Total number of nucleophiles from the following is:
\[
{NH}_3, {PhSH}, {(H}_3{C}_2{S})_2, {H}_2{C}= {C}{H}_2, \, {OH}^-, \, {H}_3{O}^+, \, ({CH}_3)_2 {CO}, \, {NCH}_3
\]
Show Hint
- 6
- 5
- 4
- 7
The Correct Option is A
Solution and Explanation
A nucleophile is a species that donates an electron pair to form a chemical bond in a reaction. The given species that can act as nucleophiles are: 1. NH3 (Ammonia)
2. PhSH (Phenyl thiol)
3. \({(H}_3{C}_2{S})_2\) (Dimethyl disulfide)
4. \({OH}^-\) (Hydroxide ion)
5. \({H}_3{O}^+\) (Hydronium ion)
6. \({NCH}_3\) (Methylamine) Thus, there are 6 nucleophiles in total.
Top Questions on Analytical Chemistry
- Given below are two statements:
Statement I: In the oxalic acid vs KMnO$_4$ (in the presence of dil H$_2$SO$_4$) titration the solution needs to be heated initially to 60°C, but no heating is required in Ferrous ammonium sulphate (FAS) vs KMnO$_4$ titration (in the presence of dil H$_2$SO$_4$).
Statement II: In oxalic acid vs KMnO$_4$ titration, the initial formation of MnSO$_4$ takes place at high temperature, which then acts as catalyst for further reaction. In the case of FAS vs KMnO$_4$, heating oxidizes Fe$^{2+}$ into Fe$^{3+}$ by oxygen of air and error may be introduced in the experiment.
In the light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
- Chemistry
- Analytical Chemistry
- Based on the data given below: \[ E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33 \, \text{V}, \quad E^\circ_{\text{Cl}_2/\text{Cl}^-} = 1.36 \, \text{V}, \quad E^\circ_{\text{MnO}_4^-/\text{Mn}^{2+}} = 1.51 \, \text{V}, \quad E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \, \text{V}. \] The strongest reducing agent is:
- JEE Main - 2025
- Chemistry
- Analytical Chemistry
- Based on the data given below: \[ E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33 \, \text{V}, \quad E^\circ_{\text{Cl}_2/\text{Cl}^-} = 1.36 \, \text{V}, \quad E^\circ_{\text{MnO}_4^-/\text{Mn}^{2+}} = 1.51 \, \text{V}, \quad E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \, \text{V}. \] The strongest reducing agent is:
- JEE Main - 2025
- Chemistry
- Analytical Chemistry
Given below are two statements:
Statement I: In the oxalic acid vs KMnO$_4$ (in the presence of dil H$_2$SO$_4$) titration the solution needs to be heated initially to 60°C, but no heating is required in Ferrous ammonium sulphate (FAS) vs KMnO$_4$ titration (in the presence of dil H$_2$SO$_4$).
Statement II: In oxalic acid vs KMnO$_4$ titration, the initial formation of MnSO$_4$ takes place at high temperature, which then acts as catalyst for further reaction. In the case of FAS vs KMnO$_4$, heating oxidizes Fe$^{2+}$ into Fe$^{3+}$ by oxygen of air and error may be introduced in the experiment.
In the light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
- Chemistry
- Analytical Chemistry
- Identify the incorrect statements regarding primary standard of titrimetric analysis
(A) It should be purely available in dry form.
(B) It should not undergo chemical change in air.
(C) It should be hygroscopic and should react with another chemical instantaneously and stoichiometrically.
(D) It should be readily soluble in water.
(E) KMnO4 & NaOH can be used as primary standard.
Choose the correct answer from the options given below :- JEE Main - 2024
- Chemistry
- Analytical Chemistry
Questions Asked in JEE Main exam
- Let \( \langle a_n \rangle \) be a sequence such that \( a_0 = 0 \), \( a_1 = \frac{1}{2} \), and \( 2a_{n+2} = 5a_{n+1} - 3a_n \).n= 0,1,2,3.... Then \( \sum_{k=1}^{100} a_k \) is equal to:
- JEE Main - 2025
- Sequences and Series
- Given below are two statements:
Statement (I): NaCl is added to the ice at \( 0^\circ C \), present in the ice cream box to prevent the melting of ice cream.
Statement (II): On addition of NaCl to ice at \( 0^\circ C \), there is a depression in freezing point.
In the light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
- Colligative Properties
- Let P be the foot of the perpendicular from the point \( (1, 2, 2) \) on the line \[ \frac{x-1}{1} = \frac{y + 1}{-1} = \frac{z - 2}{2} \] Let the line \( \mathbf{r} = (-\hat{i} + \hat{j} - 2\hat{k}) + \lambda (\hat{i} - \hat{j} + \hat{k})\), \( \lambda \in \mathbb{R} \), intersect the line \(L\) at \(Q\). Then \( 2(PQ)^2 \) is equal to:
- JEE Main - 2025
- 3D Geometry
- The electric field of an electromagnetic wave in free space is \[ \vec{E} = 57 \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] \left( 4\hat{i} - 3\hat{j} \right) \, \text{N/C}. \] The associated magnetic field in Tesla is:
- JEE Main - 2025
- Electromagnetic Induction and Inductance
- The area of the region bounded by the curves \( x(1 + y^2) = 1 \) and \( y^2 = 2x \) is:
- JEE Main - 2025
- Calculus