Thank you Write the full and shorthand electronic configurations for the followi

Question

Thank you Write the full and shorthand electronic configurations for the following ions: a) O^2- b) Ca^2+ c) Fe^2+ Draw a properly labeled Born-Haber cycle for the formation of NaCl(s) from Na(s) and Cl_2(g) Use you cycle from Q1 above and the following data to calculate the lattice energy (U) of NaCl: Enthalpy of formation (delta H_f) for NaCl(s) = -411 kJ/mol Enthalpy of sublimation (delta H_subl) of Na = 107.3 kJ/mol The first ionization energy of Na (E_il) = 495.8 kJ/mol The bond dissociation energy (D) of Cl_2 = 243 kJ/mol The electron affinity of Cl (E_ea) = -348.6 kJ/mol.

Explanation / Answer

1. a) Atomic number of oxygen (O) is 8, if we add two electrons it becomes O2- and atomic number will be 8+2 =10

   Then full electronic configuration of O2- = 1s22s22p6

when we write shorthand electronic configuration of any atom, first write the chemical symbol of nearest noble gas in brackets then carry on with the electron configuration for the following orbital sets.

Then shorthand electronic configuration of O2- = [Ne]

because atomic number of nearest noble gas Ne is 10 and electronic configurationis 1s22s22p6 .

b) Atomic number of Ca is 20, if we remove two electrons it becomes Ca2+ and atomic number will be 20-2 =18
   Then full electronic configuration of Ca2+ = 1s22s22p6 3s23p6

For shorthand electronic configuration, replace the electronic configuration of Ca2+ with the nearest noble gas argon (Ar) where 1s2 2s2 2p6 3s2 3p6 is the configuration for Argon.

Thus shorthand electronic configuration for Ca2+ = [Ar]

c) Atomic number of Fe is 26, if we remove two electrons it becomes Fe2+ and atomic number will be 26-2 =24
   Then full electronic configuration of Fe = 1s22s22p6 3s23p64s23d6

   electronic configuration of Fe2+ = 1s22s22p6 3s23p63d6

We have remove two electrons from 4s orbital because 4s is the outer most orbital it will first filled up then fill 3d orbital.

For shorthand electronic configuration, replace the portion of electronic configuration of Fe2+ with the nearest noble gas argon (Ar) where 1s2 2s2 2p6 3s2 3p6 is the configuration for Argon.

Thus shorthand electronic configuration for Fe2+ = [Ar]3d6

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