The number of 3 d electrons in manganese, given that it has two 4 s electrons and its atomic number is 25 , also explain the answer.

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Chapter 28, Problem 52P

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The number of 3d electrons in manganese, given that it has two 4s electrons and its atomic number is 25, also explain the answer.

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1. Question 1: a. Draw a schematic of the hydrogen atom with the single proton in the nucleus, and the n=1, n=2, n=3, and n=4 energy level options for the electron. Put the electron in the lowest energy configuration. b. Justify why this electron configuration represents the lowest energy state for the electron. (hint: compare the energy of the electron in both cases and be mindful of negative signs)

3. Calculate the probability to find an electron in the following regions if the electron resides in a 2p: orbital of a hydrogen atom. The "z" in the equation below is a Cartesian coordinate, not the atomic number "2" which equals 1 for the hydrogen atom. Use whatever information below is more convenient. 32 n až ao 13 I and Y(0,4) =: 24 až ao Vzp, = R(r)Y(0, 4) where R(r) = (a) Calculate the probability to find an electron in a 1.0 pm' cube centered at the Cartesian point (200, 0, 0). (b) Calculate the probability to find an electron in a 1.0 pm' cube centered at the Cartesian point (0, 2ao, 0). (c) Calculate the probability to find an electron in a 1.0 pm' cube centered at the Cartesian point (0, 0, 2a0). (d) Draw a 2p, orbital in 3D space and include the points of (a), (b), and (c). Explain how your diagram qualitatively supports the probabilities you calculated. (e) Calculate the probability to find an electron in a 1.0 pm-thick shell at a radius 2ao from the nucleus.

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