Question
Consider $|x|$ is the difference in oxidation states of Mn in highest manganese fluoride and highest manganese oxide. The ions with $|x|$ number of unpaired electrons from the following are:
A. ${Sc}^{3 +}$
B. ${Zn}^{2 +}$
C. $V^{2 +}$
D. ${Fe}^{2 +}$
E. ${Co}^{2 +}$
Choose the correct answer from the options given below:
Options
Solution
Sol. [Key Concept: Highest Mn fluoride is MnF$4$ (+4) and highest oxide Mn$2$O$7$ (+7), so $|x| = 3$; pick the ions with exactly 3 unpaired electrons.]
Oxide of Mn in its highest O.S. $= {Mn}_{2}O_{7}\left\lbrack {Mn}^{+ 7} \right\rbrack$ Fluoride of Mn in its highest O.S. $= {MnF}_{4}\left\lbrack {Mn}^{+ 4} \right\rbrack$
⇒ Difference in O.S. $= |x| = 3$
${Sc}^{+ 3} \Rightarrow \lbrack Ar\rbrack;0$ unpaired $e^{-}$
${Zn}^{2 +} \Rightarrow \lbrack Ar\rbrack 3{\text{ }d}^{10};0$ unpaired electrons
${Fe}^{+ 2} \Rightarrow \lbrack Ar\rbrack 3{\text{ }d}^{6};4$ unpaired electrons
${Co}^{+ 2} \Rightarrow \lbrack Ar\rbrack 3{\text{ }d}^{7};3$ unpaired electrons
$V^{+ 2} \Rightarrow \lbrack Ar\rbrack 3{\text{ }d}^{3};3$ unpaired electrons
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