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Standard XII

Chemistry

Valency and Oxidation State

Assign oxidat...

Question

Assign oxidation numbers to each atom in $N H_{4} N O_{2}$ :

N = 4, - 4; H = 1; O = - 2

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N = - 3; H = 1; O = 3

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N = - 1; H = 1; O = - 2

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N = 3, - 3; H = 1; O = - 2

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Solution

The correct option is C $N = 3, - 3; H = 1; O = - 2$
$N H_{4} N O_{2}$
Oxidation number of $N : x + 4 - 1 = 0$
$\Rightarrow x + 3 = 0$
$\Rightarrow x = - 3$ .
Oxidation number of $H : - 3 + 4 x - 1 = 0$
$\Rightarrow 4 x = 3 + 1$
$\Rightarrow x = 1$
Oxidation number of $O : - 2$

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Similar questions

Q. Find

A, B

and

C

in the following reactions:

$N H_{3} + A + C O_{2} \to {(N H_{4})}_{2} C O_{3}$

${(N H_{4})}_{2} C O_{3} + H_{2} O + B \to N H_{4} H C O_{3}$

$N H_{4} H C O_{3} + N a C l \to N H_{4} C l + C$

Assign oxidation numbers to the underlined elements in each of the following species:
$N a H_{2} P O_{4}$

$N a H S - - O_{4}$

$H_{4} {P - -}_{2} O_{7}$

$K_{2} M n - -- - O_{4}$

$C a {O - -}_{2}$

$N a B - - H_{4}$

$H_{2} {S - -}_{2} O_{7}$

$K A l S - - O_{4})_{2} .12 H_{2} O$

Q. Given a reaction:

N O_{3}^{⊖} + Z n ⟶ {Z n}^{2 +} + N H_{4}^{\oplus}

The balanced reaction is

N O_{3}^{⊖} + 7 H_{2} O + 4 Z n ⟶ N H_{4}^{\oplus} + 10 ⊖ O H + 4 {Z n}^{2 +}

.
If true enter 1, else enter 0.

H_{2} O + H_{3} P O_{4} + H_{3} O^{+} + H_{2} P O_{4}^{-}, p^{k_{1}} = 2.15

H_{2} O + H_{2} P O_{4}^{-} + H_{3} O^{+} + H P O_{4}^{2 -}, p^{k_{2}} = 7.20

p^{H}

of 0.01 M

N a H_{2} P O_{4}

is:

$H_{3} P O_{4} + H_{2} O ⇌ H_{3} O^{+} + H_{2} P O_{4}^{-}$ $p K_{1} = 2.15$

$H_{2} P O_{4}^{-} + H_{2} O ⇌ H_{3} O^{+} + H P O_{4}^{2} -$ $p K_{2} = 7.20$

Hence pH of 0.01 M $N a H_{2} P O_{4}$ is:

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Assign oxidation numbers to each atom in NH4NO2:

The correct option is C N=3,−3;H=1;O=−2NH4NO2Oxidation number of N:x+4−1=0⇒x+3=0⇒x=−3.Oxidation number of H:−3+4x−1=0⇒4x=3+1⇒x=1Oxidation number of O:−2

Assign oxidation numbers to each atom in $N H_{4} N O_{2}$ :

The correct option is C $N = 3, - 3; H = 1; O = - 2$
$N H_{4} N O_{2}$
Oxidation number of $N : x + 4 - 1 = 0$
$\Rightarrow x + 3 = 0$
$\Rightarrow x = - 3$ .
Oxidation number of $H : - 3 + 4 x - 1 = 0$
$\Rightarrow 4 x = 3 + 1$
$\Rightarrow x = 1$
Oxidation number of $O : - 2$