# Tag Info

10

The reason d-orbitals make a difference is that electrons in d-orbitals do not screen nuclear charge as effectively as those in s and p orbitals. This is because of something called penetration . The mathematical shapes of d-orbitals prevent them from allowing electrons to penetrate very closely to the nucleus, compared with electrons in s or p-orbitals. ...

8

This statement is relevant for the initially [incorrectly] determined structure of $\ce{B2O3}$ by Berger [1, 2] who used powder x-ray diffraction data. You can still access the structure and visualize it in 3D (ICSD 24047), but it was shown later by several researchers and summarized in paper by Gurr et al. [3] that this structure has been erroneously ...

4

It depends on what you define as a bond. A three-center two electron bond is in general stronger than an ordinary two-center bond because delocalizing the bond over three atoms instead of two makes the bonding MO more stable. But, the bridge bond is also shared between two linkages so each individual linkage has less bonding than a single two-center bond ...

3

Naively, one should expect the atomic radii to increase in a proportional fashion from boron to aluminium, gallium, indium and thallium, because each of these atoms has one more populated shell. However, the difference is not proportional because there is not a constant proton difference between the elements. From boron to aluminium 8 protons are added to ...

3

In the equation $\ce{H2O + H3BO3 -> H+ + B(OH)4-}$, one negative charge on the anion is shared among 4 electronegative oxygens. Yes, one positive charge on the proton is also shared with some water molecules, so the orthoborate ion probably shares some negative charge with some waters, but let's keep it simple. Boric acid is known to complex with sugars ...

2

It isn't just boron. The same Wikipedia list gives a -5 oxidation state for all the Group 13 elements except nihonium (which is only very recently known) and, more curiously, aluminum. For thallium Wikipedia's article on the element gives this reference, where a compound has thallium in various clusters and individual atoms bonded to alkali metals. The ...

1

Since both are metaloids, or semiconductors, look for the criteria that define them. In particular, electronegativity or ionization energy are important.

1

Going down any group of the Periodic Table the general trend is that electronegativity decreases. The relative electronegativities of In and Tl are simply a reflection of this fact. That's the end of the story; you don't need to bring in anything more complicated. You are correct that the 4f electrons are poorly shielding, and thus the 6p elements (Tl, Pb, ...

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