${X C l}_{2}$ is the chloride of metal $X$ . State the formula of the sulphate and the hydroxide of the metal $X$ respectively.

{X S O}_{3}, X ({O H)}_{3}

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{X S O}_{3}, X O H

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{X S O}_{4}, X (O H)_{2}

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{X S O}_{4}, X O H

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Solution

The correct option is C ${X S O}_{4}, X (O H)_{2}$
$X C l_{2} => x + 2 (- 1) => x = + 2$

Oxidation state of metal = +2

metal ion $X^{+ 2}$

Metal sulphate formula=> $X S O_{4}$

Metal Hydroxide formula => $X (O H)_{2}$

Option C is correct.

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Q. XCl2 is the chloride of a metal X. State the formula of its hydroxide and sulphate

THIS IS THE QUESTION TAKEN FROM 1985 BOARD ICSE PAPER OF 10TH XCL2 is the chloride of a metal X.state the formula of the sulphate and hydroxide of the metal X

Q. A metal X forms a salt XSO₄. The salt XSO₄ forms a clear solution in water which reacts with sodium hydroxide solution to form a blue precipitate Y. Metal X is used in making electric wire and alloys like brass.
(a) What do you think metal X could be?
(b) Write the name, formula and colour of salt XSO₄.
(c) What is the blue precipitate Y?
(d) Write a chemical equation of the reaction which takes place when salt XSO₄reacts with sodium hydroxide solution. Give the state symbols of all the reactants and products which occur in the above equation.

Q. Two metals X and Y form the salts XSO₄ and Y₂SO₄, respectively. The solution of salt XSO₄ is blue in colour whereas that of Y₂SO₄ is colourless. When barium chloride solution is added to XSO₄ solution, then a white precipitate Z is formed alongwith a salt which turns the solution green. And when barium chloride solution is added to Y₂SO₄ solution, then the same white precipitate Z is formed alongwith colourless common salt solution.
(a) What could the metals X and Y be?
(b) Write the name and formula of salt XSO₄.
(c) Write the name and formula of salt Y₂SO₄.
(d) What is the name and formula of white precipitate Z?
(e) Write the name and formula of the salt which turns the solution green in the first case.

Two metals X and Y form the salts $X S O_{4}$ and $Y_{2} S O_{4}$ , respectively. The solution of salt $X S O_{4}$ is blue in colour whereas that of $Y_{2} S O_{4}$ is colourless. When barium chloride solution is added to $X S O_{4}$ solution, then a white precipitate Z is formed along with a salt which turns the solution green. And when barium chloride solution is added to $Y_{2} S O_{4}$ solution, then the same white precipitate Z is formed along with colourless common salt solution.

(a) What could the metal X and Y be ?

(b) Write the name and formula of salt $X S O_{4}$ .

(d) What is the name and formla of white precipitate Z ?

(e) Write the name and formula of the salt which turns the solution green in the first case.

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XCl2 is the chloride of metal X. State the formula of the sulphate and the hydroxide of the metal X respectively.

The correct option is C XSO4,X(OH)2XCl2=>x+2(−1)=>x=+2Oxidation state of metal = +2metal ion X+2Metal sulphate formula=> XSO4Metal Hydroxide formula => X(OH)2Option C is correct.

${X C l}_{2}$ is the chloride of metal $X$ . State the formula of the sulphate and the hydroxide of the metal $X$ respectively.

The correct option is C ${X S O}_{4}, X (O H)_{2}$
$X C l_{2} => x + 2 (- 1) => x = + 2$

Oxidation state of metal = +2

metal ion $X^{+ 2}$

Metal sulphate formula=> $X S O_{4}$

Metal Hydroxide formula => $X (O H)_{2}$

Option C is correct.