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JEE Main 2023 · Chemistry · Atomic Structure · hard

The number of electrons in Fe^3+ (Z=26) in its ground state that have magnetic quantum number ml=0 is

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The number of electrons in Fe3+\mathrm{Fe^{3+}} (Z=26Z=26) in its ground state that have magnetic quantum number ml=0m_l=0 is
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Answer: 11

Fe3+\mathrm{Fe^{3+}} is 1s22s22p63s23p63d51s^22s^22p^63s^23p^63d^5. Electrons with ml=0m_l=0: every s orbital (1s,2s,3s1s,2s,3s) has ml=0m_l=0, giving 2+2+2=62+2+2=6. In each p subshell only the pzp_z orbital has ml=0m_l=0, giving 2 (2p)+2 (3p)=42\ (2p)+2\ (3p)=4. In 3d53d^5, each of the five d orbitals has one electron (Hund's rule), so exactly one electron has ml=0m_l=0. Total =6+4+1=11=6+4+1=11.
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