HBr and HI can reduce sulphurie acid, HCI can reduced `KMnO_(4)` and H – InterviewSolution

InterviewSolution

1.	HBr and HI can reduce sulphurie acid, HCI can reduced `KMnO_(4)` and HF can reduce…………….A. `H_(2)SO_(4)`B. `KMnO_(4)`C. `K_(2)Cr_(2)O_(7)`D. None of these
Answer» Correct Answer - D HF does not behave as a reducing agent. Therefore, it cannot be oxidised by any of these.

Discussion

No Comment Found

Related InterviewSolutions

Nitric oxide, though an odd electron molecule, is diamagenetic in liquid state.
In the reaction: `3Br_(2)+6OH^(Theta)rarr 5Br^(Theta)+BrO_(3)^(Theta)+3H_(2)O,Br_(2)` isA. OxidisedB. ReducedC. DisintegratedD. Disproportionted
Give balanced equations for the following : Phosphorous reacts with nitric acid to give equimolar ratio of nitric oxide and nitrogen dioxide.
(a) Though nitrogen exhibits `+5` oxidation state, it does not form pentahalide. Give reason. (b) `PH_3` has lower boiling point than `NH_3`. Why ?
In white phosphorous `/_ PPP = 90^@`.
Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen ?A. `HNO_3, NO, NH_4 Cl, N_2`B. `HNO_3, NO, N_2, NH_4 Cl`C. `HNO_3, NH_4 Cl, NO,N_2`D. `NO,HNO_3, NH_4 Cl, N_2`
Why commercial `HNO_3` is generally yellow in colour ?
The product formed in the reaction of `SOCl_2` (thionyl chloride) with white phosphorous is.A. `PCl_3`B. `SO_2 Cl_2`C. `SCl_2`D. `POCl_3`
Concentrated `HNO_3`, upon long standing, turns yellow-brown due to the formation ofA. `NO`B. `NO_2`C. `N_2 O`D. `N_2 O_4`
When sulphur is boiled is boiled with `Na_2 SO_3`, a compound `(X)` is produced, `(X)` with excess of `AgNO_3` solution gives a compound `(Y)` which is soluble in water and produces a black coloured sulphide `(Z)`. Identify compounds `(X),(Y)` and `(Z)`.