Q. Identify the correct trend given below (Atomic number, $Ti =22, Cr =24$ and $Mo =42$ )
Coordination Compounds
Solution:
For $\left[ Cr \left( H _{2} O \right)_{6}\right]^{2+}$ and $Cr ^{2+}=[ Ar ], 3 d ^{4}$
As $H _{2} O$ is a weak field lignad, so pairing of electrons does not occur.
$CFSE$ for $\left[ Cr \left( H _{2} O \right)_{6}\right]^{2+}=3\left(-0.4 \Delta_{o}\right)+1\left(0.6 \Delta_{o}\right)=-0.6 \Delta_{o}$ Similarly,
Complex
Metal ion
Electronic configuration
$CFSE$
$\left[ Mo \left( H _{2} O \right)_{6}\right]^{2+}$
$Mo ^{2+}$
$ t_{2 g}^{3}, e_{g}^{1} $
$-0.6 \Delta_{o}$
$\left[ Ti \left( H _{2} O \right)_{6}\right]^{3+}$
$Ti ^{3+}$
$t_{2 g}^{1}, e_{g}^{0}$
$-0.4 \Delta_{o}$
$\left[ Ti \left( H _{2} O \right)_{6}\right]^{2+}$
$Ti ^{2+}$
$t_{2 g}^{2}, e_{g}^{0}$
$-0.8 \Delta_{o}$
$\Delta_{o} \propto$ Crystal Field Stabilisation Energy $(CFSE)$
$\Delta_{o}$ depends on $Z_{ eff }$ and for $3 d$-series, $Z_{ eff }$ is less than $4 d$-series.
Hence, $\Delta_{o}$ of $\left[ Cr \left( H _{2} O \right)_{6}\right]^{2+}>\left[ Mo \left( H _{2} O \right)_{6}\right]^{2+}$
From above table,
$\Delta_{o} \text { of }\left[ Ti \left( H _{2} O \right)_{6}\right]^{3+}>\Delta_{o} \text { of }\left[ Ti \left( H _{2} O \right)_{6}\right]^{2+}$
| Complex | Metal ion | Electronic configuration | $CFSE$ |
|---|---|---|---|
| $\left[ Mo \left( H _{2} O \right)_{6}\right]^{2+}$ | $Mo ^{2+}$ | $ t_{2 g}^{3}, e_{g}^{1} $ | $-0.6 \Delta_{o}$ |
| $\left[ Ti \left( H _{2} O \right)_{6}\right]^{3+}$ | $Ti ^{3+}$ | $t_{2 g}^{1}, e_{g}^{0}$ | $-0.4 \Delta_{o}$ |
| $\left[ Ti \left( H _{2} O \right)_{6}\right]^{2+}$ | $Ti ^{2+}$ | $t_{2 g}^{2}, e_{g}^{0}$ | $-0.8 \Delta_{o}$ |