A colourless solution gives a white precipitate with H2S in ammoniacal medium that dissolves in dilute HCl. The ion is colourless because

it is Zn^2+ with a filled d^10 configuration, so no d--d transition is possible

it is Co^2+ with d^7, which has no d electrons

it is Mn^2+ with d^5, so all transitions are allowed but absorb in the UV

it is Ni^2+ with d^8, which has no unpaired electrons

it is Zn^2+ with a filled d^10 configuration, so no d--d transition is possible

JEE Main levelSingle correct optionDifficulty 2 of 3qc086

A colourless solution gives a white precipitate with $H X_{2} S$ in ammoniacal medium that dissolves in dilute $HCl$ . The ion is colourless because

Ait is $Co X^{2 +}$ with $d^{7}$ , which has no $d$ electrons
Bit is $Mn X^{2 +}$ with $d^{5}$ , so all transitions are allowed but absorb in the UV
Cit is $Ni X^{2 +}$ with $d^{8}$ , which has no unpaired electrons
Dit is $Zn X^{2 +}$ with a filled $d^{10}$ configuration, so no $d$ -- $d$ transition is possible

A white sulphide soluble in dilute HCl and a colourless ion point to

Zn X^{2 +}

(

3 d^{10}

). With a full

d

subshell no

d

d

excitation exists.

Mn X^{2 +}

is pale pink,

Ni X^{2 +}

green and

Co X^{2 +}

pink.