The central atom in BrF5 has 5 bonding pairs of electrons and 1 non-bonding pair of electrons.

Explanation:
To determine the bonding and non-bonding pairs of electrons in BrF5, we need to consider its Lewis structure and the octet rule.

1. Lewis Structure:
The Lewis structure of BrF5 can be determined by following the steps:
a) Calculate the total number of valence electrons:
Bromine (Br) has 7 valence electrons.
Fluorine (F) has 7 valence electrons each, and there are 5 fluorine atoms in BrF5.
Total valence electrons = 7 + (7 x 5) = 42

b) Determine the central atom:
In BrF5, bromine (Br) is the central atom as it is less electronegative than fluorine (F).

c) Distribute the valence electrons:
Place one electron pair (a bonding pair) between bromine (Br) and each fluorine (F) atom.
This accounts for 10 electrons (5 bonding pairs).
Place the remaining 32 electrons as lone pairs around bromine (Br).

d) Check the octet rule:
After distributing all the valence electrons, check if each atom (except hydrogen) has an octet (or duet for hydrogen) of electrons.
In BrF5, each fluorine (F) atom has an octet of electrons, and bromine (Br) has 5 bonding pairs and 1 non-bonding pair of electrons.
The non-bonding pair of electrons is represented as a lone pair.

2. Bonding pairs and non-bonding pairs:
In BrF5, there are 5 bonding pairs of electrons because there are 5 bonds between bromine (Br) and fluorine (F) atoms.
Additionally, there is 1 non-bonding pair of electrons (a lone pair) on bromine (Br).
Therefore, the central atom in BrF5 has 5 bonding pairs of electrons and 1 non-bonding pair of electrons.

Hence, the correct answer is option 'C': 5 and 1.