Match List-I with List-II: List-I (Salt) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Alkali metal salts impart characteristic colors to the flame due to the excitation of electrons to higher energy levels and their subsequent retur

Vedclass · Accepted Answer

$(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)$

Vedclass · Answer

(D) Alkali metal salts impart characteristic colors to the flame due to the excitation of electrons to higher energy levels and their subsequent return to the ground state.The characteristic wavelengths for these metals are:$Li$: $670.8 	ext{ nm}$ (Crimson Red)$Na$: $589.2 	ext{ nm}$ (Yellow)$Rb$: $780.0 	ext{ nm}$ (Red-violet)$Cs$: $455.5 	ext{ nm}$ (Blue)Therefore,the correct matching is: $(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)$.

List-$I$ (Salt)	List-$II$ (Flame colour wavelength)
$(a)$ $LiCl$	$(i)$ $455.5 \text{ nm}$
$(b)$ $NaCl$	$(ii)$ $670.8 \text{ nm}$
$(c)$ $RbCl$	$(iii)$ $780.0 \text{ nm}$
$(d)$ $CsCl$	$(iv)$ $589.2 \text{ nm}$

Match List-$I$ with List-$II$:

List-$I$ (Salt) List-$II$ (Flame colour wavelength)

$(a)$ $LiCl$ $(i)$ $455.5 \text{ nm}$

$(b)$ $NaCl$ $(ii)$ $670.8 \text{ nm}$

$(c)$ $RbCl$ $(iii)$ $780.0 \text{ nm}$

$(d)$ $CsCl$ $(iv)$ $589.2 \text{ nm}$

Choose the correct answer from the options given below:

Similar Questions

Sodium carbonate is manufactured by the Solvay process. The products that are recycled are:

Which of the following will release hydrogen gas upon reaction with hydrochloric acid?

The blue colour of the solution (metal dissolved in liquid ammonia) is due to

How does the ionization energy of Group $1$ elements change down the group?

How many water molecules are present in the formula of crystalline lithium chloride?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Match List-$I$ with List-$II$: List-$I$ (Salt) List-$II$ (Flame colour wavelength) $(a)$ $LiCl$ $(i)$ $455.5 \text{ nm}$ $(b)$ $NaCl$ $(ii)$ $670.8 \text{ nm}$ $(c)$ $RbCl$ $(iii)$ $780.0 \text{ nm}$ $(d)$ $CsCl$ $(iv)$ $589.2 \text{ nm}$ Choose the correct answer from the options given below: