Question

Match the column:
$\begin{array}{cccc}& & & \\ & Column-I& & Column-II\\ \left(a\right)& \left[Ag\right(N{H}_3{)}_2{]}^{+}& \left(p\right)& Octahedral\\ \left(b\right)& \left[Cu\right(N{H}_3{)}_4{]}^{2+}& \left(q\right)& Tetrahedral\\ \left(c\right)& \left[Cr\right(H_{2}O{)}_6{]}^{2+}& \left(r\right)& Trigonalbipyramidal\\ \left(d\right)& \left[Fe\right(CO{)}_5]& \left(s\right)& Squareplanar\\ & & \left(t\right)& Linear\end{array}$

• A. (a) → t; (b) → s; (c) → r; (d) → p.
• B. (a) → t; (b) → q; (c) → r; (d) → s.
• C. (a) → t; (b) → q ; (c) → s; (d) → r.
• D. (a) → t; (b) → s; (c) → p; (d) → r.

Answer 1

The correct option is D

(a) → t; (b) → s; (c) → p; (d) → r.

Use CFT (or even VBT would suffice):

This above is the simplest of all, except maybe that $\left[Ag\right(N{H}_3{)}_2{]}^{+}$ is linear!

The only complex with 6 ligands is the $\left[Cr\right(H_{2}O{)}_6{]}^{2+}$ - which is clearly octahedral.

$C{u}^{2+}$ is $3d^9$ and thus has to be square planar! There are no tetrahedral complexes from the given 4.