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Standard XII

Chemistry

Ammonia

When excess o...

Question

When excess of ammonia is used:

$3 C l_{2} + 8 N H_{3}$ $⟶$ $A + B$

When excess of chlorine is used:

$3 C l_{2} + 4 N H_{3}$ $⟶$ $C + D$

The oxidation state of nitrogen in compound $A$ is:

- 3

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Solution

The correct option is A $- 3$
$8 N H_{3} + 3 C l_{2} \to 6 N H_{4} C l + N_{2}$

Compound A is $N H_{4} C l$
$1 (x) + 4 (1) + 1 (- 1) = 0$
$x + 4 - 1 = 0$
$x = - 3$
The oxidation state of compound A is $- 3$
Hence option A is correct.

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Similar questions

Q. When excess of ammonia is used:

3 C l_{2} + 8 N H_{3}

⟶

A + B

When excess of chlorine is used:

3 C l_{2} + 4 N H_{3}

⟶

C + D

The oxidation state of nitrogen in compound

D

is:

Q. When excess of ammonia is used:

3 C l_{2} + 8 N H_{3}

⟶

A + B

When access of chlorine is used:

3 C l_{2} + 4 N H_{3}

⟶

C + D

Oxidation state of nitrogen in compound

B

is:

The weight of chlorine gas undergoing oxidation when excess of chlorine is sent into hot concentrated caustic soda solution containing 24 grams of solute in it
$3 C l_{2} + 6 N a O H \to 5 N a C l + N a C l O_{3} + 3 H_{2} O$

Q. Copy and complete the following equations:
(i) Mg₃N₂ + 6H₂O

\overset{}{\to}

(ii) 2NH₃ + 3CuO

\overset{}{\to}

(iii) 8NH₃ + 3Cl₂

\overset{}{\to}

(iv) 4NH₃ + 5O₂

\overset{}{\to}

Question 15
Ozone-layer is getting depleted because of

(a) excessive use of automobiles
(b) excessive formation of industrial units
(c) excessive use of man-made compounds containing both fluorine and chlorine
(d) excessive deforestation.

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When excess of ammonia is used: 3Cl2+8NH3⟶A+BWhen excess of chlorine is used:3Cl2+4NH3⟶C+DThe oxidation state of nitrogen in compound A is:

The correct option is A −38NH3+3Cl2→6NH4Cl+N2 Compound A is NH4Cl 1(x)+4(1)+1(−1)=0x+4−1=0x=−3The oxidation state of compound A is −3Hence option A is correct.

When excess of ammonia is used:

$3 C l_{2} + 8 N H_{3}$ $⟶$ $A + B$

When excess of chlorine is used:

$3 C l_{2} + 4 N H_{3}$ $⟶$ $C + D$

The oxidation state of nitrogen in compound $A$ is:

The correct option is A $- 3$
$8 N H_{3} + 3 C l_{2} \to 6 N H_{4} C l + N_{2}$

Compound A is $N H_{4} C l$
$1 (x) + 4 (1) + 1 (- 1) = 0$
$x + 4 - 1 = 0$
$x = - 3$
The oxidation state of compound A is $- 3$
Hence option A is correct.