Calculate the n factor of $N H_{4} C l$ for the following change:
$N H_{4} C l + N a N O_{2} \to N_{2} + 2 H_{2} O + N a C l$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 3
Steps for finding valency factor/n-factor:
1. Find the oxidation state (O.S.) of metal in reactant side.
2. Find the oxidation state of metal in product side.
3. Take the magnitude of difference of two oxidation state $| O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} |$ for the valency factor

$(N H_{4} C l + N a N O_{2} \to N_{2} + H_{2} O + N a C l$
O.S. of N in reactant side
$\Rightarrow x + 1 \times (- 1) + 4 \times (+ 1) = 0$
$\Rightarrow x = - 3$

O.S. of N in product side
$\Rightarrow x = 0$ as nitrogen is present in natural state.

$valency factor = | O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} |$
$valency factor = | 0 - (-) 3 |$
$valency factor = 3$

n-factor of $N H_{4} C l = number of N atom \times | O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} | = 1 \times 3 = 3$

Suggest Corrections

Similar questions

N H_{4} C l

N H_{4} C l + N a N O_{2} \to N_{2} + 2 H_{2} O + N a C l

$N H_{4} C l$ + $N a N O_{2} \to$ ____ + ____

{N O}_{2}^{-}

{N O}_{3}^{-}

{N O}_{2}^{-}

N H_{4} C l

{N O}_{2}^{-}

N a N O_{2} + N H_{4} C l △ ⟶ N a C l + N_{2} ↑ + 2 H_{2} O

S_{2} O_{3}

2 N a_{2} S_{2} O_{3} \to N a_{2} S_{4} O_{6}

Calculate n-factor of $N a_{3} P O_{4}$ in the following reaction.

$2 Na_3PO_4$ +$3 BaCl_2$ $\to$ 6NaCl + $B a_{3} (P O_{4})_{2}$

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program