Given :
a. [tex]BH_x[/tex] b. [tex]CH_x[/tex] c. [tex]NH_x[/tex] d. [tex]CH_2Cl_x[/tex] .
To Find :
Find the most likely vale of x for each one .
Solution :
a . [tex]BH_x[/tex]
Because boron have valency of 3 .
So , x = 3 .
b . [tex]CH_x[/tex]
Valency of carbon is 4 .
x = 4 .
c . [tex]NH_x[/tex]
Valency of nitrogen is 3 .
Therefore , x = 3 .
d . [tex]CH_2Cl_x[/tex]
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 .
