What is the oxidation number of phosphorus in the compound #"PCl"_5"#?

Oxidation numbers are all about electronegativity difference between covalently-bonded atoms.

You can assign oxidation numbers to the atoms that are a part of a covalent compound by assuming that the more electronegative atom will take both bonding electrons.

In the case of phosphorus pentachloride, #"PCl"_5#, you have one phosphorus atom that forms covalent bonds with five chlorine atoms.

Since chlorine is more electronegative than phosphorus, you can assign its oxidation number by assuming that it takes all the bonding electrons used in the molecule.

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Since each single bond contains #2# bonding electrons, one from chlorine and one from phosphorus, it follows that the more electronegative chlorine will take the electron it contributed to the bond, shown in the above image in blue, and the electron phosphorus contributed to the bond, shown in red.

So, for each bond chlorine has with phosphorus, it gains one electron; at the same time, phosphorus loses one electron. This means that the oxidation state of each chlorine atom will be #color(blue)(-1)#.

The phosphorus atom loses a total of #5# electrons, one to each chlorine atom, so its oxidation state will be #color(blue)(+5)#.

#stackrel(color(blue)(+5))("P")stackrel(color(blue)(-1))("Cl")_5#

The sum of oxidation numbers in a compound must be zero. The oxidation number of chlorine in #"PCl"_5"# is #"-1"#, so there is an overall oxidation number of #"Cl"# is #"-5"#. In order for the sum of the oxidation numbers to equal zero, the oxidation number of #"P"# is #"+5"#.

#(1xx+5)+(5xx-1)=0#

#stackrel(+5)("P")stackrel(-1)("Cl"_5")#