Question

Match the examples given in column I with the shapes of the compounds given in column II and mark the appropriate choice.

Column I		Column II
$(A)$	$[Ag(NH_{3})_{2}]^{+}$	(i)	$d^{2} sp^{3}$, octahedral
(B)	$[Ni(CN)_{4}]^{2-}$	(ii)	$dsp^{2}$, square planar
(C)	$[Ni(CO)_{4}]$	(iii)	$sp$, linear
(D)	$[Fe(CN)_{6}]^{3-}$	(iv)	$sp^{3}$, tetrahedral

• A. $(A) \to (i), (B) \to (ii), (C) \to (iii), (D) \to (iv)$
• B. $(A) \to (iii), (B) \to (ii), (C) \to (iv), (D) \to (i)$
• C. $(A) \to (iv), (B) \to (iii), (C) \to (ii), (D) \to (i)$
• D. $(A) \to (ii), (B) \to (i), (C) \to (iii), (D) \to (iv)$

Answer 1

Answer: (B)

(I) $Ag \left( NH _{3}\right)_{2}^{+} \Rightarrow Ag ^{+}=$Linear hybridisation $Sp$
(II) $\left[ Ni ( CN )_{4}\right]^{2-}= Ni ^{2+}=[ Ar ] 3 d ^{6}=$ Strong Field ligands.
$= dsp ^{2}$ Hybridisation.
$= $ Square planner.
(III) $Ni [ CO ]_{4}= Ni =[ Ar ] 3 d ^{8}= Sp ^{3}$ Hybridisation
$=$ Tetrahedral.
(IV) $\left[ Fe ( CN )_{6}\right]^{3-}= Fe ^{3+}=[ Ar ] 3 d ^{5}=$ Strong Field ligand.
$= d ^{2} sp ^{3}$ hybridise.
$=$ Octahedral.