Why is radius of Fe2+ less than that of Mn2+ ?
in paramagnetic ions all the bonds formed between Mn and O are covalent,why ?
Mn has 25+ and 23- for +2 net.
Fe has 26+ and 24- for +2 net.
Therefore, Fe^2+ has a higher + charge pulling on those outside electrons.
The reason why the radius of Fe2+ is less than that of Mn2+ is primarily due to the difference in their electronic configurations.
Fe2+ has an electronic configuration of [Ar] 3d^6, which means it has 24 electrons in total. On the other hand, Mn2+ has an electronic configuration of [Ar] 3d^5, which corresponds to 23 electrons.
In transition metals like iron (Fe) and manganese (Mn), the 3d subshell plays a significant role in determining atomic and ionic sizes. The 3d subshell is located closer to the nucleus than the 4s subshell and is shielded less effectively by the outer electrons.
With Fe2+, the 3d subshell is fully half-filled, which increases electron-electron repulsion and causes the electrons to spread out more to minimize their interaction. As a result, the extra electron in the 3d subshell of Mn2+ experiences less repulsion from other electrons compared to Fe2+, leading to a larger size for Mn2+.
Therefore, due to the additional repulsion in Fe2+, the radius of Fe2+ is smaller than that of Mn2+.
The reason why the radius of Fe2+ is less than that of Mn2+ can be explained by considering the electronic configuration and the charge of the ions.
To start, we need to take a look at the electronic configuration of both Fe2+ and Mn2+ ions. Fe2+ has lost two electrons from its neutral state (Fe), so it has an electronic configuration of [Ar]3d6. On the other hand, Mn2+ has also lost two electrons from its neutral state (Mn), so its electronic configuration is [Ar]3d5.
Now, let's consider the effect of the charge of the ions on their radii. When an ion loses electrons and becomes positively charged, there is an increase in effective nuclear charge. This means that the positive charge of the nucleus is experienced by the remaining electrons more strongly, leading to a pull that causes the electron cloud to contract.
In the case of Fe2+, it has one more electron in its 3d orbitals compared to Mn2+. The additional electron in Fe2+ experiences a greater effective nuclear charge, as it is closer to the nucleus. This stronger pull from the increased positive charge causes the electron cloud to shrink, resulting in a smaller ionic radius for Fe2+ compared to Mn2+.
In summary, the radius of Fe2+ is smaller than that of Mn2+ due to the higher effective nuclear charge experienced by the additional electron in the 3d orbital of Fe2+, causing the electron cloud to contract.