Question

which metal ion has highest stability in the Irving-Williams Order of stability of metal ions?
A.$$M{n_2}^{+}$$
B.$F{e_2}^{+}$
C.$Z{n_2}^{+}$
D.$C{u_2}^{+}$

Answer 1

Hint:The complexes or the structures formed by the elements of the first row transition element increases the stability while going from left to right in the period until the element copper after then does not apply.

Complete step by step answer:
The complexes or structures formed by the transition elements of period first, their stability series or order was proposed by Harry Irving and Robert Williams in 1953.This series or order was given the name Irving-Williams series after both its inventors.
Now, they observed that while moving left to right in the first row where we have the divalent transition elements, the stability of the complex formed by these elements increases from left to right in the period till we reach the element copper.
The series is;
$$M{n_2}^{+}<F{e_2}^{+}<C{o_2}^{+}<N{i_2}^{+}<C{u_2}^{+}>Z{n_2}^{+}$$
We can see in the above series that after copper the stability decreases of a complex made of zinc.
Now, the ionic radius of these divalent ions reduces from going left to right, which states the reason for their increase in stability of their complex.
Now, learning this series we can easily say that the answer to the above question is copper.
Hence, copper is the metal ion which has higher stability in the Irving-Williams Order of Stability of metal ions.

Hence, option D is correct.

Note:
There is a trick we can use to remember this Irving-Williams order of stability of metal ions that can be remembered as feco-niqu , which is the short form of this series. Remember it is an increasing order.