Question

1. Consider the table of reduction potentials for several transition metals (values refer to the reaction M +xe M, where x 1,

0 0
Add a comment Improve this question Transcribed image text
Answer #1

The irregularities in the trend of reduction potentials is because of the irregularity in ionisation enthalpies and hydration energies of these ions. The Mn(II) and Zn(II) ions have half filled (d​​​​​​5) and fully filled configurations (d​​​​​​10), so they have anomolous reduction potentials.

Trend in atomic or ionic radii: The radii of the transition elements should decrease with the increase in atomic no. because of increasing nuclear charge but this trend has a few anomalies. It decreases rapidly from group 3-6 (ex: Sc to Cr), then remains constant for from group 7-10 (Mn, Fe, Co and Ni) and again increases for group 11-12 (Cu,Zn). The first decrease is due to general trend i.e. increase in nuclear charge and poor shielding of d-orbitals. Then this increasing effective nuclear charge is balancd by the repulsion from increasing d electrons. And hence their reduction potentials also deviate.

While moving from 3d to 4d series, there is increase in atomic radii due to increase in no. Of shells. But while going from 4d to 5d, there is no significant change in atomic radii due to lanthanide contraction. Below is the graph showing this trend. 8.00 3- Sc da 9.00 10.00 Sc 11.00 No raduus Cl 12.00 A RL Te Ru 2n 100 CLA Mn Co 2.00 Inerrarisg at 3.00

Trend in ionisation enthalpy:

There is an increase in the ionisation enthalpies on moving from left to right because of decreasing ionic radii but there are some variations. Although the increase in second and third ionisation energies while moving along the group is much higher as compared to the increase in first ionisation energy. The irregular trend in FIRST IONSIATION ENTHALPIES(removal of one electron) is due to the alteration of relative energies of 3d and 4s orbitals with the removal of electrons. The ionisation enthalpy increases to some extent due to increase in nuclear charge. But the value of Cr is lower due to stable half filled (d5) configuration after the removal of one electron and the value for Zn is higher due to filled stable configurations(4s2 3d10).

To form M(II) ions, the sum of first and second I.E. is required. The second I.E. terms are very high for Cr and Cu where d5 and d10 configurations are stable, hence have less negative reduction potentials i.e. can easily be oxidised. This value for Zn is lower because loss of a second electron from 4s is easy.

The trend in third ionisation enthalpies is almost regular but the Mn(d5) and Zn(d10) shows difficulty in ionisation.

Add a comment
Know the answer?
Add Answer to:
1. Consider the table of reduction potentials for several transition metals (values refer to the reaction M +xe M, w...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT