Question

Three molecules X, Y and Z have a contracted octet, an octet and an expanded octet. Identity them.

• A. $C{H}_4,S{F}_6,CC{l}_4$
• B. $CC{l}_4,SC{l}_4,B{F}_3$
• C. $BeC{l}_2,CC{l}_4,S{F}_6$
• D. $Xe{F}_6,C{H}_4,I{F}_7$

Answer 1

Answer: (C)

$BeC{l}_2,CC{l}_4,S{F}_6$

According to the octet rule "two similar or dissimilar non-metallic atoms form a covalent bond to attain an octet or inert gas configuration."

In $BeC{l}_2$ Be atom (central atom) has only four electrons in its outermost shell - two of its own and the other two from the two covalently bonded $Cl$ atoms. So, the octet remains incomplete hence is a contracted octet.

For carbon tetrachloride, $\left(CC{l}_4\right)$, carbon is placed at the centre. The carbon atom then forms a single bond with each of the four chlorine atoms. Hence, the octet is complete.

In the $S{F}_6$ molecule the number of electrons in the outermost shell of the sulfur atom, the central atom is twelve - six electrons of its own and six electrons gained from six $F$ atoms. Hence, it has an expanded octet.

Hence, the correct option is $\text{C}$