Question
Given below are two statements:
Statement I : The number of pairs among $\left\lbrack {Ti}^{4 +},V^{2 +} \right\rbrack,\left\lbrack V^{2 +},{Mn}^{2 +} \right\rbrack,\left\lbrack {Mn}^{2 +},{Fe}^{3 +} \right\rbrack$ and $\left\lbrack V^{2 +},{Cr}^{2 +} \right\rbrack$ in which both ions are coloured is 3 .
Statement II : The number pairs among [ ${La}^{3 +}$, $\left. \ {Yb}^{2 +} \right\rbrack,\left\lbrack {Lu}^{3 +},{Ce}^{4 +} \right\rbrack$ and $\left\lbrack {Ac}^{3 +},{Lr}^{3 +} \right\rbrack$ ions in which both are diamagnetic is 3 .
In the light of the above statements, choose the correct answer from the option given below :
Options
Solution
Sol. [Key Concept: Among the given ion pairs exactly $3$ have both ions coloured (Statement I true), and among the $f$-block pairs exactly $3$ are both diamagnetic (${Ti}^{4 +}$ is $d^{0}$; ${La}^{3 +},{Yb}^{2 +},{Lu}^{3 +},{Ce}^{4 +},{Al}^{3 +},{Lr}^{3 +}$ diamagnetic), so both statements are correct.]
$\ {Ti}^{+ 4}$ is colourless due to $d^{0}$ configuration
∴ statement 1 is correct
${La}^{+ 3},{Yb}^{+ 2},{Lu}^{+ 3},{Ce}^{+ 4},{Al}^{+ 3}\&{Lr}^{+ 3}$ all are diamagnetic
∴ Statement 2 is correct.
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