The correct order of decreasing second ionisation enthalpy of Ti(22), V(23), Cr(24) and Mn(25) is [CBSE AIPMT 2008]

Cr > Mn > V > Ti

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V > Mn > Cr > Ti

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Mn > Cr > Ti > V

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Ti > V > Cr > Mn

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Solution

The correct option is A
Cr > Mn > V > Ti

The amount of energy required to remove an electron from unipositive ion is referred as second ionisation potential.
In Ti, V, Cr and Mn, generally second ionisation energy increases with increase in atomic number but second ionisation potential of Cr is greater than that of Mn due to the presence of exactly half-filled d-subshell in Cr.
Thus, the order of second ionisation enthalpy is Cr > Mn > V > Ti

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Similar questions

The correct order of decreasing second ionisation enthalpy of Ti(22), V(23), Cr(24) and Mn(25) is [CBSE AIPMT 2008]

(T i = 22, V = 23, C r = 24, M n = 25)

Which one of the following ions is the most stable in aqueous solution?
(Atomic number. Ti = 22, V = 23, Cr = 24, Mn = 25)

M^{2 +} / M

M^{3 +} / M^{2 +}

E^{\circ}

3 d

T i^{4 +}, V^{2 +}, M n^{3 +}, C r^{3 +}

T i = 22, V = 23, M n = 25, C r = 24)

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