A) \[TiC{{l}_{4}},{{[Ni{{({{H}_{2}}O)}_{4}}]}^{2+}}\]
B) \[TiC{{l}_{4}},{{O}_{2}}\]
C) \[TiC{{l}_{4}},{{[Ni{{(CN)}_{4}}]}^{2-}}\]
D) \[{{[Ni{{(CN)}_{4}}]}^{2-}}{{[Ni{{({{H}_{2}}O)}_{4}}]}^{2+}}\]
Correct Answer: C
Solution :
[c]| \[TiC{{l}_{4}}\] Oxidation state of \[Ti=+\,4,\] the electronic configuration is \[[Ar]\,3{{d}^{0}}.\] |
| Thus, all electrons are paired and hence, diamagnetic. |
| \[{{O}_{2}}\] No. of electron in \[{{O}_{2}}=16.\] |
| Electronic configuration: |
| \[\sigma 1{{s}^{2}}\sigma *1s\sigma 2{{s}^{2}}\sigma *2{{s}^{2}}\sigma 2{{p}_{z}}^{2}\] |
| \[\pi 2{{p}_{x}}^{2}=\pi 2{{p}_{y}}^{2}*2{{p}_{x}}^{1}=\pi *2{{p}_{y}}^{1}\] |
| Two electrons are unpaired (by MO theory). |
| Thus, paramagnetic. |
| \[{{[Ni{{(CN)}_{4}}]}^{2-}}\]Electronic configuration of \[N{{i}^{2+}}=3{{d}^{8}}4{{s}^{0}}\] |
| As \[C{{N}^{-}}\]is a strong ligand, pairing of electrons occurs. |
 |
| Hybridisation of \[{{[Ni{{(CN)}_{4}}]}^{2-}}\] is \[{{d}_{s{{p}^{2}}}}\] and shape is square plannar. |
| As all electrons paired thus, diamagnetic |
| Thus, \[[Ni{{({{H}_{2}}O)}_{4}}]\] No. pairing occurs as \[{{H}_{2}}O\] is a weak ligand. |
 |
| \[s{{p}^{3}}\text{-}\]hybridisation as two electrons are unpaired. Thus, paramagnetic in nature. |
| Therefore, option [c] is correct. |
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