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B
IF+2
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C
OF2
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D
CdBr2
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Solution
The correct options are BICl−2 DCdBr2
In ICl−2, I form sigma bonds and have three lone pair, it's hybridisation is sp3d in which lone pairs are placed in equatorial position and sigma bonds are formed along with axial position which gives it linear structure.
In CdBr2 , Cd forms 2 sigma bonds and have 0 lone pair, it's hybridisation is sp and thus is linear in shape.