1.

[Cr(H_(2)O)_(6)]Cl_(3) (at. No. of Cr = 24) has a magnetic moment of 3.83 B.M. the correct distribution of 3d electrons in the chromium of the complex is

Answer»

`3d_(xy)^(1), 3d_(yz)^(1), 3d_(xz)^(1)`
`3d_(xy)^(1), 3d_(yz)^(1), 3d_(z^(2))^(1)`
`3d_((X^(2)-y^(2)))^(1), 3d_(z^(2))^(1), 3d_(xz)^(1)`
`3d_(xy)^(1), 3d_((x^(2)-y^(2)))^(1), 3d_(yz)^(1)`

Solution :Magnetic moment `=sqrt(n(n+2))` B.M.= 3.83 B.M. (GIVEN). Hence, n = 3, i.e. there are three unpaired electrons. Thus, we have

In `d^(2)sp^(3)` HYBRIDISATION, the ORBITALS taking part are `d_(x^(2)-y^(2)) and d_(z^(2))`. Hence, unpaired electrons are present in `3d_(xy), 3d_(yz), 3d_(xz)`.


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