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Chapter 18: Q18.10-95 E (page 1047)

Oxygen forms double bonds in\({{\rm{O}}_2}\), but sulfur forms single bonds in \({{\rm{S}}_8}\). Why?

Short Answer

Expert verified

Due to its size, S-S \(\pi \) bonds are not possible.

Step by step solution

01

To explain the oxygen forms double bonds in \({{\rm{O}}_2}\), but sulfur forms single bonds in \({{\rm{S}}_8}\). why?

Sulfur atoms are larger than the oxygen atoms and have an empty d orbital. Because of this reasons, sulfur atoms are not able to form S-S double bonds or\(\pi \) bonds which oxygen atoms.

The double bond formed by Sulfur is due to the size of it as an atom which is bigger than that of an oxygen atom.

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Most popular questions from this chapter

Silicon reacts with sulfur at elevated temperatures. If \({\bf{0}}.{\bf{0923}}{\rm{ }}{\bf{g}}\) of silicon reacts with sulfur to give \({\bf{0}}.{\bf{3030}}{\rm{ }}{\bf{g}}\) of silicon sulfide, determine the empirical formula of silicon sulfide.

Write equations showing the stepwise ionization of phosphorous acid.

What is the hybridization of iodine in\({\rm{I}}{{\rm{F}}_3}\)and \({\rm{I}}{{\rm{F}}_5}\)?

Consider the production of \({\bf{100}}\) kg of sodium metal using a current of \({\bf{50}},{\bf{000}}{\rm{ }}{\bf{A}},\) assuming a \({\bf{100}}\% \) yield.

(a) How long will it take to produce the 100 kg of sodium metal?

(b) What volume of chlorine at \({\bf{25}}{\rm{ }}^\circ {\bf{C}}\)and \({\bf{1}}.{\bf{00}}\) atm forms?

Assign an oxidation state to phosphorus in each of the following:

(a) \({\rm{Na}}{{\rm{H}}_2}{\rm{P}}{{\rm{O}}_3}\)

(b) \({\rm{P}}{{\rm{F}}_5}\)

(c) \({{\rm{P}}_4}{{\rm{O}}_6}\)

(d) \({{\rm{K}}_3}{\rm{P}}{{\rm{O}}_4}\)

(e) \({\rm{N}}{{\rm{a}}_3}{\rm{P}}\)

(f) \({\rm{N}}{{\rm{a}}_4}{{\rm{P}}_2}{{\rm{O}}_7}\)
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