The maximum 2nd I.E is of _________.

M n

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S c

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C r

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T i

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Solution

The correct option is D $C r$
The $2^{n d}$ IE increases along a period towards right. So it is expected that Mn has highest second IE. However, ${C r}^{+}$ has $3 d^{5}$ configuration, which gives it added stability hence its IE increases and thus highest second IE of Cr. Thus answer is option C.

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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]

C u^{+}, F e^{2 +}, M n^{2 +}, C r^{3 +}, S c^{3 +}, T i^{4 +}

V^{2 +} < C r^{2 +} < M n^{2} < F e^{2 +}

N i^{2 +} < C o^{2 +} < F e^{2 +} < M n^{2 +}

C o^{3 +} < F e^{3 +} < C r^{3 +} < S c^{3 +}

S c < T i < C r < M n

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 +}

Z n^{2 +} > S c^{2 +} = C u^{2 +} > T i^{2 +} = N i^{2 +} > V^{2 +} = C o^{2 +} > C r^{2 +} = F e^{2 +} > M n^{2 +}

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