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
$(A)$	$[Ag(N{H}_3{)}_2{]}^{+}$	(i)	$d^{2}s{p}^3$, octahedral
(B)	$[Ni(CN{)}_4{]}^{2-}$	(ii)	$ds{p}^2$, square planar
(C)	$[Ni(CO{)}_4]$	(iii)	$sp$, linear
(D)	$[Fe(CN{)}_6{]}^{3-}$	(iv)	$s{p}^3$, 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) $Ag{\left(N{H}_3\right)}_2^{+}\Rightarrow A{g}^{+}=$Linear hybridisation $Sp$
(II) ${\left[Ni(CN{)}_4\right]}^{2-}=N{i}^{2+}=[Ar]3d^6=$ Strong Field ligands.
$=ds{p}^2$ Hybridisation.
$=$ Square planner.
(III) $Ni[CO{]}_4=Ni=[Ar]3d^8=S{p}^3$ Hybridisation
$=$ Tetrahedral.
(IV) ${\left[Fe(CN{)}_6\right]}^{3-}=F{e}^{3+}=[Ar]3d^5=$ Strong Field ligand.
$=d^{2}s{p}^3$ hybridise.
$=$ Octahedral.