1
JEE Main 2026 (Online) 28th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Identify the correct statements :

The presence of –NO2 group in benzene ring

A. activates the ring towards electrophilic substitutions.

B. deactivates the ring towards electrophilic substitutions.

C. activates the ring towards nucleophilic substitutions.

D. deactivates the ring towards nucleophilic substitutions.

Choose the correct answer from the options given below :

A

A and D Only

B

B and D Only

C

C and A Only

D

B and C Only

2
JEE Main 2026 (Online) 28th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The cyclic cations having the same number of hyperconjugation are :

JEE Main 2026 (Online) 28th January Evening Shift Chemistry - Basics of Organic Chemistry Question 24 English

Choose the correct answer from the options given below :

A

B and C Only

B

A and B Only

C

A and C Only

D

A, C and D only

3
JEE Main 2026 (Online) 28th January Evening Shift
Numerical
+4
-1
Change Language

The number of isoelectronic species among $Sc^{3+}, Cr^{2+}, Mn^{3+}, Co^{3+}$ and $Fe^{3+}$ is ‘n’. If ‘n’ moles of AgCl is formed during the reaction of complex with formula $CoCl_3(en)_2NH_3$ with excess of $AgNO_3$ solution, then the number of electrons present in the $t_{2g}$ orbital of the complex is ________.

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4
JEE Main 2026 (Online) 28th January Evening Shift
Numerical
+4
-1
Change Language

A volume of x mL of 5 M NaHCO3 solution was mixed with 10 mL of 2 M H2CO3 solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of 235.3 mV, then the value of x = ______ mL (nearest integer).

Sn(s) | Sn(OH)62− (0.5 M) | HSnO2 (0.05 M) | OH | Bi2O3(s) | Bi(s)

Consider up to one place of decimal for intermediate calculations


$\left[\begin{array}{ll}\text { Given: } & E_{Sn\left( {OH} \right)_6^{2 - } |HSnO_2^ -}^o = - 0.9V \\ & \mathrm{E}^{\mathrm{o}}{ }_{\mathrm{Bi}_2 \mathrm{O}_3 \mid \mathrm{Bi}}=-0.44 \mathrm{~V} \\ & \mathrm{pKa}_{\left(\mathrm{H}_2 \mathrm{CO}_3\right)}=6.11 \\ & \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V} \\ & \text { Antilog }(1.29)=19.5\end{array}\right]$
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