Number of electrons in this transition metal complex A chromium atom is singly b

Question

Number of electrons in this transition metal complex

A chromium atom is singly bonded to an azide (N3-), triply bonded to a nitrido (N3-), and coordinated to a ring of 4 nitrogen atoms. The complex has an overall charge of +1.

How many valence electrons does this complex have? What is the oxidation state of the metal?

The way I'm counting, I get that the ring of nitrogen atoms donate 8 electrons in all, and the ring itself is neutral. The azide is a 2 electron donor, and the nitrido is a 3 electron donor. Thus, the chromium must be +5, d1. This seems wrong, and seeing as how the complex has only 14 electrons this way, I'm sure it's wrong. But I don't know where I messed up. Any advice?

Explanation / Answer

[Cr(N3-)(N^-3)]+1

x-1-3 = +1
x = +3

Cr+3

Cr+3 = [Ar]3d^3 4S^0

3 unpaired electrons

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