MCQ
Identify the correct trend given below

(Atomic No. $= Ti : 22,$ $Cr : 24$ and  $Mo : 42$ )

  • A
    ${\Delta _0}$ of ${{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}}>\,{{[Mo{{({{H}_{2}}O)}_{6}}]}^{2+}}$ and ${\Delta _0}$ of ${[Ti{({H_2}O)_6}]^{3 + }} > {[Ti{({H_2}O)_6}]^{2 + }}$
  • B
    ${\Delta _0}$ of ${[Cr{({H_2}O)_6}]^{2 + }} > {[Mo{({H_2}O)_6}]^{2 + }}$ and ${\Delta _0}$ of ${[Ti{({H_2}O)_6}]^{3 + }} < {[Ti{({H_2}O)_6}]^{2 + }}$
  • ${\Delta _0}$ of ${[Cr{({H_2}O)_6}]^{2 + }} < {[Mo{({H_2}O)_6}]^{2 + }}$ and ${\Delta _0}$ of ${[Ti{({H_2}O)_6}]^{3 + }} > {[Ti{({H_2}O)_6}]^{2 + }}$
  • D
    ${\Delta _0}$ of ${[Cr{({H_2}O)_6}]^{2 + }} <  {[Mo{({H_2}O)_6}]^{2 + }}$ and ${\Delta _0}$ of ${[Ti{({H_2}O)_6}]^{3 + }} < {[Ti{({H_2}O)_6}]^{2 + }}$

Answer

Correct option: C.
${\Delta _0}$ of ${[Cr{({H_2}O)_6}]^{2 + }} < {[Mo{({H_2}O)_6}]^{2 + }}$ and ${\Delta _0}$ of ${[Ti{({H_2}O)_6}]^{3 + }} > {[Ti{({H_2}O)_6}]^{2 + }}$
c
The splitting is affected by the oxidation state of the central metal ion. A higher oxidation state leads to larger spitting hence.

${\Delta _0}$ of ${{[Ti{{({{H}_{2}}O)}_{6}}]}^{3+}}>$ ${\Delta _0}$ of ${{[Ti{{({{H}_{2}}O)}_{6}}]}^{2+}}$ Further ${\Delta _0}$ also depends of $Z_{eff.}$ and $Z_{eff.}$ of $4d$ series is more than $3d$ series.

Hence

${\Delta _0}$ of  ${{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}}<$ ${\Delta _0}$ of  $\,{{[Mo{{({{H}_{2}}O)}_{6}}]}^{2+}}$

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