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(B) To determine the shape,we calculate the hybridization and lone pairs:$1$. $ICl_2^+$: Central atom $I$ has $7$ valence electrons. It forms $2$ bond

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only $III$

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(B) To determine the shape,we calculate the hybridization and lone pairs:$1$. $ICl_2^+$: Central atom $I$ has $7$ valence electrons. It forms $2$ bonds and has $2$ lone pairs. Hybridization is $sp^3$,shape is angular (bent).$2$. $NH_2^-$: Central atom $N$ has $5$ valence electrons. It forms $2$ bonds and has $2$ lone pairs. Hybridization is $sp^3$,shape is angular (bent).$3$. $CS_2$: Central atom $C$ has $4$ valence electrons. It forms $2$ double bonds with $S$ and has $0$ lone pairs. Hybridization is $sp$,shape is linear.$4$. $H_2Se$: Central atom $Se$ has $6$ valence electrons. It forms $2$ bonds and has $2$ lone pairs. Hybridization is $sp^3$,shape is angular (bent).Therefore,only $CS_2$ $(III)$ is linear and does not have an angular shape.

Which among the following species does not have an angular shape?
$I.$ $ICl_2^+$
$II.$ $NH_2^-$
$III.$ $CS_2$
$IV.$ $H_2Se$

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Which among the following species does not have an angular shape?$I.$ $ICl_2^+$$II.$ $NH_2^-$$III.$ $CS_2$$IV.$ $H_2Se$