Which of the following shows the correct quantum numbers (n; ℓ; mℓ) and electron configuration for the highest energy electron(s) in the given atoms? 1. rb: [kr]5s 1 ; n = 5; ℓ = 0; mℓ = 0, 1 2. n: [he]2s 2 2p 3 ; n = 2; ℓ = 2; mℓ = −2, −1, 0, 1, 2 3. fe: [ar]4s 2 3d 6 ; n = 4; ℓ = 0; mℓ = 0 4. ca: [ar]4s 2 ; n = 4; ℓ = 1; mℓ = −1, 0+1 5. co: [ar]4s 2 3d7 ; n = 3; ℓ = 2; mℓ = −2, −1, 0, 1, 2

Co: [Ar]4s 2 3d7 ; n = 3; ℓ = 2; mℓ = −2, −1, 0, 1, 2

Explanation:

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Quantum numbers for the highest energy level of cobalt are correct, because n matches with the considered sub-orbital 3, l matches with the d region that has a value of 2 based on the rule: if region=s, l=0; if region=p, l=1;if region=d, l=2 and if region=f, l=3 and the ml goes from -2 to 2 including zero by taking into account the l quantum number. The rest of the options are wrong since the quantum numbers don't match with the corresponding orbitals.

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it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

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the answer is a and have a great day