For the complex `[Cr(H_(2)O)_(6)]^(2+)` calculate the total pairing energey in high spin and low spin state Given the mean pairing energey `=23500cm^(-1)` .

Explanation

In complex `[Cr(H_(2)O)_(6)]^(2+)` oxidation state of Cr is`+2 I .e Cr^(2+)` is pressent which has `3d^(4)` configuration The configuration of `d^(4)` ion in high spin state is `t_(2kg)^(3) eg^(1)` since no electron is paired total pairing energy `[Cr(H_(2)O)_(6)]^(2+)` in high spin state is `0 xx p =0`

`The configuration of `d^(4)` ion in low spin state is `t_(2g) ^(4)eg^(0)` since pairing of an electron takes place total pairing energey of `[Cr(H_(2)O)_(6)]^(2+)` in low spin state is `1 xx P = 1 xx 23500 cm^(-1) =23500cm^(-1)`
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