When hydrogen sulphide burns in fluorine,the oxidation state of sulphur changes from $- 2$ to

0

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+ 2

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+ 4

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+ 6

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Solution

The correct option is D $+ 6$
$H_{2} S^{- 2} + F^{0} ⟶ S^{+ 6} F$
It changes from $+ 2$ to $+ 6$

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When $H_{2} S O_{3}$ is converted into $H_{2} S O_{4}$ the change in the oxidation state of sulphur is from:

Q. When sulphur dioxide is passed in an acidified

K_{2} {C r}_{2} O_{7}

solution, the oxidation state of sulphur is changed from:

When the gases sulphur dioxide and hydrogen sulphide mix in the presence of water the reaction is :

SO₂+2H₂S=2H₂O+3S

Here hydrogen sulphide is acting as :

A. Oxidizing agent

B. Reducing agent

Translate the following statements into chemical equations and then balance the equations :

(a) Hydrogen sulphide gas burns in air to give water and sulphur dioxide.

(b) Phosphorus burns in oxygen to give phosphorus pentoxide.

(d) Aluminium metal replaces iron from ferric oxide, $F e_{2} O_{3}$ , giving aluminium oxide and iron.

(e) Barium chloride reacts with zinc sulphate to give zine chloride and barium sulphate.

Hydrogen sulphide gas burns in oxygen to yield 12.8 g of sulphur dioxide gas as under:
$2 H_{2} S + 3 O_{2} \to 2 H_{2} O + 2 S O_{2}$
(i). calculate the volume of hydrogen suulphide at STP.
Also, (ii). calculate the volume of oxygen required at STP. Which will complete the combustion of hydrogen sulphide determined in litres.

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When hydrogen sulphide burns in fluorine,the oxidation state of sulphur changes from −2 to

The correct option is D +6H2S−2+F0⟶S+6FIt changes from +2 to +6

When hydrogen sulphide burns in fluorine,the oxidation state of sulphur changes from $- 2$ to

The correct option is D $+ 6$
$H_{2} S^{- 2} + F^{0} ⟶ S^{+ 6} F$
It changes from $+ 2$ to $+ 6$