I wanna ask why the first ionization enthalpy of titanium is higher than vanadium?
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The effective nuclear charge of titanium is lower than vanadium Titanium 3d orbital is not half-filled nor full-filled. So why?
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Answer:
Data from: http://www.webelements.com/ IP α Zeff/r4s Zeff is the effective nuclear charge on the 4s e⁻ and r4s is the average radius of the 4s e⁻ Ti [Ar] 3d^2 4s^2 IP = 658 kJ mol^-1 Zeff = 4.82 r4s = 162.2 pm Zeff/r4s = 4.82/162.2 = 0.030 V [Ar] 3d^3 4s^2 IP = 650.9 kJ mol^-1 Zeff = 4.98 r4s = 154.4 pm Zeff/r4s = 4.98/154.4 = 0.032 So as you can see the Zeff on V4s^2 > Zeff on Ti 4s^2 and hence the V4s^2 is contracted as well and more difficult to ionize. The reason for this is that the extra e⁻ added on going from Ti to V enters a d AO. Zeff = Z - S where S is the screening of all the other e⁻s. Screening follows the order s>p>d>f . So the extra e⁻ does not screen as efficiently as a s or p e⁻. It is a variation on the observation that IP increases going across the PT (main group elements): S drops going from say Li to F.
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