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(I) An atom of an element contains 35 electrons and 45 neutrons. Deduce1. The number of protons 2. The electronic configuration for the element 3. All the four quantum numbers for the last electron(II) How many unpaired electrons are present in the ground state of Fe2+ (z = 26), Mn2+ (z = 25) and argon (z=18)? |
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Answer» (I) An element X contains 35 electrons and 45 neutrons 1. The number of protons must be equal to the number of electrons. So the number of protons = 35. 2. Number of electrons = 35. So the electronic configuration is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5. 3. The last electron i.e. 5th electron in 4p orbital has the following quantum numbers. n = 4, 1 = 1, m= +1, s = -\(\frac{1}{2}\) (II) Fe → Fe2+ + 3e- Fe (Z = 26) Fe3+ = number of electrons = 23 1s2 2s2 2p6 3s2 3p6 3d6 4s2 for Fe atom. 1s2 2s2 2p6 3s2 3p6 3d5 for Fe3+ ion. So, it contain 5 unpaired electrons. Mn (Z = 25). Electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d5 Mn → Mn2+ + 2e- Number of unpaired electrons in Mn2+ = 5 Ar (Z = 18). Electronic configuration is 1s2 2s2 2p6 3s2 3p6 . All orbitals are completely filled. So, no unpaired electrons in it. |
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