Solve this:
Q.1. Which of the species follows octet rule:
(1)
(2)
(3)
(4)
Dear student,
N has 5 valence electrons, its configuration is 1s22s22p3, hence N-3 has 3 more electrons hence its outermost shell contains 8 electrons and its octet is complete.
In IBr5, Iodine has seven valence electrons and in addition has 5 electrons shared from Bromine hence it now has more than 8 electrons in valence shell, hence octet is expanded and octet rule is not followed.
In SF4, Sulphur has 6 valence electrons and in addition has 4 electrons shared from fluorine hence it has more than 8 electrons now in its valence shell, hence octet is expanded and octet rule is not followed.
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).
Hope it helps
Regards
N has 5 valence electrons, its configuration is 1s22s22p3, hence N-3 has 3 more electrons hence its outermost shell contains 8 electrons and its octet is complete.
In IBr5, Iodine has seven valence electrons and in addition has 5 electrons shared from Bromine hence it now has more than 8 electrons in valence shell, hence octet is expanded and octet rule is not followed.
In SF4, Sulphur has 6 valence electrons and in addition has 4 electrons shared from fluorine hence it has more than 8 electrons now in its valence shell, hence octet is expanded and octet rule is not followed.
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).
Hope it helps
Regards