Question

Which of the species follow octet rule n-3 IBr5 SF4 Pb+4

Answer 1

Answer:

In Pb+4 the outermost shell now is 5 and its configuration is [Xe]5s25p65d10 which has more than 8 electrons hence octet rule is not followed. Hence the correct answer is (2).

Answer 2

Answer:

The required answer is an option (2) $N^{-3}$

Explanation:

Since N has 5 valence electrons and the configuration is $1s^{2} 2s^{2} 2p^{3}$, $N^{-3}$requires 3 extra electrons, bringing the total number of electrons in its outermost shell to 8, completing the octet.

Iodine has seven valence electrons in IBr5 and an additional five electrons that it shares with bromine, giving it more than eight electrons in the valence shell and breaking the rule of the octet.

The valence shell of sulphur  $SF_{4}$ has more than 8 electrons since it has 6 valence electrons and 4 shared electrons with fluorine. As a result, the octet rule is not obeyed and the octet is enlarged.

The octet rule is not obeyed in $Pb^{+4}$ because the outermost shell is now 5 and its configuration is [Xe]$5s^{2} 5p^{6} 5d^{10}$, which has more than 8 electrons.

Therefore the correct option is (2) $N^{-3}$

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