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Iron loses the 4s electrons first during ionization

Iron loses the 4s electrons first during ionization

Grade:12th pass

1 Answers

Arun
25750 Points
5 years ago
 
Dear Kajal
 

An example provided in Slater's original paper is for the iron atom which has nuclear charge 26 and electronic configuration 1s22s22p63s23p63d64s2. The screening constant, and subsequently the shielded (or effective) nuclear charge for each electron is deduced as:[1]

{\displaystyle {\begin{matrix}4s&:0.35\times 1&+&0.85\times 14&+&1.00\times 10&=&22.25&\Rightarrow &Z_{\mathrm {eff} }(4s)=26.00-22.25=3.75\\3d&:0.35\times 5&&&+&1.00\times 18&=&19.75&\Rightarrow &Z_{\mathrm {eff} }(3d)=26.00-19.75=6.25\\3s,3p&:0.35\times 7&+&0.85\times 8&+&1.00\times 2&=&11.25&\Rightarrow &Z_{\mathrm {eff} }(3s,3p)=26.00-11.25=14.75\\2s,2p&:0.35\times 7&+&0.85\times 2&&&=&4.15&\Rightarrow &Z_{\mathrm {eff} }(2s,2p)=26.00-4.15=21.85\\1s&:0.30\times 1&&&&&=&0.30&\Rightarrow &Z_{\mathrm {eff} }(1s)=26.00-0.30=25.70\end{matrix}}}

Note that the effective nuclear charge is calculated by subtracting the screening constant from the atomic number, 26.

hope it helps

Regards

Arun

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