Question

Which of these compounds do not follow the octet rule?

`NO`
`BF_3`
`(ICl_2)^-`
`OPBr_3`
`XeF_4`

Answer 1

NO , `ICl_2^-1`, `` XeF_4#

Explanation:

`BF_3` follows the octet rule because the Boron has three electrons so that it can give one electron to each of the three fluorine atoms. this gives each of the florine ions eight electrons.
The Boron ends up with no electrons in the outer shell, The complete inner shell is the same as group VIIIA helium.

NO does not follow the octet rule Nitrogen starts with five electrons gains two from Oxygen but ends with only 7

# ICl_2^-1 There are 22 electrons shared between 3 atoms. It would take 24 electrons for all three atoms to achieve the octet rule.

`OPBr_3` follows the octet rule The phosphorus shares its three unpaired electrons with Bromine providing each of the Bromine ions with eight outer elections. The phosphorus then hybridizes its two S electrons with d orbitals to enable the phosphorus atom to share two electrons with Oxygen. This gives Oxygen 8 electrons for an octet. Phosphorus has now lost all five of its outer electrons, leaving it with 8 electrons in its inner shell the same as Neon.

`Xe F_4` does not follow the octet rule. Xeon start with an octet sharing electrons with Flourine only disrupts the octet it started with
.

Answer 2

All of them.

Explanation:

We must examine the Lewis structures of these compounds to see if they follow the octet rule.

`bb"NO"`

The `"N"` atom has an only seven valence electrons, so it does not obey the octet rule.

`bb"BF"_3`

`"B"` has only six valence electrons, so it violates the octet rule.

It is pssible to draw a structure in which every atom has an octet:

However, this puts a formal positive charge on an `"F"` atom. Since `"F"` is the most electronegative atom in the Periodic Table, the second contributor is relatively unimportant.

`bb"ICl"_2^"-"`

The `"I"` atom has 10 valence electrons, so it violates the octet rule.

`bb"OPBr"_3`

The `"P"` atom has 10 valence electrons, so it violates the octet rule.

`bb"XeF"_4`

The `"Xe"` atom has 12 valence electrons, so it violates the octet rule.