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Which of the following has the shortest bond length?
(A) O2
(B) O–2
(C) O+2
(D) O-22
Correct answer is (D) O-22O2+ (BO = 2.5; maximum in available options)So, it will have shortest bond length.
Correct answer is (D) O-22
O2+ (BO = 2.5; maximum in available options)
So, it will have shortest bond length.
See lessWhich of the following cannot act as an oxidising agent ?
(A) S2–
(B) Br2
(C) HSO–4
(D) 2SO3
Correct answer is (A) S2–As sulphide (S2– ) is in its lowest oxidation state. Hence it cannot act as a oxidising agent.
Correct answer is (A) S2–
As sulphide (S2– ) is in its lowest oxidation state. Hence it cannot act as a oxidising agent.
See lessThe complex ion that does not have d electrons in the metal atom is
(A) [MnO4]–
(B) [Co(NH3)6]3+
(C) [Fe(CN)6]3–
(D) Cr(H2O)6]3+
Correct answer is (A) [MnO4]–MnO4– ⇒ Mn7+(3d0)
Correct answer is (A) [MnO4]–
MnO4– ⇒ Mn7+(3d0)
See lessThe colour changes of an indicator HIn in acid base titrations is given below
HIn (aq)⇋ H+ (aq) + In– (aq)
Colour X Colour Y
Which of the following statements is correct?
(A) In a strong alkaline solution colour Y will be observed
(B) In a strong acidic solution colour Y will be observed
(C) Concentration of in– is higher than that of HIn at the equivalence point
(D) In a strong alkaline solution colour X is observed
Correct answer is (A) In a strong alkaline solution colour Y will be observedIt is an acidic indicator therefore will remain in ionized form in strong alkaline solution (opposite ion effect).
Correct answer is (A) In a strong alkaline solution colour Y will be observed
It is an acidic indicator therefore will remain in ionized form in strong alkaline solution (opposite ion effect).
See lessEllingham diagrams are plots of ΔGº vs temperature which have applications in metallurgy
2H2 + O2 = 2H2O ΔG(J) = –247500 +55.85 T …….. (I)
2CO + O2 = 2CO2 ΔG(J) = –282400 +86.81 T………. (II)
The Ellingham diagrams for the oxidation of H2(I) and CO (II) are given below.
The two lines intersect (TE) at 1125 K.
Which of the following is correct ?
I. Δ Gº for reaction (i) is more negative at T < 1125K
II. Δ G° for the reduction of CO is more negative at T < 1125 K
III. H2 is a better reducing agent at T > 1125 K
IV. H2 is a better reducing agent at T < 1125 K
(A) I and II
(B) I and III
(C) III only
(D) I and IV
Correct answer is (C) III onlyAccording to the given data, Plot I is for CO as it has more negative ΔG value Plot II is for H2. And as we know in the Ellingham diagram, compound for which, the Plot lies below acts as a better reducing agent. So, at T > 1125 K, H2 is a better reducing agent. Note:Read more
Correct answer is (C) III only
According to the given data, Plot I is for CO as it has more negative ΔG value Plot II is for H2. And as we know in the Ellingham diagram, compound for which, the Plot lies below acts as a better reducing agent. So, at T > 1125 K, H2 is a better reducing agent.
Note: The given graph in the question is not according to given data in the question.
See less