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1

Pentanes are heavier than butanes, whatever their geometry. So the boiling point of pentanes are higher. This is a general tendency in the series of hydrocarbons. Pentanes and butanes are good examples to show that the chain length has not the most important effect on the boiling point. n-Pentane boils at $\pu{36°C}$, 2,2-dimethylpropane at $\pu{9.5°C}$. ...


-2

It is due to the difference in their second and third ionisation enthalpies. Due to the large difference in the second and third ionisation enthalpies arising from the disturbal of the stable half filled electronic configuration, +2 oxidation state is more favourable.


0

When an atom $M$ looses one or more electrons, it becomes positively charged and becomes the ion ${M^{z+}}$. In aqueous solution this positive charge attracts the unused doublets of the Oxygen atom from $H_2O$, and it repells the proton of $H^+$. If $Z$ = $1$, the ion $M^+$ attracts several molecules water around it, and it makes a Debye layer of oriented $...


2

In the lower part of the periodic table, the nucleus charge becomes so high that the electron must rotate at a relativist speed to stay in orbit in the Bohr's model. And even if the Bohr's model is outdated, it can be used to explain the strange chemistry happening in the heaviest atoms. Relativity says that when an object moves at a speed not far from c, ...


2

Both complexes (and many more such as permanganate, sulphate, …) share the same general MO scheme which I am going to shamelessly copy from my older answer: Figure 1: Qualitative MO scheme of a tetrahedric complex with σ and π bonding between metal and ligands. Double vertical lines represent electron pairs. The intense colour of permanganate ...


0

Shakespeare apparently said "What’s in a name? That which we call a rose, by any other word would smell as sweet." Rule no. 1, don't trust the internet until and unless you have some idea about the credentials of the website (and this comes with experience). The usage of the term transition metal is nothing but historical. You can stick to IUPACs definition ...


2

Compounds with Mn in formal oxidation state +1 and -1 are well known. They simply are not stable on air and in water. In general, air-stable compounds of more active 3d-metals are ionic compounds. Their stability is a result of fine balance of energy of formation of relevant ions and energy of stabilization of said ions due to ion-ion and/or ion-dipole ...


0

$\ce{Mn+}$ has configuration $\mathrm{(4s)^0(3d)^6},$ while on the other hand $\ce{Mn^2+}$ $\mathrm{(3d)^5(4s)^0}$ and also, $\ce{Mn}$ in ground state is $\mathrm{(4s)^2(3d)^5}.$ Therefore you can see in the oxidation state +2 orbitals are half-filled, and in +1 they are not half-filled. As half-filled is more stable, therefore +2 and 0 are more stable than ...


2

If you inspect Table II in Ref. 1, which you will find referenced in the link you provide to the WebElements website, you'll see that AU units are used (also in the linked page), not pm, so the radius of the 3d Cu orbital in Ref. 1 is actually $$ r_{\textrm{Mann}} = 0.613 \times 0.529 Å = 0.324 Å $$ or 32.4 pm. As for Slaters approach, it is not enough ...


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