Question

For the following, identify the most likely value for x.a. BHx b. CHx c. NHx d. CH2Clx

Answer 1

Given :

a. $BH_x$ b. $CH_x$ c. $NH_x$ d. $CH_2Cl_x$ .

To Find :

Find the most likely vale of x for each one .

Solution :

a . $BH_x$

Because boron have valency of 3 .

So , x = 3 .

b . $CH_x$

Valency of carbon is 4 .

x = 4 .

c . $NH_x$

Valency of nitrogen is 3 .

Therefore , x = 3 .

d . $CH_2Cl_x$

Now ,we know valency of carbon is 4 and hydrogen is 1 .

Also , two hydrogen are already there .

So , only 2 electrons left to share .

Since , chlorine have valency of 1 .

Therefore , only 2 electrons of chlorine can connect .

x = 2 .

Hence , this is the required solution .