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When N2 is obtained from NH; using oxide of copper, the change in the oxidation state of metal is 1) +2 to 0 2) +2 to +1 3) +1 to +2 4) 0 to +2\red\bigstar★ Explanation \red\bigstar★
\leadsto⇝ Solution:-
i) HNO
Let the valency of N be x
As it is not an ion therefore the sum of the all valencies of each element in that atom/molecule will be equal to 0
Therefore,
+ 1 + x - 2 = 0
[Here the general valencues of H and O are +1 and -2 respectively]
x - 1 = 0
Therefore
\boxed{Valency\:of\:Nitrogen\:is\:+1}
ValencyofNitrogenis+1
ii) NO
Let the valency of N be x
As it is not an ion therefore the sum of the all valencies of each element in that atom/molecule will be equal to 0
Therefore,
x - 2 = 0
Therefore
\boxed{Valency\:of\:Nitrogen\:is\:+2}
ValencyofNitrogenis+2
iii) NO2
Let the valency of N be x
As it is not an ion therefore the sum of the all valencies of each element in that atom/molecule will be equal to 0
Therefore,
x + 2(-2) = 0
x - 4 = 0
Therefore
\boxed{Valency\:of\:Nitrogen\:is\:+4}
ValencyofNitrogenis+4
iv) NH
Let the valency of N be x
As it is not an ion therefore the sum of the all valencies of each element in that atom/molecule will be equal to 0
Therefore,
x + 1 = 0
Therefore
\boxed{Valency\:of\:Nitrogen\:is\:-1}
ValencyofNitrogenis−1
When N2 is obtained from NH; using oxide of copper, the change in the oxidation state of metal is 1) +2 to 0 2) +2 to +1 3) +1 to +2 4) 0 to +2