Among $C a H_{2}, B e H_{2}, B a H_{2}$ , the order of ionic character is _______.

B a H_{2} < B e H_{2} < C a H_{2}

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B e H_{2} < C a H_{2} < B a H_{2}

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B e H_{2} < B a H_{2} < C a H_{2}

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C a H_{2} < B e H_{2} < B a H_{2}

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Solution

The correct option is B $B e H_{2} < C a H_{2} < B a H_{2}$
The ionic character of a bond is dependent on the electronegativities of the atoms involved. The higher the difference between the electronegativities of atoms, the more is the ionic character.

Electronegativity decreases down the group from Beryllium to Barium. Hence, the ionic character of their hydrides increases down the group as the difference becomes more.

Therefore, the order of ionic character is as follows :

$B e H_{2} < C a H_{2} < B a H_{2}$

Hence, the correct option is B.

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