The compound Y $B a_{2} C u_{3} O_{7}$ which shows superconductivity has copper in oxidation state ........ Assume that the rare earth element Yttrium is in its usual +3 oxidation state

3/4

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

7/3

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

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

Open in App

Solution

The correct option is B
7/3

$Y B a_{2} C u_{3} O_{7}$

$3 + 2 \times 2 + 3 x - (2 \times 7) = 0$

$3 + 4 + 3 x - 14 = 0$

$3 x = 7$

$x = \frac{7}{3}$

Suggest Corrections

Similar questions

The compound Y $B a_{2} C u_{3} O_{7}$ which shows superconductivity has copper in oxidation state ........ Assume that the rare earth element Yttrium is in its usual +3 oxidation state

Q. What is the oxidation state of copper in the compound,

Y B a_{2} C u_{3} O_{7}

which shows super conductivity? Assume that the rare earth element, yttrium is in its usual

+ 3

oxidation state.

Q. In the compound

Y {B a}_{2} {C u}_{3} O_{7}

which shows superconductivity, what is the oxidation state of

C u

?
Assume that the rare earth element yttrium is in its usual +3 oxidation state.

Q. The compound

{Y b B a}_{2} {C u}_{3} O_{7}

, which shows superconductivity has copper in oxidation state

x

, then

x

is :
(Assume that the rare earth element ytterbium is in the usual

+ 3

oxidation state)

Q. The compound

Y B a_{2} C u_{3} O_{7}

, which shows super conductivity has copper in oxidation state of

\frac{7}{3}

.
Assume that the rare earth element ytterbium is in its usual

+ 3

oxidation state.
If true enter 1, else enter 0.

Join BYJU'S Learning Program

The compound YBa2Cu3O7 which shows superconductivity has copper in oxidation state ........ Assume that the rare earth element Yttrium is in its usual +3 oxidation state

The correct option is B 7/3 YBa2Cu3O7 3+2×2+3x−(2×7)=0 3+4+3x−14=0 3x=7 x=73

The compound Y $B a_{2} C u_{3} O_{7}$ which shows superconductivity has copper in oxidation state ........ Assume that the rare earth element Yttrium is in its usual +3 oxidation state

The correct option is B
7/3

$Y B a_{2} C u_{3} O_{7}$

$3 + 2 \times 2 + 3 x - (2 \times 7) = 0$

$3 + 4 + 3 x - 14 = 0$

$3 x = 7$

$x = \frac{7}{3}$