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(A) To determine the paramagnetic behaviour,we calculate the number of unpaired electrons $(n)$ for each complex. Since $H_2O$ is a weak field ligand,

Vedclass · Accepted Answer

$[Co(H_2O)_6]^{2+}$

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(A) To determine the paramagnetic behaviour,we calculate the number of unpaired electrons $(n)$ for each complex. Since $H_2O$ is a weak field ligand,the electrons remain unpaired in the $d$-orbitals.

	
		
			Complex
			Electronic Configuration and Unpaired Electrons $(n)$
		
		
			$[Co(H_2O)_6]^{2+}$
			$Co^{2+} (d^7): t_{2g}^5 e_g^2, n=3$
		
		
			$[Fe(H_2O)_6]^{2+}$
			$Fe^{2+} (d^6): t_{2g}^4 e_g^2, n=4$
		
		
			$[Mn(H_2O)_6]^{2+}$
			$Mn^{2+} (d^5): t_{2g}^3 e_g^2, n=5$
		
		
			$[Cr(H_2O)_6]^{2+}$
			$Cr^{2+} (d^4): t_{2g}^3 e_g^1, n=4$
		
	


Paramagnetic behaviour is directly proportional to the number of unpaired electrons $(n)$.
Since $[Co(H_2O)_6]^{2+}$ has the minimum number of unpaired electrons $(n=3)$,it exhibits the least paramagnetic behaviour.

Which one of the following complexes will exhibit the least paramagnetic behaviour?
[Atomic number: $Cr=24, Mn=25, Fe=26, Co=27$]

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Which one of the following complexes will exhibit the least paramagnetic behaviour? [Atomic number: $Cr=24, Mn=25, Fe=26, Co=27$]