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$\ce{Ti}$ and $\ce{Cr^2+}$ are isoelectronic, yet their electron configurations are different, with the former's being $\ce{[Ar] 4s^2 3d^2}$ and the latter's being $\ce{[Ar] 3d^4}$.
How can that be explained?

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When you fill electrons in different orbitals, you keep in mind the $n+l$ rule, but when you remove electrons you have to consider the principal quantum number. Between 3d and 4s, 4s has a lower value of $n+l$, so electrons will be first filled in it, that's why titanium metal has the atomic configuration $\ce{[Ar] 4s^2 3d^2}$.
In case of $\ce{Cr^2+}$ you have to first write configuration of $\ce{Cr}$, which is $\ce{[Ar] 4s^2 3d^4}$ and then remove two electrons from the orbital having highest value of $n$ (principal quantum number) which is 4s. after removing 2 electrons from 4s you will get configuration of $\ce{Cr^2+}$ which is $\ce{[Ar] 3d^4}$.

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  • $\begingroup$ for n+l rule(google.co.in/…) $\endgroup$ Sep 10 '16 at 13:25
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    $\begingroup$ This would produce the wrong result for $\ce{V+}$ which should be $\ce{[Ar] 3d^4}$ but would be $\ce{[Ar] 4s^1 3d^3}$ instead if we use your method. $\endgroup$
    – DHMO
    Sep 10 '16 at 13:38
  • $\begingroup$ The reality is more complicated than a simple rule as you present, as is stated in the first sentence of my answer. $\endgroup$
    – DHMO
    Sep 10 '16 at 13:39
  • $\begingroup$ If not completely and utterly wrong, then at least that is halfhearted and incomplete. $\endgroup$ Nov 27 '17 at 12:41
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The general configuration $\ce{Sc}$ to $\ce{Zn}$ is $\ce{[Ar] 3d^n 4s^2}$, except for $\ce{Cr^2+}$, $\ce{[Ar] 3d^5 4s^1}$ and $\ce{Cu}$, $\ce{[Ar] 3d^{10} 4s^1}$ and which are attributed to the stability of half-filled and filled $\ce{d}$ shells respectively. In the half filled $\ce{3d}$ shell there is no spin pairing energy needed (one electron per orbital) compared to that energy needed to put two electrons into the $\ce{4s}$ orbital.

Titanium has the configuration $\ce{[Ar] 3d^2 4s^2}$ and chromium $\ce{[Ar] 3d^5 4s^1}$, thus loosing two electrons to make $\ce{Cr^2+}$ produces the configuration $\ce{[Ar] 3d^4 4s^0}$ (although $\ce{4s}$ and $\ce{3d}$ orbitals are close in first row transition elements) then the configuration is as one would expect.

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The reason is very complicated, but in short:
The energy gap between $\ce{3d}$ and $\ce{4s}$ is wider in $\ce{Cr^2+}$ than in $\ce{Ti}$ because $\ce{Cr^2+}$ has more protons.

When you promote an electron from $\ce{3d}$ to $\ce{4s}$, the electron gained the energy thanks to orbitals, but you also lost the energy thanks to repulsion (of electrons in the same orbital). Remember that electrons like to be in states of less energy, not more.

The energy gap between $\ce{3d}$ and $\ce{4s}$ is less in $\ce{Ti}$, so the electron, by promoting, lost more energy thanks to repulsion than it gained thanks to the orbital, so its configuration is $\ce{[Ar] 4s^2 3d^2}$.

In $\ce{Cr^2+}$, the enrgy gap widened, so the electron would gain more energy than its lost by promoting, making its configuration $\ce{[Ar] 3d^4}$.

See also chemguide.co.uk.

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