The second ionization energies in kJ/mol $^{- 1}$ of some elements of period of 4 are:
Ca Sc Ti
1145 1235 1310
V Cr Mn
1365 1592 1509 kJ/mol $^{- 1}$
Account for the trend in values.

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Solution

$(20) C a ⟶ 3 p^{6} 4 s^{2}$
$C a^{+} ⟶ 3 p^{6} 4 s^{1}$
$C a^{2 +} ⟶ 3 p^{6} 4 s^{0}$

$(21) S c ⟶ 4 s^{2} 3 d^{1}$
$S c^{+} ⟶ 4 s^{1} 3 d^{1}$
$S c^{2 +} ⟶ 4 s^{0} 3 d^{1}$

$(22) T i ⟶ 4 s^{2} 3 d^{2}$
$T i^{+} ⟶ 4 s^{1} 3 d^{2}$
$T i^{2 +} ⟶ 4 s^{0} 3 d^{2}$

$(23) V ⟶ 4 s^{2} 3 d^{3}$
$V ⟶ 4 s^{1} 3 d^{3}$
$V ⟶ 4 s^{0} 3 d^{3}$

$(24) C r ⟶ 4 s^{2} 3 d^{5}$
$C r ⟶ 4 s^{0} 3 d^{5}$
$C r ⟶ 4 s^{0} 3 d^{4}$

$(25) M n ⟶ 4 s^{2} 3 d^{5}$
$M n ⟶ 4 s^{1} 3 d^{5}$
$M n ⟶ 4 s^{0} 3 d^{5}$
$C r$ is having highest second ionisation energy value because it is present in its' extra stable half filled state which requires high amount of energy to remove an $e^{-}$ from stable half-filled state.

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k J m o l^{- 1}

k J m o l^{- 1}

I E_{1}

250 k J m o l^{- 1}

I E_{2}

820 k J m o l^{- 1}

I E_{3}

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I E_{1} = 250 {kJ mol}^{- 1}

I E_{2} = 820 {kJ mol}^{- 1}

I E_{3} = 1100 {kJ mol}^{- 1}

I E_{4} = 1400 {kJ mol}^{- 1}

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(K J {m o l}^{- 1})

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I E_{1} = 250 {kJ mol}^{- 1}

I E_{2} = 820 {kJ mol}^{- 1}

I E_{3} = 1100 {kJ mol}^{- 1}

I E_{4} = 1400 {kJ mol}^{- 1}

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