1. Which of the following ground-state electron configurations corresponds to an

Question

1. Which of the following ground-state electron configurations corresponds to an atom that has the most negative value of the electron affinity?

A.1s22s22p63s23p63d54s2

B.1s22s22p63s23p5

C.1s22s22p6

D.1s22s22p63s1

E.1s22s22p63s23p2

2. The change in energy for which of the following processes corresponds to the electron affinity of Fluorine?

A. F2(g)=2F(g)

B. F+(g) + F-(g)= F2(g)

C. F(g) +e- =F-(g)

D. F-(g) = F(g) +e-

E. F(g) = F+(g)+e-

3. What is the total number of electrons in p orbitals in a ground-state Potassium atom?

A.30

B.12

C.18

D.24

E.6

Explanation / Answer

1. The electron affinity (EA) is the energy required/released to add one electron to an atom. Halogens have the most negative EA values because addition of an e- leads to noble gas configuration,thus very favorable. From the given options, A is Cr, B is Cl, C is ne, D is Na and E is Si.

So ,the answer is B. Cl as it a halogen.

2. The electron affinity (EA) is the energy required/released to add one electron to an atom.
so, an e- is added to to give F-. So the correct option is C. F(g) +e- =F-(g)

3. K has 19 electrons. So its electronic configuration is 1s2 2s2 2p6 3s2 3p6 4s1.

So, no. of p electrons = 6 + 6 = 12 electrons.

So, B.12

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