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Which obeys the octet rule PF5 Cs2 BBr3 CO3 2-?
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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Cs2 and Co3 2 obey the octet rule. the others don't
Yes it is. It has 10 electrons (five valence electrons from phosphorus and five from five fluorine).
no it does not follow octet rule
Yes, it does.
No, Beryllium is an exception to the octet rule.
No it is not fully obeying the octet rule. Boron has only 6 electrons (3 own + 3 from each F atom), lacking two for the octet. Fluorine is 3x satisfied, each with 8 electr…ons (each has 7 own plus 1 from boron).
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 hydrogen BH3 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.
Yes. One of the "canonical" forms is O=O+-O- Just found a picture that will help see wikipedia article for ozone
No. It obeys the octet rule for the fluorine atoms but not for Se (which has 12 electrons in its valence shell.)
Yes both of carbon and oxygen obey the octet rule in CO 2
Yes, it obeys the octet rule. (Of coarse not an octet for the H atoms: they are saturated with a duplet (=2)). To show you HOW this is done, you will need the structural f…ormula: If it is a monosaccharide there are already 32 different isomeres. And even one of these and the most common one: alpha-D-gluco-pyranose (which is glucose) is too difficult to draw it in this simple 'html-text editor' layout. Just as example some explaining calculation about electron bonding pairs and free pairs. Valence electrons counting: 6 C = 6x4 = 24 e 12 H = 12x2 = 24 e 6 O = 6x6 = 36 e Total of 84 electrons are available for free or covalent (= bonding) electron pairs (total 84/2 = 42 pairs can be formed) Needed for octets and H-duplets 6 C = 6x8 = 48 e 6 O = 6x8 = 48 e 12 H = 12x2 = 24 e Total needed for octets/duplets = 120. These are 60 'needed to see' pairs C has no free pairs, all 4 are covalent pairs (= 1 octet): 6x4 = 24 covalent bondsO has 2 free and 2 covalent pairs (= 1 octet): 6x2 = 12 covalent bondsH has no free pairs and 1 covalent pair (= 1 duplet): 12x1 = 12 covalent bonds Total of 24 double counted bonding pairs (48 pairs 'seen' from one atom AND 'seen' from the other atom bound to it, called covalent*) electron pairs) One possible structure (which is not glucose) fitting the above calculation (1, 2, 3) is described here: 6 C atoms in a regular (bee hive) ring: 6 covalent pairs 6 H atoms directly to each C: 6 cov. pairs 6 O atom bonded at one side to each C atom: 6 cov. pairs 6 H atoms bonded to the other side of those 6 O atoms: 6 cov. pairs Total of 24 double counted bonding pairs*) between all atoms. This leaves 6x2 = 12 free electron pairs, two per O atom to fulfill their 'octet need' And now you can conclude: 24 double counted = 48 'seen' electron pairs which are covalent*) 12 free electron pairs (2 on each O atom) Total of 60 'structural drawn' pairs, Thus this, being the same as the 'needed to see' pairs (60, see above), means that ALL atoms are obeying the octet (+ duplet) rule. *) covalent means: 'seen by two' atoms which are bound together. (The name of this formula would be 1,2,3,4,5,6-cyclo-hexane-hexa-ol, it isn't even a monosaccharide)
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 corre…ctly!): .......O .......| O -- S -- O .......|| .......O Hope this helped!