It what happens when elements don't obey the octect rule, i.e. elements that have more than eight electrons in their outermost shell. It what happens when elements don't obey the octect rule, i.e. elements that have more than eight electrons in their outermost shell.
CF4 and XeF4 do not violate the octet rule. In CF4, carbon forms four covalent bonds with fluorine, fulfilling the octet rule. In XeF4, xenon forms four covalent bonds with fluorine and has two lone pairs, also satisfying the octet rule.
Yes, both Cu+ and Cu2+ violate the octet rule. Copper (Cu) is an exception to the octet rule due to its electron configuration, which allows it to have a partially filled d orbital. This leads to Cu forming compounds where it does not achieve a full octet of electrons.
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.
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.
Yes, the element sulfur in SO3 violates the octet rule because it has more than 8 electrons in its valence shell.
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 and XeF4 do not violate the octet rule. In CF4, carbon forms four covalent bonds with fluorine, fulfilling the octet rule. In XeF4, xenon forms four covalent bonds with fluorine and has two lone pairs, also satisfying the octet rule.
Yes, both Cu+ and Cu2+ violate the octet rule. Copper (Cu) is an exception to the octet rule due to its electron configuration, which allows it to have a partially filled d orbital. This leads to Cu forming compounds where it does not achieve a full octet of electrons.
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.
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.
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The octet rule is the tendency of many chemical elements to have eight electrons in the valence shell.
Yes, the element sulfur in SO3 violates the octet rule because it has more than 8 electrons in its valence shell.
The elements can be described by dot structure. Metals combine with other elements to make its octet complete.
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.
octet rule
Some elements that are known to violate the octet rule are: Hydrogen, Helium and Lithium (two electrons) Aluminum and Boron (less than octet but will form an octet if possible), Period 3 elements with p orbitals (more than an octet using empty d orbitals), noble gas compounds (more than an octet), and elements like nitrogen with an odd number of electrons (form free radicals when octets are not possible).