Question

How many ions are produced in an aqueous solution of \[~[Co{{({{H}_{2}}O)}_{6}}]C{{l}_{2}}\]?
(A) 2
(B) 3
(C) 4
(D) 6

Answer 1

Hint: Ions are of two types positive (action) and negative (anion) ions. To get the total number of ions we need to add all the positive and negative ions. Given compound is ionic with a positive ion at the first position and a negative ion in the second position. In an aqueous solution, compounds break down into positive and negative ions.

Complete Step by Step Solution:
Ions formation takes place with the release and gain of an electron. If any element tends to give electrons then it will attain a positive charge or can say electron donor. On the other hand, if any element tends to accept electrons then it will attain a negative charge or can say, electron acceptor.

In the given complex compound\[~[Co{{({{H}_{2}}O)}_{6}}]C{{l}_{2}}\], cobalt which is a central metal surrounded by six molecules of water (neutral ligands), tends to release electron (+2 and +3 oxidation states as per its electronic configuration). On the other hand, chlorine is an electronegative element thus, it tends to accept electrons (-1 oxidation state as per electronics configuration) such as \[C{{l}^{-}}\].

In the given compound there are two chlorine atoms, thus, both chlorine need one electron from this, we can indicate the oxidation state of cobalt which will be +2 for a given 2 electrons to two chlorine (one to each chlorine) such as \[C{{o}^{+2}}~\] .

So, in an aqueous solution, two positive ions (+2 oxidation state of cobalt with water neutral ligands) and one negative ion will produce (-1 oxidation state of chlorine).

Thus, the correct option is B.

Note: To get to know about the total number of ions, firstly know which part of the compound is positive and which part of the compound is negative. The first part (non-ionizable part) of a compound or salt is always positive and the second one is negative. Write down both separately such as \[{{[Co{{({{H}_{2}}O)}_{6}}]}^{+}}\] and \[C{{l}^{-}}\]. Now cross multiply with both charges number of an element means the number of chlorine atom (2) will multiply with a positive charge of the first part of compound and number of the first part (1) will multiply with the negative sign of chlorine atom such as \[{{[Co{{({{H}_{2}}O)}_{6}}]}^{+2}}\] and \[C{{l}^{-}}\]. As water legend is neutral thus the oxidation state of central metal (Co) will be the same as the charge on the first part of the compound.