Question

Match the symbols of the radicals with their correct combining power or capacity, (valency).
(A)- ${\text{OH}}$
(B)- ${\text{S}}{{\text{O}}_{\text{4}}}$
(C)- ${\text{N}}{{\text{O}}_{\text{3}}}$
(D)- ${\text{C}}{{\text{O}}_{\text{3}}}$
Combining power or capacity - ${\text{A:3,B:2,C:1}}$. (negative valencies)

Answer 1

Valency, capacity or combining power of any radical depends upon the outermost shell electronic configuration species. On the basis of no. of electrons present in the outermost shell on that species, we can calculate the no. of gaining electrons to the nucleus or no. of removing electrons from the nucleus; which in other terms is also known as valency.

Complete answer: Here in this question we have to find out the negative valencies of the given radicals. So if we know the positive valency of any radical which is present with the given negative radicals in any compound then we easily calculate the negative valencies because positive valencies is always equal to the negative valencies of any compound.
-In the option (A) hydroxyl radical is given, which is present in the water molecule in the attachment with hydrogen i.e. ${{\text{H}}_{\text{2}}}{\text{O = H - OH}}$. As we know hydrogen shows +1 valency, so contrary to this negative valency of ${\text{OH}}$ is -1.
-In the option (B) sulphate radical is given, which is present in the sulphuric acid in the attachment with 2 hydrogens i.e. ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_4}$. As we know hydrogen shows +1 valency, so 2 hydrogens combinedly shows +2 valency and contrary to this negative valency of ${\text{S}}{{\text{O}}_{\text{4}}}$ is -2.
-In the option (C) nitrate radical is given, which is present in the salts like sodium nitrate in the attachment with 1 sodium i.e. ${\text{NaN}}{{\text{O}}_3}$. As we know sodium shows +1 valency and contrary to this negative valency of ${\text{N}}{{\text{O}}_{\text{3}}}$ is -1.
-In the option (D) carbonate radical is given, which is present in the salts like sodium carbonate in the attachment with 2 sodium i.e. ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$. As we know sodium shows +1 valency, so 2 sodium combinedly shows +2 valency and contrary to this negative valency of ${\text{C}}{{\text{O}}_{\text{3}}}$ is -2.

Hence, the correct match of the symbols of the radicals with their correct combining power or capacity is as follow:
(A)- ${\text{OH}}$ - C
(B)- ${\text{S}}{{\text{O}}_{\text{4}}}$ - B
(C)- ${\text{N}}{{\text{O}}_{\text{3}}}$ - C
(D)- ${\text{C}}{{\text{O}}_{\text{3}}}$ - B

Note: During calculating the valency of any radical never correlate the valency of central atom present in that radical, because if you are doing so then you have to add all the valencies of the attaching atoms to the center atom to get the right answer. So, for ignoring that type of mistakes always follow the above method which explains the complete solution.