JEE Main 2024 (Online) 6th April Evening Shift | s-Block Elements Question 1 | Chemistry

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JEE Main 2024 (Online) 6th April Evening Shift

MCQ (Single Correct Answer)

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Out of Syllabus

Match List I with List II.

	LIST I Alkali Metal		LIST II Emission Wavelength in nm
A.	Li	I.	589.2
B.	Na	II.	455.5
C.	Rb	III.	670.8
D.	Cs	IV.	780.0

Choose the correct answer from the options given below :

(A)-(II), (B)-(IV), (C)-(III), (D)-(I)

(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

(A)-(I), (B)-(IV), (C)-(III), (D)-(II)

(A)-(IV), (B)-(II), (C)-(I), (D)-(III)

JEE Main 2023 (Online) 15th April Morning Shift

MCQ (Single Correct Answer)

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Out of Syllabus

Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R :

Assertion (A) : $\mathrm{BeCl}_{2}$ and $\mathrm{MgCl}_{2}$ produce characteristic flame

Reason (R) : The excitation energy is high in $\mathrm{BeCl}_{2}$ and $\mathrm{MgCl}_{2}$

In the light of the above statements, choose the correct answer from the options given below :

$\mathrm{(A)}$ is true but $(\mathrm{R})$ is false

Both (A) and $(\mathrm{R})$ are true and $(\mathrm{R})$ is the correct explanation of $(A)$

(A) is false but (R) is true

Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

JEE Main 2023 (Online) 13th April Evening Shift

MCQ (Single Correct Answer)

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Out of Syllabus

Better method for preparation of $$\mathrm{BeF}_{2}$$, among the following is :

$$\mathrm{BeH}_{2}+\mathrm{F}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2}$$

$$\left(\mathrm{NH}_{4}\right)_{2} \mathrm{BeF}_{4} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2}$$

$$\mathrm{Be}+\mathrm{F}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2}$$

$$\mathrm{BeO}+\mathrm{C}+\mathrm{F}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2}$$

JEE Main 2023 (Online) 13th April Evening Shift

MCQ (Single Correct Answer)

-1

Out of Syllabus

Identify the correct order of standard enthalpy of formation of sodium halides :

$$\mathrm{NaI}<\mathrm{NaBr}<\mathrm{NaCl}<\mathrm{NaF}$$

$$\mathrm{NaI}<\mathrm{NaBr}<\mathrm{NaF}<\mathrm{NaCl}$$

$$\mathrm{NaF}<\mathrm{NaCl}<\mathrm{NaBr}<\mathrm{NaI}$$

$$\mathrm{NaCl}<\mathrm{NaF}<\mathrm{NaBr}<\mathrm{NaI}$$

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