Match each coordination compound in List-$I$ with an appropriate pair of characteristics from List-$II$ and select the correct answer using the code given below the lists.
$\{ en = H_2NCH_2CH_2NH_2 \}$; atomic numbers: $\{Ti = 22; Cr = 24; Co = 27; Pt = 78\}$
List-$I$ List-$II$
$P.$ $[Cr(NH_3)_4Cl_2]Cl$ $1.$ Paramagnetic and exhibits ionisation isomerism
$Q.$ $[Ti(H_2O)_5Cl](NO_3)_2$ $2.$ Diamagnetic and exhibits cis-trans isomerism
$R.$ $[Pt(en)(NH_3)_2Cl_2]NO_3$ $3.$ Paramagnetic and exhibits cis-trans isomerism
$S.$ $[Co(NH_3)_4(NO_3)_2]NO_3$ $4.$ Diamagnetic and exhibits ionisation isomerism
Codes: $P, Q, R, S$

  • A
    $3, 1, 4, 2$
  • B
    $3, 1, 2, 4$
  • C
    $2, 1, 3, 4$
  • D
    $1, 3, 4, 2$

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Similar Questions

Given below are two statements:
Statement $I$ : Dimethyl glyoxime forms a six-membered covalent chelate when treated with $NiCl_2$ solution in presence of $NH_4OH$.
Statement $II$ : Prussian blue precipitate contains iron both in $(+2)$ and $(+3)$ oxidation states.
In the light of the above statements,choose the most appropriate answer from the options given below:

The reaction of $K_3[Fe(CN)_6]$ with freshly prepared $FeSO_4$ solution produces a dark blue precipitate called Turnbull's blue. The reaction of $K_4[Fe(CN)_6]$ with $FeSO_4$ solution in the complete absence of air produces a white precipitate $X$,which turns blue in air. Mixing the $FeSO_4$ solution with $NaNO_3$,followed by a slow addition of concentrated $H_2SO_4$ through the side of the test tube,produces a brown ring.
Precipitate $X$ is:
$(A)$ $Fe_4[Fe(CN)_6]_3$
$(B)$ $Fe[Fe(CN)_6]$
$(C)$ $K_2Fe[Fe(CN)_6]$
$(D)$ $KFe[Fe(CN)_6]$
Among the following,the brown ring is due to the formation of:
$(A)$ $[Fe(NO)_2(SO_4)_2]^{2-}$
$(B)$ $[Fe(NO)_2(H_2O)_4]^{3+}$
$(C)$ $[Fe(NO)_4(SO_4)_2]$
$(D)$ $[Fe(H_2O)_5(NO)]SO_4$

The equation which is balanced and represents the correct product$(s)$ is:

Total number of molecules/species from the following which will be paramagnetic is $........$
$O_2, O_2^{+}, O_2^{-}, NO, NO_2, CO, K_2[NiCl_4], [Co(NH_3)_6]Cl_3, K_2[Ni(CN)_4]$

The correct option$(s)$ regarding the complex $[Co(en)(NH_3)_3(H_2O)]^{3+}$ $(en = H_2NCH_2CH_2NH_2)$ is (are):
$A$. It has two geometrical isomers
$B$. It will have three geometrical isomers if bidentate 'en' is replaced by two cyanide ligands
$C$. It is paramagnetic
$D$. It absorbs light at longer wavelength as compared to $[Co(en)(NH_3)_4]^{3+}$

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