I think two elements that does not follow the octet rule are boron and beryllium. These elements always have one electron pair less than normal configuration. Boron is an important element for our body.
Hydrogen and helium follow a modified version of the octet rule: they only have room for two electrons in their outermost shell. Helium, as a noble gas, has a full outer shell and is therefore nonreactive.
The heavier noble gases (starting with argon) can be induced to create chemical compounds with the most reactive nonmetals, fluorine and oxygen; such compounds violate the octet rule, as noble gases (other than helium) start with a full outer shell of eight electrons.
There are numerous other examples as well; please reference the attached link.
Yes, there are many molecules, such as sulfur hexafluoride, which do not follow the octet rule.
Most molecules do not violate the octet rule.
all molecules except the group 2 and group 3 molecules of the elements
things like Sodium (Na+) and Hydrogen... elements that do not have 8 electrons to make an "octet"
BF3 and other similar Lewis Acids
A Lewis acid such as BF3
CF4 Doesn't violate the Octet Rule, the rest do.
Yes
The octet rule does not apply to transition metals.
The octet rule only applies to elements that are heavy enough to have reached the second shell of electrons. In the first shell, the octet rule does not apply because the first shell is completed with only two electrons, not eight. So no, the octet rule does not apply to beryllium hydride.
There is ClO2 and ClO2^-. For the chlorite anion (ClO2^-) the Cl will have 10 electrons and will violate the octet rule. For ClO2, all elements will have 8 electrons.
F2ccf2
The octet rule cannot be satisfied in molecules whose total number of valence electrons is an odd number.There are also molecules in which an atom has fewer, or more, than an octet of valence electrons.
CF4 Doesn't violate the Octet Rule, the rest do.
Yes
The octet rule does not apply to transition metals.
The octet rule only applies to elements that are heavy enough to have reached the second shell of electrons. In the first shell, the octet rule does not apply because the first shell is completed with only two electrons, not eight. So no, the octet rule does not apply to beryllium hydride.
There is ClO2 and ClO2^-. For the chlorite anion (ClO2^-) the Cl will have 10 electrons and will violate the octet rule. For ClO2, all elements will have 8 electrons.
The octet rule. I have the same worksheet.
In general, boron will form 3 covalent bonds, using each of its 3 valence shell electrons (sharing them). This will of course violate the octet rule, but obeys the sextet rule, and this is what makes boron stable. It (along with aluminum, eg.) do not obey the octet rule.
It does follow the octet rule!
Az important rule: any octet has to have eight parts, otherwise it is not an octet.
No, CH4 follows the octet rule.