How does boron get these two different oxidation states?

Boron has two different oxidation states because it can exist in two different forms, with different numbers of electrons. In one form, it has three electrons (boron III), and in the other form, it has five electrons (boron V). When it donates an electron to another molecule, it becomes boron III, and when it accepts an electron from another molecule, it becomes boron V.

Ytterbium is one of the lanthanide element and has the configuration [Xe]4f145d16s2. When it looses the two electrons on the 6s orbital its oxidation state is +2. When three electrons, 1 in 5d and 2 in the 6s orbital are lost, the oxidation state is +3 which appears to be the common one for the lanthanides. The oxidation state +2 is not common possibly due to lanthanide contraction.

Due to inert pair effect.

In the elements which are at lower position in group, there are d and f orbitals which have poor shielding effect. So outermost s electrons feel more attractions with nucleus and become more difficult to loose. So as 2 electrons are not getting lost, oxidation state decreases from +3 to +1.

Boron typically has an oxidation state of +3 in compounds, but it can also have an oxidation state of +5. The reason for this is that boron has three unpaired electrons in its outer shell, and these unpaired electrons are very reactive.

When boron is reacted with oxygen, it can gain or lose one of its unpaired electrons to form either BO 3 3- or BO 5 5-. BO 3 3- is the more common form of boron oxide, and it has an oxidation state of +3. BO 5 5- is the less common form of boron oxide, and it has an oxidation state of +5

There are 5 electrons in the valence (outermost) shell of Nitrogen atom: 2s2 2p3. All atoms lose or gain electrons to obtain the stable valence shell configuration of inert gases like Neon, Xenon etc. Nitrogen can obtain inert gas configuration by either gaining 3 electrons (2s2 2p6), thereby attaining oxidation state of -3, or losing all the 5 valence shell electrons, thus becoming +5. Hence, Nitrogen can show all oxidation states from -3 to +5, the most common ones however being -3, +3 or +5, and other intermediate states during gradual losing or gaining electrons.

That's wrong. In most boron compounds, the formal oxidation state is +III, so it is chemically very important…

K is always +1, so the charge of complex anion is 2-. NH3 and (O2) are neutral ligands, whereas CN is 1- and O is 2-: 2x(-1)+2x(-2)=-6. From here we conclude that chromium must be +4, which is indeed a possible oxidation state for that element.

EDIT: Sam Yan did a better observation than me so please refer to his answer.

bad question.

Mish match word salad, why not wright it in an understandable form?

And more complete.

And maybe with a line of context.

What is “ a” that appears to have a -2 charge ?

Well, yes … but we would normally use Roman numerals to represent oxidation states. Magnesium has 2 valence electrons, which it typically loses to form [math]Mg^{2+}[/math] or [math]Mg(+II)[/math] ion. And we got FORMALLY [math]B(-III)[/math]. And as always the weighted sum of the oxidation numbers gives the charge on the parent ion/molecule, here NEUTRAL.

oxidation no of Na is +1 (since oxidation no of alkali metal is +1)

oxidation no of H is -1 (hydrogen in hydrides will have -1)

sum of oxidation numbers in a compound is 0, so let oxidation no of boron be x

(+1)+x+ 4(-1)=0

1+ x -4 =0

solving it we get

x=+3

oxidation no of B is +3

Thanks for reading

I hope it helps you…

Boron can have two different oxidation states (-3 and +3) because it has 5 electrons in its outer shell. When boron has 3 of its electrons in the outer shell filled, it becomes stable and has a -3 oxidation state. When it has 4 of its electrons in the outer shell filled, it becomes unstable and has a +3 oxidation state.