Question

[co(nh3)6]2+ is low spin or high spin complex??

Answer 1

low spin complex as NH3 is a strong ligand

Answer 2

$[Co(NH_3)_6]^{2+}$ is a low spin complex.

Explanation:

• To identify whether a complex exhibits low spin or high spin, its $\mu_{0}$ value must be identified.
• The formula used to calculate $\mu_{0}$ is as follows:

$\mu_{0} = \sqrt{n(n+2)}$

where n = Number of unpaired electrons.

• The free cobalt atom has electronic configuration of $[Ar] 3d^74s^2$. It loses electrons and becomes $Co^{2+}$ and its electronic configuration becomes $[Ar] 3d^7$.
• In the compound, 6 molecules of $NH_3$ is attached to $Co^{2+}$. Thus it has 1 pair of unpaired electron.
• Thus the $\mu_{0}$ value is calculated as follows:

$\mu_{0} = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{1(3)} = \sqrt{3} = 1.73\mu_B$

• Due to presence of one pair of unpaired electron, and low $\mu_0$ value, the given compound is inferred to have low spin.

Learn more about such concept

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