Among $L i C l$ , $B e C l_{2}$ , $B C l_{3}$ and $C C l_{4}$ , the covalent bond character follows the order:

B e C l_{2} > B C l_{3} > C C l_{4} > L i C l

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B e C l_{2} < B C l_{3} < C C l_{4} < L i C l

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L i C l < B e C l_{2} < B C l_{3} < C C l_{4}

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L i C l > B e C l_{2} > B C l_{3} > C C l_{4}

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Solution

The correct option is C $L i C l < B e C l_{2} < B C l_{3} < C C l_{4}$
Along the period, as we move from $L i \to B e \to B \to C$ , the electronegativity increases and hence the E.N. difference between the element and $C l$ decreases and accordingly, the covalent character increases. Thus, $L i C l < B e C l_{2} < B C l_{3} < C C l_{4}$ is the correct order of covalent bond character.

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