The pair having similar geometry is :

P C l_{3}, N H_{4}^{+}

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B e C l_{2}, H_{2} O

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C H_{4}, C C l_{4}

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I F_{5}, P F_{5}

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Solution

The correct options are
A $P C l_{3}, N H_{4}^{+}$
C $C H_{4}, C C l_{4}$
$C H_{4}$ has 8 electrons around the central carbon atom. This means there are four electron pairs arranged in a tetrahedral shape. There are no lone pairs.

$C C l_{4}$ has no lone pairs on the central carbon and is therefore of tetrahedral geometry.

Both $C H_{4}$ and $C C l_{4}$ have tetrahedral geometry.

Hence, option $C$ is correct.

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Similar questions

{S F}_{4}, C {C l}_{4}, I F_{5}, B r F_{3}, {S O}_{2}, {O F}_{2}, {S F}_{6}, {B F}_{3}

I F_{5}

H_{2} O

B F_{3}

C C l_{4}

I

I F_{5}

H_{2} O

A l C I_{3}

C H_{4}

N F_{3}

H_{2} O

A l C l_{3}

C H_{4}

N F_{3}

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