Question

Match the following :

List I (Hybridisation)		List II (Compound/ion)
A.	$sp^3 d$	I.	$[PtCl_4]^{2-}$
B	$sp^3 d^2$	II.	$SF_6$
C.	$dsp^2$	III.	$BCl_3$
D.	$dsp^3$	IV.	$PCl_5$
		V.	$ClF_3$

The correct match is

• A. (A)- (IV), (B)-(II), (C)-(I), (D)-(V)
• B. (A)- (IV), (B)-(V), (C)-(II), (D)-(I)
• C. (A)- (II), (B)-(IV), (C)-(I), (D)-(III)
• D. (A)- (I), (B)-(III), (C)-(IV), (D)-(V)

Answer 1

Answer: (A)

(A-IV)



$5 \sigma =5-$ hybrid orbital

$ = sp^{3} d -$ hybridisation

(B-II)



$ 6 \sigma \,=6 $ - hybrid orbital

$=s p^{3} d^{2} -$ hybridisation

(C-I)



$4 \sigma =4 -$ hybrid orbital }

$=d s p^{2}-$ hybridisation

(D-V)



$ 3 \sigma+2l p =5$ hybrid orbital

$=s p^{3} d$ or $d s p^{3} $