1
JEE Main 2020 (Online) 7th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The correct order of stability for the following alkoxides is :

JEE Main 2020 (Online) 7th January Evening Slot Chemistry - Alcohols, Phenols and Ethers Question 116 English
A
(C) > (B) > (A)
B
(C) > (A) > (B)
C
(B) > (C) > (A)
D
(B) > (A) > (C)
2
JEE Main 2020 (Online) 7th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
In the following reaction, products (A) and (B) respectively, are :

NaOH + Cl2 $$ \to $$ (A) + side products
(hot and conc.)

Ca(OH)2 + Cl2 $$ \to $$ (B) + side products (dry)
A
NaOCl and Ca(OCl)2
B
NaClO3 and Ca(ClO3)2
C
NaOCl and Ca(ClO3)2
D
NaClO3 and Ca(OCl)2
3
JEE Main 2020 (Online) 7th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The bond order and the magnetic characteristics of CN- are :
A
3, paramagnetic
B
$$2{1 \over 2}$$, paramagnetic
C
3, diamagnetic
D
$$2{1 \over 2}$$, diamagnetic
4
JEE Main 2020 (Online) 7th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The equation that is incorrect is :
A
$${\left( {\Lambda _m^0} \right)_{KCl}} - {\left( {\Lambda _m^0} \right)_{NaCl}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{NaBr}}$$
B
$${\left( {\Lambda _m^0} \right)_{NaBr}} - {\left( {\Lambda _m^0} \right)_{NaI}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{NaBr}}$$
C
$${\left( {\Lambda _m^0} \right)_{NaBr}} - {\left( {\Lambda _m^0} \right)_{NaCl}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{KCl}}$$
D
$${\left( {\Lambda _m^0} \right)_{{H_2}O}} = {\left( {\Lambda _m^0} \right)_{HCl}} + {\left( {\Lambda _m^0} \right)_{NaOH}} - {\left( {\Lambda _m^0} \right)_{NaCl}}$$
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