Total number of molecules/species from the fo | vedclass

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(A) Paramagnetic species contain at least one unpaired electron. Let us analyze each species: $1. O_2$: Has $2$ unpaired electrons in $\pi^* 2p$ orbit

Vedclass · Accepted Answer

$6$

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(A) Paramagnetic species contain at least one unpaired electron. Let us analyze each species: $1. O_2$: Has $2$ unpaired electrons in $\pi^* 2p$ orbitals. (Paramagnetic) $2. O_2^{+}$: Has $1$ unpaired electron. (Paramagnetic) $3. O_2^{-}$: Has $1$ unpaired electron. (Paramagnetic) $4. NO$: Has $15$ electrons (odd electron species). (Paramagnetic) $5. NO_2$: Has $23$ electrons (odd electron species). (Paramagnetic) $6. CO$: Has $14$ electrons,all paired. (Diamagnetic) $7. K_2[NiCl_4]$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand,so it forms a tetrahedral complex with $2$ unpaired electrons. (Paramagnetic) $8. [Co(NH_3)_6]Cl_3$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand,so it forms an octahedral complex with all electrons paired. (Diamagnetic) $9. K_2[Ni(CN)_4]$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,so it forms a square planar complex with all electrons paired. (Diamagnetic) The paramagnetic species are $O_2, O_2^{+}, O_2^{-}, NO, NO_2, K_2[NiCl_4]$. The total count is $6$.

$[Ni(H_2O)_6]^{2+} : en$ molar ratio	Colour of the product
$A. 1:1$	$I. \text{Pale blue}$
$B. 1:2$	$II. \text{Blue/Purple}$
$C. 1:3$	$III. \text{Violet}$

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

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