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$C O_{2}$ has the same Geometry as
$A) H g C l_{2}$ $b) N O_{2}$ $c) S n C l_{4}$ $D) C_{2} H_{2}$

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An Intiative by Sri Chaitanya

A and C

B and D

A and D

C and D

answer is C.

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Detailed Solution

$O = C = O$ $s p -$ hybrid orbitals; linear $H g C l_{2} \to C l - H g - C l : μ = 0,$ linear

$μ = 0, s p^{3}$ hybridization trigonal planar
$H - C = C H; μ = 0,$ linear

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