Consider the addition of an electron to the following atoms from the third period. Rank the atoms in order from the most negative to the least negative electron affinity values based on their electron configurations. Atom or ion Electron configuration
Br 1s22s22p63s23p64s23d104p5
Ge 1s22s22p63s23p64s23d104p2
Kr 1s22s22p63s23p64s23d104p6

The ranks is

Ge: 3d10 4s2 4p2 (6 electrons in the outer shell)

Br: 3d10 4s2 4p5 (7 electrons in the outer shell)

Kr: 3d10 4s2 4p6 (8 electrons in the outer shell)

Explanation:

Electronic configuration reffers to the distribution of electrons of an atom or molecule in atomic or molecular orbitals. It gives us the understanding of the shape and energy of its electrons. The electronic configuration explain the The electron affinity or propensity to attract electrons

It Should be noted that the most stable configuration in an electronic configuration is attributed to when the last shell is full, i.e. when the last shell has 8 electrons.

When an atom is closer to reach the 8 electrons in the outer shell, then it's electron affinity big.

Considering the given three configuration of the elements above, we can see that "Br"needs requires only 1 electron to have 8 electrons in the outer shell, therefore, it is considered to have the biggest electron affinity among them which is reffers to as the LEAST NEGATIVE.

Ge: with the electronic configuration 3d10 4s2 4p2 has 6 electrons in the outer shell which means it still requires 2 electrons to complete 8 electrons in its outer shell, so it can be deducted that it posses an atom that is more negative than Br.

Kr: with the electronic configuration 3d10 4s2 4p6 which is a noble gas has 8 electrons in the outer shell cannot add more electrons to its outer shell because the 8 electrons is complete posses the least electron affinity among the three elements and it is the MOST NEGATIVE

the answer is b & c i believe : )