The complex $[C r {(H_{2} O)}_{4} B r_{2}] C l$ gives the test for:

B r^{-}

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

C l^{-}

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

C r^{3 +}

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

C r^{3 +} a n d C l^{-}

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

Open in App

Solution

The correct option is B $C l^{-}$
Since $C l^{-}$ is out of the coordination sphere, it is free and hence, the complex $[C r (H_{2} O)_{4} B r_{2}] C l$ gives the test for $C l^{-}$ .

Suggest Corrections

Similar questions

E_{C l_{2} / C l^{-}}^{\circ} = 1.36 V, E_{C r^{3 +} / C r} = - 0.74 V

E_{C r_{2} O_{7}^{2 -} / C r^{3 +}} = 1.33 V, E_{M n O_{4}^{-} / M n^{2 +}}^{\circ} = 1.51 V

E_{C r^{3 +} / C r}^{\circ} = - 0.74 V; E_{M n O_{4}^{-} / M n^{2 +}}^{\circ} = + 1.51 V

E_{C r_{2} O_{7}^{2 -} / C r^{3 +}}^{\circ} = + 1.33 V; E_{C l / C l^{-}}^{\circ} = + 1.36 V

E_{C r^{(} 3 +) / C r}^{0} = - 0.74 V; E_{M n O_{4}^{-} / M n^{2 +}}^{0} = 1.51 V

E_{C r_{2} O_{7}^{2 -} / C r^{3 +}}^{0} = 1.33 V; E_{C l / C l^{-}}^{0} = 1.36 V

E_{C r^{(} 3 +) / C r}^{0} = - 0.74 V; E_{M n O_{4}^{-} / M n^{2 +}}^{0} = 1.51 V

E_{C r_{2} O_{7}^{2 -} / C r^{3 +}}^{0} = 1.33 V; E_{C l / C l^{-}}^{0} = 1.36 V

E_{C l_{2} / C l^{-}}^{\circ} = 1.36 V, E_{C r^{3 +} / C r}^{\circ} = - 0.74 V E_{C r_{2} O_{7}^{2 -} / C r^{3 +}}^{\circ} = 1.33 V, E_{M n O_{4}^{-} / M n^{2 +}}^{\circ} = 1.51 V

Join BYJU'S Learning Program