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Reactivity with Air and Water (Group 2)

Two metals ...

Question

Two metals $X$ and $Y$ belong to the second group of the periodic table. $X$ forms insoluble oxide but soluble sulphate. $Y$ forms a soluble oxide but insoluble sulphate. The hydroxide of metal $X$ is soluble in $N a O H$ while that of metal $Y$ is insoluble in $N a O H$ . What are metals $X$ and $Y$ ?

X = B e, Y = B a

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X = M g, Y = C a

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X = C a, Y = S r

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X = B a, Y = M g

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Solution

The correct option is C $X = B e, Y = B a$
$B e (x) \to B e O (i n s o l u b l e) / B e S O_{4} (s o l u b l e)$
$B a (y) \to B a O (s o l u b l e) / B a S O_{4} (i n s o l u b l e)$
$B e (O H)_{2}$ soluble in $N a O H$
$B a (O H)_{2}$ insoluble in $N a O H$

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Two metals X and Y belong to the second group of the periodic table. X forms insoluble oxide but soluble sulphate. Y forms a soluble oxide but insoluble sulphate. The hydroxide of metal X is soluble in NaOH while that of metal Y is insoluble in NaOH. What are metals X and Y?

The correct option is C X=Be, Y=BaBe(x)→BeO(insoluble)/BeSO4(soluble)Ba(y)→BaO(soluble)/BaSO4(insoluble)Be(OH)2 soluble in NaOHBa(OH)2 insoluble in NaOH

The correct option is C $X = B e, Y = B a$
$B e (x) \to B e O (i n s o l u b l e) / B e S O_{4} (s o l u b l e)$
$B a (y) \to B a O (s o l u b l e) / B a S O_{4} (i n s o l u b l e)$
$B e (O H)_{2}$ soluble in $N a O H$
$B a (O H)_{2}$ insoluble in $N a O H$