{\mathrm{NO}^{+} \mathrm{ClO}_{4}^{-}}\) For \(\mathrm{\mathrm{ClO}_{4}^{-}}\), Let Ox. No.of \(\mathrm{\mathrm{Cl}=a}\) \({\rm{a + 4x( - 2) = - 1}}\) \({\rm{a = + 7}}\) Hence, the oxidation no. of \(\mathrm{\mathrm{Cl}}\) in \(\mathrm{\mathrm{NOClO}_{4}}\) is +7 .
