Question

Which of the following compound does not obey the octet rule?

• A. $S{F}_6$
• B. $B{H}_3$
• C. $B{F}_3$
• D. $PC{l}_5$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $S{F}_6$
B $B{H}_3$
C $B{F}_3$
D $PC{l}_5$

The octet rule states that atoms tend to form compounds in ways that give them eight valence electrons and thus the electron configuration of a noble gas. Compounds following the octet rule must have 8 electrons in the valence shell of their central atom.

A. In $S{F}_6$ sulphur has 6 valance electrons and 6 fluorine atoms also donate 6 electrons. So total electrons in the valence shell of sulphur are 12.

B. In $B{H}_3$ Boron has 3 valence electrons and 3 electrons are shared by 3 hydrogen atoms. So total valence electrons of boron are 6.

C. the above calculation is also valid for $B{F}_3$.

D. For $PC{l}_5$ phosphorus has 5 valence electrons and 5 electrons are also donated by the 5 chlorine atoms. So total valence shell electrons $=(5+5)=10.$

So correct options are A,B,C,D