JEE MAIN - Chemistry (2016 - 9th April Morning Slot - No. 13)
The group of molecules having identical shape is :
SF4 , XeF4 , CCl4
ClF3 , XeOF2 , XeF$$_3^ + $$
BF3 , PCl3 , XeO3
PCl5 , IF5 , XeO2F2
Explanation
H = $${1 \over 2}$$[VE + MA – c + a]
ClF3 , H = $${1 \over 2}$$ (7 + 3 – 0 + 0) = 5 (sp3d)
XeOF2 , H = $${1 \over 2}$$(8 + 2 – 0 + 0) = 5 (sp3d)
XeF3+ , H = $${1 \over 2}$$ (8 + 3 – 1 + 0 ) = 5 (sp3d)
All molecules have sp3d hybridization and 3 bond pair + 2 lone pairs. Hence all have identical stucture (T-shape).
ClF3 , H = $${1 \over 2}$$ (7 + 3 – 0 + 0) = 5 (sp3d)
XeOF2 , H = $${1 \over 2}$$(8 + 2 – 0 + 0) = 5 (sp3d)
XeF3+ , H = $${1 \over 2}$$ (8 + 3 – 1 + 0 ) = 5 (sp3d)
All molecules have sp3d hybridization and 3 bond pair + 2 lone pairs. Hence all have identical stucture (T-shape).
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