Question

which of the following elements could be e?
a. s
b. f
c. p
d. i
e. o

Explanation:

Step1: Analyze the Lewis - structure

The central atom E has 3 bonding pairs (with 1 F and 2 Cl) and 2 lone - pairs.

Step2: Consider the valence electrons requirement

Elements that can form such a structure should have enough valence electrons to form 3 bonds and have 2 lone - pairs.

Step3: Evaluate each option

• Sulfur (S) has 6 valence electrons. It can form 2 bonds and have 2 lone - pairs in a normal situation, not suitable here.
• Fluorine (F) has 7 valence electrons and usually forms only 1 bond, not suitable.
• Phosphorus (P) has 5 valence electrons. It can form 3 bonds and have 2 lone - pairs. For example, in $PCl_3F_2$ (a trigonal bipyramidal structure with 3 bonding and 2 non - bonding pairs), the central P atom can have this Lewis - structure.
• Iodine (I) has 7 valence electrons and usually forms 1, 3, 5 or 7 bonds, not suitable for this structure with 3 bonds and 2 lone - pairs in a simple case.
• Oxygen (O) has 6 valence electrons and usually forms 2 bonds and has 2 lone - pairs, not suitable for forming 3 bonds.

Answer:

c. P