The following facts are available-2 A - circleddash - B - 2 rightarrow 2 B - -circleddash - A - 2 -Which of the following statement is correct- A- E C - C 2- E B - B 2- E A - A 2- E D - D 2-B- E C - C 2- E B - B 2- E D - D 2-

The correct option is C E^∘_C^⊝|C_2 lt; E^∘_B^⊝|B_2 lt; E^∘_A^⊝|A_2 lt; E^∘_D^⊝|D_2 nbsp;Given examples are of non metals. A non metal having higher r ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Eccentricity of Hyperbola

The following...

Question

The following facts are available:
$2 A^{⊝} + B_{2} \to 2 B^{⊝} + A_{2}; 2 C^{⊝} + B_{2} \to N o r e a c t i o n; 2 D^{⊝} + A_{2} \to 2 A^{⊝} + D_{2}$
Which of the following statement is correct?

E_{C^{⊝} | C_{2}}^{\circ} > E_{B^{⊝} | B_{2}}^{\circ} > E_{A^{⊝} | A_{2}}^{\circ} > E_{D^{⊝} | D_{2}}^{\circ}

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

E_{C^{⊝} | C_{2}}^{\circ} < E_{B^{⊝} | B_{2}}^{\circ} < E_{A^{⊝} | A_{2}}^{\circ} < E_{D^{⊝} | D_{2}}^{\circ}

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

E_{C^{⊝} | C_{2}}^{\circ} < E_{B^{⊝} | B_{2}}^{\circ} > E_{A^{⊝} | A_{2}}^{\circ} > E_{D^{⊝} | D_{2}}^{\circ}

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

E_{C^{⊝} | C_{2}}^{\circ} > E_{B^{⊝} | B_{2}}^{\circ} < E_{A^{⊝} | A_{2}}^{\circ} < E_{D^{⊝} | D_{2}}^{\circ}

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

Open in App

Solution

The correct option is B $E_{C^{⊝} | C_{2}}^{\circ} < E_{B^{⊝} | B_{2}}^{\circ} < E_{A^{⊝} | A_{2}}^{\circ} < E_{D^{⊝} | D_{2}}^{\circ}$

Given examples are of non-metals. A non-metal having higher reduction potential replaces the other from its solution.
Thus, $E_{R P D}^{\circ} < E_{R P A}^{\circ} < E_{R P B}^{\circ}$
Thus, $E_{O P D}^{\circ} > E_{O P A}^{\circ} > E_{O P B}^{\circ}$

Suggest Corrections

Similar questions

If $E_{A}, E_{B}, E_{C} a n d E_{D}$ represent the magnitudes of electric fields at A, B, C and D. Which of the options are correct?

If a, b, c, d are in G.p., prove that :

(i) $(a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d)^{2}$ are in G.P.

(ii) $(a^{2} - b^{2}), (b^{2} - c^{2}), (c^{2} - d)^{2}$ are in G.P.

(iii) $\frac{1}{a^{2} + b^{2}}, \frac{1}{b^{2} + c^{2}}, \frac{1}{c^{2} + d^{2}}$ are in G.P.

(iv) $(a^{2} + b^{2} + c^{2}), (a b + b c + c d), (b^{2} + c^{2} + d^{2})$

If $E_{A}, E_{B}, E_{C} a n d E_{D}$ represent the magnitudes of electric fields at A, B, C and D. Which of the options are correct?

Q. Given a,b,c,d are in G.P then

a^{2} - b^{2}, b^{2} - c^{2}, c^{2} - d^{2}

are in ........

Q. Prove that;

a^{2} - b^{2}, b^{2} - c^{2}, c^{2} - d^{2}

are in

G . P .

Join BYJU'S Learning Program

The following facts are available: 2A⊝+B2→2B⊝+A2;2C⊝+B2→ No reaction;2D⊝+A2→2A⊝+D2 Which of the following statement is correct?

The correct option is B E∘C⊝|C2<E∘B⊝|B2<E∘A⊝|A2<E∘D⊝|D2 Given examples are of non-metals. A non-metal having higher reduction potential replaces the other from its solution. Thus, E∘RPD<E∘RPA<E∘RPB Thus,E∘OPD>E∘OPA>E∘OPB

The following facts are available:
$2 A^{⊝} + B_{2} \to 2 B^{⊝} + A_{2}; 2 C^{⊝} + B_{2} \to N o r e a c t i o n; 2 D^{⊝} + A_{2} \to 2 A^{⊝} + D_{2}$
Which of the following statement is correct?