The number of moles of KMnO₄ that will be completely reduced by one mole of ferrous oxalate in acidic medium –

Solution

The correct name for NO₂ using stock notation is :

Nitrogen dioxide
Nitrogen (iv) oxide
Nitrogen peroxide
All of these

Solution

The method of representing oxidation number by a Roman numeral within the parenthesis represents Stock notation.

FeS₂ + O₂ ⟶ Fe₂O₃+ SO₂ In the above equation the number of electrons lost by one molecule of FeS₂ are –

Solution

In FeS₂, Fe²⁺ is converting into Fe³⁺ and sulphur is changing from –1 oxidation state to +4 oxidation state.There are two S–and one Fe²⁺ in FeS₂. Thus total no.of electrons lost in the given reaction are 11.

Solution

Standard electrode potentials of redox couples A²⁺/A, B²⁺/B, C/C²⁺ and D²⁺/D are 0.3V, – 0.5V, – 0.75V and 0.9V respectively. Which of these is best oxidising agent and reducing agent respectively –

D²⁺/D and B²⁺/B
B²⁺/B and D²⁺/D
D²⁺/D and C²⁺/C
C²⁺/C and D²⁺/D

Solution

The redox couple with maximum reduction potential will be best oxidising agent and with minimum reduction potential will be best reducing agent.

The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution is:

Solution

The correct decreasing order of oxidation number of oxygen in compounds BaF₂, O₃, KO₂ and OF₂ is :

BaO₂ > KO₂ > O₃ > OF₂
OF₂ > O₃ > KO₂ > BaO₂
KO₂ > OF₂ > O₃ > BaO₂
BaO₂ > O₃ > OF₂ > KO₂

Solution

Oxidation no. of O are + 2, 0, – 1/2 and – 1 respectively.

Among NH₃, HNO₃, NaN₃ and Mg₃N₂ the number of molecules having nitrogen in negative oxidation state is

Solution

Calculating the oxidation state of nitrogen in given molecules;Oxidation state of N in NH₃ is x + 3 ×(+ 1) =0 or x = – 3 Oxidation state on N in NaNO₃ is 1 +x+ 3 × (– 2) = 0 or x = + 5 Oxidation state of N in NaN₃ is+ 1 + 3x= 0 or x= –13 Oxidation state of N in Mg₃N₂ is 3 × 2 + 2x= 0 or x = –3 Thus 3 molecules (i.e. NH₃, NaN₃ and Mg₃N₂ have nitrogen in negative oxidation state.

Solution

\(CrO_{4}^{2-}\) ⟶ \(Cr_{2}O_{7}^{2-}\) O.N of Cr does not change . It remains +6 on both sides . Hence there is no oxidation or reduction.

One mole of N₂H₄ loses 10 moles of electrons to form a new compound, y. Assuming that all nitrogen appear in the new compound, what is the oxidation state of nitrogen in y (There is no change in the oxidation state of hydrogen )

–1
–3
+3
+5

Solution

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