Question

The aqueous solution containing which one of the following ions will be colourless? (Atomic no. $Sc = 21, Fe = 26, Ti = 22, Mn = 25$)

• A. $ Sc^{3+} $
• B. $ Fe^{2+}$
• C. $ Ti^{3+} $
• D. $ Mn^{2+}$

Answer 1

The Correct Option is A

${ }_{21} Sc =1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{1}, 4 s^{2}$ So, $Sc ^{3+}=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6}$ (It is colourless due to abs�nce of unpaired electrons in $d$-stubshell) ${ }_{26} Fe =1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{6}, 4 s^{2} $ $Fe ^{2+}=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{6}$

(It is coloured due to presence of four unpaired electrons in $d$-subshell.) ${ }_{22} Ti =1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{2}, 4 s^{2}$ $Ti ^{3+}=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2}, 3 s^{6} 3 d^{1}$

(It is coloured due to presence of one unpaired electron in $d$-subshell). ${ }_{25} Mn =1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{5}, 4 s^{2} $ $Mn ^{2+}=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{5}$

(It is coloured due to presence of five unpaired electrons in $d$-subshell.)