The equivalent weights of two hydrides of nitrogen are $5.66$ and $8$ .
Briefly, explain the formulae of hydrides are:

(I) hydrides appear $N H_{2}$ (unstable),
(II) actual hydride is $N_{2} H_{4}$ (hydrazine).

Open in App

Solution

Hydrides :
(I) $N_{x} H_{y}$
Molar mass $= (14 x + y)$
$y$ $g H = (14 x + y) g$ $N_{x} H_{y}$

Using eq. concept:
$\frac{14 x + y}{y} = 5.66$ (given)
$\frac{14 x}{y} = 4.66$
$\frac{x}{y} = \frac{1}{3} = \frac{n u m b e r o f n i t r o g e n a t o m s}{n u m b e r o f h y d r o g e n a t o m s}$
Hence, hydrogen is $N H_{3}$
(II) $\frac{14 x + y}{y} = 8$ (given)
$\frac{14 x}{y} = 7$
$\frac{x}{y} = \frac{1}{2} = \frac{n u m b e r o f n i t r o g e n a t o m s}{n u m b e r o f h y d r o g e n a t o m s}$
Thus, hydrides appears $N H_{2}$ (unstable).
Thus, actual hydride is $N_{2} H_{4}$ (hydrazine).

Suggest Corrections

Similar questions

Write the names and formulae of (a) a metal hydride, and (b) a non-metal hydride.

Join BYJU'S Learning Program