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!
Yes, SiH4 obeys the octet rule
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The bolded statements are true:1. The octet rule can be violated. 2. Atoms are most stable when their atomic number is divisible by 8.3. All free atoms contribute eight valence electrons to form molecules.4. Only the oxygen atom can have an expandable octet.5. In order to obey the octet rule, some atoms have to share more than one pair of electrons.
The element tellurium would be expected to form 2covalent bonds in order to obey the octet rule.Te is a nonmetal in group 6A, and therefore has 6 valence electrons. In order to obey the octet rule, it needs to gain 2 electrons. It can do this by forming 2single covalent bonds.
BH3 does not obey octet rule.it has a total of six electrons only.boron has three electrons in the valence shell and it accepts one electron from each hydrogenBH3 is a planar molecule and is found only in the gaseous state.BH3 dimerises to form B2H6 with 4 terminal hydrogens attached by normal covalent bonds and 2 hydrogen bridges , 3 centre 2 electron bonds. Once again in the dimer it does not obey the octet rule.
more than sure that whole group is covalent bonding such as Iodine is written I2 or Bromine is Br2. Fluorine is also in this, example F2, the 2 outer electrons as each is 19, 2,8,8,1 is shared between the 2 atoms.
I know for sure BBr# & PF5 do not obey the octet rule, but i can't remember the rule of isotopes so I can't say for sure whether or not CO3 -2 obeys it or not.
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because Be has only 4 electrons and has 2 valence electron
No. Hydrogen needs only 2 electrons for a full outer shell rather than the usual 8.
The bolded statements are true:1. The octet rule can be violated. 2. Atoms are most stable when their atomic number is divisible by 8.3. All free atoms contribute eight valence electrons to form molecules.4. Only the oxygen atom can have an expandable octet.5. In order to obey the octet rule, some atoms have to share more than one pair of electrons.
The element tellurium would be expected to form 2covalent bonds in order to obey the octet rule.Te is a nonmetal in group 6A, and therefore has 6 valence electrons. In order to obey the octet rule, it needs to gain 2 electrons. It can do this by forming 2single covalent bonds.
BH3 does not obey octet rule.it has a total of six electrons only.boron has three electrons in the valence shell and it accepts one electron from each hydrogenBH3 is a planar molecule and is found only in the gaseous state.BH3 dimerises to form B2H6 with 4 terminal hydrogens attached by normal covalent bonds and 2 hydrogen bridges , 3 centre 2 electron bonds. Once again in the dimer it does not obey the octet rule.
more than sure that whole group is covalent bonding such as Iodine is written I2 or Bromine is Br2. Fluorine is also in this, example F2, the 2 outer electrons as each is 19, 2,8,8,1 is shared between the 2 atoms.
ICl2- is 10, if you are in Ap Chem on WebAssign then this is probably what you are looking for....-I already got it right on WebAssign:)
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).
Beryllium (Be) is an element in group 2 of the periodic table. It has 2 energy levels and 2 valence electrons in the outermost energy level. It reacts by losing 2 electrons to attain a noble gas structure.
H2S does follow the octet rule. When you draw the Lewis Structure for H2S, it looks like this: If you count up the lone pairs and sigma bonds (each worth 2), there are 8, thus, H2S follows the octet rule.