Which of the following is a correct Irving-Williams order (tendency of complex formation)?

M n^{2 +} < F e^{2 +} < C o^{2 +} < N i^{2 +}

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N i^{2 +} < C o^{2 +} < F e^{2 +} < M n^{2 +}

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F e^{2 +} < M n^{2 +} < N i^{2 +} < C o^{2 +}

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C o^{2 +} < M n^{2 +} < F e^{2 +} < N i^{2 +}

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Solution

The correct option is B $M n^{2 +} < F e^{2 +} < C o^{2 +} < N i^{2 +}$
For a given metal and ligand, the greater the stability the greater the charge on the metal ion. Thus, stability of coordination entities of ions of charge +3 is greater than the entities of +2 ions. For the divalent ions of the first row transition elements, irrespective of the ligand involved, the stabilities vary in the Irving-Williams order:

$M n^{+ 2} < F e^{2 +} < C O^{+ 2} < N i^{2 +} < C u^{+ 2} > Z n^{+ 2}$

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Similar questions

C u^{2 +}, N i^{2 +}, C o^{2 +}, F e^{2 +}

M n^{2 +}

V^{2 +} < C r^{2 +} < M n^{2} < F e^{2 +}

N i^{2 +} < C o^{2 +} < F e^{2 +} < M n^{2 +}

C o^{3 +} < F e^{3 +} < C r^{3 +} < S c^{3 +}

S c < T i < C r < M n

M^{2 +}

Z n^{2 +} > S c^{2 +} = C u^{2 +} > T i^{2 +} = N i^{2 +} > V^{2 +} = C o^{2 +} > C r^{2 +} = F e^{2 +} > M n^{2 +}

Magnetic moment of which ion is 2.88 BM ?

M n^{2 +}, C o^{2 +}, Z n^{2 +} & N i^{2 +}

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