Question

Match the complexes in Column-I with their properties listed in Column-II

$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{(a)}& \left[Co\right(N{H}_3{)}_4\left(H_{2}O{)}_2\right]C{l}_2& \text{(p)}& \text{Geometrical isomers}\\ \text{(b)}& \left[Pt\right(N{H}_3{)}_{2}C{l}_2]& \text{(q)}& \text{Paramagnetic}\\ \text{(c)}& \left[Co\right(H_{2}O{)}_{5}Cl]Cl& \text{(r)}& \text{Diamagnetic}\\ \text{(d)}& \left[Ni\right(H_{2}O{)}_6]C{l}_2& \text{(s)}& \text{Metal ion with + 2 oxidation state}\end{array}$

• A. (A-p, q, s) (B-p, r, s) (C-q, s) (D-q, s)
• B. (A-p, q, r) (B-p, r, s) (C-q, s) (D-q, s)
• C. (A-p, q, s) (B-p, q, s) (C-r, s) (D-q, s)
• D. (A-p, q, s) (B-p, q, s) (C-r, s) (D-r, s)

Answer 1

The correct option is A

(A-p, q, s) (B-p, r, s) (C-q, s) (D-q, s)

In all these compounds, the central metal atom is in the +2 oxidation state.

$\left[Co\right(N{H}_3{)}_4\left(H_{2}O{)}_2\right]C{l}_2$ is of the type $M{a}_4{b}_2$ so it can definitely exhibit geometrical isomerism. It also has 3 unpaired electrons and is hence paramagnetic.

$\left[Pt\right(N{H}_3{)}_{2}C{l}_2]$ can also exhibit geometrical isomerism (the chlorides can be on the same side or opposite side). It is diamagnetic.

$\left[Co\right(H_{2}O{)}_{5}Cl]$ Cl is paramagnetic with three unpaired electrons. Since it is $M{a}_{5}b$ type so there can be no geometrical isomers.

$\left[Ni\right(H_{2}O{)}_6]C{l}_2$ is paramagnetic with two unpaired electrons. Since it is $M{a}_6$ type so there can be no geometrical isomers.