The aqueous solution containing which one of the following ions will be colourless?
[Atomic number of Sc=21, Fe=26, Ti=22, Mn=25]

S c^{3 +}

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F e^{2 +}

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T i^{3 +}

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M n^{2 +}

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Solution

The correct option is A $S c^{3 +}$
(a) $_{21} S c^{3 +} = 1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6}$
It is colourless due to absence of unpaired electrons in d-subshell.

(b) $_{26} F e^{2 +} = 1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6} 3 d^{6}$
It has colour due to presence of four unpaired electrons in d-subshell.

(c) $_{22} T i^{3 +} = 1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6} 3 d^{1}$
It has colour due to presence of one unpaired electrons in d-subshell.

(d) $_{25} M n^{2 +} = 1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6} 3 d^{5}$
It has colour due to presence of 5 unpaired electrons in d-subshell.

Hence, option $A$ is correct.

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Similar questions

S c^{3 +}

{T i}^{3 +}

S c = 21, T i = 22

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

S c = 21, T i = 22, V = 23, C r = 24

3 d

+ 1

S c^{3 +}, V^{3 +}, T i^{4 +}, M n^{2 +}

S c = 21, V = 23, T i = 22, M n = 25

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

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