Question

Which of the following is a correct Irving-Williams order (tendency of complex formation)?
A) $M{n}^{2+}$<$F{e}^{2+}$<$C{o}^{2+}$<$N{i}^{2+}$
B) $N{i}^{2+}$<$C{o}^{2+}$<$F{e}^{2+}$<$M{n}^{2+}$
C) $F{e}^{2+}$<$M{n}^{2+}$<$N{i}^{2+}$<$C{o}^{2+}$
D) $C{o}^{2+}$<$M{n}^{2+}$<$F{e}^{2+}$<$N{i}^{2+}$

Answer 1

Hint: The answer to this question is based on the capability of the formation for complexes that is dependent on the sequence given by the Irving-Williams order and the sequence is to be memorised.

Complete Solution :
The concept of inorganic chemistry that deals with the chapters based on the transition complexes that is included in the inorganic part of chemistry is familiar to us.
Let us now recollect the sequence order which was given by Irving and Williams and is well known as Irving-Williams order.
- Now, this order was given which refers to the stability of the complexes that are formed by the divalent transition metals.
- The main application of this series or order is to empirically suggest an order of stability which is within the first row transition metal complexes.
- Specifically, Irving-Williams order refers to the exchange of water ligands for any other ligand (L) within that of a metal complex.
- This series increase generally across the period to the maximum stability that is possessed by the copper and the series is given by $M{n}^{2+}$<$F{e}^{2+}$<$C{o}^{2+}$<$N{i}^{2+}$<$C{u}^{2+}$>$Z{n}^{2+}$
Since, in the given options the maximum stability is possessed here will be of nickel and among the two options that is option A) and D) the correct order is present in option A)
$M{n}^{2+}$<$F{e}^{2+}$<$C{o}^{2+}$<$N{i}^{2+}$
So, the correct answer is “Option A”.

Note: Note that the sequences and some of the rules relating to the transition metal complexes are to be remembered and here in this case the rule is applicable only for divalent compounds that are metal with${}^{\prime }+2^{\prime }$ oxidation state.