Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Both $[Ni(CO)_4]$ and $[Ni(CN)_4]^{2-}$ are diamagnetic.The hybridisation of nicked in these complexes ,respectively are

$\begin{array}{1 1}(a)\;sp^3,sp^3&(b)\; sp^3,dsp^2\\(c)\;dsp^2,sp^3&(d)\;dsp^2,dsp^2\end{array}$

Can you answer this question?

1 Answer

0 votes
According to valency bond theory (VBT),in $[Ni(CO)_4]$ Ni is in zero oxidation state due to CO strong ligand,the electrons are paired up.
$Ni :3d^84s^2$
In $[Ni(CN)_4]^{2-}$ Ni is in $+2$ oxidation state.
$Ni^{+2} : 3d^84s^0$
$\therefore [Ni(Co)_4]\rightarrow sp^3$ hybridisation
$[Ni(Co)_4]\rightarrow dsp^2$ hybridisation
Hence (b) is the correct option.
answered Mar 6, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App