Single Choice

The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is :

A$$Fe^{2+}$$
B$$Co^{2+}$$
Correct Answer
C$$Mn^{2+}$$
D$$Ni^{2+}$$

Solution

The difference in the number of unpaired electrons of Metal ion in its high-spin and low-spin octahedral complexes is 2. For the metal $${Co}^{2+}\ ([Ar]3d^7)$$ ion, the difference of unpaired electrons is $$3 - 1 = 2$$ since the number of unpaired electrons of a metal ion in its high-spin complex is 3 and low-spin complex is 1. Option B is correct

SIMILAR QUESTIONS

Chemical Bonding

The correct molecular geometry of $$Fe\left ( CO \right )_{5}$$ is: (Z=26 for Fe)

Chemical Bonding

Cuprous ion is colourless while cupric ion is coloured because

Chemical Bonding

Atomic number of Mn, Fe, and Co are $$25, 26$$, and $$27$$ respectively. Which of the following inner orbital octahedral complex ions are diamagnetic?

Chemical Bonding

A solution of $$[Ni(H_{2}O)_{6}]^{2+}$$ is green but a solution of $$[Ni(CN)_{4}]^{2-}$$ is colourless, Explain.

Chemical Bonding

$$[Fe(CN)_{6}]^{4-}$$ and $$[Fe(H_{2}O)_{6}]^{2+}$$ are of different colours in dilute solutions. Why?

Chemical Bonding

Atomic number of Mn, Fe, and Co are $$25, 26$$, and $$27$$ respectively. Which of the following inner orbital octahedral complex ions are diamagnetic?

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