Match List-I with List-II and select the corr | vedclass

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

Question

(A) $(i) \, Ni^{2+}$ reacts with Dimethylglyoxime $(DMG)$ to form a rosy red precipitate of $[Ni(DMG)_2]$. Thus,$(i) - D$.$(ii) \, Ag^{+}$ reacts with

Vedclass · Accepted Answer

$D-A-C-B$

Vedclass · Answer

(A) $(i) \, Ni^{2+}$ reacts with Dimethylglyoxime $(DMG)$ to form a rosy red precipitate of $[Ni(DMG)_2]$. Thus,$(i) - D$.$(ii) \, Ag^{+}$ reacts with Sodium thiosulphate $(Na_2S_2O_3)$ to form a soluble complex $Na_3[Ag(S_2O_3)_2]$. Thus,$(ii) - A$.$(iii) \, Cu^{2+}$ reacts with Ammonia $(NH_3)$ to form a deep blue complex $[Cu(NH_3)_4]^{2+}$. Thus,$(iii) - C$.$(iv) \, S^{2-}$ reacts with Sodium nitroprusside $(Na_2[Fe(CN)_5NO])$ to form a purple colored complex $Na_4[Fe(CN)_5(NOS)]$. Thus,$(iv) - B$.Therefore,the correct matching is $(i)-D, (ii)-A, (iii)-C, (iv)-B$.

List-$I$ (Ion involved)	List-$II$ (Agent)
$(i) \, Ni^{2+}$	$(A).$ Sodium thiosulphate
$(ii) \, Ag^{+}$	$(B).$ Sodium nitroprusside
$(iii) \, Cu^{2+}$	$(C).$ Ammonia
$(iv) \, S^{2-}$	$(D).$ Dimethylglyoxime

Match List-$I$ with List-$II$ and select the correct answer using the codes given below.

List-$I$ (Ion involved) List-$II$ (Agent)

$(i) \, Ni^{2+}$ $(A).$ Sodium thiosulphate

$(ii) \, Ag^{+}$ $(B).$ Sodium nitroprusside

$(iii) \, Cu^{2+}$ $(C).$ Ammonia

$(iv) \, S^{2-}$ $(D).$ Dimethylglyoxime

Similar Questions

Concentrated aqueous $NaOH$ can separate a mixture of

In the group analysis of cations, $Ba^{2+}$ and $Ca^{2+}$ are precipitated respectively as

$H_2S$ gas is passed through an acidified solution containing $Ag^+$,$Cu^{2+}$,and $Zn^{2+}$ ions. Which of these will form a precipitate?

Three separate samples of a solution of a single salt gave these test results: one formed a white precipitate with excess ammonia solution,one formed a white precipitate with dilute $NaCl$ solution,and one formed a black precipitate with $H_2S$. The salt could be:

Three different samples of a salt solution give the following test results: One sample gives a white precipitate with ammonia solution. One sample gives a white precipitate with dilute $NaCl$ solution. One sample gives a black precipitate with $H_2S$. Identify the salt.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Match List-$I$ with List-$II$ and select the correct answer using the codes given below. List-$I$ (Ion involved) List-$II$ (Agent) $(i) \, Ni^{2+}$ $(A).$ Sodium thiosulphate $(ii) \, Ag^{+}$ $(B).$ Sodium nitroprusside $(iii) \, Cu^{2+}$ $(C).$ Ammonia $(iv) \, S^{2-}$ $(D).$ Dimethylglyoxime