18654
Max. number of moles of electrons taken up by one mole of \(NO_3^ - \) when it is reduced to
1 \(N{H_3}\)
2 \(N{H_2}OH\)
3 \(NO\)
4 \(N{O_2}\)
Explanation:
Oxidation state of \(N\) in the following compounds: \(NO _3{ }^{-} \rightarrow+5\) \(NH _3 \rightarrow-3\) [ Number of moles of electrons taken \(=5-(-3)=8\) ] \(NH _2 OH \rightarrow-1\) [ Number of moles of electrons taken \(=5-(-1)=6\) ] \(NO \rightarrow+2\) [ Number of moles of electrons taken \(=5-(+2)=3]\) \(NO _2 \rightarrow+4\) [ Number of moles of electrons taken \(\left.=5-(4)=1\right]\)
Redox Reactions
18655
In the reaction \(3Mg + {N_2} \to M{g_3}{N_2}\)
1 Magnesium is reduced
2 Magnesium is oxidized
3 Nitrogen is oxidized
4 None of these
Explanation:
(b) In the given reaction oxidation state of \(Mg\) is changing from \(0\) to \(+2\) while in nitrogen it is changing from \(0\) to \(-3\). So oxidation of \(Mg\) and reduction of nitrogen takes place.
Redox Reactions
18656
When sodium metal is dissolved in liquid ammonia, blue colour solution is formed. The blue colour is due to
1 Solvated \(N{a^ + }\) ions
2 Solvated electrons
3 Solvated \(NH_2^ - \) ions
4 Solvated protons
Explanation:
(b) When sodium metal is dissolved in liquid ammonia to form coloured solution. Dilute solutions are bright blue in colour due to the presence of solvated electrons. \(Na + (x + y)N{H_3} \to {[Na{(N{H_3})_x}]^ + } + \mathop {{{[e{{(N{H_3})}_y}]}^ - }}\limits_{{\rm{Blue \,Colour}}} \)
Redox Reactions
18657
\(SnC{l_2}\) gives a precipitate with a solution of \(HgC{l_2}.\) In this process \(HgC{l_2}\) is
1 Reduced
2 Oxidised
3 Converted into a complex compound containing both \(Sn\) and \(Hg\)
4 Converted into a chloro complex of \(Hg\)
Explanation:
(a) In this reaction \(HgC{l_2}\) is reduced in \(Hg\).
18654
Max. number of moles of electrons taken up by one mole of \(NO_3^ - \) when it is reduced to
1 \(N{H_3}\)
2 \(N{H_2}OH\)
3 \(NO\)
4 \(N{O_2}\)
Explanation:
Oxidation state of \(N\) in the following compounds: \(NO _3{ }^{-} \rightarrow+5\) \(NH _3 \rightarrow-3\) [ Number of moles of electrons taken \(=5-(-3)=8\) ] \(NH _2 OH \rightarrow-1\) [ Number of moles of electrons taken \(=5-(-1)=6\) ] \(NO \rightarrow+2\) [ Number of moles of electrons taken \(=5-(+2)=3]\) \(NO _2 \rightarrow+4\) [ Number of moles of electrons taken \(\left.=5-(4)=1\right]\)
Redox Reactions
18655
In the reaction \(3Mg + {N_2} \to M{g_3}{N_2}\)
1 Magnesium is reduced
2 Magnesium is oxidized
3 Nitrogen is oxidized
4 None of these
Explanation:
(b) In the given reaction oxidation state of \(Mg\) is changing from \(0\) to \(+2\) while in nitrogen it is changing from \(0\) to \(-3\). So oxidation of \(Mg\) and reduction of nitrogen takes place.
Redox Reactions
18656
When sodium metal is dissolved in liquid ammonia, blue colour solution is formed. The blue colour is due to
1 Solvated \(N{a^ + }\) ions
2 Solvated electrons
3 Solvated \(NH_2^ - \) ions
4 Solvated protons
Explanation:
(b) When sodium metal is dissolved in liquid ammonia to form coloured solution. Dilute solutions are bright blue in colour due to the presence of solvated electrons. \(Na + (x + y)N{H_3} \to {[Na{(N{H_3})_x}]^ + } + \mathop {{{[e{{(N{H_3})}_y}]}^ - }}\limits_{{\rm{Blue \,Colour}}} \)
Redox Reactions
18657
\(SnC{l_2}\) gives a precipitate with a solution of \(HgC{l_2}.\) In this process \(HgC{l_2}\) is
1 Reduced
2 Oxidised
3 Converted into a complex compound containing both \(Sn\) and \(Hg\)
4 Converted into a chloro complex of \(Hg\)
Explanation:
(a) In this reaction \(HgC{l_2}\) is reduced in \(Hg\).
