In \[IC{l_3}\] the oxidation number of iodine and chlorine are, respectively
A.0 and 0
B.+3 and -1
C.-1 and +3
D.-3 and +1

Hint: Oxidation number or oxidation state of an element can be defined as the degree of oxidation of an element in a given compound. In simpler terms, it can be understood as the number of the electrons gained or lost by an atom while forming a compound. This results in forming a net charge over this element, which is referred to as the oxidation state.

Complete step by step answer:
Depending on the number of electrons present in the valence shell of an atom, an element may exhibit a single or in some cases multiple oxidation states, depending on the atoms they are combining with.
Now, forming the equation for calculating the oxidation states of iodine and chlorine in \[IC{l_3}\] :
Net charge on \[IC{l_3}\] = (O.S. of iodine) (no. of atoms of iodine) + (O.S. of chlorine) (no. of atoms of Cl)
 \[0 = \left( x \right)\left( 1 \right) + \left( { - 1} \right)\left( 3 \right)\]
 \[0 = x-3\]
 \[x = + 3\]
Hence, the oxidation states of iodine and chlorine in \[IC{l_3}\] are +3 and -1 respectively.

Hence, Option B is the correct option.

Note:
The oxidation number of all halogens will always be equal to (-1). This is because of the number of valence electrons present in them. They have only 7 valence electrons, which means that they are one electron short of completing their octets. Hence, all halogens, generally just prefer to accept one electron and hence their oxidation state is (-1)