Given the unbalanced equation showing the reactants and product of a reaction occurring at 298 K and 100. kPa: P4(s) + Cl2(g) PCl3(l) + energy
Balance the equation below for the reaction, using the smallest whole number coefficients.

P4(s) + Cl2(g) PCl3(l) + energy

Balanced equation

P4(s) +6Cl2(g) ⇒4PCl3(l) + energy

Given

Reaction(unbalnced)

P4(s) +Cl2(g) ⇒PCl3(l) + energy

Required

Balanced equation

Solution

aP4(s) +bCl2(g) ⇒PCl3(l) + energy

P, left=4a, right=1⇒4a=1⇒a=1/4

Cl, left=2b, right=3⇒2b=3⇒b=3/2

The equation becomes:

1/4P4(s) +3/2Cl2(g) ⇒PCl3(l) + energy x 4

P4(s) +6Cl2(g) ⇒4PCl3(l) + energy

the answer to your question is c

substituting a hydrogen atom with a halogen in a hydrocarbon:

- the single bond remains single. a is wrong.

- the bond with hydrogen is as strong as with a halogen. c is wrong

- a hydrocarbon is saturated if there is no double bond. we do not know that so d is wrong.

- carbon will always have four valence electrons. e is wrong.

by elimination, the correct answer is b. the boiling point of the new compound increases.