Yes, CS2 Obeys the Octet rule. C double bonds with both Sulfur atoms and then each S has two pair of free electrons.
A carbon ion that obeys the octet rule typically has a charge of 4-. This is because carbon has four valence electrons and typically gains four electrons to achieve a full valence shell of eight electrons, conforming to the octet rule.
BCl3 and BEH2 obey the octet rule because Boron and Beryllium are exceptions to the octet rule and can have stable electron configurations with less than 8 electrons. Cl3CF, NO, and SbF5 do not obey the octet rule because they have incomplete or expanded valence shells.
Yes, PF5 is an exception to the Lewis octet rule. Phosphorus has 10 electrons around it in PF5, exceeding the octet rule. This is due to the availability of d-orbitals in the valence shell of phosphorus for accommodating extra electrons.
Yes, SO4 2- can be drawn without violating the octet rule. It is also a resonance structure. Here's an illustration below (ignore the dots, it was the only way it posted correctly!): .......O .......| O -- S -- O ....... .......O Hope this helped!
No, SF6 doesn't. If you draw out the Lewis structure, all 6 fluorine atoms have to connect to the sulfur.
PF5 obeys the octet rule as it has 5 bonding pairs of electrons around the central phosphorus atom, satisfying the octet. Cs2 does not follow the octet rule as Cs is in Group 1 and can only form ionic bonds. BBr3 is an exception to the octet rule as boron has only 6 electrons around it due to the empty d orbital. CO3 2- also obeys the octet rule as each oxygen atom has a complete octet.
A carbon ion that obeys the octet rule typically has a charge of 4-. This is because carbon has four valence electrons and typically gains four electrons to achieve a full valence shell of eight electrons, conforming to the octet rule.
Cs2, NO3, and PO43 do not obey the octet rule because they have an odd number of valence electrons. CI4 and SCI4 do not obey the octet rule because the central atom (Cl or S) exceeds the octet by having more than eight valence electrons.
No, Transition metals do not obey the octet rule in simple compounds , Am is an 'f' block transition metal.
BCl3 and BEH2 obey the octet rule because Boron and Beryllium are exceptions to the octet rule and can have stable electron configurations with less than 8 electrons. Cl3CF, NO, and SbF5 do not obey the octet rule because they have incomplete or expanded valence shells.
Xenon obeys octet rule and has a stable electronic configuration. So, xenon does not form any anion.
Yes, PF5 is an exception to the Lewis octet rule. Phosphorus has 10 electrons around it in PF5, exceeding the octet rule. This is due to the availability of d-orbitals in the valence shell of phosphorus for accommodating extra electrons.
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.
Yes, SO4 2- can be drawn without violating the octet rule. It is also a resonance structure. Here's an illustration below (ignore the dots, it was the only way it posted correctly!): .......O .......| O -- S -- O ....... .......O Hope this helped!
No, SF6 doesn't. If you draw out the Lewis structure, all 6 fluorine atoms have to connect to the sulfur.
Az important rule: any octet has to have eight parts, otherwise it is not an octet.
Neon does not typically form ionic compounds because it already has a full valence shell with 8 electrons, satisfying the octet rule. Its electron configuration (1s^2 2s^2 2p^6) makes it very stable and unreactive with other elements.