Identify the correct trend given below (Atomic No. = Ti : 22, Cr …

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 $\,{{[Mo{{({{H}_{2}}O)}_{6}}]}^{2+}}$