Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

The d-electron configurations of $Cr^{2+}$ , $Mn^{2+}$ , $Fe^{2+}$ and $Co^{2+}$ are $d^4$ , $d^5$ , $d^6$ and $d^7$ respectively. Which one of the following will exhibit minimum paramagnetic behaviour? (At. no. Cr = 24, Mn = 25, Fe = 26, Co = 27)

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

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

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

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

Step-by-Step Solution

Paramagnetic behaviour depends on the number of unpaired electrons ( $n$ ). The greater the number of unpaired electrons, the higher the paramagnetic behaviour. In all the given complexes, $H_2O$ is a weak field ligand, so they form high-spin complexes without pairing of electrons against Hund's rule.

Let's find the number of unpaired electrons for each:

$Fe^{2+}$ ( $3d^6$ ): High spin configuration is $t_{2g}^4 e_g^2$ . It has 4 unpaired electrons.
$Co^{2+}$ ( $3d^7$ ): High spin configuration is $t_{2g}^5 e_g^2$ . It has 3 unpaired electrons.
$Cr^{2+}$ ( $3d^4$ ): High spin configuration is $t_{2g}^3 e_g^1$ . It has 4 unpaired electrons.
$Mn^{2+}$ ( $3d^5$ ): High spin configuration is $t_{2g}^3 e_g^2$ . It has 5 unpaired electrons.

Since $[Co(H_2O)_6]^{2+}$ has the minimum number of unpaired electrons ( $n=3$ ), it will exhibit the minimum paramagnetic behaviour.

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