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The number of unpaired electrons in
$\quad\left[\mathrm{NiCl}_{4}\right]^{2-}, \mathrm{Ni}(\mathrm{CO})_{4}$ and $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2}$
respectively are
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$\quad\left[\mathrm{NiCl}_{4}\right]^{2-}, \mathrm{Ni}(\mathrm{CO})_{4}$ and $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2}$
respectively are
Solution:
2881 Upvotes
Verified Answer
The correct answer is:
2,0,1
Configuration for metal (M) in various complex.
$\mathrm{Ni}^{2 *}$ has 2 unpaired electrons
$\mathrm{Ni}^{0}$ has zero unpaired electrons with
$s p^{3}$ -hybridisation.
$\mathrm{Cu}^{2}$ "has one unpaired electron with $d s p^{2}$ -hybridisation.
$\therefore$ (b) is the correct answer, i.e.
Number of unpaired electron (2,0,1)
$\mathrm{Ni}^{2 *}$ has 2 unpaired electrons
$\mathrm{Ni}^{0}$ has zero unpaired electrons with
$s p^{3}$ -hybridisation.
$\mathrm{Cu}^{2}$ "has one unpaired electron with $d s p^{2}$ -hybridisation.
$\therefore$ (b) is the correct answer, i.e.
Number of unpaired electron (2,0,1)
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