Which is more stable in the below given pair?
$C r^{2 +}$ _____ $C r^{3 +}$

C r^{2 +} = C r^{3 +}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C r^{2 +} < C r^{3 +}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C r^{2 +} > C r^{3 +}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $C r^{2 +} < C r^{3 +}$
The outer d orbital electronic configuration of $C r^{3 +}$ is as follows:
$t_{2 g} : d_{x y}^{1} d_{y z}^{1} d_{z x}^{1}$
$e_{g} : d_{(x^{2} - y^{2})}^{0} d_{z^{2}}^{0}$ .
Whereas $C r^{2 +}$ has :
$t_{2 g} : d_{x y}^{2} d_{y z}^{1} d_{z x}^{1}$
$e_{g} : d_{(x^{2} - y^{2})}^{0} d_{(z^{2})}^{0}$ .
Now, as half filled orbital are more stable than other.
Hence , $t_{2 g}$ half filled orbital found in $C r^{3 +}$ , that is why $C r^{3 +}$ is more stable.

Suggest Corrections

Similar questions

{C r}_{2} O_{7}^{2 -} H^{+} - - \to {C r}^{3}

E q . w t o f {C r}_{2} O_{7}^{2 -}

{C r}_{2} O_{7}^{2 -} + {F e}^{2 +} ⟶ {F e}^{3 +} + {C r}^{3 +}

{C r}_{2} O_{7}^{2 -}

{C r}^{3 +}

{C r}_{2} O_{7}^{2 -} = x \times

x

H_{2} O_{2} (a q) + {C r}_{2} O_{7}^{2 -} ⟶ O_{2} (g) + {C r}^{3 +} (a q)

F e^{2 +}, F e^{3 +}, C r^{2 +}, C r^{3 +}, M n^{2 +}, M n^{3 +}

Join BYJU'S Learning Program