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)	$\left[Ag\right(N{H}_3{)}_2{]}^{+}$	(i)	$d^{2}s{p}^3,$ octahedral
(B)	$\left[Ni\right(CN{)}_4{]}^{2-}$	(ii)	$ds{p}^2$, square planar
(C)	$\left[Ni\right(CO{)}_4]$	(iii)	$sp$, linear
(D)	$\left[Fe\right(CN{)}_6{]}^{3-}$	(iv)	$s{p}^3$, tetrahedral

• A. (A) - (i), (B) - (ii), (C) - (iii), (D) - (iv)
• B. (A) - (iii), (B) - (ii), (C) - (iv), (D) - (i)
• C. (A) - (iv), (B) - (iii), (C) - (ii), (D) - (i)
• D. (A) - (ii), (B) - (i), (C) - (iii), (D) - (iv)

Answer 1

Answer: (B)

(A) - (iii), (B) - (ii), (C) - (iv), (D) - (i)

$\left(I\right)Ag(N{H}_3{)}_2^{+}\Rightarrow A{g}^{+}=$ Linear hybridisation Sp

$\left(II\right)\left[Ni\right(CN{)}_4{]}^{2-}=N{i}^{2+}=\left[Ar\right]3d^6=$ Strong Field ligands.

= $ds{p}^2$ Hybridisation.

$=$ Square planner.

$\left(III\right)Ni[CO{]}_4=Ni=[Ar]3d^8=$ $S{p}^3$ Hybridisation

$=$ Tetrahedral.

$\left(IV\right)\left[Fe\right(CN{)}_6{]}^{3-}=F{e}^{3+}=\left[Ar\right]3d^5=$ Strong Field ligand.

$=d^{2}s{p}^3$ hybridise.

$=$ Octahedral.