18654
Max. number of moles of electrons taken up by one mole of \(NO_3^ - \) when it is reduced to
1 \(N{H_3}\)
2 \(N{H_2}OH\)
3 \(NO\)
4 \(N{O_2}\)
Explanation:
Oxidation state of \(N\) in the following compounds: \(NO _3{ }^{-} \rightarrow+5\) \(NH _3 \rightarrow-3\) [ Number of moles of electrons taken \(=5-(-3)=8\) ] \(NH _2 OH \rightarrow-1\) [ Number of moles of electrons taken \(=5-(-1)=6\) ] \(NO \rightarrow+2\) [ Number of moles of electrons taken \(=5-(+2)=3]\) \(NO _2 \rightarrow+4\) [ Number of moles of electrons taken \(\left.=5-(4)=1\right]\)
Redox Reactions
18655
In the reaction \(3Mg + {N_2} \to M{g_3}{N_2}\)
1 Magnesium is reduced
2 Magnesium is oxidized
3 Nitrogen is oxidized
4 None of these
Explanation:
(b) In the given reaction oxidation state of \(Mg\) is changing from \(0\) to \(+2\) while in nitrogen it is changing from \(0\) to \(-3\). So oxidation of \(Mg\) and reduction of nitrogen takes place.
Redox Reactions
18656
When sodium metal is dissolved in liquid ammonia, blue colour solution is formed. The blue colour is due to
1 Solvated \(N{a^ + }\) ions
2 Solvated electrons
3 Solvated \(NH_2^ - \) ions
4 Solvated protons
Explanation:
(b) When sodium metal is dissolved in liquid ammonia to form coloured solution. Dilute solutions are bright blue in colour due to the presence of solvated electrons. \(Na + (x + y)N{H_3} \to {[Na{(N{H_3})_x}]^ + } + \mathop {{{[e{{(N{H_3})}_y}]}^ - }}\limits_{{\rm{Blue \,Colour}}} \)
Redox Reactions
18657
\(SnC{l_2}\) gives a precipitate with a solution of \(HgC{l_2}.\) In this process \(HgC{l_2}\) is
1 Reduced
2 Oxidised
3 Converted into a complex compound containing both \(Sn\) and \(Hg\)
4 Converted into a chloro complex of \(Hg\)
Explanation:
(a) In this reaction \(HgC{l_2}\) is reduced in \(Hg\).
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Redox Reactions
18654
Max. number of moles of electrons taken up by one mole of \(NO_3^ - \) when it is reduced to
1 \(N{H_3}\)
2 \(N{H_2}OH\)
3 \(NO\)
4 \(N{O_2}\)
Explanation:
Oxidation state of \(N\) in the following compounds: \(NO _3{ }^{-} \rightarrow+5\) \(NH _3 \rightarrow-3\) [ Number of moles of electrons taken \(=5-(-3)=8\) ] \(NH _2 OH \rightarrow-1\) [ Number of moles of electrons taken \(=5-(-1)=6\) ] \(NO \rightarrow+2\) [ Number of moles of electrons taken \(=5-(+2)=3]\) \(NO _2 \rightarrow+4\) [ Number of moles of electrons taken \(\left.=5-(4)=1\right]\)
Redox Reactions
18655
In the reaction \(3Mg + {N_2} \to M{g_3}{N_2}\)
1 Magnesium is reduced
2 Magnesium is oxidized
3 Nitrogen is oxidized
4 None of these
Explanation:
(b) In the given reaction oxidation state of \(Mg\) is changing from \(0\) to \(+2\) while in nitrogen it is changing from \(0\) to \(-3\). So oxidation of \(Mg\) and reduction of nitrogen takes place.
Redox Reactions
18656
When sodium metal is dissolved in liquid ammonia, blue colour solution is formed. The blue colour is due to
1 Solvated \(N{a^ + }\) ions
2 Solvated electrons
3 Solvated \(NH_2^ - \) ions
4 Solvated protons
Explanation:
(b) When sodium metal is dissolved in liquid ammonia to form coloured solution. Dilute solutions are bright blue in colour due to the presence of solvated electrons. \(Na + (x + y)N{H_3} \to {[Na{(N{H_3})_x}]^ + } + \mathop {{{[e{{(N{H_3})}_y}]}^ - }}\limits_{{\rm{Blue \,Colour}}} \)
Redox Reactions
18657
\(SnC{l_2}\) gives a precipitate with a solution of \(HgC{l_2}.\) In this process \(HgC{l_2}\) is
1 Reduced
2 Oxidised
3 Converted into a complex compound containing both \(Sn\) and \(Hg\)
4 Converted into a chloro complex of \(Hg\)
Explanation:
(a) In this reaction \(HgC{l_2}\) is reduced in \(Hg\).
