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(Valance Bond Theory) Magnetic measurements indicate that $ [Co(OH_2)_6]^{2+}$ has 3 unpaired electrons. Therefore, the hybridization of the metal's orbitals in $[Co(OH_2)_6]^{2+}$ is :

$(a)\;sp^3\qquad(b)\;sp^2d\qquad(c)\;dsp^2\qquad(d)\;sp^3d^2$

1 Answer

the hybridization of the metal's orbitals in $[Co(OH_2)_6]^{2+}$ is $sp^3d^2$
Hence (d) is the correct option.
answered Feb 3, 2014 by sreemathi.v
 

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