The compound having more covalent nature is:

B a C l_{2}

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M g C l_{2}

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S r C l_{2}

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B e C l_{2}

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Solution

The correct option is D $B e C l_{2}$
$B e, M g, S r$ and $B a$ belong to the same group. Also, we know that down the group, the size of atom increases. So, $B e$ will be the smallest in size and $B a$ will be largest. Now, from Fajan's rule of polarization, smaller the size, more is the polarizing power, so more is the covalent character.
Since, $B e$ has the smallest size, $B e C l_{2}$ will have maximum covalent character.

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Similar questions

R a C l_{2} < B a C l_{2} < S r C l_{2} < C a C l_{2} < M g C l_{2} < B e C l_{2}

B e C l_{2} < M g C l_{2} < C a C l_{2} < S r C l_{2} < B a C l_{2} < R a C l_{2}

R a C l_{2} < B a C l_{2} < S r C l_{2} < C a C l_{2} < M g C l_{2} < B e C l_{2}

R a C l_{2} < B a C l_{2} < S r C l_{2} < C a C l_{2} < M g C l_{2} < B e C l_{2}

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B e C l_{2}

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