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1.	The colour changes of an indicator HIn in acid base titrations is given belowHIn (aq)⇋ H+ (aq) + In– (aq)Colour X Colour YWhich 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
Answer» 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).

The colour changes of an indicator HIn in acid base titrations is given belowHIn (aq)⇋ H+ (aq) + In– (aq)Colour X Colour YWhich 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

Answer»

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).

Discussion

2.	Which of the following has the shortest bond length? (A) O2 (B) O-2(C) O+2(D) O-22
Answer» Correct answer is (D) O^-2₂ O₂⁺ (BO = 2.5; maximum in available options) So, it will have shortest bond length.

Which of the following has the shortest bond length? (A) O2 (B) O-2(C) O+2(D) O-22

Answer»

Correct answer is (D) O^-2₂

O₂⁺ (BO = 2.5; maximum in available options)

So, it will have shortest bond length.

Discussion

3.	The 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+
Answer» Correct answer is (A) [MnO₄]^– MnO₄^– ⇒ Mn⁷⁺(3d0)

The 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+

Answer»

Correct answer is (A) [MnO₄]^–

MnO₄^– ⇒ Mn⁷⁺(3d0)

Discussion

4.	Which of the following cannot act as an oxidising agent ? (A) S2– (B) Br2 (C) HSO-4 (D) 2SO3
Answer» Correct answer is (A) S^2– As sulphide (S^2– ) is in its lowest oxidation state. Hence it cannot act as a oxidising agent.

Which of the following cannot act as an oxidising agent ? (A) S2– (B) Br2 (C) HSO-4 (D) 2SO3

Answer»

Correct answer is (A) S^2–

As sulphide (S^2– ) is in its lowest oxidation state. Hence it cannot act as a oxidising agent.

Discussion

5.	Ellingham diagrams are plots of ΔGº vs temperature which have applications in metallurgy2H2 + 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 < 1125KII. Δ G° for the reduction of CO is more negative at T < 1125 KIII. H2 is a better reducing agent at T > 1125 KIV. 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
Answer» 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.

Ellingham diagrams are plots of ΔGº vs temperature which have applications in metallurgy2H2 + 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 < 1125KII. Δ G° for the reduction of CO is more negative at T < 1125 KIII. H2 is a better reducing agent at T > 1125 KIV. 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

Answer»

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.

Discussion